3.1.56 \(\int x^{3/2} (a+b \sec (c+d \sqrt {x}))^2 \, dx\) [56]

Optimal. Leaf size=451 \[ -\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \text {ArcTan}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 i b^2 \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d} \]

[Out]

-96*I*a*b*polylog(4,-I*exp(I*(c+d*x^(1/2))))*x^(1/2)/d^4+2/5*a^2*x^(5/2)+16*I*a*b*x^(3/2)*polylog(2,-I*exp(I*(
c+d*x^(1/2))))/d^2+8*b^2*x^(3/2)*ln(1+exp(2*I*(c+d*x^(1/2))))/d^2-8*I*a*b*x^2*arctan(exp(I*(c+d*x^(1/2))))/d-1
2*I*b^2*x*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^3+6*I*b^2*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^5-48*a*b*x*polyl
og(3,-I*exp(I*(c+d*x^(1/2))))/d^3+48*a*b*x*polylog(3,I*exp(I*(c+d*x^(1/2))))/d^3+96*I*a*b*polylog(4,I*exp(I*(c
+d*x^(1/2))))*x^(1/2)/d^4+96*a*b*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5-96*a*b*polylog(5,I*exp(I*(c+d*x^(1/2))
))/d^5+12*b^2*polylog(3,-exp(2*I*(c+d*x^(1/2))))*x^(1/2)/d^4-2*I*b^2*x^2/d-16*I*a*b*x^(3/2)*polylog(2,I*exp(I*
(c+d*x^(1/2))))/d^2+2*b^2*x^2*tan(c+d*x^(1/2))/d

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 451, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {4289, 4275, 4266, 2611, 6744, 2320, 6724, 4269, 3800, 2221} \begin {gather*} \frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \text {ArcTan}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {96 a b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {6 i b^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {12 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}-\frac {2 i b^2 x^2}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*Sec[c + d*Sqrt[x]])^2,x]

[Out]

((-2*I)*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - ((8*I)*a*b*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (8*b^2*x^(3/2)*Log[
1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((16*I)*a*b*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((16*I
)*a*b*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((12*I)*b^2*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])
/d^3 - (48*a*b*x*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (48*a*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/
d^3 + (12*b^2*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((96*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(I*(
c + d*Sqrt[x]))])/d^4 + ((96*I)*a*b*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((6*I)*b^2*PolyLog[4, -
E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (96*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (96*a*b*PolyLog[5, I*E
^(I*(c + d*Sqrt[x]))])/d^5 + (2*b^2*x^2*Tan[c + d*Sqrt[x]])/d

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4266

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

Rule 4269

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 4275

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 4289

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[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 6724

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 6744

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^{3/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \text {Subst}\left (\int x^4 (a+b \sec (c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (a^2 x^4+2 a b x^4 \sec (c+d x)+b^2 x^4 \sec ^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}+(4 a b) \text {Subst}\left (\int x^4 \sec (c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^4 \sec ^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d}-\frac {\left (8 b^2\right ) \text {Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}-\frac {(48 i a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {(48 i a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (16 i b^2\right ) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^3}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (24 b^2\right ) \text {Subst}\left (\int x^2 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(96 i a b) \text {Subst}\left (\int \text {Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(96 i a b) \text {Subst}\left (\int \text {Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}+\frac {\left (24 i b^2\right ) \text {Subst}\left (\int x \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}+\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 a b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}+\frac {\left (6 i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ &=-\frac {2 i b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {16 i a b x^{3/2} \text {Li}_2\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 i a b x^{3/2} \text {Li}_2\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {12 i b^2 x \text {Li}_2\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {12 b^2 \sqrt {x} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 i a b \sqrt {x} \text {Li}_4\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {96 i a b \sqrt {x} \text {Li}_4\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {6 i b^2 \text {Li}_4\left (-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {Li}_5\left (-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {2 b^2 x^2 \tan \left (c+d \sqrt {x}\right )}{d}\\ \end {align*}

