3.6.85 \(\int \frac {(f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x))^2}{\sqrt {d+i c d x}} \, dx\) [585]

Optimal. Leaf size=615 \[ -\frac {68 i b^2 f^3 \left (1+c^2 x^2\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 b^2 f^3 x \left (1+c^2 x^2\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f^3 \left (1+c^2 x^2\right )^2}{27 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b^2 f^3 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {22 i b f^3 x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

-68/9*I*b^2*f^3*(c^2*x^2+1)/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-3/4*b^2*f^3*x*(c^2*x^2+1)/(d+I*c*d*x)^(1/2)/
(f-I*c*f*x)^(1/2)+2/27*I*b^2*f^3*(c^2*x^2+1)^2/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-11/3*I*f^3*(c^2*x^2+1)*(a
+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-3/2*f^3*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x
)^(1/2)/(f-I*c*f*x)^(1/2)+1/3*I*c*f^3*x^2*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)
+3/4*b^2*f^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+22/3*I*b*f^3*x*(a+b*arcsinh(
c*x))*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+3/2*b*c*f^3*x^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/
2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-2/9*I*b*c^2*f^3*x^3*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1
/2)/(f-I*c*f*x)^(1/2)+5/6*f^3*(a+b*arcsinh(c*x))^3*(c^2*x^2+1)^(1/2)/b/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.50, antiderivative size = 615, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 10, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {5796, 5843, 3398, 3377, 2718, 3392, 32, 2715, 8, 2713} \begin {gather*} \frac {5 f^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {22 i b f^3 x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f^3 \left (c^2 x^2+1\right )^2}{27 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 b^2 f^3 x \left (c^2 x^2+1\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {68 i b^2 f^3 \left (c^2 x^2+1\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b^2 f^3 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-68*I)/9)*b^2*f^3*(1 + c^2*x^2))/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - (3*b^2*f^3*x*(1 + c^2*x^2))/(4*S
qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (((2*I)/27)*b^2*f^3*(1 + c^2*x^2)^2)/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f
*x]) + (3*b^2*f^3*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(4*c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (((22*I)/3)*b*f^
3*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (3*b*c*f^3*x^2*Sqrt[1 + c^
2*x^2]*(a + b*ArcSinh[c*x]))/(2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - (((2*I)/9)*b*c^2*f^3*x^3*Sqrt[1 + c^2*x
^2]*(a + b*ArcSinh[c*x]))/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - (((11*I)/3)*f^3*(1 + c^2*x^2)*(a + b*ArcSinh
[c*x])^2)/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - (3*f^3*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(2*Sqrt[d +
I*c*d*x]*Sqrt[f - I*c*f*x]) + ((I/3)*c*f^3*x^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(Sqrt[d + I*c*d*x]*Sqrt[
f - I*c*f*x]) + (5*f^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^3)/(6*b*c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3392

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*m*(c + d*x)^(m - 1)*((
b*Sin[e + f*x])^n/(f^2*n^2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[d^2*m*((m - 1)/(f^2*n^2)), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[b*(c + d*x)^m*Cos[e + f
*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3398

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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 5843

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

Rubi steps

\begin {align*} \int \frac {(f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {d+i c d x}} \, dx &=\frac {\sqrt {1+c^2 x^2} \int \frac {(f-i c f x)^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {\sqrt {1+c^2 x^2} \text {Subst}\left (\int (a+b x)^2 (c f-i c f \sinh (x))^3 \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {\sqrt {1+c^2 x^2} \text {Subst}\left (\int \left (c^3 f^3 (a+b x)^2-3 i c^3 f^3 (a+b x)^2 \sinh (x)-3 c^3 f^3 (a+b x)^2 \sinh ^2(x)+i c^3 f^3 (a+b x)^2 \sinh ^3(x)\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (i f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x)^2 \sinh ^3(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (3 i f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x)^2 \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (3 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x)^2 \sinh ^2(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (2 i f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x)^2 \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (3 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x)^2 \, dx,x,\sinh ^{-1}(c x)\right )}{2 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (6 i b f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \cosh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (2 i b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \sinh ^3(x) \, dx,x,\sinh ^{-1}(c x)\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (3 b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \sinh ^2(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=-\frac {3 b^2 f^3 x \left (1+c^2 x^2\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {6 i b f^3 x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (4 i b f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \cosh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (2 i b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\sqrt {1+c^2 x^2}\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (6 i b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (3 b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int 1 \, dx,x,\sinh ^{-1}(c x)\right )}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=-\frac {56 i b^2 f^3 \left (1+c^2 x^2\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 b^2 f^3 x \left (1+c^2 x^2\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f^3 \left (1+c^2 x^2\right )^2}{27 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b^2 f^3 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {22 i b f^3 x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (4 i b^2 f^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \sinh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=-\frac {68 i b^2 f^3 \left (1+c^2 x^2\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 b^2 f^3 x \left (1+c^2 x^2\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f^3 \left (1+c^2 x^2\right )^2}{27 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b^2 f^3 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {22 i b f^3 x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ \end {align*}

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Mathematica [A]
time = 2.11, size = 723, normalized size = 1.18 \begin {gather*} \frac {1620 i a b c f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x}-792 i a^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-1620 i b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-324 a^2 c f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+72 i a^2 c^2 f^2 x^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+180 b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^3+162 a b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )+4 i b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (3 \sinh ^{-1}(c x)\right )+6 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (27 b \cosh \left (2 \sinh ^{-1}(c x)\right )+2 i \left (-4 b c x \left (-33+c^2 x^2\right )+27 a (-5+2 i c x) \sqrt {1+c^2 x^2}+3 a \cosh \left (3 \sinh ^{-1}(c x)\right )\right )\right )+540 a^2 \sqrt {d} f^{5/2} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )-81 b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (2 \sinh ^{-1}(c x)\right )+18 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2 \left (30 a-45 i b \sqrt {1+c^2 x^2}+i b \cosh \left (3 \sinh ^{-1}(c x)\right )-9 b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )-12 i a b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (3 \sinh ^{-1}(c x)\right )}{216 c d \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1620*I)*a*b*c*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (792*I)*a^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x
]*Sqrt[1 + c^2*x^2] - (1620*I)*b^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 324*a^2*c*f^2*x
*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + (72*I)*a^2*c^2*f^2*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c
*f*x]*Sqrt[1 + c^2*x^2] + 180*b^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^3 + 162*a*b*f^2*Sqrt[d
+ I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + (4*I)*b^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[3*A
rcSinh[c*x]] + 6*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]*(27*b*Cosh[2*ArcSinh[c*x]] + (2*I)*(-4
*b*c*x*(-33 + c^2*x^2) + 27*a*(-5 + (2*I)*c*x)*Sqrt[1 + c^2*x^2] + 3*a*Cosh[3*ArcSinh[c*x]])) + 540*a^2*Sqrt[d
]*f^(5/2)*Sqrt[1 + c^2*x^2]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] - 81*b^2*f^2*Sq
rt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[2*ArcSinh[c*x]] + 18*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[
c*x]^2*(30*a - (45*I)*b*Sqrt[1 + c^2*x^2] + I*b*Cosh[3*ArcSinh[c*x]] - 9*b*Sinh[2*ArcSinh[c*x]]) - (12*I)*a*b*
f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[3*ArcSinh[c*x]])/(216*c*d*Sqrt[1 + c^2*x^2])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\sqrt {i c d x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6189 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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