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

Optimal. Leaf size=45 \[ \frac{b \tan ^{-1}(c x)}{\pi ^{3/2} c^2}-\frac{a+b \sinh ^{-1}(c x)}{\pi c^2 \sqrt{\pi c^2 x^2+\pi }} \]

[Out]

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

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Rubi [A]  time = 0.0723462, antiderivative size = 70, normalized size of antiderivative = 1.56, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5717, 203} \[ \frac{b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{\pi c^2 \sqrt{\pi c^2 x^2+\pi }}-\frac{a+b \sinh ^{-1}(c x)}{\pi c^2 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5717

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.107267, size = 52, normalized size = 1.16 \[ \frac{-a+b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)-b \sinh ^{-1}(c x)}{\pi ^{3/2} c^2 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.086, size = 103, normalized size = 2.3 \begin{align*} -{\frac{a}{\pi \,{c}^{2}}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}{c}^{2}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{2}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ) }-{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{2}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}-i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b{\left (\frac{-\operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right )}{\pi ^{\frac{3}{2}} c^{2}} - \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{\pi ^{\frac{3}{2}} \sqrt{c^{2} x^{2} + 1} c^{2}} - \int \frac{1}{\pi ^{\frac{3}{2}} c^{5} x^{4} + \pi ^{\frac{3}{2}} c^{3} x^{2} +{\left (\pi ^{\frac{3}{2}} c^{4} x^{3} + \pi ^{\frac{3}{2}} c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x}\right )} - \frac{a}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.88157, size = 305, normalized size = 6.78 \begin{align*} -\frac{\sqrt{\pi }{\left (b c^{2} x^{2} + b\right )} \arctan \left (-\frac{2 \, \sqrt{\pi } \sqrt{\pi + \pi c^{2} x^{2}} \sqrt{c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 2 \, \sqrt{\pi + \pi c^{2} x^{2}} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \, \sqrt{\pi + \pi c^{2} x^{2}} a}{2 \,{\left (\pi ^{2} c^{4} x^{2} + \pi ^{2} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(pi)*(b*c^2*x^2 + b)*arctan(-2*sqrt(pi)*sqrt(pi + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*c*x/(pi - pi*c^4*x^4
)) + 2*sqrt(pi + pi*c^2*x^2)*b*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(pi + pi*c^2*x^2)*a)/(pi^2*c^4*x^2 + pi^2*
c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b x \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b*x*asinh(c*x)/(c**2*x**2*s
qrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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