3.185 \(\int \frac{x^3}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=383 \[ -\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{\sqrt{a^2-b^2}+a}\right )}{\sqrt{a^2-b^2}} \]

[Out]

((-I)*x^3*Log[1 + (b*E^(I*x))/(a - Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2] + (I*x^3*Log[1 + (b*E^(I*x))/(a + Sqrt[a
^2 - b^2])])/Sqrt[a^2 - b^2] - (3*x^2*PolyLog[2, -((b*E^(I*x))/(a - Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] + (3*x
^2*PolyLog[2, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] - ((6*I)*x*PolyLog[3, -((b*E^(I*x))/(a -
Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] + ((6*I)*x*PolyLog[3, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^
2] + (6*PolyLog[4, -((b*E^(I*x))/(a - Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] - (6*PolyLog[4, -((b*E^(I*x))/(a + S
qrt[a^2 - b^2]))])/Sqrt[a^2 - b^2]

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Rubi [A]  time = 0.558972, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {3321, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{\sqrt{a^2-b^2}+a}\right )}{\sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Cos[x]),x]

[Out]

((-I)*x^3*Log[1 + (b*E^(I*x))/(a - Sqrt[a^2 - b^2])])/Sqrt[a^2 - b^2] + (I*x^3*Log[1 + (b*E^(I*x))/(a + Sqrt[a
^2 - b^2])])/Sqrt[a^2 - b^2] - (3*x^2*PolyLog[2, -((b*E^(I*x))/(a - Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] + (3*x
^2*PolyLog[2, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] - ((6*I)*x*PolyLog[3, -((b*E^(I*x))/(a -
Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] + ((6*I)*x*PolyLog[3, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^
2] + (6*PolyLog[4, -((b*E^(I*x))/(a - Sqrt[a^2 - b^2]))])/Sqrt[a^2 - b^2] - (6*PolyLog[4, -((b*E^(I*x))/(a + S
qrt[a^2 - b^2]))])/Sqrt[a^2 - b^2]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && 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 6609

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]

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^3}{a+b \cos (x)} \, dx &=2 \int \frac{e^{i x} x^3}{b+2 a e^{i x}+b e^{2 i x}} \, dx\\ &=\frac{(2 b) \int \frac{e^{i x} x^3}{2 a-2 \sqrt{a^2-b^2}+2 b e^{i x}} \, dx}{\sqrt{a^2-b^2}}-\frac{(2 b) \int \frac{e^{i x} x^3}{2 a+2 \sqrt{a^2-b^2}+2 b e^{i x}} \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{(3 i) \int x^2 \log \left (1+\frac{2 b e^{i x}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}-\frac{(3 i) \int x^2 \log \left (1+\frac{2 b e^{i x}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 \int x \text{Li}_2\left (-\frac{2 b e^{i x}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}-\frac{6 \int x \text{Li}_2\left (-\frac{2 b e^{i x}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{(6 i) \int \text{Li}_3\left (-\frac{2 b e^{i x}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}-\frac{(6 i) \int \text{Li}_3\left (-\frac{2 b e^{i x}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{b x}{-a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i x}\right )}{\sqrt{a^2-b^2}}-\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i x}\right )}{\sqrt{a^2-b^2}}\\ &=-\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{i x^3 \log \left (1+\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{6 \text{Li}_4\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}\\ \end{align*}

Mathematica [A]  time = 0.898692, size = 290, normalized size = 0.76 \[ \frac{-3 x^2 \text{Li}_2\left (\frac{b e^{i x}}{\sqrt{a^2-b^2}-a}\right )+3 x^2 \text{Li}_2\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )-6 i x \text{Li}_3\left (\frac{b e^{i x}}{\sqrt{a^2-b^2}-a}\right )+6 i x \text{Li}_3\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )+6 \text{Li}_4\left (\frac{b e^{i x}}{\sqrt{a^2-b^2}-a}\right )-6 \text{Li}_4\left (-\frac{b e^{i x}}{a+\sqrt{a^2-b^2}}\right )-i x^3 \log \left (1+\frac{b e^{i x}}{a-\sqrt{a^2-b^2}}\right )+i x^3 \log \left (1+\frac{b e^{i x}}{\sqrt{a^2-b^2}+a}\right )}{\sqrt{a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b*Cos[x]),x]

[Out]

((-I)*x^3*Log[1 + (b*E^(I*x))/(a - Sqrt[a^2 - b^2])] + I*x^3*Log[1 + (b*E^(I*x))/(a + Sqrt[a^2 - b^2])] - 3*x^
2*PolyLog[2, (b*E^(I*x))/(-a + Sqrt[a^2 - b^2])] + 3*x^2*PolyLog[2, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))] - (6
*I)*x*PolyLog[3, (b*E^(I*x))/(-a + Sqrt[a^2 - b^2])] + (6*I)*x*PolyLog[3, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))
] + 6*PolyLog[4, (b*E^(I*x))/(-a + Sqrt[a^2 - b^2])] - 6*PolyLog[4, -((b*E^(I*x))/(a + Sqrt[a^2 - b^2]))])/Sqr
t[a^2 - b^2]

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Maple [F]  time = 0.231, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{a+b\cos \left ( x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+b*cos(x)),x)

[Out]

int(x^3/(a+b*cos(x)),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^3/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.53652, size = 2719, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*cos(x)),x, algorithm="fricas")

[Out]

-1/4*(2*I*b*x^3*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(x) + 2*I*a*sin(x) + 2*(b*cos(x) + I*b*sin(x))*sqrt((a^2
- b^2)/b^2) + 2*b)/b) - 2*I*b*x^3*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(x) + 2*I*a*sin(x) - 2*(b*cos(x) + I*
b*sin(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) - 2*I*b*x^3*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*cos(x) - 2*I*a*sin(x)
+ 2*(b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 2*I*b*x^3*sqrt((a^2 - b^2)/b^2)*log(1/2*(2*a*co
s(x) - 2*I*a*sin(x) - 2*(b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b) + 6*b*x^2*sqrt((a^2 - b^2)/b^2
)*dilog(-1/2*(2*a*cos(x) + 2*I*a*sin(x) + 2*(b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 6*b*
x^2*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(x) + 2*I*a*sin(x) - 2*(b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b
^2) + 2*b)/b + 1) + 6*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(x) - 2*I*a*sin(x) + 2*(b*cos(x) - I*b*si
n(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) - 6*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-1/2*(2*a*cos(x) - 2*I*a*sin(x
) - 2*(b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + 2*b)/b + 1) + 12*I*b*x*sqrt((a^2 - b^2)/b^2)*polylog(3,
-(a*cos(x) + I*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b) - 12*I*b*x*sqrt((a^2 - b^2)/b^2)*p
olylog(3, -(a*cos(x) + I*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b) - 12*I*b*x*sqrt((a^2 - b
^2)/b^2)*polylog(3, -(a*cos(x) - I*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b) + 12*I*b*x*sqr
t((a^2 - b^2)/b^2)*polylog(3, -(a*cos(x) - I*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b) - 12
*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cos(x) + I*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b
) + 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cos(x) + I*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b
^2))/b) - 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cos(x) - I*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt((a^2 -
b^2)/b^2))/b) + 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cos(x) - I*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt((
a^2 - b^2)/b^2))/b))/(a^2 - b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+b*cos(x)),x)

[Out]

Integral(x**3/(a + b*cos(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{b \cos \left (x\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*cos(x)),x, algorithm="giac")

[Out]

integrate(x^3/(b*cos(x) + a), x)