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

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Rubi [A]  time = 0.21, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.45, 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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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ \frac {4 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b x \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 4 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a x + {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac {b c^{2} x^{4} + \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d f x^{2} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} + b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) - {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac {b c^{2} x^{4} - \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d f x^{2} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} + b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) - 2 \, {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac {b c^{2} x^{3} + \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d f x \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} + b x}{b c^{2} x^{2} + b}\right ) + 2 \, {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} \log \left (\frac {b c^{2} x^{3} - \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d f x \sqrt {\frac {b^{2}}{c^{2} d^{3} f^{3}}} + b x}{b c^{2} x^{2} + b}\right ) + 4 \, {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )} {\rm integral}\left (-\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x}{c^{4} d^{2} f^{2} x^{4} + 2 \, c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}}, x\right )}{4 \, {\left (c^{2} d^{2} f^{2} x^{2} + d^{2} f^{2}\right )}} \]

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

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]

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

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsinh \left (c x \right )}{\left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {3}{2}}}\, dx \]

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 = 0.58, size = 82, normalized size = 0.80 \[ \frac {b x \operatorname {arsinh}\left (c x\right )}{\sqrt {c^{2} d f x^{2} + d f} d f} + \frac {a x}{\sqrt {c^{2} d f x^{2} + d f} d f} - \frac {b \sqrt {\frac {1}{d f}} \log \left (x^{2} + \frac {1}{c^{2}}\right )}{2 \, c d f} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

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

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]

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

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