3.6.48 \(\int \sqrt {d+i c d x} (f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [548]

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

[Out]

3/8*f^2*x*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1/4*c^2*f^2*x^3*(a+b*arcsinh(c*x))*(d+I*c*d*x
)^(1/2)*(f-I*c*f*x)^(1/2)-2/3*I*f^2*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/c+2/3*I
*b*f^2*x*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)-3/16*b*c*f^2*x^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^
(1/2)/(c^2*x^2+1)^(1/2)+2/9*I*b*c^2*f^2*x^3*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+1/16*b*c^3*f
^2*x^4*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+5/16*f^2*(a+b*arcsinh(c*x))^2*(d+I*c*d*x)^(1/2)*(
f-I*c*f*x)^(1/2)/b/c/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.40, antiderivative size = 416, normalized size of antiderivative = 1.00, 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 = {5796, 5838, 5785, 5783, 30, 5798, 5806, 5812} \begin {gather*} -\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 {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}}+\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 {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}} \end {gather*}

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 30

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

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 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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 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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5806

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[(1/(m + 2))*Simp[Sqrt[d + e*x^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/(f*(m + 2)))
*Simp[Sqrt[d + e*x^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] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])

Rule 5812

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

Rule 5838

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]))

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 = 0.73, size = 565, normalized size = 1.36 \begin {gather*} \frac {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} \sqrt {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 \sqrt {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} \sqrt {1+c^2 x^2}+360 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2-144 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 )+720 a \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 )+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 )+12 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (-48 i \sqrt {1+c^2 x^2}-16 i \cosh \left (3 \sinh ^{-1}(c x)\right )+24 \sinh \left (2 \sinh ^{-1}(c x)\right )-3 \sinh \left (4 \sinh ^{-1}(c x)\right )\right )}{1152 c \sqrt {1+c^2 x^2}} \end {gather*}

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])

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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))*(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.

________________________________________________________________________________________

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))*(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)

________________________________________________________________________________________

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))*(d+I*c*d*x)**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6188 deep

________________________________________________________________________________________

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))*(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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________