3.6.47 \(\int (d+i c d x)^{3/2} (f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [547]

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

[Out]

1/5*I*b*f*x*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)/(c^2*x^2+1)^(3/2)-5/16*b*c*f*x^2*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)
^(3/2)/(c^2*x^2+1)^(3/2)+2/15*I*b*c^2*f*x^3*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)/(c^2*x^2+1)^(3/2)-1/16*b*c^3*f
*x^4*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)/(c^2*x^2+1)^(3/2)+1/25*I*b*c^4*f*x^5*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3
/2)/(c^2*x^2+1)^(3/2)+1/4*f*x*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))+3/8*f*x*(d+I*c*d*x)^(3/2)
*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))/(c^2*x^2+1)-1/5*I*f*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(c^2*x^2+1)*(a+b
*arcsinh(c*x))/c+3/16*f*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))^2/b/c/(c^2*x^2+1)^(3/2)

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Rubi [A]
time = 0.29, antiderivative size = 459, normalized size of antiderivative = 1.00, 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 = {5796, 5838, 5786, 5785, 5783, 30, 14, 5798, 200} \begin {gather*} \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 {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}}+\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 {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}} \end {gather*}

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 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 30

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

Rule 200

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]

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 5786

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2)^(p - 1/2)*(a + b*A
rcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 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 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 (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*}

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Mathematica [A]
time = 1.03, size = 683, normalized size = 1.49 \begin {gather*} \frac {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} \sinh ^{-1}(c x)^2-1200 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 )+3600 a d^{3/2} 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 )+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 )+60 b d f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (-10 i \cosh \left (3 \sinh ^{-1}(c x)\right )-2 i \cosh \left (5 \sinh ^{-1}(c x)\right )+5 \left (-4 i \sqrt {1+c^2 x^2}+8 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )\right )\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 )}{9600 c \sqrt {1+c^2 x^2}} \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

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 = 0.00, 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((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: 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((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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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((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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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