3.8 \(\int x^5 (a+b \csc (c+d x^2))^2 \, dx\)

Optimal. Leaf size=228 \[ \frac{2 i a b x^2 \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 a b \text{PolyLog}\left (3,-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{PolyLog}\left (3,e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac{i b^2 \text{PolyLog}\left (2,e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 x^2 \log \left (1-e^{2 i \left (c+d x^2\right )}\right )}{d^2}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}-\frac{i b^2 x^4}{2 d} \]

[Out]

((-I/2)*b^2*x^4)/d + (a^2*x^6)/6 - (2*a*b*x^4*ArcTanh[E^(I*(c + d*x^2))])/d - (b^2*x^4*Cot[c + d*x^2])/(2*d) +
(b^2*x^2*Log[1 - E^((2*I)*(c + d*x^2))])/d^2 + ((2*I)*a*b*x^2*PolyLog[2, -E^(I*(c + d*x^2))])/d^2 - ((2*I)*a*
b*x^2*PolyLog[2, E^(I*(c + d*x^2))])/d^2 - ((I/2)*b^2*PolyLog[2, E^((2*I)*(c + d*x^2))])/d^3 - (2*a*b*PolyLog[
3, -E^(I*(c + d*x^2))])/d^3 + (2*a*b*PolyLog[3, E^(I*(c + d*x^2))])/d^3

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Rubi [A]  time = 0.368607, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {4205, 4190, 4183, 2531, 2282, 6589, 4184, 3717, 2190, 2279, 2391} \[ \frac{2 i a b x^2 \text{PolyLog}\left (2,-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{PolyLog}\left (2,e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 a b \text{PolyLog}\left (3,-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{PolyLog}\left (3,e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac{i b^2 \text{PolyLog}\left (2,e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 x^2 \log \left (1-e^{2 i \left (c+d x^2\right )}\right )}{d^2}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}-\frac{i b^2 x^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Csc[c + d*x^2])^2,x]

[Out]

((-I/2)*b^2*x^4)/d + (a^2*x^6)/6 - (2*a*b*x^4*ArcTanh[E^(I*(c + d*x^2))])/d - (b^2*x^4*Cot[c + d*x^2])/(2*d) +
(b^2*x^2*Log[1 - E^((2*I)*(c + d*x^2))])/d^2 + ((2*I)*a*b*x^2*PolyLog[2, -E^(I*(c + d*x^2))])/d^2 - ((2*I)*a*
b*x^2*PolyLog[2, E^(I*(c + d*x^2))])/d^2 - ((I/2)*b^2*PolyLog[2, E^((2*I)*(c + d*x^2))])/d^3 - (2*a*b*PolyLog[
3, -E^(I*(c + d*x^2))])/d^3 + (2*a*b*PolyLog[3, E^(I*(c + d*x^2))])/d^3

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 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 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 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 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 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 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 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 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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x^5 \left (a+b \csc \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b \csc (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x^2+2 a b x^2 \csc (c+d x)+b^2 x^2 \csc ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}+(a b) \operatorname{Subst}\left (\int x^2 \csc (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x^2 \csc ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}-\frac{(2 a b) \operatorname{Subst}\left (\int x \log \left (1-e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac{(2 a b) \operatorname{Subst}\left (\int x \log \left (1+e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int x \cot (c+d x) \, dx,x,x^2\right )}{d}\\ &=-\frac{i b^2 x^4}{2 d}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}+\frac{2 i a b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{(2 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}+\frac{(2 i a b) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}-\frac{\left (2 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x}{1-e^{2 i (c+d x)}} \, dx,x,x^2\right )}{d}\\ &=-\frac{i b^2 x^4}{2 d}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}+\frac{b^2 x^2 \log \left (1-e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac{b^2 \operatorname{Subst}\left (\int \log \left (1-e^{2 i (c+d x)}\right ) \, dx,x,x^2\right )}{d^2}\\ &=-\frac{i b^2 x^4}{2 d}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}+\frac{b^2 x^2 \log \left (1-e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 a b \text{Li}_3\left (-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{Li}_3\left (e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}\\ &=-\frac{i b^2 x^4}{2 d}+\frac{a^2 x^6}{6}-\frac{2 a b x^4 \tanh ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac{b^2 x^4 \cot \left (c+d x^2\right )}{2 d}+\frac{b^2 x^2 \log \left (1-e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac{2 i a b x^2 \text{Li}_2\left (-e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{Li}_2\left (e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i b^2 \text{Li}_2\left (e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}-\frac{2 a b \text{Li}_3\left (-e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{Li}_3\left (e^{i \left (c+d x^2\right )}\right )}{d^3}\\ \end{align*}

