3.175 \(\int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx\)

Optimal. Leaf size=235 \[ \frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {96 i \text {Li}_4\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {96 i \text {Li}_4\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {4 x^3 \sin \left (\frac {x}{2}\right ) \tanh ^{-1}\left (e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}} \]

[Out]

-4*x^3*arctanh(exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)+12*I*x^2*polylog(2,-exp(1/2*I*x))*sin(1/2*x)/(a-a*c
os(x))^(1/2)-12*I*x^2*polylog(2,exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)-48*x*polylog(3,-exp(1/2*I*x))*sin(
1/2*x)/(a-a*cos(x))^(1/2)+48*x*polylog(3,exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)-96*I*polylog(4,-exp(1/2*I
*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)+96*I*polylog(4,exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3319, 4183, 2531, 6609, 2282, 6589} \[ \frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {96 i \text {Li}_4\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {96 i \text {Li}_4\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {4 x^3 \sin \left (\frac {x}{2}\right ) \tanh ^{-1}\left (e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a - a*Cos[x]],x]

[Out]

(-4*x^3*ArcTanh[E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] + ((12*I)*x^2*PolyLog[2, -E^((I/2)*x)]*Sin[x/2])/Sqr
t[a - a*Cos[x]] - ((12*I)*x^2*PolyLog[2, E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] - (48*x*PolyLog[3, -E^((I/2
)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] + (48*x*PolyLog[3, E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] - ((96*I)*Poly
Log[4, -E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] + ((96*I)*PolyLog[4, E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x
]]

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 3319

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((2*a)^IntPart[n
]*(a + b*Sin[e + f*x])^FracPart[n])/Sin[e/2 + (a*Pi)/(4*b) + (f*x)/2]^(2*FracPart[n]), Int[(c + d*x)^m*Sin[e/2
+ (a*Pi)/(4*b) + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[n
+ 1/2] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && 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}{\sqrt {a-a \cos (x)}} \, dx &=\frac {\sin \left (\frac {x}{2}\right ) \int x^3 \csc \left (\frac {x}{2}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (6 \sin \left (\frac {x}{2}\right )\right ) \int x^2 \log \left (1-e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}+\frac {\left (6 \sin \left (\frac {x}{2}\right )\right ) \int x^2 \log \left (1+e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (24 i \sin \left (\frac {x}{2}\right )\right ) \int x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}+\frac {\left (24 i \sin \left (\frac {x}{2}\right )\right ) \int x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {\left (48 \sin \left (\frac {x}{2}\right )\right ) \int \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}-\frac {\left (48 \sin \left (\frac {x}{2}\right )\right ) \int \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (96 i \sin \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}}+\frac {\left (96 i \sin \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^3 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {12 i x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {96 i \text {Li}_4\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {96 i \text {Li}_4\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 170, normalized size = 0.72 \[ -\frac {i \sin \left (\frac {x}{2}\right ) \left (-48 x^2 \text {Li}_2\left (e^{-\frac {i x}{2}}\right )-48 x^2 \text {Li}_2\left (-e^{\frac {i x}{2}}\right )+192 i x \text {Li}_3\left (e^{-\frac {i x}{2}}\right )-192 i x \text {Li}_3\left (-e^{\frac {i x}{2}}\right )+384 \text {Li}_4\left (e^{-\frac {i x}{2}}\right )+384 \text {Li}_4\left (-e^{\frac {i x}{2}}\right )-x^4+8 i x^3 \log \left (1-e^{-\frac {i x}{2}}\right )-8 i x^3 \log \left (1+e^{\frac {i x}{2}}\right )+8 \pi ^4\right )}{4 \sqrt {a-a \cos (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a - a*Cos[x]],x]

[Out]

((-1/4*I)*(8*Pi^4 - x^4 + (8*I)*x^3*Log[1 - E^((-1/2*I)*x)] - (8*I)*x^3*Log[1 + E^((I/2)*x)] - 48*x^2*PolyLog[
2, E^((-1/2*I)*x)] - 48*x^2*PolyLog[2, -E^((I/2)*x)] + (192*I)*x*PolyLog[3, E^((-1/2*I)*x)] - (192*I)*x*PolyLo
g[3, -E^((I/2)*x)] + 384*PolyLog[4, E^((-1/2*I)*x)] + 384*PolyLog[4, -E^((I/2)*x)])*Sin[x/2])/Sqrt[a - a*Cos[x
]]

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a \cos \relax (x) + a} x^{3}}{a \cos \relax (x) - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a-a*cos(x))^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a*cos(x) + a)*x^3/(a*cos(x) - a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a-a*cos(x))^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/sqrt(-a*cos(x) + a), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {a -a \cos \relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a-a*cos(x))^(1/2),x)

[Out]

int(x^3/(a-a*cos(x))^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a-a*cos(x))^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/sqrt(-a*cos(x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a - a*cos(x))^(1/2),x)

[Out]

int(x^3/(a - a*cos(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\sqrt {- a \left (\cos {\relax (x )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a-a*cos(x))**(1/2),x)

[Out]

Integral(x**3/sqrt(-a*(cos(x) - 1)), x)

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