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]

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

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Rubi [A]  time = 0.11, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.22, 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]

-1/2*a^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])*ArcSi
nh[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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fricas [B]  time = 0.61, size = 185, normalized size = 2.18 \[ \frac {2 \, a b c^{2} x^{2} + 2 \, \sqrt {c^{2} x^{2} + 1} a b c x - b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} - a^{2} + 2 \, a b - {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) + 2 \, {\left (a b c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c x\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{2} x^{2} + a b\right )} \log \left (-c x + \sqrt {c^{2} x^{2} + 1}\right )}{2 \, {\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} \]

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

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)

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maple [B]  time = 0.10, size = 222, normalized size = 2.61 \[ -\frac {a^{2}}{2 c^{2} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {2 b^{2} \arcsinh \left (c x \right )}{c^{2} d^{2}}+\frac {b^{2} \arcsinh \left (c x \right ) x}{c \,d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {b^{2} \arcsinh \left (c x \right ) x^{2}}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )^{2}}{2 c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \arcsinh \left (c x \right )}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{2} d^{2}}-\frac {a b \arcsinh \left (c x \right )}{c^{2} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {a b x}{c \,d^{2} \sqrt {c^{2} x^{2}+1}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{2 \, {\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} - \frac {a^{2}}{2 \, {\left (c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}} + \int \frac {{\left ({\left (2 \, a b c^{2} + b^{2} c^{2}\right )} x^{2} + \sqrt {c^{2} x^{2} + 1} {\left (2 \, a b c + b^{2} c\right )} x + b^{2}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{2} x^{5} + 2 \, c^{4} d^{2} x^{3} + c^{2} d^{2} x + {\left (c^{5} d^{2} x^{4} + 2 \, c^{3} d^{2} x^{2} + c d^{2}\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

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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