\(\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx\) [234]

Optimal result

Integrand size = 35, antiderivative size = 294 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {i b d^2 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^2 x \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d^2 \left (1+c^2 x^2\right )^{5/2} \arctan (c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \] Output:

1/3*I*b*d^2*(c^2*x^2+1)^(5/2)/c/(I+c*x)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2 
)-2/3*I*d^2*(1+I*c*x)*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/( 
f-I*c*f*x)^(5/2)+1/3*d^2*x*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(5 
/2)/(f-I*c*f*x)^(5/2)+1/3*I*b*d^2*(c^2*x^2+1)^(5/2)*arctan(c*x)/c/(d+I*c*d 
*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/6*b*d^2*(c^2*x^2+1)^(5/2)*ln(c^2*x^2+1)/c/(d 
+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)
 

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.47 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {\sqrt {f-i c f x} \left ((2 i+c x) \left (a+i a c x+i b \sqrt {1+c^2 x^2}\right )+i b \left (2+i c x+c^2 x^2\right ) \text {arcsinh}(c x)+b (1-i c x) \sqrt {1+c^2 x^2} \log (d (-1+i c x))\right )}{3 c f^3 (i+c x)^2 \sqrt {d+i c d x}} \] Input:

Integrate[(a + b*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(5/2)),x]
 

Output:

(Sqrt[f - I*c*f*x]*((2*I + c*x)*(a + I*a*c*x + I*b*Sqrt[1 + c^2*x^2]) + I* 
b*(2 + I*c*x + c^2*x^2)*ArcSinh[c*x] + b*(1 - I*c*x)*Sqrt[1 + c^2*x^2]*Log 
[d*(-1 + I*c*x)]))/(3*c*f^3*(I + c*x)^2*Sqrt[d + I*c*d*x])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.58, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6211, 27, 6252, 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.

\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx\)

\(\Big \downarrow \) 6211

\(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^2 (i c x+1)^2 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^2 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\)

\(\Big \downarrow \) 6252

\(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \left (-b c \int \left (\frac {x}{3 \left (c^2 x^2+1\right )}-\frac {2 i (i c x+1)}{3 c \left (c^2 x^2+1\right )^2}\right )dx+\frac {x (a+b \text {arcsinh}(c x))}{3 \sqrt {c^2 x^2+1}}-\frac {2 i (1+i c x) (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \left (\frac {x (a+b \text {arcsinh}(c x))}{3 \sqrt {c^2 x^2+1}}-\frac {2 i (1+i c x) (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}-b c \left (-\frac {i \arctan (c x)}{3 c^2}+\frac {i (-c x+i)}{3 c^2 \left (c^2 x^2+1\right )}+\frac {\log \left (c^2 x^2+1\right )}{6 c^2}\right )\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\)

Input:

Int[(a + b*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(5/2)),x]
 

Output:

(d^2*(1 + c^2*x^2)^(5/2)*((((-2*I)/3)*(1 + I*c*x)*(a + b*ArcSinh[c*x]))/(c 
*(1 + c^2*x^2)^(3/2)) + (x*(a + b*ArcSinh[c*x]))/(3*Sqrt[1 + c^2*x^2]) - b 
*c*(((I/3)*(I - c*x))/(c^2*(1 + c^2*x^2)) - ((I/3)*ArcTan[c*x])/c^2 + Log[ 
1 + c^2*x^2]/(6*c^2))))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 6211

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_ 
) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 + c^2*x 
^2)^q)   Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n, x], 
x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^ 
2 + e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
 

rule 6252

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + ( 
e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p 
, x]}, Simp[(a + b*ArcSinh[c*x])   u, x] - Simp[b*c   Int[1/Sqrt[1 + c^2*x^ 
2]   u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ 
[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])
 

Maple [A] (verified)

