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

Optimal. Leaf size=362 \[ -\frac{x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{i \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}+\frac{i \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \sqrt{a^2-b^2}}-\frac{i x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \sqrt{a^2-b^2}}+\frac{i x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \sqrt{a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.878822, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3379, 3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac{x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^2 \sqrt{a^2-b^2}}+\frac{x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^2 \sqrt{a^2-b^2}}-\frac{i \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{d^3 \sqrt{a^2-b^2}}+\frac{i \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{d^3 \sqrt{a^2-b^2}}-\frac{i x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )}{2 d \sqrt{a^2-b^2}}+\frac{i x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )}{2 d \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sin[c + d*x^2]),x]

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.209812, size = 289, normalized size = 0.8 \[ \frac{-i \left (2 i d x^2 \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )+2 \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{a-\sqrt{a^2-b^2}}\right )-2 \text{PolyLog}\left (3,\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )+d^2 x^4 \log \left (1+\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )-d^2 x^4 \log \left (1-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}+a}\right )\right )-2 d x^2 \text{PolyLog}\left (2,-\frac{i b e^{i \left (c+d x^2\right )}}{\sqrt{a^2-b^2}-a}\right )}{2 d^3 \sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{a+b\sin \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [C]  time = 3.32103, size = 3376, normalized size = 9.33 \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)),x, algorithm="fricas")

[Out]

-1/8*(-4*I*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*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*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*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))*sq
rt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - 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*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*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*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*(b*d^2*x^4 - b*c^2)*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*(b*d^2*x^4 - b*c^2)*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*(b*d^2*x^4 - b*c^2)*s
qrt(-(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*(b*d^2*x^4 - b*c^2)*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) + 4
*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*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*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*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))/((a^2 - b^2)*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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