3.37 \(\int x^2 (a+b \csc (c+d \sqrt {x}))^2 \, dx\)

Optimal. Leaf size=513 \[ \frac {a^2 x^3}{3}+\frac {480 i a b \text {Li}_6\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {15 b^2 \text {Li}_5\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {30 b^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{5/2}}{d} \]

[Out]

-20*I*a*b*x^2*polylog(2,exp(I*(c+d*x^(1/2))))/d^2+1/3*a^2*x^3-8*a*b*x^(5/2)*arctanh(exp(I*(c+d*x^(1/2))))/d-2*
b^2*x^(5/2)*cot(c+d*x^(1/2))/d+10*b^2*x^2*ln(1-exp(2*I*(c+d*x^(1/2))))/d^2-20*I*b^2*x^(3/2)*polylog(2,exp(2*I*
(c+d*x^(1/2))))/d^3+240*I*a*b*x*polylog(4,exp(I*(c+d*x^(1/2))))/d^4-480*I*a*b*polylog(6,exp(I*(c+d*x^(1/2))))/
d^6-80*a*b*x^(3/2)*polylog(3,-exp(I*(c+d*x^(1/2))))/d^3+80*a*b*x^(3/2)*polylog(3,exp(I*(c+d*x^(1/2))))/d^3+30*
b^2*x*polylog(3,exp(2*I*(c+d*x^(1/2))))/d^4-2*I*b^2*x^(5/2)/d+480*I*a*b*polylog(6,-exp(I*(c+d*x^(1/2))))/d^6-1
5*b^2*polylog(5,exp(2*I*(c+d*x^(1/2))))/d^6+30*I*b^2*polylog(4,exp(2*I*(c+d*x^(1/2))))*x^(1/2)/d^5+20*I*a*b*x^
2*polylog(2,-exp(I*(c+d*x^(1/2))))/d^2-240*I*a*b*x*polylog(4,-exp(I*(c+d*x^(1/2))))/d^4+480*a*b*polylog(5,-exp
(I*(c+d*x^(1/2))))*x^(1/2)/d^5-480*a*b*polylog(5,exp(I*(c+d*x^(1/2))))*x^(1/2)/d^5

________________________________________________________________________________________

Rubi [A]  time = 0.61, antiderivative size = 513, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4205, 4190, 4183, 2531, 6609, 2282, 6589, 4184, 3717, 2190} \[ \frac {20 i a b x^2 \text {PolyLog}\left (2,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {PolyLog}\left (2,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {80 a b x^{3/2} \text {PolyLog}\left (3,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {PolyLog}\left (3,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {240 i a b x \text {PolyLog}\left (4,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {PolyLog}\left (4,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 i a b \text {PolyLog}\left (6,-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {PolyLog}\left (6,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {20 i b^2 x^{3/2} \text {PolyLog}\left (2,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {PolyLog}\left (3,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {PolyLog}\left (4,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 b^2 \text {PolyLog}\left (5,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Csc[c + d*Sqrt[x]])^2,x]

[Out]

((-2*I)*b^2*x^(5/2))/d + (a^2*x^3)/3 - (8*a*b*x^(5/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d - (2*b^2*x^(5/2)*Cot[c
+ d*Sqrt[x]])/d + (10*b^2*x^2*Log[1 - E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((20*I)*a*b*x^2*PolyLog[2, -E^(I*(c +
d*Sqrt[x]))])/d^2 - ((20*I)*a*b*x^2*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - ((20*I)*b^2*x^(3/2)*PolyLog[2, E
^((2*I)*(c + d*Sqrt[x]))])/d^3 - (80*a*b*x^(3/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))])/d^3 + (80*a*b*x^(3/2)*Pol
yLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 + (30*b^2*x*PolyLog[3, E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((240*I)*a*b*x*Po
lyLog[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + ((240*I)*a*b*x*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4 + ((30*I)*b^2*Sq
rt[x]*PolyLog[4, E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (480*a*b*Sqrt[x]*PolyLog[5, -E^(I*(c + d*Sqrt[x]))])/d^5 -
(480*a*b*Sqrt[x]*PolyLog[5, E^(I*(c + d*Sqrt[x]))])/d^5 - (15*b^2*PolyLog[5, E^((2*I)*(c + d*Sqrt[x]))])/d^6 +
((480*I)*a*b*PolyLog[6, -E^(I*(c + d*Sqrt[x]))])/d^6 - ((480*I)*a*b*PolyLog[6, E^(I*(c + d*Sqrt[x]))])/d^6

