3.181 \(\int \frac {x^2}{(a+a \cos (x))^{3/2}} \, dx\)

Optimal. Leaf size=257 \[ \frac {2 i x \text {Li}_2\left (-i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {2 i x \text {Li}_2\left (i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {4 \text {Li}_3\left (-i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}+\frac {4 \text {Li}_3\left (i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {i x^2 \cos \left (\frac {x}{2}\right ) \tan ^{-1}\left (e^{\frac {i x}{2}}\right )}{a \sqrt {a \cos (x)+a}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a \cos (x)+a}}-\frac {2 x}{a \sqrt {a \cos (x)+a}}+\frac {4 \cos \left (\frac {x}{2}\right ) \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right )}{a \sqrt {a \cos (x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3319, 4186, 3770, 4181, 2531, 2282, 6589} \[ \frac {2 i x \text {Li}_2\left (-i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {2 i x \text {Li}_2\left (i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {4 \text {Li}_3\left (-i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}+\frac {4 \text {Li}_3\left (i e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a \cos (x)+a}}-\frac {i x^2 \cos \left (\frac {x}{2}\right ) \tan ^{-1}\left (e^{\frac {i x}{2}}\right )}{a \sqrt {a \cos (x)+a}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a \cos (x)+a}}-\frac {2 x}{a \sqrt {a \cos (x)+a}}+\frac {4 \cos \left (\frac {x}{2}\right ) \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right )}{a \sqrt {a \cos (x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + a*Cos[x])^(3/2),x]

[Out]

(-2*x)/(a*Sqrt[a + a*Cos[x]]) - (I*x^2*ArcTan[E^((I/2)*x)]*Cos[x/2])/(a*Sqrt[a + a*Cos[x]]) + (4*ArcTanh[Sin[x
/2]]*Cos[x/2])/(a*Sqrt[a + a*Cos[x]]) + ((2*I)*x*Cos[x/2]*PolyLog[2, (-I)*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]])
- ((2*I)*x*Cos[x/2]*PolyLog[2, I*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) - (4*Cos[x/2]*PolyLog[3, (-I)*E^((I/2)*
x)])/(a*Sqrt[a + a*Cos[x]]) + (4*Cos[x/2]*PolyLog[3, I*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) + (x^2*Tan[x/2])/(
2*a*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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

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}{(a+a \cos (x))^{3/2}} \, dx &=\frac {\cos \left (\frac {x}{2}\right ) \int x^2 \sec ^3\left (\frac {x}{2}\right ) \, dx}{2 a \sqrt {a+a \cos (x)}}\\ &=-\frac {2 x}{a \sqrt {a+a \cos (x)}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}+\frac {\cos \left (\frac {x}{2}\right ) \int x^2 \sec \left (\frac {x}{2}\right ) \, dx}{4 a \sqrt {a+a \cos (x)}}+\frac {\left (2 \cos \left (\frac {x}{2}\right )\right ) \int \sec \left (\frac {x}{2}\right ) \, dx}{a \sqrt {a+a \cos (x)}}\\ &=-\frac {2 x}{a \sqrt {a+a \cos (x)}}-\frac {i x^2 \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {4 \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}-\frac {\cos \left (\frac {x}{2}\right ) \int x \log \left (1-i e^{\frac {i x}{2}}\right ) \, dx}{a \sqrt {a+a \cos (x)}}+\frac {\cos \left (\frac {x}{2}\right ) \int x \log \left (1+i e^{\frac {i x}{2}}\right ) \, dx}{a \sqrt {a+a \cos (x)}}\\ &=-\frac {2 x}{a \sqrt {a+a \cos (x)}}-\frac {i x^2 \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {4 \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}-\frac {\left (2 i \cos \left (\frac {x}{2}\right )\right ) \int \text {Li}_2\left (-i e^{\frac {i x}{2}}\right ) \, dx}{a \sqrt {a+a \cos (x)}}+\frac {\left (2 i \cos \left (\frac {x}{2}\right )\right ) \int \text {Li}_2\left (i e^{\frac {i x}{2}}\right ) \, dx}{a \sqrt {a+a \cos (x)}}\\ &=-\frac {2 x}{a \sqrt {a+a \cos (x)}}-\frac {i x^2 \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {4 \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}-\frac {\left (4 \cos \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {\left (4 \cos \left (\frac {x}{2}\right )\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}\\ &=-\frac {2 x}{a \sqrt {a+a \cos (x)}}-\frac {i x^2 \tan ^{-1}\left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {4 \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {2 i x \cos \left (\frac {x}{2}\right ) \text {Li}_2\left (i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {4 \cos \left (\frac {x}{2}\right ) \text {Li}_3\left (-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {4 \cos \left (\frac {x}{2}\right ) \text {Li}_3\left (i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x^2 \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 185, normalized size = 0.72 \[ \frac {\cos \left (\frac {x}{2}\right ) \left (4 i x \text {Li}_2\left (-i e^{\frac {i x}{2}}\right ) \cos ^2\left (\frac {x}{2}\right )-4 i x \text {Li}_2\left (i e^{\frac {i x}{2}}\right ) \cos ^2\left (\frac {x}{2}\right )-8 \text {Li}_3\left (-i e^{\frac {i x}{2}}\right ) \cos ^2\left (\frac {x}{2}\right )+8 \text {Li}_3\left (i e^{\frac {i x}{2}}\right ) \cos ^2\left (\frac {x}{2}\right )+x^2 \sin \left (\frac {x}{2}\right )-2 i x^2 \cos ^2\left (\frac {x}{2}\right ) \tan ^{-1}\left (e^{\frac {i x}{2}}\right )-4 x \cos \left (\frac {x}{2}\right )+8 \cos ^2\left (\frac {x}{2}\right ) \tanh ^{-1}\left (\sin \left (\frac {x}{2}\right )\right )\right )}{(a (\cos (x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + a*Cos[x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cos \relax (x) + a} x^{2}}{a^{2} \cos \relax (x)^{2} + 2 \, a^{2} \cos \relax (x) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (a \cos \relax (x) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(a*cos(x) + a)^(3/2), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a +a \cos \relax (x )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2}{{\left (a+a\,\cos \relax (x)\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + a*cos(x))^(3/2),x)

[Out]

int(x^2/(a + a*cos(x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (a \left (\cos {\relax (x )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(a+a*cos(x))**(3/2),x)

[Out]

Integral(x**2/(a*(cos(x) + 1))**(3/2), x)

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