\(\int \frac {(a+i a \sinh (e+f x))^{3/2}}{x^2} \, dx\) [129]

Optimal result

Integrand size = 21, antiderivative size = 302 \[ \int \frac {(a+i a \sinh (e+f x))^{3/2}}{x^2} \, dx=-\frac {2 a \cosh ^2\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)}}{x}-\frac {3}{4} a f \text {Chi}\left (\frac {3 f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (6 e-i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {3}{4} a f \text {Chi}\left (\frac {f x}{2}\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \sqrt {a+i a \sinh (e+f x)}+\frac {3}{4} a f \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {f x}{2}\right )-\frac {3}{4} a f \cosh \left (\frac {1}{4} (6 e-i \pi )\right ) \text {sech}\left (\frac {e}{2}+\frac {i \pi }{4}+\frac {f x}{2}\right ) \sqrt {a+i a \sinh (e+f x)} \text {Shi}\left (\frac {3 f x}{2}\right ) \] Output:

-2*a*cosh(1/2*e+1/4*I*Pi+1/2*f*x)^2*(a+I*a*sinh(f*x+e))^(1/2)/x+3/4*I*a*f* 
Chi(3/2*f*x)*sech(1/2*e+1/4*I*Pi+1/2*f*x)*cosh(3/2*e+1/4*I*Pi)*(a+I*a*sinh 
(f*x+e))^(1/2)+3/4*a*f*Chi(1/2*f*x)*sech(1/2*e+1/4*I*Pi+1/2*f*x)*sinh(1/2* 
e+1/4*I*Pi)*(a+I*a*sinh(f*x+e))^(1/2)+3/4*a*f*cosh(1/2*e+1/4*I*Pi)*sech(1/ 
2*e+1/4*I*Pi+1/2*f*x)*(a+I*a*sinh(f*x+e))^(1/2)*Shi(1/2*f*x)+3/4*I*a*f*sin 
h(3/2*e+1/4*I*Pi)*sech(1/2*e+1/4*I*Pi+1/2*f*x)*(a+I*a*sinh(f*x+e))^(1/2)*S 
hi(3/2*f*x)
 

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \sinh (e+f x))^{3/2}}{x^2} \, dx=\frac {a (-i+\sinh (e+f x)) \sqrt {a+i a \sinh (e+f x)} \left (-6 i \cosh \left (\frac {1}{2} (e+f x)\right )+2 i \cosh \left (\frac {3}{2} (e+f x)\right )-3 f x \text {Chi}\left (\frac {f x}{2}\right ) \left (\cosh \left (\frac {e}{2}\right )-i \sinh \left (\frac {e}{2}\right )\right )-3 f x \text {Chi}\left (\frac {3 f x}{2}\right ) \left (\cosh \left (\frac {3 e}{2}\right )+i \sinh \left (\frac {3 e}{2}\right )\right )+6 \sinh \left (\frac {1}{2} (e+f x)\right )+2 \sinh \left (\frac {3}{2} (e+f x)\right )+3 i f x \cosh \left (\frac {e}{2}\right ) \text {Shi}\left (\frac {f x}{2}\right )-3 f x \sinh \left (\frac {e}{2}\right ) \text {Shi}\left (\frac {f x}{2}\right )-3 i f x \cosh \left (\frac {3 e}{2}\right ) \text {Shi}\left (\frac {3 f x}{2}\right )-3 f x \sinh \left (\frac {3 e}{2}\right ) \text {Shi}\left (\frac {3 f x}{2}\right )\right )}{4 x \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

(a*(-I + Sinh[e + f*x])*Sqrt[a + I*a*Sinh[e + f*x]]*((-6*I)*Cosh[(e + f*x) 
/2] + (2*I)*Cosh[(3*(e + f*x))/2] - 3*f*x*CoshIntegral[(f*x)/2]*(Cosh[e/2] 
 - I*Sinh[e/2]) - 3*f*x*CoshIntegral[(3*f*x)/2]*(Cosh[(3*e)/2] + I*Sinh[(3 
*e)/2]) + 6*Sinh[(e + f*x)/2] + 2*Sinh[(3*(e + f*x))/2] + (3*I)*f*x*Cosh[e 
/2]*SinhIntegral[(f*x)/2] - 3*f*x*Sinh[e/2]*SinhIntegral[(f*x)/2] - (3*I)* 
f*x*Cosh[(3*e)/2]*SinhIntegral[(3*f*x)/2] - 3*f*x*Sinh[(3*e)/2]*SinhIntegr 
al[(3*f*x)/2]))/(4*x*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3800, 3042, 3794, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

\(\displaystyle 2 a \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)} \int \frac {\cosh ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{x^2}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3794

\(\displaystyle 2 a \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)} \left (\frac {3}{2} i f \int \left (\frac {\cosh \left (\frac {3 e}{2}+\frac {3 f x}{2}+\frac {i \pi }{4}\right )}{4 x}-\frac {i \sinh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{4 x}\right )dx-\frac {\cosh ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 a \text {sech}\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (e+f x)} \left (\frac {3}{2} i f \left (\frac {1}{4} i \sinh \left (\frac {1}{4} (6 e-i \pi )\right ) \text {Chi}\left (\frac {3 f x}{2}\right )-\frac {1}{4} i \sinh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {Chi}\left (\frac {f x}{2}\right )-\frac {1}{4} i \cosh \left (\frac {1}{4} (2 e+i \pi )\right ) \text {Shi}\left (\frac {f x}{2}\right )+\frac {1}{4} i \cosh \left (\frac {1}{4} (6 e-i \pi )\right ) \text {Shi}\left (\frac {3 f x}{2}\right )\right )-\frac {\cosh ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{x}\right )\)

Input:

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

Output:

2*a*Sech[e/2 + (I/4)*Pi + (f*x)/2]*Sqrt[a + I*a*Sinh[e + f*x]]*(-(Cosh[e/2 
 + (I/4)*Pi + (f*x)/2]^3/x) + ((3*I)/2)*f*((I/4)*CoshIntegral[(3*f*x)/2]*S 
inh[(6*e - I*Pi)/4] - (I/4)*CoshIntegral[(f*x)/2]*Sinh[(2*e + I*Pi)/4] - ( 
I/4)*Cosh[(2*e + I*Pi)/4]*SinhIntegral[(f*x)/2] + (I/4)*Cosh[(6*e - I*Pi)/ 
4]*SinhIntegral[(3*f*x)/2]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {(a+i a \sinh (e+f x))^{3/2}}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

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

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

Output:

Integral((I*a*(sinh(e + f*x) - I))**(3/2)/x**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \sinh (e+f x))^{3/2}}{x^2} \, dx=\int \frac {{\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{x^2} \,d x \] Input:

int((a + a*sinh(e + f*x)*1i)^(3/2)/x^2,x)
 

Output:

int((a + a*sinh(e + f*x)*1i)^(3/2)/x^2, x)
 

Reduce [F]

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

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

Output:

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