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

Optimal. Leaf size=130 \[ \frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{c^2 d x^2+d}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{c^2 d x^2+d}}+\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^3 d \sqrt{c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.168188, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5751, 5677, 5675, 260} \[ \frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt{c^2 d x^2+d}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt{c^2 d x^2+d}}+\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 c^3 d \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5751

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p
+ 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 5677

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e,
c^2*d] &&  !GtQ[d, 0]

Rule 5675

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

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

Mathematica [A]  time = 0.195132, size = 146, normalized size = 1.12 \[ -\frac{a x \sqrt{d \left (c^2 x^2+1\right )}}{c^2 d^2 \left (c^2 x^2+1\right )}+\frac{a \log \left (\sqrt{d} \sqrt{d \left (c^2 x^2+1\right )}+c d x\right )}{c^3 d^{3/2}}+\frac{b \left (\sqrt{c^2 x^2+1} \left (2 \log \left (\sqrt{c^2 x^2+1}\right )+\sinh ^{-1}(c x)^2\right )-2 c x \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d \left (c^2 x^2+1\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.151, size = 232, normalized size = 1.8 \begin{align*} -{\frac{ax}{{c}^{2}d}{\frac{1}{\sqrt{{c}^{2}d{x}^{2}+d}}}}+{\frac{a}{{c}^{2}d}\ln \left ({{c}^{2}dx{\frac{1}{\sqrt{{c}^{2}d}}}}+\sqrt{{c}^{2}d{x}^{2}+d} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{2\,{c}^{3}{d}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{c}^{3}{d}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) x}{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}+1 \right ) }\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b}{{c}^{3}{d}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}\right )}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*x^2*arcsinh(c*x) + a*x^2)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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