\(\int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx\) [134]

Optimal result

Integrand size = 21, antiderivative size = 444 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=-\frac {4 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}-\frac {5}{8} a^2 d \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {15}{8} a^2 d \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} a^2 d \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {d x}{2}\right )-\frac {15}{8} a^2 d \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {3 d x}{2}\right )-\frac {5}{8} a^2 d \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {5 d x}{2}\right ) \] Output:

-4*a^2*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^4*(a+I*a*sinh(d*x+c))^(1/2)/x-5/8*a^2* 
d*Chi(5/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(5/2*c+1/4*I*Pi)*(a+I*a*si 
nh(d*x+c))^(1/2)+15/8*I*a^2*d*Chi(3/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*co 
sh(3/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/4*a^2*d*Chi(1/2*d*x)*sech(1 
/2*c+1/4*I*Pi+1/2*d*x)*sinh(1/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/4* 
a^2*d*cosh(1/2*c+1/4*I*Pi)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*(a+I*a*sinh(d*x+c) 
)^(1/2)*Shi(1/2*d*x)+15/8*I*a^2*d*sinh(3/2*c+1/4*I*Pi)*sech(1/2*c+1/4*I*Pi 
+1/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)*Shi(3/2*d*x)-5/8*a^2*d*cosh(5/2*c+1/4* 
I*Pi)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*(a+I*a*sinh(d*x+c))^(1/2)*Shi(5/2*d*x)
 

Mathematica [A] (verified)

Time = 5.01 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\frac {a^2 (-i+\sinh (c+d x))^2 \sqrt {a+i a \sinh (c+d x)} \left (20 \cosh \left (\frac {1}{2} (c+d x)\right )-10 \cosh \left (\frac {3}{2} (c+d x)\right )-2 \cosh \left (\frac {5}{2} (c+d x)\right )+5 i d x \cosh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )-10 i d x \text {Chi}\left (\frac {d x}{2}\right ) \left (\cosh \left (\frac {c}{2}\right )-i \sinh \left (\frac {c}{2}\right )\right )+15 d x \text {Chi}\left (\frac {3 d x}{2}\right ) \left (-i \cosh \left (\frac {3 c}{2}\right )+\sinh \left (\frac {3 c}{2}\right )\right )+5 d x \text {Chi}\left (\frac {5 d x}{2}\right ) \sinh \left (\frac {5 c}{2}\right )+20 i \sinh \left (\frac {1}{2} (c+d x)\right )+10 i \sinh \left (\frac {3}{2} (c+d x)\right )-2 i \sinh \left (\frac {5}{2} (c+d x)\right )-10 d x \cosh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )-10 i d x \sinh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )+15 d x \cosh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )-15 i d x \sinh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+5 d x \cosh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )+5 i d x \sinh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )\right )}{8 x \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:

Integrate[(a + I*a*Sinh[c + d*x])^(5/2)/x^2,x]
 

Output:

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(20*Cosh[(c + d*x) 
/2] - 10*Cosh[(3*(c + d*x))/2] - 2*Cosh[(5*(c + d*x))/2] + (5*I)*d*x*Cosh[ 
(5*c)/2]*CoshIntegral[(5*d*x)/2] - (10*I)*d*x*CoshIntegral[(d*x)/2]*(Cosh[ 
c/2] - I*Sinh[c/2]) + 15*d*x*CoshIntegral[(3*d*x)/2]*((-I)*Cosh[(3*c)/2] + 
 Sinh[(3*c)/2]) + 5*d*x*CoshIntegral[(5*d*x)/2]*Sinh[(5*c)/2] + (20*I)*Sin 
h[(c + d*x)/2] + (10*I)*Sinh[(3*(c + d*x))/2] - (2*I)*Sinh[(5*(c + d*x))/2 
] - 10*d*x*Cosh[c/2]*SinhIntegral[(d*x)/2] - (10*I)*d*x*Sinh[c/2]*SinhInte 
gral[(d*x)/2] + 15*d*x*Cosh[(3*c)/2]*SinhIntegral[(3*d*x)/2] - (15*I)*d*x* 
Sinh[(3*c)/2]*SinhIntegral[(3*d*x)/2] + 5*d*x*Cosh[(5*c)/2]*SinhIntegral[( 
5*d*x)/2] + (5*I)*d*x*Sinh[(5*c)/2]*SinhIntegral[(5*d*x)/2]))/(8*x*(Cosh[( 
c + d*x)/2] + I*Sinh[(c + d*x)/2])^5)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.54, 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.

\(\displaystyle \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+a \sin (i c+i d x))^{5/2}}{x^2}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \int \frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \int \frac {\sin \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^5}{x^2}dx\)

\(\Big \downarrow \) 3794

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (\frac {5}{2} i d \int \left (\frac {3 \cosh \left (\frac {3 c}{2}+\frac {3 d x}{2}+\frac {i \pi }{4}\right )}{16 x}-\frac {i \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{8 x}+\frac {i \sinh \left (\frac {5 c}{2}+\frac {5 d x}{2}+\frac {i \pi }{4}\right )}{16 x}\right )dx-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (\frac {5}{2} i d \left (\frac {1}{16} i \sinh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )+\frac {3}{16} i \sinh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right )-\frac {1}{8} i \sinh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right )-\frac {1}{8} i \cosh \left (\frac {1}{4} (2 c+i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right )+\frac {3}{16} i \cosh \left (\frac {1}{4} (6 c-i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+\frac {1}{16} i \cosh \left (\frac {5 c}{2}+\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )\right )-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{x}\right )\)

Input:

Int[(a + I*a*Sinh[c + d*x])^(5/2)/x^2,x]
 

Output:

4*a^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*(-(Cosh[c 
/2 + (I/4)*Pi + (d*x)/2]^5/x) + ((5*I)/2)*d*((I/16)*CoshIntegral[(5*d*x)/2 
]*Sinh[(5*c)/2 + (I/4)*Pi] + ((3*I)/16)*CoshIntegral[(3*d*x)/2]*Sinh[(6*c 
- I*Pi)/4] - (I/8)*CoshIntegral[(d*x)/2]*Sinh[(2*c + I*Pi)/4] - (I/8)*Cosh 
[(2*c + I*Pi)/4]*SinhIntegral[(d*x)/2] + ((3*I)/16)*Cosh[(6*c - I*Pi)/4]*S 
inhIntegral[(3*d*x)/2] + (I/16)*Cosh[(5*c)/2 + (I/4)*Pi]*SinhIntegral[(5*d 
*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]

\[\int \frac {\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}{x^{2}}d x\]

Input:

int((a+I*a*sinh(d*x+c))^(5/2)/x^2,x)
 

Output:

int((a+I*a*sinh(d*x+c))^(5/2)/x^2,x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*sinh(d*x+c))**(5/2)/x**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\int { \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="maxima")
 

Output:

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^2, x)
 

Giac [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\int { \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{2}} \,d x } \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^2,x, algorithm="giac")
 

Output:

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\int \frac {{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{x^2} \,d x \] Input:

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^2,x)
 

Output:

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^2, x)
 

Reduce [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^2} \, dx=\sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}}{x^{2}}d x -\left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )^{2}}{x^{2}}d x \right )+2 \left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )}{x^{2}}d x \right ) i \right ) \] Input:

int((a+I*a*sinh(d*x+c))^(5/2)/x^2,x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sinh(c + d*x)*i + 1)/x**2,x) - int((sqrt(sinh(c + d 
*x)*i + 1)*sinh(c + d*x)**2)/x**2,x) + 2*int((sqrt(sinh(c + d*x)*i + 1)*si 
nh(c + d*x))/x**2,x)*i)