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int x^2 \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \operatorname {Subst}\left (\int x^5 (a+b \csc (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (a^2 x^5+2 a b x^5 \csc (c+d x)+b^2 x^5 \csc ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}+(4 a b) \operatorname {Subst}\left (\int x^5 \csc (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \operatorname {Subst}\left (\int x^5 \csc ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}-\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(20 a b) \operatorname {Subst}\left (\int x^4 \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (10 b^2\right ) \operatorname {Subst}\left (\int x^4 \cot (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {(80 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(80 i a b) \operatorname {Subst}\left (\int x^3 \text {Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (20 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^4}{1-e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(240 a b) \operatorname {Subst}\left (\int x^2 \text {Li}_3\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (40 b^2\right ) \operatorname {Subst}\left (\int x^3 \log \left (1-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {(480 i a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(480 i a b) \operatorname {Subst}\left (\int x \text {Li}_4\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (60 i b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_2\left (e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(480 a b) \operatorname {Subst}\left (\int \text {Li}_5\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}-\frac {\left (60 b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_3\left (e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {(480 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {(480 i a b) \operatorname {Subst}\left (\int \frac {\text {Li}_5(x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {\left (30 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_4\left (e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 i a b \text {Li}_6\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ &=-\frac {2 i b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {2 b^2 x^{5/2} \cot \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {20 i a b x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i a b x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 i b^2 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {30 b^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 i a b x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {240 i a b x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {30 i b^2 \sqrt {x} \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {15 b^2 \text {Li}_5\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {480 i a b \text {Li}_6\left (-e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 i a b \text {Li}_6\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ \end {align*}

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Mathematica [A]  time = 14.10, size = 779, normalized size = 1.52 \[ \frac {a^2 x^3 \sin ^2\left (c+d \sqrt {x}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2}{3 \left (a \sin \left (c+d \sqrt {x}\right )+b\right )^2}+\frac {b^2 x^{5/2} \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d \sqrt {x}}{2}\right ) \sin ^2\left (c+d \sqrt {x}\right ) \csc \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2}{d \left (a \sin \left (c+d \sqrt {x}\right )+b\right )^2}+\frac {b^2 x^{5/2} \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d \sqrt {x}}{2}\right ) \sin ^2\left (c+d \sqrt {x}\right ) \sec \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2}{d \left (a \sin \left (c+d \sqrt {x}\right )+b\right )^2}-\frac {i b \sin ^2\left (c+d \sqrt {x}\right ) \left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2 \left (i \left (4 a d^5 x^{5/2} \log \left (1-e^{i \left (c+d \sqrt {x}\right )}\right )-4 a d^5 x^{5/2} \log \left (1+e^{i \left (c+d \sqrt {x}\right )}\right )-80 a d^3 x^{3/2} \text {Li}_3\left (-e^{i \left (c+d \sqrt {x}\right )}\right )+80 a d^3 x^{3/2} \text {Li}_3\left (e^{i \left (c+d \sqrt {x}\right )}\right )-240 i a d^2 x \text {Li}_4\left (-e^{i \left (c+d \sqrt {x}\right )}\right )+240 i a d^2 x \text {Li}_4\left (e^{i \left (c+d \sqrt {x}\right )}\right )+480 a d \sqrt {x} \text {Li}_5\left (-e^{i \left (c+d \sqrt {x}\right )}\right )-480 a d \sqrt {x} \text {Li}_5\left (e^{i \left (c+d \sqrt {x}\right )}\right )+480 i a \text {Li}_6\left (-e^{i \left (c+d \sqrt {x}\right )}\right )-480 i a \text {Li}_6\left (e^{i \left (c+d \sqrt {x}\right )}\right )+10 b d^4 x^2 \log \left (1-e^{2 i \left (c+d \sqrt {x}\right )}\right )-20 i b d^3 x^{3/2} \text {Li}_2\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )+30 b d^2 x \text {Li}_3\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )+30 i b d \sqrt {x} \text {Li}_4\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )-15 b \text {Li}_5\left (e^{2 i \left (c+d \sqrt {x}\right )}\right )\right )-20 a d^4 x^2 \text {Li}_2\left (-e^{i \left (c+d \sqrt {x}\right )}\right )+20 a d^4 x^2 \text {Li}_2\left (e^{i \left (c+d \sqrt {x}\right )}\right )+\frac {4 b e^{2 i c} d^5 x^{5/2}}{-1+e^{2 i c}}\right )}{d^6 \left (a \sin \left (c+d \sqrt {x}\right )+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*Csc[c + d*Sqrt[x]])^2,x]

