\(\int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx\) [269]

Optimal result

Integrand size = 37, antiderivative size = 529 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 b d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {4 b^2 d \sqrt {1+c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 b d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 i b^2 d \sqrt {1+c^2 x^2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}} \] Output:

1/3*d*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^2/c/f^2/(d+I*c*d*x)^(1/2)/(f-I* 
c*f*x)^(1/2)+4/3*b*d*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))*ln(1+I/(c*x+(c^2 
*x^2+1)^(1/2)))/c/f^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-4/3*b^2*d*(c^2*x 
^2+1)^(1/2)*polylog(2,-I/(c*x+(c^2*x^2+1)^(1/2)))/c/f^2/(d+I*c*d*x)^(1/2)/ 
(f-I*c*f*x)^(1/2)+2/3*b*d*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))*sec(1/4*Pi+ 
1/2*I*arcsinh(c*x))^2/c/f^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+4/3*I*b^2* 
d*(c^2*x^2+1)^(1/2)*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/f^2/(d+I*c*d*x)^(1/2) 
/(f-I*c*f*x)^(1/2)+1/3*I*d*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^2*tan(1/4* 
Pi+1/2*I*arcsinh(c*x))/c/f^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-1/3*I*d*( 
c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^2*sec(1/4*Pi+1/2*I*arcsinh(c*x))^2*tan 
(1/4*Pi+1/2*I*arcsinh(c*x))/c/f^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 9.65 (sec) , antiderivative size = 788, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((2*I)/3)*a^2)/(f^3*(I + c* 
x)^2) - a^2/(3*f^3*(I + c*x))))/c - ((I/3)*a*b*Sqrt[I*((-I)*d + c*d*x)]*Sq 
rt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(Cosh[ArcSinh[c*x]/2] + 
I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcT 
an[Coth[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]])) + Cosh[ArcSinh[c*x]/ 
2]*(4*I + 3*ArcSinh[c*x] - 6*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*I)*Log[Sqrt 
[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(I*ArcSinh[c*x] + (2*I)*ArcTan[Coth 
[ArcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]) + 2*(1 + I*ArcSinh[c*x] + (2*I 
)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x 
]/2]))/(c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[Ar 
cSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) - ((I/3)*b^2*(-I + c*x)*Sqrt[I* 
((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*((-1 
 - I)*ArcSinh[c*x]^2 - (2*ArcSinh[c*x]*(2*I + ArcSinh[c*x]))/(I + c*x) - ( 
2*I)*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] - I*Pi*(3*ArcSinh 
[c*x] - 4*Log[1 + E^ArcSinh[c*x]] - 2*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4 
]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) + 4*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (4 
*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSi 
nh[c*x]/2])^3 + (2*(4 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSin 
h[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(c*f^3*Sqrt[-(((-I)*d + c*d*x)*(I*f 
+ c*f*x))]*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x...
 

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.51, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6211, 27, 6259, 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.

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

\(\Big \downarrow \) 6211

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 6259

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {(a+b \text {arcsinh}(c x))^2}{3 c}+\frac {4 b \log \left (1+i e^{-\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {2 b \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}-\frac {4 b^2 \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c}+\frac {4 i b^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\)

Input:

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

Output:

(d^3*(1 + c^2*x^2)^(5/2)*((a + b*ArcSinh[c*x])^2/(3*c) + (4*b*(a + b*ArcSi 
nh[c*x])*Log[1 + I/E^ArcSinh[c*x]])/(3*c) - (4*b^2*PolyLog[2, (-I)/E^ArcSi 
nh[c*x]])/(3*c) + (2*b*(a + b*ArcSinh[c*x])*Sec[Pi/4 + (I/2)*ArcSinh[c*x]] 
^2)/(3*c) + (((4*I)/3)*b^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/c + ((I/3)*(a + 
 b*ArcSinh[c*x])^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/c - ((I/3)*(a + b*ArcSi 
nh[c*x])^2*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]] 
)/c))/((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 6259

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d 
_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcSinh[c* 
x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{ 
a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IntegerQ[m] && ILtQ[p + 1/2, 0 
] && GtQ[d, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2756 vs. \(2 (447 ) = 894\).

