3.547 \(\int (d+i c d x)^{3/2} (f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=459 \[ \frac{3 f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (c^2 x^2+1\right )}+\frac{3 f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \left (c^2 x^2+1\right )^{3/2}}-\frac{i f \left (c^2 x^2+1\right ) (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c}+\frac{1}{4} f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{i b c^4 f x^5 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{25 \left (c^2 x^2+1\right )^{3/2}}-\frac{b c^3 f x^4 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}}+\frac{2 i b c^2 f x^3 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{15 \left (c^2 x^2+1\right )^{3/2}}-\frac{5 b c f x^2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}}+\frac{i b f x (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{5 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

((I/5)*b*f*x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(1 + c^2*x^2)^(3/2) - (5*b*c*f*x^2*(d + I*c*d*x)^(3/2)*(
f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) + (((2*I)/15)*b*c^2*f*x^3*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)
)/(1 + c^2*x^2)^(3/2) - (b*c^3*f*x^4*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) + ((I/2
5)*b*c^4*f*x^5*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(1 + c^2*x^2)^(3/2) + (f*x*(d + I*c*d*x)^(3/2)*(f - I*
c*f*x)^(3/2)*(a + b*ArcSinh[c*x]))/4 + (3*f*x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x]))/(8
*(1 + c^2*x^2)) - ((I/5)*f*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/c + (3*
f*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x])^2)/(16*b*c*(1 + c^2*x^2)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.425982, antiderivative size = 459, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {5712, 5821, 5684, 5682, 5675, 30, 14, 5717, 194} \[ \frac{3 f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (c^2 x^2+1\right )}+\frac{3 f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \left (c^2 x^2+1\right )^{3/2}}-\frac{i f \left (c^2 x^2+1\right ) (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{5 c}+\frac{1}{4} f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{i b c^4 f x^5 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{25 \left (c^2 x^2+1\right )^{3/2}}-\frac{b c^3 f x^4 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}}+\frac{2 i b c^2 f x^3 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{15 \left (c^2 x^2+1\right )^{3/2}}-\frac{5 b c f x^2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}}+\frac{i b f x (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{5 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*b*f*x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(1 + c^2*x^2)^(3/2) - (5*b*c*f*x^2*(d + I*c*d*x)^(3/2)*(
f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) + (((2*I)/15)*b*c^2*f*x^3*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)
)/(1 + c^2*x^2)^(3/2) - (b*c^3*f*x^4*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) + ((I/2
5)*b*c^4*f*x^5*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(1 + c^2*x^2)^(3/2) + (f*x*(d + I*c*d*x)^(3/2)*(f - I*
c*f*x)^(3/2)*(a + b*ArcSinh[c*x]))/4 + (3*f*x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x]))/(8
*(1 + c^2*x^2)) - ((I/5)*f*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/c + (3*
f*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x])^2)/(16*b*c*(1 + c^2*x^2)^(3/2))

Rule 5712

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 5821

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

Rule 5684

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*
(a + b*ArcSinh[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^
n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1
+ c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && Gt
Q[n, 0] && GtQ[p, 0]

Rule 5682

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*
(a + b*ArcSinh[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 + c^2*x^2]), Int[(a + b*ArcSinh[c*x])^n/Sqrt[1
+ c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 + c^2*x^2]), Int[x*(a + b*ArcSinh[c*x])^(n - 1),
x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 194