Mathematica [B]  time = 3.148, size = 639, normalized size = 2.8 \[ \frac{12 i b \left (-1+e^{2 i c}\right ) \left (b-2 a d x^2\right ) \text{PolyLog}\left (2,-e^{-i \left (c+d x^2\right )}\right )+12 i b \left (-1+e^{2 i c}\right ) \left (2 a d x^2+b\right ) \text{PolyLog}\left (2,e^{-i \left (c+d x^2\right )}\right )-24 a b e^{2 i c} \text{PolyLog}\left (3,-e^{-i \left (c+d x^2\right )}\right )+24 a b \text{PolyLog}\left (3,-e^{-i \left (c+d x^2\right )}\right )+24 a b e^{2 i c} \text{PolyLog}\left (3,e^{-i \left (c+d x^2\right )}\right )-24 a b \text{PolyLog}\left (3,e^{-i \left (c+d x^2\right )}\right )+2 a^2 e^{2 i c} d^3 x^6-2 a^2 d^3 x^6+12 a b e^{2 i c} d^2 x^4 \log \left (1-e^{-i \left (c+d x^2\right )}\right )-12 a b d^2 x^4 \log \left (1-e^{-i \left (c+d x^2\right )}\right )-12 a b e^{2 i c} d^2 x^4 \log \left (1+e^{-i \left (c+d x^2\right )}\right )+12 a b d^2 x^4 \log \left (1+e^{-i \left (c+d x^2\right )}\right )+3 b^2 e^{2 i c} d^2 x^4 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x^2}{2}\right ) \csc \left (\frac{1}{2} \left (c+d x^2\right )\right )-3 b^2 d^2 x^4 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x^2}{2}\right ) \csc \left (\frac{1}{2} \left (c+d x^2\right )\right )+3 b^2 e^{2 i c} d^2 x^4 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x^2}{2}\right ) \sec \left (\frac{1}{2} \left (c+d x^2\right )\right )-3 b^2 d^2 x^4 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x^2}{2}\right ) \sec \left (\frac{1}{2} \left (c+d x^2\right )\right )+12 b^2 e^{2 i c} d x^2 \log \left (1-e^{-i \left (c+d x^2\right )}\right )-12 b^2 d x^2 \log \left (1-e^{-i \left (c+d x^2\right )}\right )+12 b^2 e^{2 i c} d x^2 \log \left (1+e^{-i \left (c+d x^2\right )}\right )-12 b^2 d x^2 \log \left (1+e^{-i \left (c+d x^2\right )}\right )-12 i b^2 d^2 x^4}{12 \left (-1+e^{2 i c}\right ) d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*Csc[c + d*x^2])^2,x]

[Out]

