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

Optimal. Leaf size=80 \[ -\frac{a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{b x}{6 c d^3 \sqrt{c^2 x^2+1}}+\frac{b x}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

(b*x)/(12*c*d^3*(1 + c^2*x^2)^(3/2)) + (b*x)/(6*c*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])/(4*c^2*d^3*(1
+ c^2*x^2)^2)

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Rubi [A]  time = 0.053746, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {5717, 192, 191} \[ -\frac{a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}+\frac{b x}{6 c d^3 \sqrt{c^2 x^2+1}}+\frac{b x}{12 c d^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x)/(12*c*d^3*(1 + c^2*x^2)^(3/2)) + (b*x)/(6*c*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])/(4*c^2*d^3*(1
+ c^2*x^2)^2)

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 192

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^3} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{b \int \frac{1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}\\ &=\frac{b x}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac{b \int \frac{1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{6 c d^3}\\ &=\frac{b x}{12 c d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{b x}{6 c d^3 \sqrt{1+c^2 x^2}}-\frac{a+b \sinh ^{-1}(c x)}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0699482, size = 56, normalized size = 0.7 \[ \frac{-3 a+b c x \sqrt{c^2 x^2+1} \left (2 c^2 x^2+3\right )-3 b \sinh ^{-1}(c x)}{12 d^3 \left (c^3 x^2+c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a + b*c*x*Sqrt[1 + c^2*x^2]*(3 + 2*c^2*x^2) - 3*b*ArcSinh[c*x])/(12*d^3*(c + c^3*x^2)^2)

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Maple [A]  time = 0.006, size = 76, normalized size = 1. \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{a}{4\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{b}{{d}^{3}} \left ( -{\frac{{\it Arcsinh} \left ( cx \right ) }{4\, \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{cx}{12} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{cx}{6}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/4*a/d^3/(c^2*x^2+1)^2+b/d^3*(-1/4/(c^2*x^2+1)^2*arcsinh(c*x)+1/12/(c^2*x^2+1)^(3/2)*c*x+1/6*c*x/(c^2
*x^2+1)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, b{\left (\frac{4 \, \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 1}{c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}} - 16 \, \int \frac{1}{4 \,{\left (c^{8} d^{3} x^{7} + 3 \, c^{6} d^{3} x^{5} + 3 \, c^{4} d^{3} x^{3} + c^{2} d^{3} x +{\left (c^{7} d^{3} x^{6} + 3 \, c^{5} d^{3} x^{4} + 3 \, c^{3} d^{3} x^{2} + c d^{3}\right )} \sqrt{c^{2} x^{2} + 1}\right )}}\,{d x}\right )} - \frac{a}{4 \,{\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*((4*log(c*x + sqrt(c^2*x^2 + 1)) + 1)/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3) - 16*integrate(1/4/(c^8*
d^3*x^7 + 3*c^6*d^3*x^5 + 3*c^4*d^3*x^3 + c^2*d^3*x + (c^7*d^3*x^6 + 3*c^5*d^3*x^4 + 3*c^3*d^3*x^2 + c*d^3)*sq
rt(c^2*x^2 + 1)), x)) - 1/4*a/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

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Fricas [A]  time = 2.34944, size = 207, normalized size = 2.59 \begin{align*} \frac{3 \, a c^{4} x^{4} + 6 \, a c^{2} x^{2} - 3 \, b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt{c^{2} x^{2} + 1}}{12 \,{\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*x^4 + 6*a*c^2*x^2 - 3*b*log(c*x + sqrt(c^2*x^2 + 1)) + (2*b*c^3*x^3 + 3*b*c*x)*sqrt(c^2*x^2 + 1)
)/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac{b x \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b*x*asinh(c*x)/(c**6*x**6 + 3*c**4*x*
*4 + 3*c**2*x**2 + 1), x))/d**3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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