\(\int \frac {x}{(a+a \cos (x))^{3/2}} \, dx\) [182]

Optimal result

Integrand size = 12, antiderivative size = 150 \[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=-\frac {1}{a \sqrt {a+a \cos (x)}}-\frac {i x \arctan \left (e^{\frac {i x}{2}}\right ) \cos \left (\frac {x}{2}\right )}{a \sqrt {a+a \cos (x)}}+\frac {i \cos \left (\frac {x}{2}\right ) \operatorname {PolyLog}\left (2,-i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}-\frac {i \cos \left (\frac {x}{2}\right ) \operatorname {PolyLog}\left (2,i e^{\frac {i x}{2}}\right )}{a \sqrt {a+a \cos (x)}}+\frac {x \tan \left (\frac {x}{2}\right )}{2 a \sqrt {a+a \cos (x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.10 \[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\frac {\sec \left (\frac {x}{2}\right ) \left (-4 \cos \left (\frac {x}{2}\right )+x \log \left (1-i e^{\frac {i x}{2}}\right )+x \cos (x) \log \left (1-i e^{\frac {i x}{2}}\right )-x \log \left (1+i e^{\frac {i x}{2}}\right )-x \cos (x) \log \left (1+i e^{\frac {i x}{2}}\right )+2 i (1+\cos (x)) \operatorname {PolyLog}\left (2,-i e^{\frac {i x}{2}}\right )-2 i (1+\cos (x)) \operatorname {PolyLog}\left (2,i e^{\frac {i x}{2}}\right )+2 x \sin \left (\frac {x}{2}\right )\right )}{4 a \sqrt {a (1+\cos (x))}} \] Input:

Integrate[x/(a + a*Cos[x])^(3/2),x]
 

Output:

(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])*PolyLo 
g[2, I*E^((I/2)*x)] + 2*x*Sin[x/2]))/(4*a*Sqrt[a*(1 + Cos[x])])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.69, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3800, 3042, 4673, 3042, 4669, 2715, 2838}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \int x \sec ^3\left (\frac {x}{2}\right )dx}{2 a \sqrt {a \cos (x)+a}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \left (\frac {1}{2} \int x \sec \left (\frac {x}{2}\right )dx-2 \sec \left (\frac {x}{2}\right )+x \tan \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right )\right )}{2 a \sqrt {a \cos (x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \left (\frac {1}{2} \int x \csc \left (\frac {x}{2}+\frac {\pi }{2}\right )dx-2 \sec \left (\frac {x}{2}\right )+x \tan \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right )\right )}{2 a \sqrt {a \cos (x)+a}}\)

\(\Big \downarrow \) 4669

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \left (\frac {1}{2} \left (-2 \int \log \left (1-i e^{\frac {i x}{2}}\right )dx+2 \int \log \left (1+i e^{\frac {i x}{2}}\right )dx-4 i x \arctan \left (e^{\frac {i x}{2}}\right )\right )-2 \sec \left (\frac {x}{2}\right )+x \tan \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right )\right )}{2 a \sqrt {a \cos (x)+a}}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \left (\frac {1}{2} \left (4 i \int e^{-\frac {i x}{2}} \log \left (1-i e^{\frac {i x}{2}}\right )de^{\frac {i x}{2}}-4 i \int e^{-\frac {i x}{2}} \log \left (1+i e^{\frac {i x}{2}}\right )de^{\frac {i x}{2}}-4 i x \arctan \left (e^{\frac {i x}{2}}\right )\right )-2 \sec \left (\frac {x}{2}\right )+x \tan \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right )\right )}{2 a \sqrt {a \cos (x)+a}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\cos \left (\frac {x}{2}\right ) \left (\frac {1}{2} \left (-4 i x \arctan \left (e^{\frac {i x}{2}}\right )+4 i \operatorname {PolyLog}\left (2,-i e^{\frac {i x}{2}}\right )-4 i \operatorname {PolyLog}\left (2,i e^{\frac {i x}{2}}\right )\right )-2 \sec \left (\frac {x}{2}\right )+x \tan \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right )\right )}{2 a \sqrt {a \cos (x)+a}}\)

Input:

Int[x/(a + a*Cos[x])^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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]
 