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Mathematica [A]
time = 1.38, size = 443, normalized size = 0.98 \begin {gather*} \frac {2 \left (-5 i b^2 d^4 x^2+a^2 d^5 x^{5/2}-20 i a b d^4 x^2 \text {ArcTan}\left (e^{i \left (c+d \sqrt {x}\right )}\right )+20 b^2 d^3 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )+40 i a b d^3 x^{3/2} \text {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )-40 i a b d^3 x^{3/2} \text {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )-30 i b^2 d^2 x \text {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-120 a b d^2 x \text {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )+120 a b d^2 x \text {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )+30 b^2 d \sqrt {x} \text {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )-240 i a b d \sqrt {x} \text {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )+240 i a b d \sqrt {x} \text {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )+15 i b^2 \text {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )+240 a b \text {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )-240 a b \text {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )+5 b^2 d^4 x^2 \tan \left (c+d \sqrt {x}\right )\right )}{5 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(a + b*Sec[c + d*Sqrt[x]])^2,x]

[Out]

(2*((-5*I)*b^2*d^4*x^2 + a^2*d^5*x^(5/2) - (20*I)*a*b*d^4*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))] + 20*b^2*d^3*x^(3/
2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + (40*I)*a*b*d^3*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (40*I)
*a*b*d^3*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (30*I)*b^2*d^2*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))]
- 120*a*b*d^2*x*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))] + 120*a*b*d^2*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] +
30*b^2*d*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (240*I)*a*b*d*Sqrt[x]*PolyLog[4, (-I)*E^(I*(c + d*Sq
rt[x]))] + (240*I)*a*b*d*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + (15*I)*b^2*PolyLog[4, -E^((2*I)*(c + d*
Sqrt[x]))] + 240*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))] - 240*a*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))] + 5*
b^2*d^4*x^2*Tan[c + d*Sqrt[x]]))/(5*d^5)

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Maple [F]
time = 0.72, size = 0, normalized size = 0.00 \[\int x^{\frac {3}{2}} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x)

[Out]

int(x^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2869 vs. \(2 (346) = 692\).
time = 0.70, size = 2869, normalized size = 6.36 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="maxima")

[Out]

