3.569 \(\int \frac{a+b \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=203 \[ \frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

(b*(1 + c^2*x^2)^(3/2))/(6*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))
/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*x*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/
2)*(f - I*c*f*x)^(5/2)) - (b*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2
))

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Rubi [A]  time = 0.244513, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5712, 5690, 5687, 260, 261} \[ \frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)),x]

[Out]

(b*(1 + c^2*x^2)^(3/2))/(6*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))
/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*x*(1 + c^2*x^2)^2*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/
2)*(f - I*c*f*x)^(5/2)) - (b*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2
))

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 5690

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

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]

Rule 261

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

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{b \left (1+c^2 x^2\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{b \left (1+c^2 x^2\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.553006, size = 193, normalized size = 0.95 \[ \frac{i \sqrt{f-i c f x} \left (4 a c^3 x^3+6 a c x-2 b c^2 x^2 \sqrt{c^2 x^2+1} \log (d+i c d x)-2 b \left (c^2 x^2+1\right )^{3/2} \log (d (-1+i c x))-2 b \sqrt{c^2 x^2+1} \log (d+i c d x)+b \sqrt{c^2 x^2+1}+2 b c x \left (2 c^2 x^2+3\right ) \sinh ^{-1}(c x)\right )}{6 c d^2 f^3 (c x-i) (c x+i)^2 \sqrt{d+i c d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)),x]

[Out]

((I/6)*Sqrt[f - I*c*f*x]*(6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 + c^2*x^2] + 2*b*c*x*(3 + 2*c^2*x^2)*ArcSinh[c*x] -
2*b*(1 + c^2*x^2)^(3/2)*Log[d*(-1 + I*c*x)] - 2*b*Sqrt[1 + c^2*x^2]*Log[d + I*c*d*x] - 2*b*c^2*x^2*Sqrt[1 + c
^2*x^2]*Log[d + I*c*d*x]))/(c*d^2*f^3*(-I + c*x)*(I + c*x)^2*Sqrt[d + I*c*d*x])

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Maple [F]  time = 0.254, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{-{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{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)^(5/2)/(f-I*c*f*x)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.24843, size = 215, normalized size = 1.06 \begin{align*} \frac{1}{6} \, b c{\left (\frac{1}{c^{4} d^{\frac{5}{2}} f^{\frac{5}{2}} x^{2} + c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}}\right )} + \frac{1}{3} \, b{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} 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)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 + c^2*d^(5/2)*f^(5/2)) - 2*log(c^2*x^2 + 1)/(c^2*d^(5/2)*f^(5/2))) + 1/3*b
*(x/((c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))*arcsinh(c*x) + 1/3*a*(x/((c^2*d*f
*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x^2 - 2*(2*b*c^2*x^3 + 3*b*x)*sqrt(I*c*d*x +
d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) - (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/
(c^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x^2*sqrt(b^2/(c^2*d^5*f^5
)) + b*c^2*x^4 + b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) + (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(
b^2/(c^2*d^5*f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x^2*sqrt(b^2/(c^2*d^
5*f^5)) - b*c^2*x^4 - b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) + 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3
)*sqrt(b^2/(c^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x*sqrt(b^2/(c^
2*d^5*f^5)) + b*c^2*x^3 + b*x)/(b*c^2*x^2 + b)) - 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/(
c^2*d^5*f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x*sqrt(b^2/(c^2*d^5*f^5))
- b*c^2*x^3 - b*x)/(b*c^2*x^2 + b)) - 2*(2*a*c^2*x^3 + 3*a*x)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - 6*(c^4*d
^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*integral(-2/3*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)
*b*c*x/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3), x))/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)

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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)**(5/2)/(f-I*c*f*x)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: AttributeError