\(\int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx\) [233]

Optimal result

Integrand size = 35, antiderivative size = 190 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\frac {2 i b d \sqrt {1+c^2 x^2}}{3 c f^2 (i+c x) \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i d (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (1+c^2 x^2\right )}+\frac {b d \sqrt {1+c^2 x^2} \log (i+c x)}{3 c f^2 \sqrt {d+i c d x} \sqrt {f-i c f x}} \] Output:

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

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {i d \sqrt {f-i c f x} \left ((-i+c x) \left (-i a+a c x+b \sqrt {1+c^2 x^2}\right )+b (-i+c x)^2 \text {arcsinh}(c x)-b (i+c x) \sqrt {1+c^2 x^2} \log (d (-1+i c x))\right )}{3 c f^3 (i+c x)^2 \sqrt {d+i c d x}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6211, 27, 6252, 27, 456, 49, 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.

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

\(\Big \downarrow \) 6211

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 6252

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 456

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(d^3*(1 + c^2*x^2)^(5/2)*(((-1/3*I)*(1 + I*c*x)^3*(a + b*ArcSinh[c*x]))/(c 
*(1 + c^2*x^2)^(3/2)) + (I/3)*b*(2/(c*(I + c*x)) - (I*Log[I + 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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 456

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && 
EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] &&  !Integ 
erQ[n]))
 

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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (157 ) = 314\).

Time = 7.76 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.29

method	result	size
default	\(a \left (-\frac {i \sqrt {i c d x +d}}{c f \left (-i c f x +f \right )^{\frac {3}{2}}}-d \left (-\frac {i \sqrt {i c d x +d}}{3 d c f \left (-i c f x +f \right )^{\frac {3}{2}}}-\frac {i \sqrt {i c d x +d}}{3 c d \,f^{2} \sqrt {-i c f x +f}}\right )\right )-\frac {b \left (i \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{3} c^{3}-i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+3 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-3 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}-3 i \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x c -c^{2} x^{2}-i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )-2 i x c +\sqrt {c^{2} x^{2}+1}\, x c -i \sqrt {c^{2} x^{2}+1}-\operatorname {arcsinh}\left (x c \right )+\ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )+1\right ) \left (x^{3} c^{3}-3 i c^{2} x^{2}+x^{2} c^{2} \sqrt {c^{2} x^{2}+1}-3 x c +i+\sqrt {c^{2} x^{2}+1}\right ) \sqrt {-i \left (x c +i\right ) f}\, \sqrt {i \left (x c -i\right ) d}}{3 f^{3} \left (3 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )^{2} c}\)	\(435\)
parts	\(a \left (-\frac {i \sqrt {i c d x +d}}{c f \left (-i c f x +f \right )^{\frac {3}{2}}}-d \left (-\frac {i \sqrt {i c d x +d}}{3 d c f \left (-i c f x +f \right )^{\frac {3}{2}}}-\frac {i \sqrt {i c d x +d}}{3 c d \,f^{2} \sqrt {-i c f x +f}}\right )\right )-\frac {b \left (i \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{3} c^{3}-i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}+3 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-3 \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x^{2} c^{2}-3 i \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right ) x c -c^{2} x^{2}-i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )-2 i x c +\sqrt {c^{2} x^{2}+1}\, x c -i \sqrt {c^{2} x^{2}+1}-\operatorname {arcsinh}\left (x c \right )+\ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )+1\right ) \left (x^{3} c^{3}-3 i c^{2} x^{2}+x^{2} c^{2} \sqrt {c^{2} x^{2}+1}-3 x c +i+\sqrt {c^{2} x^{2}+1}\right ) \sqrt {-i \left (x c +i\right ) f}\, \sqrt {i \left (x c -i\right ) d}}{3 f^{3} \left (3 c^{2} x^{2}-1\right ) \left (c^{2} x^{2}+1\right )^{2} c}\)	\(435\)

Input:

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

Output:

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

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. 548 vs. \(2 (146) = 292\).

Time = 0.16 (sec) , antiderivative size = 548, normalized size of antiderivative = 2.88 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {4 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x + 2 \, {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (i \, c^{9} f^{3} x^{4} - 2 \, c^{8} f^{3} x^{3} + i \, c^{7} f^{3} x^{2} - 2 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) - {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (-i \, c^{9} f^{3} x^{4} + 2 \, c^{8} f^{3} x^{3} - i \, c^{7} f^{3} x^{2} + 2 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 i \, a c x - a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{6 \, {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {6 \, \sqrt {d}}{3 i \, c^{3} f^{\frac {5}{2}} x - 3 \, c^{2} f^{\frac {5}{2}}} - \frac {\sqrt {d} \log \left (c x + i\right )}{c^{2} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (-\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} + 2 i \, c^{2} f^{3} x - c f^{3}} - \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (-\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} + 2 i \, c^{2} f^{3} x - c f^{3}} - \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(sqrt(d)*( - 3*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)*b*c**2*i*x + 3*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)*b*c - a*c**2*i*x**2 - 2*a*c*x + a*i) 
)/(3*sqrt(f)*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*c*f**2*(c*i*x - 1))