[Out]

(a^2*x^3*(a + b*Csc[c + d*Sqrt[x]])^2*Sin[c + d*Sqrt[x]]^2)/(3*(b + a*Sin[c + d*Sqrt[x]])^2) - (I*b*(a + b*Csc
[c + d*Sqrt[x]])^2*((4*b*d^5*E^((2*I)*c)*x^(5/2))/(-1 + E^((2*I)*c)) - 20*a*d^4*x^2*PolyLog[2, -E^(I*(c + d*Sq
rt[x]))] + 20*a*d^4*x^2*PolyLog[2, E^(I*(c + d*Sqrt[x]))] + I*(4*a*d^5*x^(5/2)*Log[1 - E^(I*(c + d*Sqrt[x]))]
- 4*a*d^5*x^(5/2)*Log[1 + E^(I*(c + d*Sqrt[x]))] + 10*b*d^4*x^2*Log[1 - E^((2*I)*(c + d*Sqrt[x]))] - (20*I)*b*
d^3*x^(3/2)*PolyLog[2, E^((2*I)*(c + d*Sqrt[x]))] - 80*a*d^3*x^(3/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))] + 80*a
*d^3*x^(3/2)*PolyLog[3, E^(I*(c + d*Sqrt[x]))] + 30*b*d^2*x*PolyLog[3, E^((2*I)*(c + d*Sqrt[x]))] - (240*I)*a*
d^2*x*PolyLog[4, -E^(I*(c + d*Sqrt[x]))] + (240*I)*a*d^2*x*PolyLog[4, E^(I*(c + d*Sqrt[x]))] + (30*I)*b*d*Sqrt
[x]*PolyLog[4, E^((2*I)*(c + d*Sqrt[x]))] + 480*a*d*Sqrt[x]*PolyLog[5, -E^(I*(c + d*Sqrt[x]))] - 480*a*d*Sqrt[
x]*PolyLog[5, E^(I*(c + d*Sqrt[x]))] - 15*b*PolyLog[5, E^((2*I)*(c + d*Sqrt[x]))] + (480*I)*a*PolyLog[6, -E^(I
*(c + d*Sqrt[x]))] - (480*I)*a*PolyLog[6, E^(I*(c + d*Sqrt[x]))]))*Sin[c + d*Sqrt[x]]^2)/(d^6*(b + a*Sin[c + d
*Sqrt[x]])^2) + (b^2*x^(5/2)*Csc[c/2]*Csc[c/2 + (d*Sqrt[x])/2]*(a + b*Csc[c + d*Sqrt[x]])^2*Sin[c + d*Sqrt[x]]
^2*Sin[(d*Sqrt[x])/2])/(d*(b + a*Sin[c + d*Sqrt[x]])^2) + (b^2*x^(5/2)*(a + b*Csc[c + d*Sqrt[x]])^2*Sec[c/2]*S
ec[c/2 + (d*Sqrt[x])/2]*Sin[c + d*Sqrt[x]]^2*Sin[(d*Sqrt[x])/2])/(d*(b + a*Sin[c + d*Sqrt[x]])^2)

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{2} \csc \left (d \sqrt {x} + c\right )^{2} + 2 \, a b x^{2} \csc \left (d \sqrt {x} + c\right ) + a^{2} x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*csc(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^2*csc(d*sqrt(x) + c)^2 + 2*a*b*x^2*csc(d*sqrt(x) + c) + a^2*x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \csc \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*csc(c+d*x^(1/2)))^2,x, algorithm="giac")

[Out]

integrate((b*csc(d*sqrt(x) + c) + a)^2*x^2, x)

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maple [F]  time = 4.13, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +b \csc \left (c +d \sqrt {x}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*csc(c+d*x^(1/2)))^2,x)

[Out]

int(x^2*(a+b*csc(c+d*x^(1/2)))^2,x)

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maxima [B]  time = 1.60, size = 3856, normalized size = 7.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*csc(c+d*x^(1/2)))^2,x, algorithm="maxima")

[Out]

