3.104 \(\int \frac{x^2 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=80 \[ \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{b}{6 \pi ^{5/2} c^3 \left (c^2 x^2+1\right )}-\frac{b \log \left (c^2 x^2+1\right )}{6 \pi ^{5/2} c^3} \]

[Out]

-b/(6*c^3*Pi^(5/2)*(1 + c^2*x^2)) + (x^3*(a + b*ArcSinh[c*x]))/(3*Pi*(Pi + c^2*Pi*x^2)^(3/2)) - (b*Log[1 + c^2
*x^2])/(6*c^3*Pi^(5/2))

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Rubi [A]  time = 0.128864, antiderivative size = 119, normalized size of antiderivative = 1.49, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {5723, 266, 43} \[ \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{b}{6 \pi ^2 c^3 \sqrt{c^2 x^2+1} \sqrt{\pi c^2 x^2+\pi }}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 \pi ^2 c^3 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

-b/(6*c^3*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*Pi*x^2]) + (x^3*(a + b*ArcSinh[c*x]))/(3*Pi*(Pi + c^2*Pi*x^2)^(
3/2)) - (b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^3*Pi^2*Sqrt[Pi + c^2*Pi*x^2])

Rule 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \int \frac{x^3}{\left (1+c^2 x^2\right )^2} \, dx}{3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2 \left (1+c^2 x\right )^2}+\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b}{6 c^3 \pi ^2 \sqrt{1+c^2 x^2} \sqrt{\pi +c^2 \pi x^2}}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ \end{align*}

Mathematica [A]  time = 0.17163, size = 88, normalized size = 1.1 \[ -\frac{-2 a c^3 x^3+b \sqrt{c^2 x^2+1}+b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )-2 b c^3 x^3 \sinh ^{-1}(c x)}{6 \pi ^{5/2} c^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

-(-2*a*c^3*x^3 + b*Sqrt[1 + c^2*x^2] - 2*b*c^3*x^3*ArcSinh[c*x] + b*(1 + c^2*x^2)^(3/2)*Log[1 + c^2*x^2])/(6*c
^3*Pi^(5/2)*(1 + c^2*x^2)^(3/2))

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Maple [B]  time = 0.158, size = 707, normalized size = 8.8 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

-1/3*a/c^2/Pi*x/(Pi*c^2*x^2+Pi)^(3/2)+1/3*a/c^2/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)+2/3*b/c^3/Pi^(5/2)*arcsinh(c*x)-b
/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2*c^5*arcsinh(c*x)*x^8+b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x
^2+1)^(3/2)*c^4*arcsinh(c*x)*x^7-1/6*b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2*c^5*x^8+1/6*b/Pi^(5/2)/(
3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)*c^3*x^6-3*b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2*c^3*arcsinh(c*x)
*x^6+b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^(3/2)*c^2*arcsinh(c*x)*x^5-2/3*b/Pi^(5/2)/(3*c^4*x^4+3*c^2
*x^2+1)/(c^2*x^2+1)^2*c^3*x^6-10/3*b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2*c*arcsinh(c*x)*x^4+1/3*b/P
i^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^3-b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2
+1)^2*c*x^4-5/3*b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2/c*arcsinh(c*x)*x^2-2/3*b/Pi^(5/2)/(3*c^4*x^4+
3*c^2*x^2+1)/(c^2*x^2+1)^2/c*x^2-1/3*b/Pi^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2/c^3*arcsinh(c*x)-1/6*b/P
i^(5/2)/(3*c^4*x^4+3*c^2*x^2+1)/(c^2*x^2+1)^2/c^3-1/3*b/c^3/Pi^(5/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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Maxima [B]  time = 1.26992, size = 185, normalized size = 2.31 \begin{align*} -\frac{1}{6} \, b c{\left (\frac{1}{\pi ^{\frac{5}{2}} c^{6} x^{2} + \pi ^{\frac{5}{2}} c^{4}} + \frac{\log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac{5}{2}} c^{4}}\right )} - \frac{1}{3} \, b{\left (\frac{x}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2}} - \frac{x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} c^{2}}\right )} \operatorname{arsinh}\left (c x\right ) - \frac{1}{3} \, a{\left (\frac{x}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2}} - \frac{x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} c^{2}}\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)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(1/(pi^(5/2)*c^6*x^2 + pi^(5/2)*c^4) + log(c^2*x^2 + 1)/(pi^(5/2)*c^4)) - 1/3*b*(x/(pi*(pi + pi*c^2*x
^2)^(3/2)*c^2) - x/(pi^2*sqrt(pi + pi*c^2*x^2)*c^2))*arcsinh(c*x) - 1/3*a*(x/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2)
- x/(pi^2*sqrt(pi + pi*c^2*x^2)*c^2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\pi + \pi c^{2} x^{2}}{\left (b x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}\right )}}{\pi ^{3} c^{6} x^{6} + 3 \, \pi ^{3} c^{4} x^{4} + 3 \, \pi ^{3} c^{2} x^{2} + \pi ^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*x^2*arcsinh(c*x) + a*x^2)/(pi^3*c^6*x^6 + 3*pi^3*c^4*x^4 + 3*pi^3*c^2*x^2 +
pi^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{2}}{c^{4} x^{4} \sqrt{c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b x^{2} \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt{c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac{5}{2}}} \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)**(5/2),x)

[Out]

(Integral(a*x**2/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) +
Integral(b*x**2*asinh(c*x)/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2
+ 1)), x))/pi**(5/2)

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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 )} x^{2}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^2/(pi + pi*c^2*x^2)^(5/2), x)