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*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 = 0.56, 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 = {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 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.92, 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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fricas [C]  time = 0.75, size = 1074, normalized size = 2.80 \[ \text {result too large to display} \]

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,
-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))/b) - 12*I*b*x*sqrt((a^2 - b
^2)/b^2)*polylog(3, -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))/b) - 12
*I*b*x*sqrt((a^2 - b^2)/b^2)*polylog(3, -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))/b) + 12*I*b*x*sqrt((a^2 - b^2)/b^2)*polylog(3, -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))/b) - 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -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))/b) + 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -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))/b) - 12*b*sqrt((a^2 - b^2)/b^2)*polylog(4, -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))/b) + 12*b*sqrt((a^2 - b^2)/b^
2)*polylog(4, -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))/b))/(a^2 - b^
2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{b \cos \relax (x) + a}\,{d x} \]

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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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{a +b \cos \relax (x )}\, dx \]

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 \[ \text {Exception raised: ValueError} \]

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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{a+b\,\cos \relax (x)} \,d x \]

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{a + b \cos {\relax (x )}}\, dx \]

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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