\(\int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx\) [135]

Optimal result

Integrand size = 21, antiderivative size = 536 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=-\frac {2 a^2 \cosh ^4\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x^2}-\frac {25}{32} i a^2 d^2 \text {Chi}\left (\frac {5 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {5}{16} i a^2 d^2 \text {Chi}\left (\frac {d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}+\frac {45}{32} i a^2 d^2 \text {Chi}\left (\frac {3 d x}{2}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \sqrt {a+i a \sinh (c+d x)}-\frac {5 a^2 d \cosh ^3\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)}}{x}+\frac {5}{16} i a^2 d^2 \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {d x}{2}\right )+\frac {45}{32} i a^2 d^2 \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {3 d x}{2}\right )-\frac {25}{32} i a^2 d^2 \cosh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {sech}\left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sqrt {a+i a \sinh (c+d x)} \text {Shi}\left (\frac {5 d x}{2}\right ) \] Output:

-2*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^2-25/32* 
a^2*d^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)+5/16*a^2*d^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)+45/32*I*a^2*d^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*a^2*d*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^3*sinh(1/2*c+1/4*I*Pi+1/2*d*x 
)*(a+I*a*sinh(d*x+c))^(1/2)/x+5/16*a^2*d^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)+45/32*I*a^2*d^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)-25/32*a^2*d^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 [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(4751\) vs. \(2(536)=1072\).

Time = 10.21 (sec) , antiderivative size = 4751, normalized size of antiderivative = 8.86 \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + I*a*Sinh[c + d*x])^(5/2)/x^3,x]
 

Output:

(2*((1/128 + I/128)*Cosh[5*(c/2 + (d*x)/2)] - (1/128 + I/128)*Sinh[5*(c/2 
+ (d*x)/2)])*(a + I*a*Sinh[c + d*x])^(5/2)*((-4*I)*d^3 - (10*I)*c*d^3 + (2 
0*I)*d^3*(c/2 + (d*x)/2) + 20*d^3*Cosh[2*(c/2 + (d*x)/2)] + 30*c*d^3*Cosh[ 
2*(c/2 + (d*x)/2)] - 60*d^3*(c/2 + (d*x)/2)*Cosh[2*(c/2 + (d*x)/2)] + (40* 
I)*d^3*Cosh[4*(c/2 + (d*x)/2)] + (20*I)*c*d^3*Cosh[4*(c/2 + (d*x)/2)] - (4 
0*I)*d^3*(c/2 + (d*x)/2)*Cosh[4*(c/2 + (d*x)/2)] - 40*d^3*Cosh[6*(c/2 + (d 
*x)/2)] + 20*c*d^3*Cosh[6*(c/2 + (d*x)/2)] - 40*d^3*(c/2 + (d*x)/2)*Cosh[6 
*(c/2 + (d*x)/2)] - (20*I)*d^3*Cosh[8*(c/2 + (d*x)/2)] + (30*I)*c*d^3*Cosh 
[8*(c/2 + (d*x)/2)] - (60*I)*d^3*(c/2 + (d*x)/2)*Cosh[8*(c/2 + (d*x)/2)] + 
 4*d^3*Cosh[10*(c/2 + (d*x)/2)] - 10*c*d^3*Cosh[10*(c/2 + (d*x)/2)] + 20*d 
^3*(c/2 + (d*x)/2)*Cosh[10*(c/2 + (d*x)/2)] - (10*I)*c^2*d^3*Cosh[c/2 - 5* 
(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] + (40*I)*c*d^3*(c/2 + (d*x)/2)*Cosh 
[c/2 - 5*(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] - (40*I)*d^3*(c/2 + (d*x)/ 
2)^2*Cosh[c/2 - 5*(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] + 10*c^2*d^3*Cosh 
[c/2 + 5*(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] - 40*c*d^3*(c/2 + (d*x)/2) 
*Cosh[c/2 + 5*(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] + 40*d^3*(c/2 + (d*x) 
/2)^2*Cosh[c/2 + 5*(c/2 + (d*x)/2)]*CoshIntegral[(d*x)/2] - 45*c^2*d^3*Cos 
h[(3*c)/2 - 5*(c/2 + (d*x)/2)]*CoshIntegral[(-3*c)/2 + 3*(c/2 + (d*x)/2)] 
+ 180*c*d^3*(c/2 + (d*x)/2)*Cosh[(3*c)/2 - 5*(c/2 + (d*x)/2)]*CoshIntegral 
[(-3*c)/2 + 3*(c/2 + (d*x)/2)] - 180*d^3*(c/2 + (d*x)/2)^2*Cosh[(3*c)/2...
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 408, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3800, 3042, 3795, 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.

