\(\int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx\) [565]

Optimal result

Integrand size = 35, antiderivative size = 362 \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\frac {4 i b d^4 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 (1+i c x)^3 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^4 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {5796, 683, 667, 221, 5837, 641, 45, 31, 5783} \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\frac {2 i d^4 (1+i c x) \left (c^2 x^2+1\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 (1+i c x)^3 \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (c^2 x^2+1\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d^4 \left (c^2 x^2+1\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b d^4 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^4 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[In]

Int[((d + I*c*d*x)^(3/2)*(a + b*ArcSinh[c*x]))/(f - I*c*f*x)^(5/2),x]

[Out]

(((4*I)/3)*b*d^4*(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)) - (b*d^4*(1 + c^2*
x^2)^(5/2)*ArcSinh[c*x]^2)/(2*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((2*I)/3)*d^4*(1 + I*c*x)^3*(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)) + ((2*I)*d^4*(1 + I*c*x)*(1 + c^2*x^
2)^2*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (d^4*(1 + c^2*x^2)^(5/2)*ArcSinh[c*x]
*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (8*b*d^4*(1 + c^2*x^2)^(5/2)*Log[I + c*x]
)/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

Rule 31

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 221

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

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 683

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -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)^4 (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^4 (1+i c x)^3 \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 {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (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^4 (1+i c x)^3}{3 c \left (1+c^2 x^2\right )^2}+\frac {2 i d^4 (1+i c x)}{c \left (1+c^2 x^2\right )}+\frac {d^4 \text {arcsinh}(c x)}{c \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {2 i d^4 (1+i c x)^3 \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 {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(1+i c x)^3}{\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 (2 i b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1+i c x}{1+c^2 x^2} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\text {arcsinh}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {b d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 (1+i c x)^3 \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 {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1+i c x}{(1-i c x)^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{1-i c x} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = -\frac {b d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 (1+i c x)^3 \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 {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^4 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b d^4 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {2}{(i+c x)^2}-\frac {i}{i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ & = \frac {4 i b d^4 \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 {b d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 (1+i c x)^3 \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 {2 i d^4 (1+i c x) \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^4 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 706, normalized size of antiderivative = 1.95 \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\frac {-\frac {16 a d (i+2 c x) \sqrt {d+i c d x} \sqrt {f-i c f x}}{f^3 (i+c x)^2}+\frac {12 a d^{3/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{f^{5/2}}-\frac {2 i b d \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right ) \left (-\cosh \left (\frac {3}{2} \text {arcsinh}(c x)\right ) \left (\text {arcsinh}(c x)-2 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\frac {1}{2} i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (4 i+3 \text {arcsinh}(c x)-6 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\frac {3}{2} i \log \left (1+c^2 x^2\right )\right )+2 \left (2+2 i \text {arcsinh}(c x)+4 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (1+c^2 x^2\right )+\frac {1}{2} \sqrt {1+c^2 x^2} \left (2 i \text {arcsinh}(c x)+4 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )}{f^3 (1+i c x) \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^4}+\frac {b d \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right ) \left (\cosh \left (\frac {3}{2} \text {arcsinh}(c x)\right ) \left ((14 i-3 \text {arcsinh}(c x)) \text {arcsinh}(c x)+28 i \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )-7 \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (8+6 i \text {arcsinh}(c x)+9 \text {arcsinh}(c x)^2-84 i \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+21 \log \left (1+c^2 x^2\right )\right )-2 i \left (4+4 i \text {arcsinh}(c x)+6 \text {arcsinh}(c x)^2-56 i \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+14 \log \left (1+c^2 x^2\right )+\sqrt {1+c^2 x^2} \left (\text {arcsinh}(c x) (14 i+3 \text {arcsinh}(c x))-28 i \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+7 \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )}{f^3 (1+i c x) \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^4}}{12 c} \]

[In]

Integrate[((d + I*c*d*x)^(3/2)*(a + b*ArcSinh[c*x]))/(f - I*c*f*x)^(5/2),x]

[Out]

((-16*a*d*(I + 2*c*x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(f^3*(I + c*x)^2) + (12*a*d^(3/2)*Log[c*d*f*x + Sqr
t[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/f^(5/2) - ((2*I)*b*d*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(C
osh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSin
h[c*x]/2]] + (I/2)*Log[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)/2)*Log[1 + c^2*x^2]) + 2*(2 + (2*I)*ArcSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[1 +
c^2*x^2] + (Sqrt[1 + c^2*x^2]*((2*I)*ArcSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2]))/2
)*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) + (b*d*Sqrt[d + I
*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(Cosh[(3*ArcSinh[c*x])/2]*((14*I - 3
*ArcSinh[c*x])*ArcSinh[c*x] + (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] - 7*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]
*(8 + (6*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 - (84*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 21*Log[1 + c^2*x^2]) - (2*
I)*(4 + (4*I)*ArcSinh[c*x] + 6*ArcSinh[c*x]^2 - (56*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 14*Log[1 + c^2*x^2] + Sq
rt[1 + c^2*x^2]*(ArcSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 7*Log[1 + c^2*x^
2]))*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4))/(12*c)

Maple [F]

\[\int \frac {\left (i c d x +d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]

[In]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x)

[Out]

int((d+I*c*d*x)^(3/2)*(a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2),x)

Fricas [F]

\[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((d+I*c*d*x)**(3/2)*(a+b*asinh(c*x))/(f-I*c*f*x)**(5/2),x)

[Out]

Integral((I*d*(c*x - I))**(3/2)*(a + b*asinh(c*x))/(-I*f*(c*x + I))**(5/2), x)

Maxima [F]

\[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

-1/3*a*(3*I*(c^2*d*f*x^2 + d*f)^(3/2)/(3*I*c^4*f^4*x^3 - 9*c^3*f^4*x^2 - 9*I*c^2*f^4*x + 3*c*f^4) - 2*I*sqrt(c
^2*d*f*x^2 + d*f)*d/(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3) - 21*I*sqrt(c^2*d*f*x^2 + d*f)*d/(-3*I*c^2*f^3*x + 3
*c*f^3) - 3*d^2*arcsinh(c*x)/(c*f^3*sqrt(d/f))) + b*integrate((I*c*d*x + d)^(3/2)*log(c*x + sqrt(c^2*x^2 + 1))
/(-I*c*f*x + f)^(5/2), x)

Giac [F]

\[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate((I*c*d*x + d)^(3/2)*(b*arcsinh(c*x) + a)/(-I*c*f*x + f)^(5/2), x)

Mupad [F(-1)]

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

[In]

int(((a + b*asinh(c*x))*(d + c*d*x*1i)^(3/2))/(f - c*f*x*1i)^(5/2),x)

[Out]

int(((a + b*asinh(c*x))*(d + c*d*x*1i)^(3/2))/(f - c*f*x*1i)^(5/2), x)