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

Optimal. Leaf size=283 \[ -\frac {3 d^3 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {i d^3 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 i d^3 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {i b d^3 x \left (c^2 x^2+1\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 b d^3 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

I*b*d^3*x*(c^2*x^2+1)^(3/2)/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-4*I*d^3*(1+I*c*x)*(c^2*x^2+1)*(a+b*arcsinh(c*x
))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-I*d^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x
)^(3/2)-3/2*d^3*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))^2/b/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-4*b*d^3*(c^2*x^
2+1)^(3/2)*ln(I+c*x)/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)

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Rubi [A]  time = 0.48, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5712, 5833, 637, 5819, 12, 627, 31, 5675, 5717, 8} \[ -\frac {3 d^3 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {i d^3 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 i d^3 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {i b d^3 x \left (c^2 x^2+1\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 b d^3 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

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 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 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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[p + 1/2, 0]
&& GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 5833

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

Rubi steps

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

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Mathematica [A]  time = 2.43, size = 515, normalized size = 1.82 \[ \frac {-\frac {6 a d^{3/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{f^{3/2}}+\frac {2 a d (5-i c x) \sqrt {d+i c d x} \sqrt {f-i c f x}}{f^2 (c x+i)}+\frac {b d \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (4 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (c^2 x^2+1\right )\right )-\left (\sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )+4 \sinh ^{-1}(c x) \left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{f^2 \sqrt {c^2 x^2+1} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}+\frac {2 b d \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\sinh ^{-1}(c x) \left (-\left (\sqrt {c^2 x^2+1}-2\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \left (\sqrt {c^2 x^2+1}+2\right ) \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\left (\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (i \log \left (c^2 x^2+1\right )+c x-4 \tan ^{-1}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )-\left (\sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )}{f^2 \sqrt {c^2 x^2+1} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*d*(5 - I*c*x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(f^2*(I + c*x)) - (6*a*d^(3/2)*Log[c*d*f*x + Sqrt[d]*
Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/f^(3/2) + (b*d*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-(ArcSinh[c*
x]^2*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])) + 4*ArcSinh[c*x]*((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSi
nh[c*x]/2]) + 2*(4*ArcTan[Tanh[ArcSinh[c*x]/2]] + I*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c
*x]/2])))/(f^2*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])) + (2*b*d*Sqrt[d + I*c*d*x]*S
qrt[f - I*c*f*x]*(-(ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])) + (c*x - 4*ArcTan[Coth[Arc
Sinh[c*x]/2]] + I*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2]) + ArcSinh[c*x]*((-I)*(2 +
Sqrt[1 + c^2*x^2])*Cosh[ArcSinh[c*x]/2] - (-2 + Sqrt[1 + c^2*x^2])*Sinh[ArcSinh[c*x]/2])))/(f^2*Sqrt[1 + c^2*x
^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(2*c)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-i \, b c d x - b d\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (-i \, a c d x - a d\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{c^{2} f^{2} x^{2} + 2 i \, c f^{2} x - f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [c,d,t_nostep]=[30,75,38]Precision problem choosing root in common_EXT, current precision 14Prec
ision problem choosing root in common_EXT, current precision 28Precision problem choosing root in common_EXT,
current precision 56Precision problem choosing root in common_EXT, current precision 112Precision problem choo
sing root in common_EXT, current precision 224Precision problem choosing root in common_EXT, current precision
448Precision problem choosing root in common_EXT, current precision 896Unable to transpose Error: Bad Argumen
t ValueWarning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Evaluation time: 0.75sym
2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (i c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )}{\left (-i c f x +f \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (-\frac {i \, {\left (c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}}}{c^{3} f^{3} x^{2} + 2 i \, c^{2} f^{3} x - c f^{3}} - \frac {6 i \, \sqrt {c^{2} d f x^{2} + d f} d}{-i \, c^{2} f^{2} x + c f^{2}} - \frac {3 \, d^{2} \operatorname {arsinh}\left (c x\right )}{c f^{2} \sqrt {\frac {d}{f}}}\right )} + b \int \frac {{\left (i \, c d x + d\right )}^{\frac {3}{2}} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (-i \, c f x + f\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-I*(c^2*d*f*x^2 + d*f)^(3/2)/(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3) - 6*I*sqrt(c^2*d*f*x^2 + d*f)*d/(-I*c^2*
f^2*x + c*f^2) - 3*d^2*arcsinh(c*x)/(c*f^2*sqrt(d/f))) + b*integrate((I*c*d*x + d)^(3/2)*log(c*x + sqrt(c^2*x^
2 + 1))/(-I*c*f*x + f)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

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)^(3/2),x)

[Out]

int(((a + b*asinh(c*x))*(d + c*d*x*1i)^(3/2))/(f - c*f*x*1i)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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