3.89 \(\int \frac{x^5}{(a+b \sin (c+d x^3))^2} \, dx\)

Optimal. Leaf size=324 \[ -\frac{a \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^3\right )}}{a-\sqrt{a^2-b^2}}\right )}{3 d^2 \left (a^2-b^2\right )^{3/2}}+\frac{a \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{3 d^2 \left (a^2-b^2\right )^{3/2}}-\frac{\log \left (a+b \sin \left (c+d x^3\right )\right )}{3 d^2 \left (a^2-b^2\right )}-\frac{i a x^3 \log \left (1-\frac{i b e^{i \left (c+d x^3\right )}}{a-\sqrt{a^2-b^2}}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac{i a x^3 \log \left (1-\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac{b x^3 \cos \left (c+d x^3\right )}{3 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.59335, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3379, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac{a \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^3\right )}}{a-\sqrt{a^2-b^2}}\right )}{3 d^2 \left (a^2-b^2\right )^{3/2}}+\frac{a \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{3 d^2 \left (a^2-b^2\right )^{3/2}}-\frac{\log \left (a+b \sin \left (c+d x^3\right )\right )}{3 d^2 \left (a^2-b^2\right )}-\frac{i a x^3 \log \left (1-\frac{i b e^{i \left (c+d x^3\right )}}{a-\sqrt{a^2-b^2}}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac{i a x^3 \log \left (1-\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{3 d \left (a^2-b^2\right )^{3/2}}+\frac{b x^3 \cos \left (c+d x^3\right )}{3 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sin[c + d*x^3])^2,x]

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.993256, size = 302, normalized size = 0.93 \[ \frac{-\frac{a \text{PolyLog}\left (2,-\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}-a}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a \text{PolyLog}\left (2,\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{i a d x^3 \log \left (1+\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}-a}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{i a d x^3 \log \left (1-\frac{i b e^{i \left (c+d x^3\right )}}{\sqrt{a^2-b^2}+a}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{\log \left (a+b \sin \left (c+d x^3\right )\right )}{a^2-b^2}+\frac{b d x^3 \cos \left (c+d x^3\right )}{\left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*Sin[c + d*x^3])^2,x]

[Out]

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

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Maple [F]  time = 0.773, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{ \left ( a+b\sin \left ( d{x}^{3}+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^3+c))^2,x)

[Out]

int(x^5/(a+b*sin(d*x^3+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^3+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.28416, size = 3438, normalized size = 10.61 \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^3+c))^2,x, algorithm="fricas")

[Out]

1/6*(2*(a^2*b - b^3)*d*x^3*cos(d*x^3 + c) + (I*a*b^2*sin(d*x^3 + c) + I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1
/2*(2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) + 2*(b*cos(d*x^3 + c) - I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^
2) + 2*b)/b + 1) + (-I*a*b^2*sin(d*x^3 + c) - I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x^3 + c)
+ 2*a*sin(d*x^3 + c) - 2*(b*cos(d*x^3 + c) - I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-I*a
*b^2*sin(d*x^3 + c) - I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) +
2*(b*cos(d*x^3 + c) + I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (I*a*b^2*sin(d*x^3 + c) + I*
a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) - 2*(b*cos(d*x^3 + c) + I
*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (a^2*b*d*x^3 + a^2*b*c + (a*b^2*d*x^3 + a*b^2*c)*sin
(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) + 2*(b*cos(d*x^3 + c) -
I*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*x^3 + a^2*b*c + (a*b^2*d*x^3 + a*b^2*c)*sin(d
*x^3 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) - 2*(b*cos(d*x^3 + c) - I
*b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (a^2*b*d*x^3 + a^2*b*c + (a*b^2*d*x^3 + a*b^2*c)*sin(d*x
^3 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) + 2*(b*cos(d*x^3 + c) + I*
b*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*x^3 + a^2*b*c + (a*b^2*d*x^3 + a*b^2*c)*sin(d*x^
3 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x^3 + c) + 2*a*sin(d*x^3 + c) - 2*(b*cos(d*x^3 + c) + I*b
*sin(d*x^3 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x^3 + c) + (a*b^2*c*sin
(d*x^3 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(d*x^3 + c) + 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2
- b^2)/b^2) + 2*I*a) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x^3 + c) + (a*b^2*c*sin(d*x^3 + c) + a^2*b*c)*sqrt(-
(a^2 - b^2)/b^2))*log(2*b*cos(d*x^3 + c) - 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - (a^3 -
a*b^2 + (a^2*b - b^3)*sin(d*x^3 + c) - (a*b^2*c*sin(d*x^3 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*co
s(d*x^3 + c) + 2*I*b*sin(d*x^3 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d
*x^3 + c) - (a*b^2*c*sin(d*x^3 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(d*x^3 + c) - 2*I*b*sin(d*x
^3 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a))/((a^4*b - 2*a^2*b^3 + b^5)*d^2*sin(d*x^3 + c) + (a^5 - 2*a^3*b^
2 + a*b^4)*d^2)

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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**3+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^{3} + 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^3+c))^2,x, algorithm="giac")

[Out]

integrate(x^5/(b*sin(d*x^3 + c) + a)^2, x)