3.43 \(\int \frac{x^5}{(a+b \sin (c+d x^2))^2} \, dx\)

Optimal. Leaf size=663 \[ -\frac{a x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{a x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{i \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}+\frac{i \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )}-\frac{i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}+\frac{i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}-\frac{x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}-\frac{x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i a x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac{i a x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac{b x^4 \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )}+\frac{i x^4}{2 d \left (a^2-b^2\right )} \]

[Out]

((I/2)*x^4)/((a^2 - b^2)*d) - (x^2*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^2) -
((I/2)*a*x^4*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (x^2*Log[1 - (I*
b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^2) + ((I/2)*a*x^4*Log[1 - (I*b*E^(I*(c + d*x^2)))/
(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + (I*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/
((a^2 - b^2)*d^3) - (a*x^2*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2)
+ (I*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^3) + (a*x^2*PolyLog[2, (I*b*E^(
I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (I*a*PolyLog[3, (I*b*E^(I*(c + d*x^2)))/(a -
Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^3) + (I*a*PolyLog[3, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/
((a^2 - b^2)^(3/2)*d^3) + (b*x^4*Cos[c + d*x^2])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x^2]))

________________________________________________________________________________________

Rubi [A]  time = 1.30189, antiderivative size = 663, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {3379, 3324, 3323, 2264, 2190, 2531, 2282, 6589, 4519, 2279, 2391} \[ -\frac{a x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{a x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac{i \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )}+\frac{i \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )}-\frac{i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}+\frac{i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \left (a^2-b^2\right )^{3/2}}-\frac{x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )}-\frac{x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \left (a^2-b^2\right )}-\frac{i a x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac{i a x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac{b x^4 \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )}+\frac{i x^4}{2 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sin[c + d*x^2])^2,x]

[Out]

((I/2)*x^4)/((a^2 - b^2)*d) - (x^2*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^2) -
((I/2)*a*x^4*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - (x^2*Log[1 - (I*
b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^2) + ((I/2)*a*x^4*Log[1 - (I*b*E^(I*(c + d*x^2)))/
(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + (I*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/
((a^2 - b^2)*d^3) - (a*x^2*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2)
+ (I*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)*d^3) + (a*x^2*PolyLog[2, (I*b*E^(
I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^2) - (I*a*PolyLog[3, (I*b*E^(I*(c + d*x^2)))/(a -
Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d^3) + (I*a*PolyLog[3, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/
((a^2 - b^2)^(3/2)*d^3) + (b*x^4*Cos[c + d*x^2])/(2*(a^2 - b^2)*d*(a + b*Sin[c + d*x^2]))

Rule 3379

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

Rule 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 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 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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

Mathematica [A]  time = 2.3089, size = 513, normalized size = 0.77 \[ \frac{\left (-\frac{2 a d x^2}{\sqrt{a^2-b^2}}+2 i\right ) \text{PolyLog}\left (2,-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )+\left (\frac{2 a d x^2}{\sqrt{a^2-b^2}}+2 i\right ) \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )-\frac{2 i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{2 i a \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{\sqrt{a^2-b^2}}-\frac{i a d^2 x^4 \log \left (1+\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )}{\sqrt{a^2-b^2}}+\frac{i a d^2 x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{\sqrt{a^2-b^2}}-2 d x^2 \log \left (1+\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )-2 d x^2 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )+\frac{b d^2 x^4 \cos \left (c+d x^2\right )}{a+b \sin \left (c+d x^2\right )}+i d^2 x^4}{2 d^3 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*d^2*x^4 - 2*d*x^2*Log[1 + (I*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])] - (I*a*d^2*x^4*Log[1 + (I*b*E^(I*
(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2] - 2*d*x^2*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2
- b^2])] + (I*a*d^2*x^4*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2] + (2*I - (2*a
*d*x^2)/Sqrt[a^2 - b^2])*PolyLog[2, ((-I)*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])] + (2*I + (2*a*d*x^2)/Sq
rt[a^2 - b^2])*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])] - ((2*I)*a*PolyLog[3, (I*b*E^(I*(c +
d*x^2)))/(a - Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2] + ((2*I)*a*PolyLog[3, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 -
b^2])])/Sqrt[a^2 - b^2] + (b*d^2*x^4*Cos[c + d*x^2])/(a + b*Sin[c + d*x^2]))/(2*(a^2 - b^2)*d^3)

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Maple [F]  time = 0.669, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{ \left ( a+b\sin \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*sin(d*x^2+c))^2,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 4.06118, size = 5577, normalized size = 8.41 \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*sin(d*x^2+c))^2,x, algorithm="fricas")

[Out]