1/3*((d*sqrt(x) + c)^6*a^2 - 6*(d*sqrt(x) + c)^5*a^2*c + 15*(d*sqrt(x) + c)^4*a^2*c^2 - 20*(d*sqrt(x) + c)^3*a
^2*c^3 + 15*(d*sqrt(x) + c)^2*a^2*c^4 - 6*(d*sqrt(x) + c)*a^2*c^5 + 12*a*b*c^5*log(cot(d*sqrt(x) + c) + csc(d*
sqrt(x) + c)) + 6*(4*b^2*c^5 + (4*(d*sqrt(x) + c)^5*a*b - 10*b^2*c^4 - 10*(2*a*b*c + b^2)*(d*sqrt(x) + c)^4 +
40*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^3 - 20*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 20*(a*b*c^4 + 2*b^2*c^
3)*(d*sqrt(x) + c) - 2*(2*(d*sqrt(x) + c)^5*a*b - 5*b^2*c^4 - 5*(2*a*b*c + b^2)*(d*sqrt(x) + c)^4 + 20*(a*b*c^
2 + b^2*c)*(d*sqrt(x) + c)^3 - 10*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 10*(a*b*c^4 + 2*b^2*c^3)*(d*sqrt
(x) + c))*cos(2*d*sqrt(x) + 2*c) - (4*I*(d*sqrt(x) + c)^5*a*b - 10*I*b^2*c^4 + (-20*I*a*b*c - 10*I*b^2)*(d*sqr
t(x) + c)^4 + (40*I*a*b*c^2 + 40*I*b^2*c)*(d*sqrt(x) + c)^3 + (-40*I*a*b*c^3 - 60*I*b^2*c^2)*(d*sqrt(x) + c)^2
+ (20*I*a*b*c^4 + 40*I*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(d*sqrt(x) + c), cos(d*sq
rt(x) + c) + 1) + (10*b^2*c^4*cos(2*d*sqrt(x) + 2*c) + 10*I*b^2*c^4*sin(2*d*sqrt(x) + 2*c) - 10*b^2*c^4)*arcta
n2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) - 1) + (4*(d*sqrt(x) + c)^5*a*b - 10*(2*a*b*c - b^2)*(d*sqrt(x) + c)
^4 + 40*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c)^3 - 20*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 20*(a*b*c^4 - 2*b
^2*c^3)*(d*sqrt(x) + c) - 2*(2*(d*sqrt(x) + c)^5*a*b - 5*(2*a*b*c - b^2)*(d*sqrt(x) + c)^4 + 20*(a*b*c^2 - b^2
*c)*(d*sqrt(x) + c)^3 - 10*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 10*(a*b*c^4 - 2*b^2*c^3)*(d*sqrt(x) + c
))*cos(2*d*sqrt(x) + 2*c) - (4*I*(d*sqrt(x) + c)^5*a*b + (-20*I*a*b*c + 10*I*b^2)*(d*sqrt(x) + c)^4 + (40*I*a*
b*c^2 - 40*I*b^2*c)*(d*sqrt(x) + c)^3 + (-40*I*a*b*c^3 + 60*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (20*I*a*b*c^4 - 40*
I*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(d*sqrt(x) + c), -cos(d*sqrt(x) + c) + 1) - 4*(
(d*sqrt(x) + c)^5*b^2 - 5*(d*sqrt(x) + c)^4*b^2*c + 10*(d*sqrt(x) + c)^3*b^2*c^2 - 10*(d*sqrt(x) + c)^2*b^2*c^
3 + 5*(d*sqrt(x) + c)*b^2*c^4)*cos(2*d*sqrt(x) + 2*c) - (20*(d*sqrt(x) + c)^4*a*b + 20*a*b*c^4 + 40*b^2*c^3 -
40*(2*a*b*c + b^2)*(d*sqrt(x) + c)^3 + 120*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 40*(2*a*b*c^3 + 3*b^2*c^2)*(d
*sqrt(x) + c) - 20*((d*sqrt(x) + c)^4*a*b + a*b*c^4 + 2*b^2*c^3 - 2*(2*a*b*c + b^2)*(d*sqrt(x) + c)^3 + 6*(a*b
*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 2*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (-20*I*(
d*sqrt(x) + c)^4*a*b - 20*I*a*b*c^4 - 40*I*b^2*c^3 + (80*I*a*b*c + 40*I*b^2)*(d*sqrt(x) + c)^3 + (-120*I*a*b*c
^2 - 120*I*b^2*c)*(d*sqrt(x) + c)^2 + (80*I*a*b*c^3 + 120*I*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*
