3.2.38 \(\int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx\) [138]

3.2.38.1 Optimal result

Integrand size = 19, antiderivative size = 207 \[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\frac {4 i x \text {arctanh}\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {4 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {4 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}} \]

output

-4*I*x*arctanh(exp(1/2*e+3/4*I*Pi+1/2*f*x))*cosh(1/2*e+1/4*I*Pi+1/2*f*x)/f 
/(a+I*a*sinh(f*x+e))^(1/2)+4*I*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(2,exp( 
1/2*e+3/4*I*Pi+1/2*f*x))/f^2/(a+I*a*sinh(f*x+e))^(1/2)-4*I*cosh(1/2*e+1/4* 
I*Pi+1/2*f*x)*polylog(2,-exp(1/2*e+3/4*I*Pi+1/2*f*x))/f^2/(a+I*a*sinh(f*x+ 
e))^(1/2)
 

3.2.38.2 Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\frac {(2-2 i) (-1)^{3/4} \left (i e \arctan \left (\sqrt [4]{-1} e^{\frac {1}{2} (e+f x)}\right )+\frac {1}{2} (e+f x) \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-\frac {1}{2} (e+f x) \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-\operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+\operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )\right ) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}} \]

input

Integrate[x/Sqrt[a + I*a*Sinh[e + f*x]],x]
 

output

((2 - 2*I)*(-1)^(3/4)*(I*e*ArcTan[(-1)^(1/4)*E^((e + f*x)/2)] + ((e + f*x) 
*Log[1 - (-1)^(3/4)*E^((e + f*x)/2)])/2 - ((e + f*x)*Log[1 + (-1)^(3/4)*E^ 
((e + f*x)/2)])/2 - PolyLog[2, -((-1)^(3/4)*E^((e + f*x)/2))] + PolyLog[2, 
 (-1)^(3/4)*E^((e + f*x)/2)])*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2]))/( 
f^2*Sqrt[a + I*a*Sinh[e + f*x]])
 

3.2.38.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.65, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3800, 3042, 4670, 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.

\(\displaystyle \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x}{\sqrt {a+a \sin (i e+i f x)}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \int x \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx}{\sqrt {a+i a \sinh (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \int x \csc \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{4}\right )dx}{\sqrt {a+i a \sinh (e+f x)}}\)

\(\Big \downarrow \) 4670

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

input

Int[x/Sqrt[a + I*a*Sinh[e + f*x]],x]
 

output

(Cosh[e/2 + (I/4)*Pi + (f*x)/2]*(((4*I)*x*ArcTanh[E^((2*e - I*Pi)/4 + (f*x 
)/2)])/f + ((4*I)*PolyLog[2, -E^((2*e - I*Pi)/4 + (f*x)/2)])/f^2 - ((4*I)* 
PolyLog[2, E^((2*e - I*Pi)/4 + (f*x)/2)])/f^2))/Sqrt[a + I*a*Sinh[e + f*x] 
]
 

3.2.38.3.1 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 4670

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x 
_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] 
 + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*fz*x 
)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e 
+ f*fz*x)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

3.2.38.4 Maple [F]

\[\int \frac {x}{\sqrt {a +i a \sinh \left (f x +e \right )}}d x\]

input

int(x/(a+I*a*sinh(f*x+e))^(1/2),x)
 

output

int(x/(a+I*a*sinh(f*x+e))^(1/2),x)
 

3.2.38.5 Fricas [F]

\[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int { \frac {x}{\sqrt {i \, a \sinh \left (f x + e\right ) + a}} \,d x } \]

input

integrate(x/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="fricas")
 

output

integral(-2*I*sqrt(1/2*I*a*e^(-f*x - e))*x*e^(f*x + e)/(a*e^(f*x + e) - I* 
a), x)
 

3.2.38.6 Sympy [F]

\[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int \frac {x}{\sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}}\, dx \]

input

integrate(x/(a+I*a*sinh(f*x+e))**(1/2),x)
 

output

Integral(x/sqrt(I*a*(sinh(e + f*x) - I)), x)
 

3.2.38.7 Maxima [F]

\[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int { \frac {x}{\sqrt {i \, a \sinh \left (f x + e\right ) + a}} \,d x } \]

input

integrate(x/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="maxima")
 

output

integrate(x/sqrt(I*a*sinh(f*x + e) + a), x)
 

3.2.38.8 Giac [F]

\[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int { \frac {x}{\sqrt {i \, a \sinh \left (f x + e\right ) + a}} \,d x } \]

input

integrate(x/(a+I*a*sinh(f*x+e))^(1/2),x, algorithm="giac")
 

output

integrate(x/sqrt(I*a*sinh(f*x + e) + a), x)
 

3.2.38.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int \frac {x}{\sqrt {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

input

int(x/(a + a*sinh(e + f*x)*1i)^(1/2),x)
 

output

int(x/(a + a*sinh(e + f*x)*1i)^(1/2), x)