\(\int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx\) [175]

Optimal result

Integrand size = 15, antiderivative size = 235 \[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=-\frac {4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {12 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {12 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {48 x \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {48 x \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}-\frac {96 i \operatorname {PolyLog}\left (4,-e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}}+\frac {96 i \operatorname {PolyLog}\left (4,e^{\frac {i x}{2}}\right ) \sin \left (\frac {x}{2}\right )}{\sqrt {a-a \cos (x)}} \] Output:

-4*x^3*arctanh(exp(1/2*I*x))*sin(1/2*x)/(a-a*cos(x))^(1/2)+12*I*x^2*polylo 
g(2,-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*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)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.72 \[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=-\frac {i \left (8 \pi ^4-x^4+8 i x^3 \log \left (1-e^{-\frac {i x}{2}}\right )-8 i x^3 \log \left (1+e^{\frac {i x}{2}}\right )-48 x^2 \operatorname {PolyLog}\left (2,e^{-\frac {i x}{2}}\right )-48 x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )+192 i x \operatorname {PolyLog}\left (3,e^{-\frac {i x}{2}}\right )-192 i x \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )+384 \operatorname {PolyLog}\left (4,e^{-\frac {i x}{2}}\right )+384 \operatorname {PolyLog}\left (4,-e^{\frac {i x}{2}}\right )\right ) \sin \left (\frac {x}{2}\right )}{4 \sqrt {a-a \cos (x)}} \] Input:

Integrate[x^3/Sqrt[a - a*Cos[x]],x]
 

Output:

((-1/4*I)*(8*Pi^4 - x^4 + (8*I)*x^3*Log[1 - E^((-1/2*I)*x)] - (8*I)*x^3*Lo 
g[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*PolyLog[ 
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]]
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.64, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3800, 3042, 4671, 3011, 7163, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^3}{\sqrt {a-a \sin \left (x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \int x^3 \csc \left (\frac {x}{2}\right )dx}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \int x^3 \csc \left (\frac {x}{2}\right )dx}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \left (-6 \int x^2 \log \left (1-e^{\frac {i x}{2}}\right )dx+6 \int x^2 \log \left (1+e^{\frac {i x}{2}}\right )dx-4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right )\right )}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \left (6 \left (2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )-4 i \int x \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )dx\right )-6 \left (2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right )-4 i \int x \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right )dx\right )-4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right )\right )}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \left (6 \left (2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )-4 i \left (2 i \int \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )dx-2 i x \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )\right )\right )-6 \left (2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right )-4 i \left (2 i \int \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right )dx-2 i x \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right )\right )\right )-4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right )\right )}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \left (6 \left (2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )-4 i \left (4 \int e^{-\frac {i x}{2}} \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )de^{\frac {i x}{2}}-2 i x \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )\right )\right )-6 \left (2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right )-4 i \left (4 \int e^{-\frac {i x}{2}} \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right )de^{\frac {i x}{2}}-2 i x \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right )\right )\right )-4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right )\right )}{\sqrt {a-a \cos (x)}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\sin \left (\frac {x}{2}\right ) \left (-4 x^3 \text {arctanh}\left (e^{\frac {i x}{2}}\right )+6 \left (2 i x^2 \operatorname {PolyLog}\left (2,-e^{\frac {i x}{2}}\right )-4 i \left (4 \operatorname {PolyLog}\left (4,-e^{\frac {i x}{2}}\right )-2 i x \operatorname {PolyLog}\left (3,-e^{\frac {i x}{2}}\right )\right )\right )-6 \left (2 i x^2 \operatorname {PolyLog}\left (2,e^{\frac {i x}{2}}\right )-4 i \left (4 \operatorname {PolyLog}\left (4,e^{\frac {i x}{2}}\right )-2 i x \operatorname {PolyLog}\left (3,e^{\frac {i x}{2}}\right )\right )\right )\right )}{\sqrt {a-a \cos (x)}}\)

Input:

Int[x^3/Sqrt[a - a*Cos[x]],x]
 

Output:

((-4*x^3*ArcTanh[E^((I/2)*x)] + 6*((2*I)*x^2*PolyLog[2, -E^((I/2)*x)] - (4 
*I)*((-2*I)*x*PolyLog[3, -E^((I/2)*x)] + 4*PolyLog[4, -E^((I/2)*x)])) - 6* 
((2*I)*x^2*PolyLog[2, E^((I/2)*x)] - (4*I)*((-2*I)*x*PolyLog[3, E^((I/2)*x 
)] + 4*PolyLog[4, E^((I/2)*x)])))*Sin[x/2])/Sqrt[a - a*Cos[x]]
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(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 4671

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] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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 7163

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] - Simp[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]
 

Maple [F]

\[\int \frac {x^{3}}{\sqrt {a -a \cos \left (x \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=\int { \frac {x^{3}}{\sqrt {-a \cos \left (x\right ) + a}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=\int \frac {x^{3}}{\sqrt {- a \left (\cos {\left (x \right )} - 1\right )}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=\int { \frac {x^{3}}{\sqrt {-a \cos \left (x\right ) + a}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=\int { \frac {x^{3}}{\sqrt {-a \cos \left (x\right ) + a}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=\int \frac {x^3}{\sqrt {a-a\,\cos \left (x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{\sqrt {a-a \cos (x)}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\cos \left (x \right )+1}\, x^{3}}{\cos \left (x \right )-1}d x \right )}{a} \] Input:

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

Output:

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