dilog(-e^(I*d*sqrt(x) + I*c)) + (20*(d*sqrt(x) + c)^4*a*b + 20*a*b*c^4 - 40*b^2*c^3 - 40*(2*a*b*c - b^2)*(d*sq
rt(x) + c)^3 + 120*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c)^2 - 40*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c) - 20*((d*s
qrt(x) + c)^4*a*b + a*b*c^4 - 2*b^2*c^3 - 2*(2*a*b*c - b^2)*(d*sqrt(x) + c)^3 + 6*(a*b*c^2 - b^2*c)*(d*sqrt(x)
+ c)^2 - 2*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (20*I*(d*sqrt(x) + c)^4*a*b + 20
*I*a*b*c^4 - 40*I*b^2*c^3 + (-80*I*a*b*c + 40*I*b^2)*(d*sqrt(x) + c)^3 + (120*I*a*b*c^2 - 120*I*b^2*c)*(d*sqrt
(x) + c)^2 + (-80*I*a*b*c^3 + 120*I*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*dilog(e^(I*d*sqrt(x) + I
*c)) - (2*I*(d*sqrt(x) + c)^5*a*b - 5*I*b^2*c^4 + (-10*I*a*b*c - 5*I*b^2)*(d*sqrt(x) + c)^4 + (20*I*a*b*c^2 +
20*I*b^2*c)*(d*sqrt(x) + c)^3 + (-20*I*a*b*c^3 - 30*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (10*I*a*b*c^4 + 20*I*b^2*c^
3)*(d*sqrt(x) + c) + (-2*I*(d*sqrt(x) + c)^5*a*b + 5*I*b^2*c^4 + (10*I*a*b*c + 5*I*b^2)*(d*sqrt(x) + c)^4 + (-
20*I*a*b*c^2 - 20*I*b^2*c)*(d*sqrt(x) + c)^3 + (20*I*a*b*c^3 + 30*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (-10*I*a*b*c^
4 - 20*I*b^2*c^3)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (2*(d*sqrt(x) + c)^5*a*b - 5*b^2*c^4 - 5*(2*a*b*c
+ b^2)*(d*sqrt(x) + c)^4 + 20*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^3 - 10*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)
^2 + 10*(a*b*c^4 + 2*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x
) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) - (-2*I*(d*sqrt(x) + c)^5*a*b - 5*I*b^2*c^4 + (10*I*a*b*c - 5*I*b^2)*(d*s
qrt(x) + c)^4 + (-20*I*a*b*c^2 + 20*I*b^2*c)*(d*sqrt(x) + c)^3 + (20*I*a*b*c^3 - 30*I*b^2*c^2)*(d*sqrt(x) + c)
^2 + (-10*I*a*b*c^4 + 20*I*b^2*c^3)*(d*sqrt(x) + c) + (2*I*(d*sqrt(x) + c)^5*a*b + 5*I*b^2*c^4 + (-10*I*a*b*c
+ 5*I*b^2)*(d*sqrt(x) + c)^4 + (20*I*a*b*c^2 - 20*I*b^2*c)*(d*sqrt(x) + c)^3 + (-20*I*a*b*c^3 + 30*I*b^2*c^2)*
(d*sqrt(x) + c)^2 + (10*I*a*b*c^4 - 20*I*b^2*c^3)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (2*(d*sqrt(x) + c)
^5*a*b + 5*b^2*c^4 - 5*(2*a*b*c - b^2)*(d*sqrt(x) + c)^4 + 20*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c)^3 - 10*(2*a*b*
c^3 - 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 10*(a*b*c^4 - 2*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*log(cos
(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) + 480*(a*b*cos(2*d*sqrt(x) + 2*c) + I*a*b
*sin(2*d*sqrt(x) + 2*c) - a*b)*polylog(6, -e^(I*d*sqrt(x) + I*c)) - 480*(a*b*cos(2*d*sqrt(x) + 2*c) + I*a*b*si
n(2*d*sqrt(x) + 2*c) - a*b)*polylog(6, e^(I*d*sqrt(x) + I*c)) - (-480*I*(d*sqrt(x) + c)*a*b + 480*I*a*b*c + 24
0*I*b^2 + (480*I*(d*sqrt(x) + c)*a*b - 480*I*a*b*c - 240*I*b^2)*cos(2*d*sqrt(x) + 2*c) - 240*(2*(d*sqrt(x) + c
)*a*b - 2*a*b*c - b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(5, -e^(I*d*sqrt(x) + I*c)) - (480*I*(d*sqrt(x) + c)*a*b
