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

Optimal. Leaf size=416 \[ -\frac{1}{4} c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}-\frac{2 i f^2 \left (c^2 x^2+1\right ) \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{3}{8} f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac{b c^3 f^2 x^4 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{c^2 x^2+1}}+\frac{2 i b c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x}}{9 \sqrt{c^2 x^2+1}}-\frac{3 b c f^2 x^2 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{c^2 x^2+1}}+\frac{2 i b f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}}{3 \sqrt{c^2 x^2+1}} \]

[Out]

(((2*I)/3)*b*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[1 + c^2*x^2] - (3*b*c*f^2*x^2*Sqrt[d + I*c*d*x]*S
qrt[f - I*c*f*x])/(16*Sqrt[1 + c^2*x^2]) + (((2*I)/9)*b*c^2*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[
1 + c^2*x^2] + (b*c^3*f^2*x^4*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(16*Sqrt[1 + c^2*x^2]) + (3*f^2*x*Sqrt[d +
I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/8 - (c^2*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*A
rcSinh[c*x]))/4 - (((2*I)/3)*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/c + (
5*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/(16*b*c*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.571439, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {5712, 5821, 5682, 5675, 30, 5717, 5742, 5758} \[ -\frac{1}{4} c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{c^2 x^2+1}}-\frac{2 i f^2 \left (c^2 x^2+1\right ) \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{3}{8} f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac{b c^3 f^2 x^4 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{c^2 x^2+1}}+\frac{2 i b c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x}}{9 \sqrt{c^2 x^2+1}}-\frac{3 b c f^2 x^2 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{c^2 x^2+1}}+\frac{2 i b f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}}{3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/3)*b*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[1 + c^2*x^2] - (3*b*c*f^2*x^2*Sqrt[d + I*c*d*x]*S
qrt[f - I*c*f*x])/(16*Sqrt[1 + c^2*x^2]) + (((2*I)/9)*b*c^2*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[
1 + c^2*x^2] + (b*c^3*f^2*x^4*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(16*Sqrt[1 + c^2*x^2]) + (3*f^2*x*Sqrt[d +
I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/8 - (c^2*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*A
rcSinh[c*x]))/4 - (((2*I)/3)*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/c + (
5*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/(16*b*c*Sqrt[1 + c^2*x^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 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 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 5742

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

Rule 5758

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

Rubi steps

\begin{align*} \int \sqrt{d+i c d x} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left (\sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int (f-i c f x)^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{\left (\sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \left (f^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-2 i c f^2 x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-c^2 f^2 x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{\left (f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}-\frac{\left (2 i c f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}-\frac{\left (c^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{1}{2} f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{4} c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 i f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{\left (f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}+\frac{\left (2 i b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 \sqrt{1+c^2 x^2}}-\frac{\left (b c f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (c^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{4 \sqrt{1+c^2 x^2}}+\frac{\left (b c^3 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x^3 \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{2 i b f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}}{3 \sqrt{1+c^2 x^2}}-\frac{b c f^2 x^2 \sqrt{d+i c d x} \sqrt{f-i c f x}}{4 \sqrt{1+c^2 x^2}}+\frac{2 i b c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x}}{9 \sqrt{1+c^2 x^2}}+\frac{b c^3 f^2 x^4 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{1+c^2 x^2}}+\frac{3}{8} f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{4} c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 i f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{1+c^2 x^2}}+\frac{\left (f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{8 \sqrt{1+c^2 x^2}}+\frac{\left (b c f^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x \, dx}{8 \sqrt{1+c^2 x^2}}\\ &=\frac{2 i b f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}}{3 \sqrt{1+c^2 x^2}}-\frac{3 b c f^2 x^2 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{1+c^2 x^2}}+\frac{2 i b c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x}}{9 \sqrt{1+c^2 x^2}}+\frac{b c^3 f^2 x^4 \sqrt{d+i c d x} \sqrt{f-i c f x}}{16 \sqrt{1+c^2 x^2}}+\frac{3}{8} f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{1}{4} c^2 f^2 x^3 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac{2 i f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac{5 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.18699, size = 565, normalized size = 1.36 \[ \frac{-288 a c^3 f^2 x^3 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-768 i a c^2 f^2 x^2 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+432 a c f^2 x \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-768 i a f^2 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+720 a \sqrt{d} f^{5/2} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+12 b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x) \left (-48 i \sqrt{c^2 x^2+1}+24 \sinh \left (2 \sinh ^{-1}(c x)\right )-3 \sinh \left (4 \sinh ^{-1}(c x)\right )-16 i \cosh \left (3 \sinh ^{-1}(c x)\right )\right )+576 i b c f^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}+360 b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2+64 i b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh \left (3 \sinh ^{-1}(c x)\right )-144 b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )+9 b f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (4 \sinh ^{-1}(c x)\right )}{1152 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((576*I)*b*c*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (768*I)*a*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqr
t[1 + c^2*x^2] + 432*a*c*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - (768*I)*a*c^2*f^2*x^2*S
qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 288*a*c^3*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*S
qrt[1 + c^2*x^2] + 360*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 - 144*b*f^2*Sqrt[d + I*c*d*x]*
Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + 9*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[4*ArcSinh[c*x]] + 72
0*a*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]] + (64
*I)*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[3*ArcSinh[c*x]] + 12*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f
*x]*ArcSinh[c*x]*((-48*I)*Sqrt[1 + c^2*x^2] - (16*I)*Cosh[3*ArcSinh[c*x]] + 24*Sinh[2*ArcSinh[c*x]] - 3*Sinh[4
*ArcSinh[c*x]]))/(1152*c*Sqrt[1 + c^2*x^2])

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Maple [F]  time = 0.267, size = 0, normalized size = 0. \begin{align*} \int \left ( f-icfx \right ) ^{{\frac{5}{2}}} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \sqrt{d+icdx}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),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((f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),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 (b c^{2} f^{2} x^{2} + 2 i \, b c f^{2} x - b 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 (a c^{2} f^{2} x^{2} + 2 i \, a c f^{2} x - a 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((f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: TypeError