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

Optimal. Leaf size=103 \[ \frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.205475, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {5712, 5687, 260} \[ \frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b*(1 + c^2*x^2)^(3/2)*Log[
1 + c^2*x^2])/(2*c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(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 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.465136, size = 118, normalized size = 1.15 \[ \frac{i \sqrt{f-i c f x} \left (2 a c x-b \sqrt{c^2 x^2+1} \log (d (-1+i c x))-b \sqrt{c^2 x^2+1} \log (d+i c d x)+2 b c x \sinh ^{-1}(c x)\right )}{2 c d f^2 (c x+i) \sqrt{d+i c d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.243, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{-{\frac{3}{2}}} \left ( f-icfx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22233, size = 112, normalized size = 1.09 \begin{align*} -\frac{b c \sqrt{\frac{1}{c^{4} d f}} \log \left (x^{2} + \frac{1}{c^{2}}\right )}{2 \, d f} + \frac{b x \operatorname{arsinh}\left (c x\right )}{\sqrt{c^{2} d f x^{2} + d f} d f} + \frac{a x}{\sqrt{c^{2} d f x^{2} + d f} d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*c*sqrt(1/(c^4*d*f))*log(x^2 + 1/c^2)/(d*f) + b*x*arcsinh(c*x)/(sqrt(c^2*d*f*x^2 + d*f)*d*f) + a*x/(sqrt
(c^2*d*f*x^2 + d*f)*d*f)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{4 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 4 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a x +{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac{b c^{2} x^{4} + \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} c d f x^{2} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} + b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) -{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac{b c^{2} x^{4} - \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} c d f x^{2} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} + b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) - 2 \,{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac{b c^{2} x^{3} + \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} c d f x \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} + b x}{b c^{2} x^{2} + b}\right ) + 2 \,{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac{b c^{2} x^{3} - \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} c d f x \sqrt{\frac{b^{2}}{c^{2} d^{3} f^{3}}} + b x}{b c^{2} x^{2} + b}\right ) + 4 \,{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )}{\rm integral}\left (-\frac{\sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b c x}{c^{4} d^{2} f^{2} x^{4} + 2 \, c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}}, x\right )}{4 \,{\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*x*log(c*x + sqrt(c^2*x^2 + 1)) + 4*sqrt(I*c*d*x + d)*sqrt(-I*c*f
*x + f)*a*x + (c^2*d^2*f^2*x^2 + d^2*f^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^4 + sqrt(c^2*x^2 + 1)*sqrt(I*c*
d*x + d)*sqrt(-I*c*f*x + f)*c*d*f*x^2*sqrt(b^2/(c^2*d^3*f^3)) + b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) - (c^2*d
^2*f^2*x^2 + d^2*f^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^4 - sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f
*x + f)*c*d*f*x^2*sqrt(b^2/(c^2*d^3*f^3)) + b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) - 2*(c^2*d^2*f^2*x^2 + d^2*f
^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^3 + sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f*x*sq
rt(b^2/(c^2*d^3*f^3)) + b*x)/(b*c^2*x^2 + b)) + 2*(c^2*d^2*f^2*x^2 + d^2*f^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c
^2*x^3 - sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f*x*sqrt(b^2/(c^2*d^3*f^3)) + b*x)/(b*c^2*
x^2 + b)) + 4*(c^2*d^2*f^2*x^2 + d^2*f^2)*integral(-sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c
*x/(c^4*d^2*f^2*x^4 + 2*c^2*d^2*f^2*x^2 + d^2*f^2), x))/(c^2*d^2*f^2*x^2 + d^2*f^2)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError