3.2.85 \(\int \frac {x^3}{a+b \cos (x)} \, dx\) [185]

Optimal. Leaf size=383 \[ -\frac {i x^3 \log \left (1+\frac {b e^{i x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {i x^3 \log \left (1+\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {3 x^2 \text {PolyLog}\left (2,-\frac {b e^{i x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {3 x^2 \text {PolyLog}\left (2,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {6 i x \text {PolyLog}\left (3,-\frac {b e^{i x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {6 i x \text {PolyLog}\left (3,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {6 \text {PolyLog}\left (4,-\frac {b e^{i x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {6 \text {PolyLog}\left (4,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \]

[Out]

-I*x^3*ln(1+b*exp(I*x)/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)+I*x^3*ln(1+b*exp(I*x)/(a+(a^2-b^2)^(1/2)))/(a^2-b^
2)^(1/2)-3*x^2*polylog(2,-b*exp(I*x)/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)+3*x^2*polylog(2,-b*exp(I*x)/(a+(a^2-
b^2)^(1/2)))/(a^2-b^2)^(1/2)-6*I*x*polylog(3,-b*exp(I*x)/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)+6*I*x*polylog(3,
-b*exp(I*x)/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)+6*polylog(4,-b*exp(I*x)/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)-
6*polylog(4,-b*exp(I*x)/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(1/2)

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Rubi [A]
time = 0.37, antiderivative size = 383, normalized size of antiderivative = 1.00, 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 = {3402, 2296, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\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}} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3402

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 6724

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 6744

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]

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 \text {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 \text {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*}

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Mathematica [A]
time = 0.56, size = 290, normalized size = 0.76 \begin {gather*} \frac {-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}}{a+\sqrt {a^2-b^2}}\right )-3 x^2 \text {PolyLog}\left (2,\frac {b e^{i x}}{-a+\sqrt {a^2-b^2}}\right )+3 x^2 \text {PolyLog}\left (2,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )-6 i x \text {PolyLog}\left (3,\frac {b e^{i x}}{-a+\sqrt {a^2-b^2}}\right )+6 i x \text {PolyLog}\left (3,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )+6 \text {PolyLog}\left (4,\frac {b e^{i x}}{-a+\sqrt {a^2-b^2}}\right )-6 \text {PolyLog}\left (4,-\frac {b e^{i x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}} \end {gather*}

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 = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{3}}{a +b \cos \left (x \right )}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1030 vs. \(2 (317) = 634\).
time = 0.48, size = 1030, normalized size = 2.69 \begin {gather*} \frac {-i \, b x^{3} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) + i \, b x^{3} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) + i \, b x^{3} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) - i \, b x^{3} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} \log \left (\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b}\right ) - 3 \, b x^{2} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) + 3 \, b x^{2} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) - 3 \, b x^{2} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) + 3 \, b x^{2} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm Li}_2\left (-\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} + b}{b} + 1\right ) - 6 i \, b x \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 i \, b x \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 i \, b x \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 6 i \, b x \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (3, -\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (4, -\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 6 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (4, -\frac {a \cos \left (x\right ) + i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) + i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 6 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (4, -\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) + {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 6 \, b \sqrt {\frac {a^{2} - b^{2}}{b^{2}}} {\rm polylog}\left (4, -\frac {a \cos \left (x\right ) - i \, a \sin \left (x\right ) - {\left (b \cos \left (x\right ) - i \, b \sin \left (x\right )\right )} \sqrt {\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right )}{2 \, {\left (a^{2} - b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-I*b*x^3*sqrt((a^2 - b^2)/b^2)*log((a*cos(x) + I*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2)
+ b)/b) + I*b*x^3*sqrt((a^2 - b^2)/b^2)*log((a*cos(x) + I*a*sin(x) - (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)
/b^2) + b)/b) + I*b*x^3*sqrt((a^2 - b^2)/b^2)*log((a*cos(x) - I*a*sin(x) + (b*cos(x) - I*b*sin(x))*sqrt((a^2 -
b^2)/b^2) + b)/b) - I*b*x^3*sqrt((a^2 - b^2)/b^2)*log((a*cos(x) - I*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt((
a^2 - b^2)/b^2) + b)/b) - 3*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(x) + I*a*sin(x) + (b*cos(x) + I*b*sin(x)
)*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + 3*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(x) + I*a*sin(x) - (b*cos(x)
+ I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 3*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(x) - I*a*sin(x)
+ (b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) + 3*b*x^2*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cos(x) -
I*a*sin(x) - (b*cos(x) - I*b*sin(x))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - 6*I*b*x*sqrt((a^2 - b^2)/b^2)*polylo
g(3, -(a*cos(x) + I*a*sin(x) + (b*cos(x) + I*b*sin(x))*sqrt((a^2 - b^2)/b^2))/b) + 6*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) + 6*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) - 6*I*b*x*s
qrt((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) +
6*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) - 6*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) + 6*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) - 6*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3}{a+b\,\cos \left (x\right )} \,d x \end {gather*}

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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