((-12*I)*b^2*d^2*x^4 - 2*a^2*d^3*x^6 + 2*a^2*d^3*E^((2*I)*c)*x^6 - 12*b^2*d*x^2*Log[1 - E^((-I)*(c + d*x^2))]
+ 12*b^2*d*E^((2*I)*c)*x^2*Log[1 - E^((-I)*(c + d*x^2))] - 12*a*b*d^2*x^4*Log[1 - E^((-I)*(c + d*x^2))] + 12*a
*b*d^2*E^((2*I)*c)*x^4*Log[1 - E^((-I)*(c + d*x^2))] - 12*b^2*d*x^2*Log[1 + E^((-I)*(c + d*x^2))] + 12*b^2*d*E
^((2*I)*c)*x^2*Log[1 + E^((-I)*(c + d*x^2))] + 12*a*b*d^2*x^4*Log[1 + E^((-I)*(c + d*x^2))] - 12*a*b*d^2*E^((2
*I)*c)*x^4*Log[1 + E^((-I)*(c + d*x^2))] + (12*I)*b*(-1 + E^((2*I)*c))*(b - 2*a*d*x^2)*PolyLog[2, -E^((-I)*(c
+ d*x^2))] + (12*I)*b*(-1 + E^((2*I)*c))*(b + 2*a*d*x^2)*PolyLog[2, E^((-I)*(c + d*x^2))] + 24*a*b*PolyLog[3,
-E^((-I)*(c + d*x^2))] - 24*a*b*E^((2*I)*c)*PolyLog[3, -E^((-I)*(c + d*x^2))] - 24*a*b*PolyLog[3, E^((-I)*(c +
d*x^2))] + 24*a*b*E^((2*I)*c)*PolyLog[3, E^((-I)*(c + d*x^2))] - 3*b^2*d^2*x^4*Csc[c/2]*Csc[(c + d*x^2)/2]*Si
n[(d*x^2)/2] + 3*b^2*d^2*E^((2*I)*c)*x^4*Csc[c/2]*Csc[(c + d*x^2)/2]*Sin[(d*x^2)/2] - 3*b^2*d^2*x^4*Sec[c/2]*S
ec[(c + d*x^2)/2]*Sin[(d*x^2)/2] + 3*b^2*d^2*E^((2*I)*c)*x^4*Sec[c/2]*Sec[(c + d*x^2)/2]*Sin[(d*x^2)/2])/(12*d
^3*(-1 + E^((2*I)*c)))

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Maple [F]  time = 0.268, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \left ( a+b\csc \left ( d{x}^{2}+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*csc(d*x^2+c))^2,x)

[Out]

int(x^5*(a+b*csc(d*x^2+c))^2,x)

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Maxima [B]  time = 1.57885, size = 1087, normalized size = 4.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c))^2,x, algorithm="maxima")

[Out]

1/6*a^2*x^6 - (2*b^2*d^2*x^4*cos(2*d*x^2 + 2*c) + 2*I*b^2*d^2*x^4*sin(2*d*x^2 + 2*c) - (2*a*b*d^2*x^4 - 2*b^2*
d*x^2 - 2*(a*b*d^2*x^4 - b^2*d*x^2)*cos(2*d*x^2 + 2*c) - (2*I*a*b*d^2*x^4 - 2*I*b^2*d*x^2)*sin(2*d*x^2 + 2*c))
*arctan2(sin(d*x^2 + c), cos(d*x^2 + c) + 1) - (2*a*b*d^2*x^4 + 2*b^2*d*x^2 - 2*(a*b*d^2*x^4 + b^2*d*x^2)*cos(
2*d*x^2 + 2*c) - (2*I*a*b*d^2*x^4 + 2*I*b^2*d*x^2)*sin(2*d*x^2 + 2*c))*arctan2(sin(d*x^2 + c), -cos(d*x^2 + c)
+ 1) + (4*a*b*d*x^2 - 2*b^2 - 2*(2*a*b*d*x^2 - b^2)*cos(2*d*x^2 + 2*c) + (-4*I*a*b*d*x^2 + 2*I*b^2)*sin(2*d*x
^2 + 2*c))*dilog(-e^(I*d*x^2 + I*c)) - (4*a*b*d*x^2 + 2*b^2 - 2*(2*a*b*d*x^2 + b^2)*cos(2*d*x^2 + 2*c) - (4*I*
a*b*d*x^2 + 2*I*b^2)*sin(2*d*x^2 + 2*c))*dilog(e^(I*d*x^2 + I*c)) + (I*a*b*d^2*x^4 - I*b^2*d*x^2 + (-I*a*b*d^2
*x^4 + I*b^2*d*x^2)*cos(2*d*x^2 + 2*c) + (a*b*d^2*x^4 - b^2*d*x^2)*sin(2*d*x^2 + 2*c))*log(cos(d*x^2 + c)^2 +
sin(d*x^2 + c)^2 + 2*cos(d*x^2 + c) + 1) + (-I*a*b*d^2*x^4 - I*b^2*d*x^2 + (I*a*b*d^2*x^4 + I*b^2*d*x^2)*cos(2
*d*x^2 + 2*c) - (a*b*d^2*x^4 + b^2*d*x^2)*sin(2*d*x^2 + 2*c))*log(cos(d*x^2 + c)^2 + sin(d*x^2 + c)^2 - 2*cos(
d*x^2 + c) + 1) + (-4*I*a*b*cos(2*d*x^2 + 2*c) + 4*a*b*sin(2*d*x^2 + 2*c) + 4*I*a*b)*polylog(3, -e^(I*d*x^2 +
I*c)) + (4*I*a*b*cos(2*d*x^2 + 2*c) - 4*a*b*sin(2*d*x^2 + 2*c) - 4*I*a*b)*polylog(3, e^(I*d*x^2 + I*c)))/(-2*I
*d^3*cos(2*d*x^2 + 2*c) + 2*d^3*sin(2*d*x^2 + 2*c) + 2*I*d^3)