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

Rubi steps

\begin{align*} \int (d+i c d x)^{3/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left ((d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int (f-i c f x) \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\left (1+c^2 x^2\right )^{3/2}}\\ &=\frac{\left ((d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \left (f \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-i c f x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\left (1+c^2 x^2\right )^{3/2}}\\ &=\frac{\left (f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\left (1+c^2 x^2\right )^{3/2}}-\frac{\left (i c f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int x \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\left (1+c^2 x^2\right )^{3/2}}\\ &=\frac{1}{4} f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{i f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{5 c}+\frac{\left (3 f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{4 \left (1+c^2 x^2\right )^{3/2}}+\frac{\left (i b f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \left (1+c^2 x^2\right )^2 \, dx}{5 \left (1+c^2 x^2\right )^{3/2}}-\frac{\left (b c f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{4 \left (1+c^2 x^2\right )^{3/2}}\\ &=\frac{1}{4} f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (1+c^2 x^2\right )}-\frac{i f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{5 c}+\frac{\left (3 f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{8 \left (1+c^2 x^2\right )^{3/2}}+\frac{\left (i b f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \left (1+2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \left (1+c^2 x^2\right )^{3/2}}-\frac{\left (b c f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{4 \left (1+c^2 x^2\right )^{3/2}}-\frac{\left (3 b c f (d+i c d x)^{3/2} (f-i c f x)^{3/2}\right ) \int x \, dx}{8 \left (1+c^2 x^2\right )^{3/2}}\\ &=\frac{i b f x (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{5 \left (1+c^2 x^2\right )^{3/2}}-\frac{5 b c f x^2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 i b c^2 f x^3 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{15 \left (1+c^2 x^2\right )^{3/2}}-\frac{b c^3 f x^4 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (1+c^2 x^2\right )^{3/2}}+\frac{i b c^4 f x^5 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{25 \left (1+c^2 x^2\right )^{3/2}}+\frac{1}{4} f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{3 f x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (1+c^2 x^2\right )}-\frac{i f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{5 c}+\frac{3 f (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \left (1+c^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.59681, size = 683, normalized size = 1.49 \[ \frac{3600 a d^{3/2} f^{5/2} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )-1920 i a c^4 d f^2 x^4 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+2400 a c^3 d f^2 x^3 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-3840 i a c^2 d f^2 x^2 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+6000 a c d f^2 x \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-1920 i a d f^2 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+60 b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x) \left (5 \left (-4 i \sqrt{c^2 x^2+1}+8 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )\right )-10 i \cosh \left (3 \sinh ^{-1}(c x)\right )-2 i \cosh \left (5 \sinh ^{-1}(c x)\right )\right )+1200 i b c d f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}+1800 b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2+200 i b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh \left (3 \sinh ^{-1}(c x)\right )+24 i b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh \left (5 \sinh ^{-1}(c x)\right )-1200 b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )-75 b d f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (4 \sinh ^{-1}(c x)\right )}{9600 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1200*I)*b*c*d*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (1920*I)*a*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*
x]*Sqrt[1 + c^2*x^2] + 6000*a*c*d*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - (3840*I)*a*c^2
*d*f^2*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 2400*a*c^3*d*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - (1920*I)*a*c^4*d*f^2*x^4*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^
2] + 1800*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 - 1200*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f -
I*c*f*x]*Cosh[2*ArcSinh[c*x]] - 75*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[4*ArcSinh[c*x]] + 3600*a*
d^(3/2)*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]] + (200*I)
*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[3*ArcSinh[c*x]] + 60*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*
f*x]*ArcSinh[c*x]*((-10*I)*Cosh[3*ArcSinh[c*x]] - (2*I)*Cosh[5*ArcSinh[c*x]] + 5*((-4*I)*Sqrt[1 + c^2*x^2] + 8
*Sinh[2*ArcSinh[c*x]] + Sinh[4*ArcSinh[c*x]])) + (24*I)*b*d*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[5*Arc
Sinh[c*x]])/(9600*c*Sqrt[1 + c^2*x^2])

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Maple [F]  time = 0.253, size = 0, normalized size = 0. \begin{align*} \int \left ( d+icdx \right ) ^{{\frac{3}{2}}} \left ( f-icfx \right ) ^{{\frac{5}{2}}} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-i \, b c^{3} d f^{2} x^{3} + b c^{2} d f^{2} x^{2} - i \, b c d f^{2} x + b d f^{2}\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 (-i \, a c^{3} d f^{2} x^{3} + a c^{2} d f^{2} x^{2} - i \, a c d f^{2} x + a d f^{2}\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError