3.83 \(\int \frac{x^2 (a+b \sinh ^{-1}(c x))}{\sqrt{\pi +c^2 \pi x^2}} \, dx\)

Optimal. Leaf size=75 \[ \frac{x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi c^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 \sqrt{\pi } b c^3}-\frac{b x^2}{4 \sqrt{\pi } c} \]

[Out]

-(b*x^2)/(4*c*Sqrt[Pi]) + (x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(2*c^2*Pi) - (a + b*ArcSinh[c*x])^2/(
4*b*c^3*Sqrt[Pi])

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Rubi [A]  time = 0.121491, antiderivative size = 97, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {5758, 5675, 30} \[ \frac{x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi c^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{4 \sqrt{\pi } b c^3}-\frac{b x^2 \sqrt{c^2 x^2+1}}{4 c \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5758

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.175136, size = 69, normalized size = 0.92 \[ \frac{\sinh ^{-1}(c x) \left (2 b \sinh \left (2 \sinh ^{-1}(c x)\right )-4 a\right )+4 a c x \sqrt{c^2 x^2+1}-2 b \sinh ^{-1}(c x)^2-b \cosh \left (2 \sinh ^{-1}(c x)\right )}{8 \sqrt{\pi } c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*c*x*Sqrt[1 + c^2*x^2] - 2*b*ArcSinh[c*x]^2 - b*Cosh[2*ArcSinh[c*x]] + ArcSinh[c*x]*(-4*a + 2*b*Sinh[2*Arc
Sinh[c*x]]))/(8*c^3*Sqrt[Pi])

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Maple [A]  time = 0.055, size = 125, normalized size = 1.7 \begin{align*}{\frac{ax}{2\,\pi \,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{a}{2\,{c}^{2}}\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) x}{2\,{c}^{2}\sqrt{\pi }}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{x}^{2}}{4\,c\sqrt{\pi }}}-{\frac{b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{4\,{c}^{3}\sqrt{\pi }}}-{\frac{b}{4\,{c}^{3}\sqrt{\pi }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x)

[Out]

1/2*a*x/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-1/2*a/c^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2
)+1/2*b/c^2/Pi^(1/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-1/4*b*x^2/c/Pi^(1/2)-1/4*b/c^3/Pi^(1/2)*arcsinh(c*x)^2-1
/4*b/c^3/Pi^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.64037, size = 92, normalized size = 1.23 \begin{align*} \frac{a x \sqrt{c^{2} x^{2} + 1}}{2 \sqrt{\pi } c^{2}} - \frac{a \operatorname{asinh}{\left (c x \right )}}{2 \sqrt{\pi } c^{3}} + \frac{b \left (\begin{cases} - \frac{x^{2}}{4 c} + \frac{x \sqrt{c^{2} x^{2} + 1} \operatorname{asinh}{\left (c x \right )}}{2 c^{2}} - \frac{\operatorname{asinh}^{2}{\left (c x \right )}}{4 c^{3}} & \text{for}\: c \neq 0 \\0 & \text{otherwise} \end{cases}\right )}{\sqrt{\pi }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*x*sqrt(c**2*x**2 + 1)/(2*sqrt(pi)*c**2) - a*asinh(c*x)/(2*sqrt(pi)*c**3) + b*Piecewise((-x**2/(4*c) + x*sqrt
(c**2*x**2 + 1)*asinh(c*x)/(2*c**2) - asinh(c*x)**2/(4*c**3), Ne(c, 0)), (0, True))/sqrt(pi)

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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}}{\sqrt{\pi + \pi c^{2} x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

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