- 480*I*a*b*c + 240*I*b^2 + (-480*I*(d*sqrt(x) + c)*a*b + 480*I*a*b*c - 240*I*b^2)*cos(2*d*sqrt(x) + 2*c) + 2
40*(2*(d*sqrt(x) + c)*a*b - 2*a*b*c + b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(5, e^(I*d*sqrt(x) + I*c)) + (240*(d
*sqrt(x) + c)^2*a*b + 240*a*b*c^2 + 240*b^2*c - 240*(2*a*b*c + b^2)*(d*sqrt(x) + c) - 240*((d*sqrt(x) + c)^2*a
*b + a*b*c^2 + b^2*c - (2*a*b*c + b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (240*I*(d*sqrt(x) + c)^2*a*b
+ 240*I*a*b*c^2 + 240*I*b^2*c + (-480*I*a*b*c - 240*I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*polylog(4,
-e^(I*d*sqrt(x) + I*c)) - (240*(d*sqrt(x) + c)^2*a*b + 240*a*b*c^2 - 240*b^2*c - 240*(2*a*b*c - b^2)*(d*sqrt(
x) + c) - 240*((d*sqrt(x) + c)^2*a*b + a*b*c^2 - b^2*c - (2*a*b*c - b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*
c) + (-240*I*(d*sqrt(x) + c)^2*a*b - 240*I*a*b*c^2 + 240*I*b^2*c + (480*I*a*b*c - 240*I*b^2)*(d*sqrt(x) + c))*
sin(2*d*sqrt(x) + 2*c))*polylog(4, e^(I*d*sqrt(x) + I*c)) - (80*I*(d*sqrt(x) + c)^3*a*b - 80*I*a*b*c^3 - 120*I
*b^2*c^2 + (-240*I*a*b*c - 120*I*b^2)*(d*sqrt(x) + c)^2 + (240*I*a*b*c^2 + 240*I*b^2*c)*(d*sqrt(x) + c) + (-80
*I*(d*sqrt(x) + c)^3*a*b + 80*I*a*b*c^3 + 120*I*b^2*c^2 + (240*I*a*b*c + 120*I*b^2)*(d*sqrt(x) + c)^2 + (-240*
I*a*b*c^2 - 240*I*b^2*c)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + 40*(2*(d*sqrt(x) + c)^3*a*b - 2*a*b*c^3 - 3
*b^2*c^2 - 3*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 6*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*
polylog(3, -e^(I*d*sqrt(x) + I*c)) - (-80*I*(d*sqrt(x) + c)^3*a*b + 80*I*a*b*c^3 - 120*I*b^2*c^2 + (240*I*a*b*
c - 120*I*b^2)*(d*sqrt(x) + c)^2 + (-240*I*a*b*c^2 + 240*I*b^2*c)*(d*sqrt(x) + c) + (80*I*(d*sqrt(x) + c)^3*a*
b - 80*I*a*b*c^3 + 120*I*b^2*c^2 + (-240*I*a*b*c + 120*I*b^2)*(d*sqrt(x) + c)^2 + (240*I*a*b*c^2 - 240*I*b^2*c
)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - 40*(2*(d*sqrt(x) + c)^3*a*b - 2*a*b*c^3 + 3*b^2*c^2 - 3*(2*a*b*c -
b^2)*(d*sqrt(x) + c)^2 + 6*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*polylog(3, e^(I*d*sqrt(
x) + I*c)) - (4*I*(d*sqrt(x) + c)^5*b^2 - 20*I*(d*sqrt(x) + c)^4*b^2*c + 40*I*(d*sqrt(x) + c)^3*b^2*c^2 - 40*I
*(d*sqrt(x) + c)^2*b^2*c^3 + 20*I*(d*sqrt(x) + c)*b^2*c^4)*sin(2*d*sqrt(x) + 2*c))/(-2*I*cos(2*d*sqrt(x) + 2*c
) + 2*sin(2*d*sqrt(x) + 2*c) + 2*I))/d^6

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a+\frac {b}{\sin \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b/sin(c + d*x^(1/2)))^2,x)

[Out]

int(x^2*(a + b/sin(c + d*x^(1/2)))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \csc {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*csc(c+d*x**(1/2)))**2,x)

[Out]

Integral(x**2*(a + b*csc(c + d*sqrt(x)))**2, x)

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