Time = 6.38 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.01

method	result	size
default	\(a \left (-\frac {i \sqrt {i c d x +d}}{3 d c f \left (-i c f x +f \right )^{\frac {3}{2}}}-\frac {i \sqrt {i c d x +d}}{3 c d \,f^{2} \sqrt {-i c f x +f}}\right )+\frac {b \left (\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-2 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{4} c^{4}+\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+i x^{3} c^{3}+2 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-4 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+3 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}-2 i \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )+i x c +\operatorname {arcsinh}\left (x c \right )-2 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )+1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{3 \left (c^{2} x^{2}+1\right )^{\frac {5}{2}} f^{3} c d}\)	\(297\)
parts	\(a \left (-\frac {i \sqrt {i c d x +d}}{3 d c f \left (-i c f x +f \right )^{\frac {3}{2}}}-\frac {i \sqrt {i c d x +d}}{3 c d \,f^{2} \sqrt {-i c f x +f}}\right )+\frac {b \left (\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-2 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{4} c^{4}+\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+i x^{3} c^{3}+2 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-4 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+3 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +c^{2} x^{2}-2 i \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )+i x c +\operatorname {arcsinh}\left (x c \right )-2 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )+1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{3 \left (c^{2} x^{2}+1\right )^{\frac {5}{2}} f^{3} c d}\)	\(297\)

Input:

int((a+b*arcsinh(x*c))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

a*(-1/3*I/d/c/f/(f-I*c*f*x)^(3/2)*(d+I*c*d*x)^(1/2)-1/3*I/c/d/f^2/(f-I*c*f 
*x)^(1/2)*(d+I*c*d*x)^(1/2))+1/3*b*(arcsinh(x*c)*c^4*x^4-2*ln(x*c+(c^2*x^2 
+1)^(1/2)+I)*x^4*c^4+arcsinh(x*c)*(c^2*x^2+1)^(1/2)*x^3*c^3+I*x^3*c^3+2*ar 
csinh(x*c)*c^2*x^2-4*ln(x*c+(c^2*x^2+1)^(1/2)+I)*x^2*c^2+3*arcsinh(x*c)*(c 
^2*x^2+1)^(1/2)*x*c+c^2*x^2-2*I*(c^2*x^2+1)^(1/2)*arcsinh(x*c)+I*x*c+arcsi 
nh(x*c)-2*ln(x*c+(c^2*x^2+1)^(1/2)+I)+1)*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f 
)^(1/2)/(c^2*x^2+1)^(5/2)/f^3/c/d
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (228) = 456\).

Time = 0.20 (sec) , antiderivative size = 576, normalized size of antiderivative = 1.96 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=-\frac {2 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x - 2 \, {\left (b c^{2} x^{2} + i \, b c x + 2 \, b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (i \, c^{9} d f^{3} x^{4} - 2 \, c^{8} d f^{3} x^{3} + i \, c^{7} d f^{3} x^{2} - 2 \, c^{6} d f^{3} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) + {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (-i \, c^{9} d f^{3} x^{4} + 2 \, c^{8} d f^{3} x^{3} - i \, c^{7} d f^{3} x^{2} + 2 \, c^{6} d f^{3} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) - 2 \, {\left (a c^{2} x^{2} + i \, a c x + 2 \, a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{6 \, {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )}} \] Input:

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x, algori 
thm="fricas")
 

Output:

-1/6*(2*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x - 2*( 
b*c^2*x^2 + I*b*c*x + 2*b)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + 
sqrt(c^2*x^2 + 1)) - (c^4*d*f^3*x^3 + I*c^3*d*f^3*x^2 + c^2*d*f^3*x + I*c* 
d*f^3)*sqrt(b^2/(c^2*d*f^5))*log(-1/8*((-I*b*c^6*x^2 + 2*b*c^5*x + 2*I*b*c 
^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) + (I*c^9*d*f^3* 
x^4 - 2*c^8*d*f^3*x^3 + I*c^7*d*f^3*x^2 - 2*c^6*d*f^3*x)*sqrt(b^2/(c^2*d*f 
^5)))/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + I*b)) + (c^4*d*f^3*x^3 + I*c^3*d* 
f^3*x^2 + c^2*d*f^3*x + I*c*d*f^3)*sqrt(b^2/(c^2*d*f^5))*log(-1/8*((-I*b*c 
^6*x^2 + 2*b*c^5*x + 2*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(- 
I*c*f*x + f) + (-I*c^9*d*f^3*x^4 + 2*c^8*d*f^3*x^3 - I*c^7*d*f^3*x^2 + 2*c 
^6*d*f^3*x)*sqrt(b^2/(c^2*d*f^5)))/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + I*b) 
) - 2*(a*c^2*x^2 + I*a*c*x + 2*a)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))/(c 
^4*d*f^3*x^3 + I*c^3*d*f^3*x^2 + c^2*d*f^3*x + I*c*d*f^3)
 