2/5*((d*sqrt(x) + c)^5*a^2 - 5*(d*sqrt(x) + c)^4*a^2*c + 10*(d*sqrt(x) + c)^3*a^2*c^2 - 10*(d*sqrt(x) + c)^2*a
^2*c^3 + 5*(d*sqrt(x) + c)*a^2*c^4 + 10*a*b*c^4*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 5*(6*b^2*c^4 -
6*((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3
+ ((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^
3)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a*b*c + 6*I*(d*sqrt(x) + c)^2*a*b
*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + c) + 1
) - 6*((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b
*c^3 + ((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*
b*c^3)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a*b*c + 6*I*(d*sqrt(x) + c)^2
*a*b*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c
) + 1) + 4*(4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c)*b^2*c^2 - 3*b^2*c^3 + (4*(
d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c)*b^2*c^2 - 3*b^2*c^3)*cos(2*d*sqrt(x) + 2*
c) - (-4*I*(d*sqrt(x) + c)^3*b^2 + 9*I*(d*sqrt(x) + c)^2*b^2*c - 9*I*(d*sqrt(x) + c)*b^2*c^2 + 3*I*b^2*c^3)*si
n(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) - 6*((d*sqrt(x) + c)^4*b^2 -
4*(d*sqrt(x) + c)^3*b^2*c + 6*(d*sqrt(x) + c)^2*b^2*c^2 - 4*(d*sqrt(x) + c)*b^2*c^3)*cos(2*d*sqrt(x) + 2*c) -
6*(4*(d*sqrt(x) + c)^2*b^2 - 6*(d*sqrt(x) + c)*b^2*c + 3*b^2*c^2 + (4*(d*sqrt(x) + c)^2*b^2 - 6*(d*sqrt(x) +
c)*b^2*c + 3*b^2*c^2)*cos(2*d*sqrt(x) + 2*c) + (4*I*(d*sqrt(x) + c)^2*b^2 - 6*I*(d*sqrt(x) + c)*b^2*c + 3*I*b^
2*c^2)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - 24*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) +
c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 - a*b*c^3 + ((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*s
qrt(x) + c)*a*b*c^2 - a*b*c^3)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)^3*a*b - 3*I*(d*sqrt(x) + c)^2*a*b*c
+ 3*I*(d*sqrt(x) + c)*a*b*c^2 - I*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*dilog(I*e^(I*d*sqrt(x) + I*c)) + 24*((d*sq
rt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 - a*b*c^3 + ((d*sqrt(x) + c)^3*a*b -
3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 - a*b*c^3)*cos(2*d*sqrt(x) + 2*c) - (-I*(d*sqrt(x) + c)^
3*a*b + 3*I*(d*sqrt(x) + c)^2*a*b*c - 3*I*(d*sqrt(x) + c)*a*b*c^2 + I*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*dilog(-
I*e^(I*d*sqrt(x) + I*c)) - 2*(4*I*(d*sqrt(x) + c)^3*b^2 - 9*I*(d*sqrt(x) + c)^2*b^2*c + 9*I*(d*sqrt(x) + c)*b^
2*c^2 - 3*I*b^2*c^3 + (4*I*(d*sqrt(x) + c)^3*b^2 - 9*I*(d*sqrt(x) + c)^2*b^2*c + 9*I*(d*sqrt(x) + c)*b^2*c^2 -
3*I*b^2*c^3)*cos(2*d*sqrt(x) + 2*c) - (4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c
)*b^2*c^2 - 3*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))*log(cos(2*d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x) + 2*c)^2 + 2*cos
(2*d*sqrt(x) + 2*c) + 1) - 3*(I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a*b*c + 6*I*(d*sqrt(x) + c)^2*a*
b*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3 + (I*(d*sqrt(x) + c)^4*a*b - 4*I*(d*sqrt(x) + c)^3*a*b*c + 6*I*(d*sqrt(x)
+ c)^2*a*b*c^2 - 4*I*(d*sqrt(x) + c)*a*b*c^3)*cos(2*d*sqrt(x) + 2*c) - ((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) +
c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(
x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*sin(d*sqrt(x) + c) + 1) - 3*(-I*(d*sqrt(x) + c)^4*a*b + 4*I*(d*sqrt(x) +
c)^3*a*b*c - 6*I*(d*sqrt(x) + c)^2*a*b*c^2 + 4*I*(d*sqrt(x) + c)*a*b*c^3 + (-I*(d*sqrt(x) + c)^4*a*b + 4*I*(d*
sqrt(x) + c)^3*a*b*c - 6*I*(d*sqrt(x) + c)^2*a*b*c^2 + 4*I*(d*sqrt(x) + c)*a*b*c^3)*cos(2*d*sqrt(x) + 2*c) + (
(d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3)*s
in(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) - 144*(-I*a
*b*cos(2*d*sqrt(x) + 2*c) + a*b*sin(2*d*sqrt(x) + 2*c) - I*a*b)*polylog(5, I*e^(I*d*sqrt(x) + I*c)) - 144*(I*a
*b*cos(2*d*sqrt(x) + 2*c) - a*b*sin(2*d*sqrt(x) + 2*c) + I*a*b)*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) + 12*(b^2
*cos(2*d*sqrt(x) + 2*c) + I*b^2*sin(2*d*sqrt(x) + 2*c) + b^2)*polylog(4, -e^(2*I*d*sqrt(x) + 2*I*c)) + 144*((d
*sqrt(x) + c)*a*b - a*b*c + ((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) - (-I*(d*sqrt(x) + c)*a*b + I
*a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(4, I*e^(I*d*sqrt(x) + I*c)) - 144*((d*sqrt(x) + c)*a*b - a*b*c + ((d*s
qrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) + (I*(d*sqrt(x) + c)*a*b - I*a*b*c)*sin(2*d*sqrt(x) + 2*c))*po
lylog(4, -I*e^(I*d*sqrt(x) + I*c)) - 6*(4*I*(d*sqrt(x) + c)*b^2 - 3*I*b^2*c + (4*I*(d*sqrt(x) + c)*b^2 - 3*I*b
^2*c)*cos(2*d*sqrt(x) + 2*c) - (4*(d*sqrt(x) + c)*b^2 - 3*b^2*c)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -e^(2*I*d*
sqrt(x) + 2*I*c)) - 72*(I*(d*sqrt(x) + c)^2*a*b...

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^(3/2)*sec(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*sec(d*sqrt(x) + c) + a^2*x^(3/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {3}{2}} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(a+b*sec(c+d*x**(1/2)))**2,x)

[Out]

Integral(x**(3/2)*(a + b*sec(c + d*sqrt(x)))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="giac")

[Out]

integrate((b*sec(d*sqrt(x) + c) + a)^2*x^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{3/2}\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b/cos(c + d*x^(1/2)))^2,x)

[Out]

int(x^(3/2)*(a + b/cos(c + d*x^(1/2)))^2, x)

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