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

Optimal. Leaf size=85 \[ \frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt{c^2 x^2+1}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.106848, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5717, 5687, 260} \[ \frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt{c^2 x^2+1}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*(a + b*ArcSinh[c*x]))/(c*d^2*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])^2/(2*c^2*d^2*(1 + c^2*x^2)) - (b^2
*Log[1 + c^2*x^2])/(2*c^2*d^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 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{x \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac{b \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{c d^2}\\ &=\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{b^2 \int \frac{x}{1+c^2 x^2} \, dx}{d^2}\\ &=\frac{b x \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac{b^2 \log \left (1+c^2 x^2\right )}{2 c^2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.218641, size = 145, normalized size = 1.71 \[ -\frac{a^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac{a b x}{c d^2 \sqrt{c^2 x^2+1}}+\frac{b \sinh ^{-1}(c x) \left (b c x \sqrt{c^2 x^2+1}-a\right )}{c^2 d^2 \left (c^2 x^2+1\right )}-\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 d^2}-\frac{b^2 \sinh ^{-1}(c x)^2}{2 c^2 d^2 \left (c^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.063, size = 222, normalized size = 2.6 \begin{align*} -{\frac{{a}^{2}}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+2\,{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{{c}^{2}{d}^{2}}}+{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) x}{c{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ){x}^{2}}{{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2}{\it Arcsinh} \left ( cx \right ) }{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{{b}^{2}}{{c}^{2}{d}^{2}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{ab{\it Arcsinh} \left ( cx \right ) }{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{abx}{c{d}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.6841, size = 397, normalized size = 4.67 \begin{align*} \frac{2 \, a b c^{2} x^{2} + 2 \, \sqrt{c^{2} x^{2} + 1} a b c x - b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} - a^{2} + 2 \, a b -{\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) + 2 \,{\left (a b c^{2} x^{2} + \sqrt{c^{2} x^{2} + 1} b^{2} c x\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (a b c^{2} x^{2} + a b\right )} \log \left (-c x + \sqrt{c^{2} x^{2} + 1}\right )}{2 \,{\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2} x}{{\left (c^{2} d x^{2} + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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