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\sqrt {i d \left (c x - i\right )} \left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+b*asinh(c*x))/(d+I*c*d*x)**(1/2)/(f-I*c*f*x)**(5/2),x)
 

Output:

Integral((a + b*asinh(c*x))/(sqrt(I*d*(c*x - I))*(-I*f*(c*x + I))**(5/2)), 
 x)
 

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {3}{3 i \, c^{3} \sqrt {d} f^{\frac {5}{2}} x - 3 \, c^{2} \sqrt {d} f^{\frac {5}{2}}} + \frac {\log \left (c x + i\right )}{c^{2} \sqrt {d} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} + 2 i \, c^{2} d f^{3} x - c d f^{3}} + \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} d f^{3} x + 3 \, c d f^{3}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} + 2 i \, c^{2} d f^{3} x - c d f^{3}} + \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} d f^{3} x + 3 \, c d f^{3}}\right )} \] Input:

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x, algori 
thm="maxima")
 

Output:

-1/3*b*c*(3/(3*I*c^3*sqrt(d)*f^(5/2)*x - 3*c^2*sqrt(d)*f^(5/2)) + log(c*x 
+ I)/(c^2*sqrt(d)*f^(5/2))) - 1/3*b*(-I*sqrt(c^2*d*f*x^2 + d*f)/(c^3*d*f^3 
*x^2 + 2*I*c^2*d*f^3*x - c*d*f^3) + 3*I*sqrt(c^2*d*f*x^2 + d*f)/(-3*I*c^2* 
d*f^3*x + 3*c*d*f^3))*arcsinh(c*x) - 1/3*a*(-I*sqrt(c^2*d*f*x^2 + d*f)/(c^ 
3*d*f^3*x^2 + 2*I*c^2*d*f^3*x - c*d*f^3) + 3*I*sqrt(c^2*d*f*x^2 + d*f)/(-3 
*I*c^2*d*f^3*x + 3*c*d*f^3))
 

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{\sqrt {i \, c d x + d} {\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x, algori 
thm="giac")
 

Output:

integrate((b*arcsinh(c*x) + a)/(sqrt(I*c*d*x + d)*(-I*c*f*x + f)^(5/2)), x 
)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int((a + b*asinh(c*x))/((d + c*d*x*1i)^(1/2)*(f - c*f*x*1i)^(5/2)),x)
 

Output:

int((a + b*asinh(c*x))/((d + c*d*x*1i)^(1/2)*(f - c*f*x*1i)^(5/2)), x)
 

Reduce [F]

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

int((a+b*asinh(c*x))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x)
 

Output:

( - 3*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int(asinh(c*x)/(sqrt(c*i*x + 1)*s 
qrt( - c*i*x + 1)*c**2*x**2 + 2*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*c*i*x - 
 sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)),x)*b*c**3*x**2 - 3*sqrt(c*i*x + 1)*sq 
rt( - c*i*x + 1)*int(asinh(c*x)/(sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*c**2*x 
**2 + 2*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*c*i*x - sqrt(c*i*x + 1)*sqrt( - 
 c*i*x + 1)),x)*b*c + a*c**3*x**3 + 3*a*c*x - 2*a*i)/(3*sqrt(f)*sqrt(d)*sq 
rt(c*i*x + 1)*sqrt( - c*i*x + 1)*c*f**2*(c**2*x**2 + 1))