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

Optimal. Leaf size=150 \[ \frac{i \text{Li}_2\left (-i e^{\frac{i x}{2}}\right ) \cos \left (\frac{x}{2}\right )}{a \sqrt{a \cos (x)+a}}-\frac{i \text{Li}_2\left (i e^{\frac{i x}{2}}\right ) \cos \left (\frac{x}{2}\right )}{a \sqrt{a \cos (x)+a}}-\frac{1}{a \sqrt{a \cos (x)+a}}-\frac{i x \cos \left (\frac{x}{2}\right ) \tan ^{-1}\left (e^{\frac{i x}{2}}\right )}{a \sqrt{a \cos (x)+a}}+\frac{x \tan \left (\frac{x}{2}\right )}{2 a \sqrt{a \cos (x)+a}} \]

[Out]

-(1/(a*Sqrt[a + a*Cos[x]])) - (I*x*ArcTan[E^((I/2)*x)]*Cos[x/2])/(a*Sqrt[a + a*Cos[x]]) + (I*Cos[x/2]*PolyLog[
2, (-I)*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) - (I*Cos[x/2]*PolyLog[2, I*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) +
(x*Tan[x/2])/(2*a*Sqrt[a + a*Cos[x]])

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Rubi [A]  time = 0.11658, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3319, 4185, 4181, 2279, 2391} \[ \frac{i \text{Li}_2\left (-i e^{\frac{i x}{2}}\right ) \cos \left (\frac{x}{2}\right )}{a \sqrt{a \cos (x)+a}}-\frac{i \text{Li}_2\left (i e^{\frac{i x}{2}}\right ) \cos \left (\frac{x}{2}\right )}{a \sqrt{a \cos (x)+a}}-\frac{1}{a \sqrt{a \cos (x)+a}}-\frac{i x \cos \left (\frac{x}{2}\right ) \tan ^{-1}\left (e^{\frac{i x}{2}}\right )}{a \sqrt{a \cos (x)+a}}+\frac{x \tan \left (\frac{x}{2}\right )}{2 a \sqrt{a \cos (x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + a*Cos[x])^(3/2),x]

[Out]

-(1/(a*Sqrt[a + a*Cos[x]])) - (I*x*ArcTan[E^((I/2)*x)]*Cos[x/2])/(a*Sqrt[a + a*Cos[x]]) + (I*Cos[x/2]*PolyLog[
2, (-I)*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) - (I*Cos[x/2]*PolyLog[2, I*E^((I/2)*x)])/(a*Sqrt[a + a*Cos[x]]) +
(x*Tan[x/2])/(2*a*Sqrt[a + a*Cos[x]])

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 4185

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.179946, size = 165, normalized size = 1.1 \[ \frac{\sec \left (\frac{x}{2}\right ) \left (2 i \text{Li}_2\left (-i e^{\frac{i x}{2}}\right ) (\cos (x)+1)-2 i \text{Li}_2\left (i e^{\frac{i x}{2}}\right ) (\cos (x)+1)+x \log \left (1-i e^{\frac{i x}{2}}\right )-x \log \left (1+i e^{\frac{i x}{2}}\right )+2 x \sin \left (\frac{x}{2}\right )-4 \cos \left (\frac{x}{2}\right )+x \log \left (1-i e^{\frac{i x}{2}}\right ) \cos (x)-x \log \left (1+i e^{\frac{i x}{2}}\right ) \cos (x)\right )}{4 a \sqrt{a (\cos (x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + a*Cos[x])^(3/2),x]

[Out]

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

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+a\cos \left ( x \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \cos \left (x\right ) + a} x}{a^{2} \cos \left (x\right )^{2} + 2 \, a^{2} \cos \left (x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a \left (\cos{\left (x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (a \cos \left (x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/(a*cos(x) + a)^(3/2), x)