\(\displaystyle \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+a \sin (i c+i d x))^{5/2}}{x^3}dx\)

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3795

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (\frac {25}{8} d^2 \int \frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{x}dx-\frac {5}{2} d^2 \int \frac {\cosh ^3\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{x}dx-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 x^2}-\frac {5 d \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (-\frac {5}{2} d^2 \int \frac {\sin \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^3}{x}dx+\frac {25}{8} d^2 \int \frac {\sin \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^5}{x}dx-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 x^2}-\frac {5 d \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 x}\right )\)

\(\Big \downarrow \) 3793

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (-\frac {5}{2} d^2 \int \left (\frac {3 i \sinh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{4 x}+\frac {i \sinh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{4 x}\right )dx+\frac {25}{8} d^2 \int \left (\frac {5 i \sinh \left (\frac {1}{4} (2 c-i \pi )+\frac {d x}{2}\right )}{8 x}+\frac {5 i \sinh \left (\frac {1}{4} (6 c+i \pi )+\frac {3 d x}{2}\right )}{16 x}-\frac {i \sinh \left (\frac {1}{4} (10 c-i \pi )+\frac {5 d x}{2}\right )}{16 x}\right )dx-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 x^2}-\frac {5 d \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 a^2 \text {sech}\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \sqrt {a+i a \sinh (c+d x)} \left (-\frac {5}{2} d^2 \left (\frac {3}{4} i \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right )+\frac {1}{4} i \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right )+\frac {3}{4} i \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right )+\frac {1}{4} i \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right )\right )+\frac {25}{8} d^2 \left (-\frac {1}{16} i \sinh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Chi}\left (\frac {5 d x}{2}\right )+\frac {5}{8} i \sinh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Chi}\left (\frac {d x}{2}\right )+\frac {5}{16} i \sinh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Chi}\left (\frac {3 d x}{2}\right )+\frac {5}{8} i \cosh \left (\frac {1}{4} (2 c-i \pi )\right ) \text {Shi}\left (\frac {d x}{2}\right )+\frac {5}{16} i \cosh \left (\frac {1}{4} (6 c+i \pi )\right ) \text {Shi}\left (\frac {3 d x}{2}\right )-\frac {1}{16} i \cosh \left (\frac {5 c}{2}-\frac {i \pi }{4}\right ) \text {Shi}\left (\frac {5 d x}{2}\right )\right )-\frac {\cosh ^5\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{2 x^2}-\frac {5 d \sinh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right ) \cosh ^4\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{4 x}\right )\)

Input:

Int[(a + I*a*Sinh[c + d*x])^(5/2)/x^3,x]
 

Output:

4*a^2*Sech[c/2 + (I/4)*Pi + (d*x)/2]*Sqrt[a + I*a*Sinh[c + d*x]]*(-1/2*Cos 
h[c/2 + (I/4)*Pi + (d*x)/2]^5/x^2 - (5*d*Cosh[c/2 + (I/4)*Pi + (d*x)/2]^4* 
Sinh[c/2 + (I/4)*Pi + (d*x)/2])/(4*x) - (5*d^2*(((3*I)/4)*CoshIntegral[(d* 
x)/2]*Sinh[(2*c - I*Pi)/4] + (I/4)*CoshIntegral[(3*d*x)/2]*Sinh[(6*c + I*P 
i)/4] + ((3*I)/4)*Cosh[(2*c - I*Pi)/4]*SinhIntegral[(d*x)/2] + (I/4)*Cosh[ 
(6*c + I*Pi)/4]*SinhIntegral[(3*d*x)/2]))/2 + (25*d^2*((-1/16*I)*CoshInteg 
ral[(5*d*x)/2]*Sinh[(5*c)/2 - (I/4)*Pi] + ((5*I)/8)*CoshIntegral[(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)*C 
osh[(6*c + I*Pi)/4]*SinhIntegral[(3*d*x)/2] - (I/16)*Cosh[(5*c)/2 - (I/4)* 
Pi]*SinhIntegral[(5*d*x)/2]))/8)
 

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 3795

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ 
b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) 
*(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2)))   Int[(c + 
d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* 
(m + 2)))   Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, 
c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
 

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]

\[\int \frac {\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}{x^{3}}d x\]

Input:

int((a+I*a*sinh(d*x+c))^(5/2)/x^3,x)
 

Output:

int((a+I*a*sinh(d*x+c))^(5/2)/x^3,x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^3,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*sinh(d*x+c))**(5/2)/x**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\int { \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{3}} \,d x } \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^3,x, algorithm="maxima")
 

Output:

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^3, x)
 

Giac [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\int { \frac {{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{x^{3}} \,d x } \] Input:

integrate((a+I*a*sinh(d*x+c))^(5/2)/x^3,x, algorithm="giac")
 

Output:

integrate((I*a*sinh(d*x + c) + a)^(5/2)/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\int \frac {{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{x^3} \,d x \] Input:

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^3,x)
 

Output:

int((a + a*sinh(c + d*x)*1i)^(5/2)/x^3, x)
 

Reduce [F]

\[ \int \frac {(a+i a \sinh (c+d x))^{5/2}}{x^3} \, dx=\sqrt {a}\, a^{2} \left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}}{x^{3}}d x -\left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )^{2}}{x^{3}}d x \right )+2 \left (\int \frac {\sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )}{x^{3}}d x \right ) i \right ) \] Input:

int((a+I*a*sinh(d*x+c))^(5/2)/x^3,x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sinh(c + d*x)*i + 1)/x**3,x) - int((sqrt(sinh(c + d 
*x)*i + 1)*sinh(c + d*x)**2)/x**3,x) + 2*int((sqrt(sinh(c + d*x)*i + 1)*si 
nh(c + d*x))/x**3,x)*i)