\(\int \frac {x}{(a+a \sin (e+f x))^{3/2}} \, dx\) [141]

Optimal result

Integrand size = 16, antiderivative size = 249 \[ \int \frac {x}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {1}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {x \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {x \text {arctanh}\left (e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f \sqrt {a+a \sin (e+f x)}}+\frac {i \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {i \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}} \] Output:

-1/a/f^2/(a+a*sin(f*x+e))^(1/2)-1/2*x*cot(1/2*e+1/4*Pi+1/2*f*x)/a/f/(a+a*s 
in(f*x+e))^(1/2)-x*arctanh(exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2 
*f*x)/a/f/(a+a*sin(f*x+e))^(1/2)+I*polylog(2,-exp(1/4*I*(2*f*x+Pi+2*e)))*s 
in(1/2*e+1/4*Pi+1/2*f*x)/a/f^2/(a+a*sin(f*x+e))^(1/2)-I*polylog(2,exp(1/4* 
I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^2/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 3.23 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.24 \[ \int \frac {x}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {2 f x \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-(2+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+\frac {\left (-\pi \text {arctanh}\left (\frac {-1+\tan \left (\frac {1}{4} (e+f x)\right )}{\sqrt {2}}\right )+\frac {1}{2} (2 e+\pi +2 f x) \left (\log \left (1-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right )-\log \left (1+e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right )-\operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{\sqrt {2}}+\frac {e \arcsin \left (\csc \left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) (1+\sin (e+f x)) \sin \left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{\sqrt {\frac {-1+\sin (e+f x)}{1+\sin (e+f x)}}}}{2 f^2 (a (1+\sin (e+f x)))^{3/2}} \] Input:

Integrate[x/(a + a*Sin[e + f*x])^(3/2),x]
 

Output:

(2*f*x*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - (2 + f*x)* 
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + ((-(Pi*ArcTanh[(-1 + Tan[(e + f* 
x)/4])/Sqrt[2]]) + ((2*e + Pi + 2*f*x)*(Log[1 - E^((I/4)*(2*e + Pi + 2*f*x 
))] - Log[1 + E^((I/4)*(2*e + Pi + 2*f*x))]))/2 + (2*I)*(PolyLog[2, -E^((I 
/4)*(2*e + Pi + 2*f*x))] - PolyLog[2, E^((I/4)*(2*e + Pi + 2*f*x))]))*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/Sqrt[2] + (e*ArcSin[Csc[(2*e + Pi + 2 
*f*x)/4]]*(1 + Sin[e + f*x])*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[(-1 + Sin[e + 
 f*x])/(1 + Sin[e + f*x])])/(2*f^2*(a*(1 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.76, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3800, 3042, 4673, 3042, 4671, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x}{(a \sin (e+f x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \int x \csc ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \int x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )^3dx}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \int x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx-\frac {2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \int x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx-\frac {2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (-\frac {2 \int \log \left (1-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}+\frac {2 \int \log \left (1+e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}-\frac {4 x \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}\right )-\frac {2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (\frac {4 i \int e^{-\frac {1}{4} i (2 e+2 f x+\pi )} \log \left (1-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )de^{\frac {1}{4} i (2 e+2 f x+\pi )}}{f^2}-\frac {4 i \int e^{-\frac {1}{4} i (2 e+2 f x+\pi )} \log \left (1+e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )de^{\frac {1}{4} i (2 e+2 f x+\pi )}}{f^2}-\frac {4 x \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}\right )-\frac {2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (-\frac {4 x \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}+\frac {4 i \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f^2}-\frac {4 i \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f^2}\right )-\frac {2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

Input:

Int[x/(a + a*Sin[e + f*x])^(3/2),x]
 

Output:

(((-2*Csc[e/2 + Pi/4 + (f*x)/2])/f^2 - (x*Cot[e/2 + Pi/4 + (f*x)/2]*Csc[e/ 
2 + Pi/4 + (f*x)/2])/f + ((-4*x*ArcTanh[E^((I/4)*(2*e + Pi + 2*f*x))])/f + 
 ((4*I)*PolyLog[2, -E^((I/4)*(2*e + Pi + 2*f*x))])/f^2 - ((4*I)*PolyLog[2, 
 E^((I/4)*(2*e + Pi + 2*f*x))])/f^2)/2)*Sin[e/2 + Pi/4 + (f*x)/2])/(2*a*Sq 
rt[a + a*Sin[e + f*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 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 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]

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

Input:

int(x/(a+a*sin(f*x+e))^(3/2),x)
 

Output:

int(x/(a+a*sin(f*x+e))^(3/2),x)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate(x/(a+a*sin(f*x+e))**(3/2),x)
 

Output:

Integral(x/(a*(sin(e + f*x) + 1))**(3/2), x)
 

Maxima [F]

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

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

Output:

integrate(x/(a*sin(f*x + e) + a)^(3/2), x)
 

Giac [F]

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

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

Output:

integrate(x/(a*sin(f*x + e) + a)^(3/2), x)
 

Mupad [F(-1)]

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

int(x/(a + a*sin(e + f*x))^(3/2),x)
 

Output:

int(x/(a + a*sin(e + f*x))^(3/2), x)
 

Reduce [F]

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

int(x/(a+a*sin(f*x+e))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*x)/(sin(e + f*x)**2 + 2*sin(e + f*x) 
+ 1),x))/a**2