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 4669

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] + (-Si 
mp[d*(m/f)   Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], 
 x] + Simp[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 4673

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] + (-Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x] + S 
imp[b^2*((n - 2)/(n - 1))   Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2), x], x]) 
 /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2]
 

Maple [F]

\[\int \frac {x}{\left (a +a \cos \left (x \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\int { \frac {x}{{\left (a \cos \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\int \frac {x}{\left (a \left (\cos {\left (x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\int { \frac {x}{{\left (a \cos \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

1/3*(8*x*cos(3/2*x)*sin(3*x)^3 - 8*x*cos(3*x)^3*sin(3/2*x) - 8*((3*x*sin(3 
/2*x) + 2*cos(3/2*x))*cos(2*x) - (3*x*sin(x) - 2*cos(x))*cos(3/2*x) - (3*x 
*cos(3/2*x) + 2*sin(3/2*x))*sin(2*x) + (3*x*cos(x) + 3*x - 2*sin(x))*sin(3 
/2*x))*cos(3*x)^2 - 48*cos(2*x)^2*cos(3/2*x) - 24*((3*x*sin(3/2*x) - 2*cos 
(3/2*x))*cos(3*x) + 3*(3*x*sin(3/2*x) - 2*cos(3/2*x))*cos(2*x) - (9*x*sin( 
x) + 6*cos(x) + 2)*cos(3/2*x) - 27*(a^2*cos(3*x)^2 + 9*a^2*cos(2*x)^2 + 9* 
a^2*cos(x)^2 + a^2*sin(3*x)^2 + 9*a^2*sin(2*x)^2 + 18*a^2*sin(2*x)*sin(x) 
+ 9*a^2*sin(x)^2 + 6*a^2*cos(x) + a^2 + 2*(3*a^2*cos(2*x) + 3*a^2*cos(x) + 
 a^2)*cos(3*x) + 6*(3*a^2*cos(x) + a^2)*cos(2*x) + 6*(a^2*sin(2*x) + a^2*s 
in(x))*sin(3*x))*integrate(1/3*(x*cos(4*x)*cos(3/2*x) + 4*x*cos(3*x)*cos(3 
/2*x) + 6*x*cos(2*x)*cos(3/2*x) + x*sin(4*x)*sin(3/2*x) + 4*x*sin(3*x)*sin 
(3/2*x) + 6*x*sin(2*x)*sin(3/2*x) + 4*x*sin(3/2*x)*sin(x) + (4*x*cos(x) + 
x)*cos(3/2*x))/(a^2*cos(4*x)^2 + 16*a^2*cos(3*x)^2 + 36*a^2*cos(2*x)^2 + 1 
6*a^2*cos(x)^2 + a^2*sin(4*x)^2 + 16*a^2*sin(3*x)^2 + 36*a^2*sin(2*x)^2 + 
48*a^2*sin(2*x)*sin(x) + 16*a^2*sin(x)^2 + 8*a^2*cos(x) + a^2 + 2*(4*a^2*c 
os(3*x) + 6*a^2*cos(2*x) + 4*a^2*cos(x) + a^2)*cos(4*x) + 8*(6*a^2*cos(2*x 
) + 4*a^2*cos(x) + a^2)*cos(3*x) + 12*(4*a^2*cos(x) + a^2)*cos(2*x) + 4*(2 
*a^2*sin(3*x) + 3*a^2*sin(2*x) + 2*a^2*sin(x))*sin(4*x) + 16*(3*a^2*sin(2* 
x) + 2*a^2*sin(x))*sin(3*x)), x) - (3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(3*x 
) - 3*(3*x*cos(3/2*x) + 2*sin(3/2*x))*sin(2*x) + 3*(3*x*cos(x) + x - 2*...
 

Giac [F]

\[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\int { \frac {x}{{\left (a \cos \left (x\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\int \frac {x}{{\left (a+a\,\cos \left (x\right )\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x}{(a+a \cos (x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (x \right )+1}\, x}{\cos \left (x \right )^{2}+2 \cos \left (x \right )+1}d x \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(cos(x) + 1)*x)/(cos(x)**2 + 2*cos(x) + 1),x))/a**2