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

Optimal. Leaf size=119 \[ \frac{1}{4} x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{\sqrt{\pi } x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\sqrt{\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3}-\frac{1}{16} \sqrt{\pi } b c x^4-\frac{\sqrt{\pi } b x^2}{16 c} \]

[Out]

-(b*Sqrt[Pi]*x^2)/(16*c) - (b*c*Sqrt[Pi]*x^4)/16 + (Sqrt[Pi]*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(8*c^2)
+ (x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/4 - (Sqrt[Pi]*(a + b*ArcSinh[c*x])^2)/(16*b*c^3)

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Rubi [A]  time = 0.197875, antiderivative size = 181, normalized size of antiderivative = 1.52, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5742, 5758, 5675, 30} \[ \frac{1}{4} x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}-\frac{\sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt{c^2 x^2+1}}-\frac{b c x^4 \sqrt{\pi c^2 x^2+\pi }}{16 \sqrt{c^2 x^2+1}}-\frac{b x^2 \sqrt{\pi c^2 x^2+\pi }}{16 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x^2*Sqrt[Pi + c^2*Pi*x^2])/(16*c*Sqrt[1 + c^2*x^2]) - (b*c*x^4*Sqrt[Pi + c^2*Pi*x^2])/(16*Sqrt[1 + c^2*x^2
]) + (x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(8*c^2) + (x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))
/4 - (Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(16*b*c^3*Sqrt[1 + c^2*x^2])

Rule 5742

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

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 x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\sqrt{\pi +c^2 \pi x^2} \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{4 \sqrt{1+c^2 x^2}}-\frac{\left (b c \sqrt{\pi +c^2 \pi x^2}\right ) \int x^3 \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=-\frac{b c x^4 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}+\frac{x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac{1}{4} x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\sqrt{\pi +c^2 \pi x^2} \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{8 c^2 \sqrt{1+c^2 x^2}}-\frac{\left (b \sqrt{\pi +c^2 \pi x^2}\right ) \int x \, dx}{8 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b x^2 \sqrt{\pi +c^2 \pi x^2}}{16 c \sqrt{1+c^2 x^2}}-\frac{b c x^4 \sqrt{\pi +c^2 \pi x^2}}{16 \sqrt{1+c^2 x^2}}+\frac{x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{8 c^2}+\frac{1}{4} x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c^3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.168271, size = 79, normalized size = 0.66 \[ \frac{\sqrt{\pi } \left (\sinh ^{-1}(c x) \left (4 b \sinh \left (4 \sinh ^{-1}(c x)\right )-16 a\right )+16 a c x \sqrt{c^2 x^2+1} \left (2 c^2 x^2+1\right )-8 b \sinh ^{-1}(c x)^2-b \cosh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Pi]*(16*a*c*x*Sqrt[1 + c^2*x^2]*(1 + 2*c^2*x^2) - 8*b*ArcSinh[c*x]^2 - b*Cosh[4*ArcSinh[c*x]] + ArcSinh[
c*x]*(-16*a + 4*b*Sinh[4*ArcSinh[c*x]])))/(128*c^3)

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Maple [A]  time = 0.055, size = 170, normalized size = 1.4 \begin{align*}{\frac{ax}{4\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}-{\frac{ax}{8\,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{a\pi }{8\,{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\sqrt{\pi }{\it Arcsinh} \left ( cx \right ){x}^{3}}{4}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{bc{x}^{4}\sqrt{\pi }}{16}}+{\frac{b\sqrt{\pi }{\it Arcsinh} \left ( cx \right ) x}{8\,{c}^{2}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{x}^{2}\sqrt{\pi }}{16\,c}}-{\frac{b\sqrt{\pi } \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{16\,{c}^{3}}} \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]

1/4*a*x*(Pi*c^2*x^2+Pi)^(3/2)/Pi/c^2-1/8*a/c^2*x*(Pi*c^2*x^2+Pi)^(1/2)-1/8*a/c^2*Pi*ln(Pi*x*c^2/(Pi*c^2)^(1/2)
+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/4*b*Pi^(1/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-1/16*b*c*x^4*Pi^(1/2)
+1/8*b*Pi^(1/2)/c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-1/16*b*x^2*Pi^(1/2)/c-1/16*b*Pi^(1/2)/c^3*arcsinh(c*x)^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 (\sqrt{\pi + \pi c^{2} x^{2}}{\left (b x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}\right )}, 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(sqrt(pi + pi*c^2*x^2)*(b*x^2*arcsinh(c*x) + a*x^2), x)

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

sqrt(pi)*(Integral(a*x**2*sqrt(c**2*x**2 + 1), x) + Integral(b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x), x))

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