\(\int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx\) [136]

Optimal result

Integrand size = 16, antiderivative size = 175 \[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {4 x \text {arctanh}\left (e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d \sqrt {a+a \sin (c+d x)}}+\frac {4 i \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}}-\frac {4 i \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right ) \sin \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{d^2 \sqrt {a+a \sin (c+d x)}} \] Output:

-4*x*arctanh(exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d/(a+a*s 
in(d*x+c))^(1/2)+4*I*polylog(2,-exp(1/4*I*(2*d*x+Pi+2*c)))*sin(1/2*c+1/4*P 
i+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)-4*I*polylog(2,exp(1/4*I*(2*d*x+Pi+2* 
c)))*sin(1/2*c+1/4*Pi+1/2*d*x)/d^2/(a+a*sin(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.18 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.32 \[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 \left (\frac {\left (-\pi \text {arctanh}\left (\frac {-1+\tan \left (\frac {1}{4} (c+d x)\right )}{\sqrt {2}}\right )+\frac {1}{2} (2 c+\pi +2 d x) \left (\log \left (1-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right )-\log \left (1+e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right )-\operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 c+\pi +2 d x)}\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {2}}+\frac {c \arcsin \left (\csc \left (\frac {1}{4} (2 c+\pi +2 d x)\right )\right ) \sin \left (\frac {1}{4} (2 c-\pi +2 d x)\right )}{\sqrt {\frac {-1+\sin (c+d x)}{1+\sin (c+d x)}}}\right )}{d^2 \sqrt {a (1+\sin (c+d x))}} \] Input:

Integrate[x/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(2*(((-(Pi*ArcTanh[(-1 + Tan[(c + d*x)/4])/Sqrt[2]]) + ((2*c + Pi + 2*d*x) 
*(Log[1 - E^((I/4)*(2*c + Pi + 2*d*x))] - Log[1 + E^((I/4)*(2*c + Pi + 2*d 
*x))]))/2 + (2*I)*(PolyLog[2, -E^((I/4)*(2*c + Pi + 2*d*x))] - PolyLog[2, 
E^((I/4)*(2*c + Pi + 2*d*x))]))*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/Sqr 
t[2] + (c*ArcSin[Csc[(2*c + Pi + 2*d*x)/4]]*Sin[(2*c - Pi + 2*d*x)/4])/Sqr 
t[(-1 + Sin[c + d*x])/(1 + Sin[c + d*x])]))/(d^2*Sqrt[a*(1 + Sin[c + d*x]) 
])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3800, 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.

\(\displaystyle \int \frac {x}{\sqrt {a \sin (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x}{\sqrt {a \sin (c+d x)+a}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \int x \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \int x \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{\sqrt {a \sin (c+d x)+a}}\)

\(\Big \downarrow \) 4671

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

Int[x/Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(((-4*x*ArcTanh[E^((I/4)*(2*c + Pi + 2*d*x))])/d + ((4*I)*PolyLog[2, -E^(( 
I/4)*(2*c + Pi + 2*d*x))])/d^2 - ((4*I)*PolyLog[2, E^((I/4)*(2*c + Pi + 2* 
d*x))])/d^2)*Sin[c/2 + Pi/4 + (d*x)/2])/Sqrt[a + a*Sin[c + d*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]
 

Maple [F]

\[\int \frac {x}{\sqrt {a +a \sin \left (d x +c \right )}}d x\]

Input:

int(x/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

int(x/(a+a*sin(d*x+c))^(1/2),x)
 

Fricas [F]

\[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {x}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(x/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(x/sqrt(a*sin(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {x}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \] Input:

integrate(x/(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(x/sqrt(a*(sin(c + d*x) + 1)), x)
 

Maxima [F]

\[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {x}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(x/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(x/sqrt(a*sin(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {x}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(x/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(x/sqrt(a*sin(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {x}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \] Input:

int(x/(a + a*sin(c + d*x))^(1/2),x)
 

Output:

int(x/(a + a*sin(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {x}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, x}{\sin \left (d x +c \right )+1}d x \right )}{a} \] Input:

int(x/(a+a*sin(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*x)/(sin(c + d*x) + 1),x))/a