3.176 \(\int \frac {x^2}{\sqrt {a-a \cos (x)}} \, dx\)

Optimal. Leaf size=163 \[ \frac {8 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {16 \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {16 \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {4 x^2 \sin \left (\frac {x}{2}\right ) \tanh ^{-1}\left (e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}} \]

[Out]

-4*x^2*arctanh(exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)+8*I*x*polylog(2,-exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(
x))^(1/2)-8*I*x*polylog(2,exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)-16*polylog(3,-exp(1/2*I*x))*sin(1/2*x)/(
a-a*cos(x))^(1/2)+16*polylog(3,exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3319, 4183, 2531, 2282, 6589} \[ \frac {8 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {16 \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {16 \text {Li}_3\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {4 x^2 \sin \left (\frac {x}{2}\right ) \tanh ^{-1}\left (e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a - a*Cos[x]],x]

[Out]

(-4*x^2*ArcTanh[E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] + ((8*I)*x*PolyLog[2, -E^((I/2)*x)]*Sin[x/2])/Sqrt[a
- a*Cos[x]] - ((8*I)*x*PolyLog[2, E^((I/2)*x)]*Sin[x/2])/Sqrt[a - a*Cos[x]] - (16*PolyLog[3, -E^((I/2)*x)]*Si
n[x/2])/Sqrt[a - a*Cos[x]] + (16*PolyLog[3, 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]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {a-a \cos (x)}} \, dx &=\frac {\sin \left (\frac {x}{2}\right ) \int x^2 \csc \left (\frac {x}{2}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (4 \sin \left (\frac {x}{2}\right )\right ) \int x \log \left (1-e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}+\frac {\left (4 \sin \left (\frac {x}{2}\right )\right ) \int x \log \left (1+e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (8 i \sin \left (\frac {x}{2}\right )\right ) \int \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}+\frac {\left (8 i \sin \left (\frac {x}{2}\right )\right ) \int \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \, dx}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {\left (16 \sin \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}}+\frac {\left (16 \sin \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{\sqrt {a-a \cos (x)}}\\ &=-\frac {4 x^2 \tanh ^{-1}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {8 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {8 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {16 \text {Li}_3\left (-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {16 \text {Li}_3\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 = 0.05, size = 117, normalized size = 0.72 \[ \frac {2 \sin \left (\frac {x}{2}\right ) \left (4 i x \text {Li}_2\left (-e^{\frac {i x}{2}}\right )-4 i x \text {Li}_2\left (e^{\frac {i x}{2}}\right )-8 \text {Li}_3\left (-e^{\frac {i x}{2}}\right )+8 \text {Li}_3\left (e^{\frac {i x}{2}}\right )+x^2 \log \left (1-e^{\frac {i x}{2}}\right )-x^2 \log \left (1+e^{\frac {i x}{2}}\right )\right )}{\sqrt {a-a \cos (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a - a*Cos[x]],x]

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a \cos \relax (x) + a} x^{2}}{a \cos \relax (x) - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a*cos(x) + a)*x^2/(a*cos(x) - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-a \cos \relax (x) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-a*cos(x) + a), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {a -a \cos \relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {-a \cos \relax (x) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(-a*cos(x) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\sqrt {a-a\,\cos \relax (x)}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a - a*cos(x))^(1/2),x)

[Out]

int(x^2/(a - a*cos(x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {- a \left (\cos {\relax (x )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/sqrt(-a*(cos(x) - 1)), x)

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