1/8*(4*(a^2*b - b^3)*d^2*x^4*cos(d*x^2 + c) - 4*(a*b^2*sin(d*x^2 + c) + a^2*b)*sqrt(-(a^2 - b^2)/b^2)*polylog(
3, 1/2*(2*I*a*cos(d*x^2 + c) - 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2
)/b^2))/b) + 4*(a*b^2*sin(d*x^2 + c) + a^2*b)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(2*I*a*cos(d*x^2 + c) - 2*
a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 4*(a*b^2*sin(d*x^2 +
c) + a^2*b)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x^2 + c) + a*sin(d*x^2 + c) + (b*cos(d*x^2 + c) - I
*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 4*(a*b^2*sin(d*x^2 + c) + a^2*b)*sqrt(-(a^2 - b^2)/b^2)*polylo
g(3, -(I*a*cos(d*x^2 + c) + a*sin(d*x^2 + c) - (b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))
/b) + (-4*I*a^3 + 4*I*a*b^2 + (-4*I*a^2*b + 4*I*b^3)*sin(d*x^2 + c) + 2*(2*I*a*b^2*d*x^2*sin(d*x^2 + c) + 2*I*
a^2*b*d*x^2)*sqrt(-(a^2 - b^2)/b^2))*dilog(-1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 +
c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-4*I*a^3 + 4*I*a*b^2 + (-4*I*a^2*b + 4*I*b^3)
*sin(d*x^2 + c) + 2*(-2*I*a*b^2*d*x^2*sin(d*x^2 + c) - 2*I*a^2*b*d*x^2)*sqrt(-(a^2 - b^2)/b^2))*dilog(-1/2*(2*
I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2
*b)/b + 1) + (4*I*a^3 - 4*I*a*b^2 + (4*I*a^2*b - 4*I*b^3)*sin(d*x^2 + c) + 2*(-2*I*a*b^2*d*x^2*sin(d*x^2 + c)
- 2*I*a^2*b*d*x^2)*sqrt(-(a^2 - b^2)/b^2))*dilog(-1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d
*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (4*I*a^3 - 4*I*a*b^2 + (4*I*a^2*b - 4*I
*b^3)*sin(d*x^2 + c) + 2*(2*I*a*b^2*d*x^2*sin(d*x^2 + c) + 2*I*a^2*b*d*x^2)*sqrt(-(a^2 - b^2)/b^2))*dilog(-1/2
*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2
) + 2*b)/b + 1) + 2*(2*(a^2*b - b^3)*c*sin(d*x^2 + c) + 2*(a^3 - a*b^2)*c + (a*b^2*c^2*sin(d*x^2 + c) + a^2*b*
c^2)*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*
a) + 2*(2*(a^2*b - b^3)*c*sin(d*x^2 + c) + 2*(a^3 - a*b^2)*c + (a*b^2*c^2*sin(d*x^2 + c) + a^2*b*c^2)*sqrt(-(a
^2 - b^2)/b^2))*log(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*(2*(a^
2*b - b^3)*c*sin(d*x^2 + c) + 2*(a^3 - a*b^2)*c - (a*b^2*c^2*sin(d*x^2 + c) + a^2*b*c^2)*sqrt(-(a^2 - b^2)/b^2
))*log(-2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*(2*(a^2*b - b^3)*c
*sin(d*x^2 + c) + 2*(a^3 - a*b^2)*c - (a*b^2*c^2*sin(d*x^2 + c) + a^2*b*c^2)*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*
cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - 2*(2*(a^3 - a*b^2)*d*x^2 + 2*(a^
3 - a*b^2)*c + 2*((a^2*b - b^3)*d*x^2 + (a^2*b - b^3)*c)*sin(d*x^2 + c) - (a^2*b*d^2*x^4 - a^2*b*c^2 + (a*b^2*
d^2*x^4 - a*b^2*c^2)*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c
) + 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(2*(a^3 - a*b^2)*d*x^2 + 2*
(a^3 - a*b^2)*c + 2*((a^2*b - b^3)*d*x^2 + (a^2*b - b^3)*c)*sin(d*x^2 + c) + (a^2*b*d^2*x^4 - a^2*b*c^2 + (a*b
^2*d^2*x^4 - a*b^2*c^2)*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2
+ c) - 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(2*(a^3 - a*b^2)*d*x^2 +
2*(a^3 - a*b^2)*c + 2*((a^2*b - b^3)*d*x^2 + (a^2*b - b^3)*c)*sin(d*x^2 + c) - (a^2*b*d^2*x^4 - a^2*b*c^2 + (
a*b^2*d^2*x^4 - a*b^2*c^2)*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))*log(1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*
x^2 + c) + 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(2*(a^3 - a*b^2)*d*x
^2 + 2*(a^3 - a*b^2)*c + 2*((a^2*b - b^3)*d*x^2 + (a^2*b - b^3)*c)*sin(d*x^2 + c) + (a^2*b*d^2*x^4 - a^2*b*c^2
+ (a*b^2*d^2*x^4 - a*b^2*c^2)*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2))*log(1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*si
n(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b))/((a^4*b - 2*a^2*b^3
+ b^5)*d^3*sin(d*x^2 + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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