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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 105, normalized size of antiderivative = 1.31, 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 = {5751, 5675, 260} \[ -\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\pi c^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 \pi ^{3/2} b c^3}+\frac {b \sqrt {c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 \pi c^3 \sqrt {\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 78, normalized size = 0.98 \[ \frac {\sinh ^{-1}(c x) \left (2 a-\frac {2 b c x}{\sqrt {c^2 x^2+1}}\right )-\frac {2 a c x}{\sqrt {c^2 x^2+1}}+b \log \left (c^2 x^2+1\right )+b \sinh ^{-1}(c x)^2}{2 \pi ^{3/2} c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b x^{2} \operatorname {arsinh}\left (c x\right ) + a x^{2}\right )}}{\pi ^{2} c^{4} x^{4} + 2 \, \pi ^{2} c^{2} x^{2} + \pi ^{2}}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.16, size = 196, normalized size = 2.45 \[ -\frac {a x}{\pi \,c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {a \ln \left (\frac {\pi x \,c^{2}}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{\pi \,c^{2} \sqrt {\pi \,c^{2}}}+\frac {b \arcsinh \left (c x \right )^{2}}{2 c^{3} \pi ^{\frac {3}{2}}}-\frac {2 b \arcsinh \left (c x \right )}{c^{3} \pi ^{\frac {3}{2}}}+\frac {b \arcsinh \left (c x \right ) x^{2}}{\pi ^{\frac {3}{2}} c \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right ) x}{\pi ^{\frac {3}{2}} c^{2} \sqrt {c^{2} x^{2}+1}}+\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} c^{3} \left (c^{2} x^{2}+1\right )}+\frac {b \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{3} \pi ^{\frac {3}{2}}} \]

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)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {x}{\pi \sqrt {\pi + \pi c^{2} x^{2}} c^{2}} - \frac {\operatorname {arsinh}\left (c x\right )}{\pi ^{\frac {3}{2}} c^{3}}\right )} + b \int \frac {x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

-a*(x/(pi*sqrt(pi + pi*c^2*x^2)*c^2) - arcsinh(c*x)/(pi^(3/2)*c^3)) + b*integrate(x^2*log(c*x + sqrt(c^2*x^2 +
1))/(pi + pi*c^2*x^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(3/2),x)

[Out]

int((x^2*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{2}}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{2} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \]

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)**(3/2),x)

[Out]

(Integral(a*x**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**2*
x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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