Time = 7.65 (sec) , antiderivative size = 2757, normalized size of antiderivative = 5.21

method	result	size
default	\(\text {Expression too large to display}\)	\(2757\)
parts	\(\text {Expression too large to display}\)	\(2757\)

Input:

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

Output:

-b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1 
)^2*arcsinh(x*c)^2*x-2/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/ 
(3*c^2*x^2-1)/(c^2*x^2+1)^2*arcsinh(x*c)*x-8/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I 
*(I+x*c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)^2*c^4*x^5+4/3*b^2*(I*(x*c- 
I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/(c^2*x^2+1)^(1/2)/c/f^3*arcsinh(x*c)*ln(1 
-I*(x*c+(c^2*x^2+1)^(1/2)))-4/3*I*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^( 
1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)/c+8/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x* 
c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*c*x^2+2/3*b^2*(I*(x*c-I)*d 
)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)*arcsinh(x*c)*x+ 
4*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+ 
1)^(3/2)*c^3*x^4-4/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c 
^2*x^2-1)/(c^2*x^2+1)*c^2*x^3-1/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^( 
1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/c*arcsinh(x*c)^2-4/3*b^2*(I*(x*c- 
I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/c*arc 
sinh(x*c)+a^2*(-I/c/f*(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(3/2)-d*(-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)))-4/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/(3*c^2* 
x^2-1)/(c^2*x^2+1)*x+8/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/f^3/ 
(3*c^2*x^2-1)/(c^2*x^2+1)^2*x-4/3*b^2*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^( 
1/2)/f^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/c-2/3*b^2*(I*(x*c-I)*d)^(1/2)*...
 

Fricas [F]

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

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

Output:

-1/3*((b^2*c*x - I*b^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqr 
t(c^2*x^2 + 1))^2 - 3*(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3)*integral(1/3*( 
-3*I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2 + 2*(sqrt(c^2*x^2 + 1)*sqrt( 
I*c*d*x + d)*sqrt(-I*c*f*x + f)*b^2 - 3*I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x 
+ f)*a*b)*log(c*x + sqrt(c^2*x^2 + 1)))/(c^3*f^3*x^3 + 3*I*c^2*f^3*x^2 - 3 
*c*f^3*x - I*f^3), x))/(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {\sqrt {d}\, \left (-6 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {c i x +1}\, \mathit {asinh} \left (c x \right )}{\sqrt {-c i x +1}\, c^{2} x^{2}+2 \sqrt {-c i x +1}\, c i x -\sqrt {-c i x +1}}d x \right ) a b \,c^{2} i x +6 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {c i x +1}\, \mathit {asinh} \left (c x \right )}{\sqrt {-c i x +1}\, c^{2} x^{2}+2 \sqrt {-c i x +1}\, c i x -\sqrt {-c i x +1}}d x \right ) a b c -3 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {c i x +1}\, \mathit {asinh} \left (c x \right )^{2}}{\sqrt {-c i x +1}\, c^{2} x^{2}+2 \sqrt {-c i x +1}\, c i x -\sqrt {-c i x +1}}d x \right ) b^{2} c^{2} i x +3 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {c i x +1}\, \mathit {asinh} \left (c x \right )^{2}}{\sqrt {-c i x +1}\, c^{2} x^{2}+2 \sqrt {-c i x +1}\, c i x -\sqrt {-c i x +1}}d x \right ) b^{2} c -a^{2} c^{2} i \,x^{2}-2 a^{2} c x +a^{2} i \right )}{3 \sqrt {f}\, \sqrt {c i x +1}\, \sqrt {-c i x +1}\, c \,f^{2} \left (c i x -1\right )} \] Input:

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

Output:

(sqrt(d)*( - 6*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int((sqrt(c*i*x + 1)*asi 
nh(c*x))/(sqrt( - c*i*x + 1)*c**2*x**2 + 2*sqrt( - c*i*x + 1)*c*i*x - sqrt 
( - c*i*x + 1)),x)*a*b*c**2*i*x + 6*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int 
((sqrt(c*i*x + 1)*asinh(c*x))/(sqrt( - c*i*x + 1)*c**2*x**2 + 2*sqrt( - c* 
i*x + 1)*c*i*x - sqrt( - c*i*x + 1)),x)*a*b*c - 3*sqrt(c*i*x + 1)*sqrt( - 
c*i*x + 1)*int((sqrt(c*i*x + 1)*asinh(c*x)**2)/(sqrt( - c*i*x + 1)*c**2*x* 
*2 + 2*sqrt( - c*i*x + 1)*c*i*x - sqrt( - c*i*x + 1)),x)*b**2*c**2*i*x + 3 
*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int((sqrt(c*i*x + 1)*asinh(c*x)**2)/(s 
qrt( - c*i*x + 1)*c**2*x**2 + 2*sqrt( - c*i*x + 1)*c*i*x - sqrt( - c*i*x + 
 1)),x)*b**2*c - a**2*c**2*i*x**2 - 2*a**2*c*x + a**2*i))/(3*sqrt(f)*sqrt( 
c*i*x + 1)*sqrt( - c*i*x + 1)*c*f**2*(c*i*x - 1))