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Fricas [C]  time = 0.61137, size = 1723, normalized size = 7.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c))^2,x, algorithm="fricas")

[Out]

1/6*(a^2*d^3*x^6*sin(d*x^2 + c) - 3*b^2*d^2*x^4*cos(d*x^2 + c) + 6*a*b*polylog(3, cos(d*x^2 + c) + I*sin(d*x^2
+ c))*sin(d*x^2 + c) + 6*a*b*polylog(3, cos(d*x^2 + c) - I*sin(d*x^2 + c))*sin(d*x^2 + c) - 6*a*b*polylog(3,
-cos(d*x^2 + c) + I*sin(d*x^2 + c))*sin(d*x^2 + c) - 6*a*b*polylog(3, -cos(d*x^2 + c) - I*sin(d*x^2 + c))*sin(
d*x^2 + c) + (-6*I*a*b*d*x^2 - 3*I*b^2)*dilog(cos(d*x^2 + c) + I*sin(d*x^2 + c))*sin(d*x^2 + c) + (6*I*a*b*d*x
^2 + 3*I*b^2)*dilog(cos(d*x^2 + c) - I*sin(d*x^2 + c))*sin(d*x^2 + c) + (-6*I*a*b*d*x^2 + 3*I*b^2)*dilog(-cos(
d*x^2 + c) + I*sin(d*x^2 + c))*sin(d*x^2 + c) + (6*I*a*b*d*x^2 - 3*I*b^2)*dilog(-cos(d*x^2 + c) - I*sin(d*x^2
+ c))*sin(d*x^2 + c) - 3*(a*b*d^2*x^4 - b^2*d*x^2)*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + 1)*sin(d*x^2 + c) -
3*(a*b*d^2*x^4 - b^2*d*x^2)*log(cos(d*x^2 + c) - I*sin(d*x^2 + c) + 1)*sin(d*x^2 + c) + 3*(a*b*c^2 - b^2*c)*l
og(-1/2*cos(d*x^2 + c) + 1/2*I*sin(d*x^2 + c) + 1/2)*sin(d*x^2 + c) + 3*(a*b*c^2 - b^2*c)*log(-1/2*cos(d*x^2 +
c) - 1/2*I*sin(d*x^2 + c) + 1/2)*sin(d*x^2 + c) + 3*(a*b*d^2*x^4 + b^2*d*x^2 - a*b*c^2 + b^2*c)*log(-cos(d*x^
2 + c) + I*sin(d*x^2 + c) + 1)*sin(d*x^2 + c) + 3*(a*b*d^2*x^4 + b^2*d*x^2 - a*b*c^2 + b^2*c)*log(-cos(d*x^2 +
c) - I*sin(d*x^2 + c) + 1)*sin(d*x^2 + c))/(d^3*sin(d*x^2 + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (a + b \csc{\left (c + d x^{2} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*csc(d*x**2+c))**2,x)

[Out]

Integral(x**5*(a + b*csc(c + d*x**2))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \csc \left (d x^{2} + c\right ) + a\right )}^{2} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*csc(d*x^2+c))^2,x, algorithm="giac")

[Out]

integrate((b*csc(d*x^2 + c) + a)^2*x^5, x)