\(\int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx\) [137]

Optimal result

Integrand size = 21, antiderivative size = 349 \[ \int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\frac {4 i x^2 \text {arctanh}\left (e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right ) \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right )}{f \sqrt {a+i a \sinh (e+f x)}}+\frac {8 i x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {8 i x \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2 \sqrt {a+i a \sinh (e+f x)}}-\frac {16 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}}+\frac {16 i \cosh \left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}} \] Output:

-4*I*x^2*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)+8*I*x*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)-8*I*x*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*sin 
h(f*x+e))^(1/2)-16*I*cosh(1/2*e+1/4*I*Pi+1/2*f*x)*polylog(3,exp(1/2*e+3/4* 
I*Pi+1/2*f*x))/f^3/(a+I*a*sinh(f*x+e))^(1/2)+16*I*cosh(1/2*e+1/4*I*Pi+1/2* 
f*x)*polylog(3,-exp(1/2*e+3/4*I*Pi+1/2*f*x))/f^3/(a+I*a*sinh(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.79 \[ \int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\frac {(1+i) (-1)^{3/4} \left (-2 i e^2 \arctan \left (\sqrt [4]{-1} e^{\frac {1}{2} (e+f x)}\right )-e^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+f^2 x^2 \log \left (1-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+e^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-f^2 x^2 \log \left (1+(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-4 f x \operatorname {PolyLog}\left (2,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+4 f x \operatorname {PolyLog}\left (2,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )+8 \operatorname {PolyLog}\left (3,-(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )-8 \operatorname {PolyLog}\left (3,(-1)^{3/4} e^{\frac {1}{2} (e+f x)}\right )\right ) \left (-i \cosh \left (\frac {1}{2} (e+f x)\right )+\sinh \left (\frac {1}{2} (e+f x)\right )\right )}{f^3 \sqrt {a+i a \sinh (e+f x)}} \] Input:

Integrate[x^2/Sqrt[a + I*a*Sinh[e + f*x]],x]
 

Output:

((1 + I)*(-1)^(3/4)*((-2*I)*e^2*ArcTan[(-1)^(1/4)*E^((e + f*x)/2)] - e^2*L 
og[1 - (-1)^(3/4)*E^((e + f*x)/2)] + f^2*x^2*Log[1 - (-1)^(3/4)*E^((e + f* 
x)/2)] + e^2*Log[1 + (-1)^(3/4)*E^((e + f*x)/2)] - f^2*x^2*Log[1 + (-1)^(3 
/4)*E^((e + f*x)/2)] - 4*f*x*PolyLog[2, -((-1)^(3/4)*E^((e + f*x)/2))] + 4 
*f*x*PolyLog[2, (-1)^(3/4)*E^((e + f*x)/2)] + 8*PolyLog[3, -((-1)^(3/4)*E^ 
((e + f*x)/2))] - 8*PolyLog[3, (-1)^(3/4)*E^((e + f*x)/2)])*((-I)*Cosh[(e 
+ f*x)/2] + Sinh[(e + f*x)/2]))/(f^3*Sqrt[a + I*a*Sinh[e + f*x]])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.60, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3800, 3042, 4670, 3011, 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.

\(\displaystyle \int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4670

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\cosh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \left (\frac {4 i x^2 \text {arctanh}\left (e^{\frac {f x}{2}+\frac {1}{4} (2 e-i \pi )}\right )}{f}-\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2}-\frac {2 x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f}\right )}{f}+\frac {4 i \left (\frac {4 \operatorname {PolyLog}\left (3,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f^2}-\frac {2 x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} (2 e-i \pi )+\frac {f x}{2}}\right )}{f}\right )}{f}\right )}{\sqrt {a+i a \sinh (e+f x)}}\)

Input:

Int[x^2/Sqrt[a + I*a*Sinh[e + f*x]],x]
 

Output:

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

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]
 

Maple [F]

\[\int \frac {x^{2}}{\sqrt {a +i a \sinh \left (f x +e \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^2/(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^2*e^(f*x + e)/(a*e^(f*x + e) - 
I*a), x)
 

Sympy [F]

\[ \int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int \frac {x^{2}}{\sqrt {i a \left (\sinh {\left (e + f x \right )} - i\right )}}\, dx \] Input:

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

Output:

Integral(x**2/sqrt(I*a*(sinh(e + f*x) - I)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+i a \sinh (e+f x)}} \, dx=\int \frac {x^2}{\sqrt {a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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