\(\int \frac {(a+i a \sinh (c+d x))^{5/2}}{x} \, dx\) [133]

Optimal result

Integrand size = 21, antiderivative size = 403 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x} \, dx=-\frac {1}{4} i a^2 \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{4} i a^2 \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{2} i a^2 \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {d x}{2}\right )+\frac {5}{4} i a^2 \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {3 d x}{2}\right )-\frac {1}{4} i a^2 \cosh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {5 d x}{2}\right ) \] Output:

-1/4*a^2*Chi(5/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)*cosh(5/2*c+1/4*I*Pi)*(a 
+I*a*sinh(d*x+c))^(1/2)+5/2*a^2*Chi(1/2*d*x)*sech(1/2*c+1/4*I*Pi+1/2*d*x)* 
cosh(1/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/4*I*a^2*Chi(3/2*d*x)*sech 
(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(3/2*c+1/4*I*Pi)*(a+I*a*sinh(d*x+c))^(1/2)+5/ 
2*a^2*sinh(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)+5/4*I*a^2*cosh(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)-1/4*a^2*sinh(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 = 4.32 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.60 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x} \, dx=\frac {a^2 (-i+\sinh (c+d x))^2 \sqrt {a+i a \sinh (c+d x)} \left (\cosh \left (\frac {5 c}{2}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )-10 \text {Chi}\left (\frac {d x}{2}\right ) \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right )+5 \text {Chi}\left (\frac {3 d x}{2}\right ) \left (\cosh \left (\frac {3 c}{2}\right )-i \sinh \left (\frac {3 c}{2}\right )\right )+i \text {Chi}\left (\frac {5 d x}{2}\right ) \sinh \left (\frac {5 c}{2}\right )-10 i \cosh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )-10 \sinh \left (\frac {c}{2}\right ) \text {Shi}\left (\frac {d x}{2}\right )-5 i \cosh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+5 \sinh \left (\frac {3 c}{2}\right ) \text {Shi}\left (\frac {3 d x}{2}\right )+i \cosh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )+\sinh \left (\frac {5 c}{2}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )\right )}{4 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:

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

Output:

(a^2*(-I + Sinh[c + d*x])^2*Sqrt[a + I*a*Sinh[c + d*x]]*(Cosh[(5*c)/2]*Cos 
hIntegral[(5*d*x)/2] - 10*CoshIntegral[(d*x)/2]*(Cosh[c/2] + I*Sinh[c/2]) 
+ 5*CoshIntegral[(3*d*x)/2]*(Cosh[(3*c)/2] - I*Sinh[(3*c)/2]) + I*CoshInte 
gral[(5*d*x)/2]*Sinh[(5*c)/2] - (10*I)*Cosh[c/2]*SinhIntegral[(d*x)/2] - 1 
0*Sinh[c/2]*SinhIntegral[(d*x)/2] - (5*I)*Cosh[(3*c)/2]*SinhIntegral[(3*d* 
x)/2] + 5*Sinh[(3*c)/2]*SinhIntegral[(3*d*x)/2] + I*Cosh[(5*c)/2]*SinhInte 
gral[(5*d*x)/2] + Sinh[(5*c)/2]*SinhIntegral[(5*d*x)/2]))/(4*(Cosh[(c + d* 
x)/2] + I*Sinh[(c + d*x)/2])^5)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.51, 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, 3793, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \int \left (\frac {5 i \sinh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{8 x}+\frac {5 i \sinh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{16 x}-\frac {i \sinh \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{16 x}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

4*a^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*((-1/16*I 
)*CoshIntegral[(5*d*x)/2]*Sinh[(5*c)/2 - (I/4)*Pi] + ((5*I)/8)*CoshIntegra 
l[(d*x)/2]*Sinh[(2*c - I*Pi)/4] + ((5*I)/16)*CoshIntegral[(3*d*x)/2]*Sinh[ 
(6*c + I*Pi)/4] + ((5*I)/8)*Cosh[(2*c - I*Pi)/4]*SinhIntegral[(d*x)/2] + ( 
(5*I)/16)*Cosh[(6*c + I*Pi)/4]*SinhIntegral[(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 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && 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]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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