\(\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx\) [567]

Optimal result

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

[Out]

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)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {5796, 667, 197, 5837, 641, 46, 209, 266} \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {d^2 x \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^2 (1+i c x) \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d^2 \left (c^2 x^2+1\right )^{5/2} \arctan (c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[In]

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

[Out]

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

Rule 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 667

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)*((a + c*x^2)^(p + 1)/(c*(p
+ 1))), x] - Dist[e^2*((p + 2)/(c*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 5796

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :>
Dist[(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 5837

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], 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])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x)^2 (a+b \text {arcsinh}(c x))}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(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 {\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {2 i d^2 (1+i c x)}{3 c \left (1+c^2 x^2\right )^2}+\frac {d^2 x}{3 \left (1+c^2 x^2\right )}\right ) \, dx}{(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 {\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1+i c x}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b c d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 (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 {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}}+\frac {\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{(1-i c x)^2 (1+i c x)} \, dx}{3 (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 {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}}+\frac {\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {1}{2 (i+c x)^2}+\frac {1}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{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}}{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 {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}}+\frac {\left (i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{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}}{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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.47 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {\sqrt {f-i c f x} \left ((2 i+c x) \left (a+i a c x+i b \sqrt {1+c^2 x^2}\right )+i b \left (2+i c x+c^2 x^2\right ) \text {arcsinh}(c x)+b (1-i c x) \sqrt {1+c^2 x^2} \log (d (-1+i c x))\right )}{3 c f^3 (i+c x)^2 \sqrt {d+i c d x}} \]

[In]

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

[Out]

(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])

Maple [F]

\[\int \frac {a +b \,\operatorname {arcsinh}\left (c x \right )}{\left (-i c f x +f \right )^{\frac {5}{2}} \sqrt {i c d x +d}}d x\]

[In]

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

[Out]

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

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 = 0.33 (sec) , antiderivative size = 576, normalized size of antiderivative = 1.96 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=-\frac {2 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x - 2 \, {\left (b c^{2} x^{2} + i \, b c x + 2 \, b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (i \, c^{9} d f^{3} x^{4} - 2 \, c^{8} d f^{3} x^{3} + i \, c^{7} d f^{3} x^{2} - 2 \, c^{6} d f^{3} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) + {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (-i \, c^{9} d f^{3} x^{4} + 2 \, c^{8} d f^{3} x^{3} - i \, c^{7} d f^{3} x^{2} + 2 \, c^{6} d f^{3} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) - 2 \, {\left (a c^{2} x^{2} + i \, a c x + 2 \, a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{6 \, {\left (c^{4} d f^{3} x^{3} + i \, c^{3} d f^{3} x^{2} + c^{2} d f^{3} x + i \, c d f^{3}\right )}} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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)

none

Time = 0.32 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {3}{3 i \, c^{3} \sqrt {d} f^{\frac {5}{2}} x - 3 \, c^{2} \sqrt {d} f^{\frac {5}{2}}} + \frac {\log \left (c x + i\right )}{c^{2} \sqrt {d} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} + 2 i \, c^{2} d f^{3} x - c d f^{3}} + \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} d f^{3} x + 3 \, c d f^{3}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} + 2 i \, c^{2} d f^{3} x - c d f^{3}} + \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} d f^{3} x + 3 \, c d f^{3}}\right )} \]

[In]

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

[Out]

-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]

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

[In]

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

[Out]

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

[In]

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

[Out]

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