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

Optimal. Leaf size=108 \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{b}{6 \pi ^{5/2} c \left (c^2 x^2+1\right )}-\frac{b \log \left (c^2 x^2+1\right )}{3 \pi ^{5/2} c} \]

[Out]

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

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Rubi [A]  time = 0.0873414, antiderivative size = 147, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {5690, 5687, 260, 261} \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{b}{6 \pi ^2 c \sqrt{c^2 x^2+1} \sqrt{\pi c^2 x^2+\pi }}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 \pi ^2 c \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5690

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

Rule 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh
[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[
c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

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 261

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

Rubi steps

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

Mathematica [A]  time = 0.146641, size = 100, normalized size = 0.93 \[ \frac{4 a c^3 x^3+6 a c x+b \sqrt{c^2 x^2+1}-2 b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )+2 b c x \left (2 c^2 x^2+3\right ) \sinh ^{-1}(c x)}{6 \pi ^{5/2} c \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 + c^2*x^2] + 2*b*c*x*(3 + 2*c^2*x^2)*ArcSinh[c*x] - 2*b*(1 + c^2*x^2)^(3/2)*
Log[1 + c^2*x^2])/(6*c*Pi^(5/2)*(1 + c^2*x^2)^(3/2))

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Maple [B]  time = 0.09, size = 618, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

1/3*a/Pi*x/(Pi*c^2*x^2+Pi)^(3/2)+2/3*a/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)+4/3*b/c/Pi^(5/2)*arcsinh(c*x)+2/3*b/Pi^(5/
2)*c^7/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^8-2/3*b/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)*x^6-2*b/Pi^(5/2)*c^5/(3*c^
2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^6+2*b/Pi^(5/2)*c^4/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^5+8/3*
b/Pi^(5/2)*c^5/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^6-2*b/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)*x^4-20/3*b/Pi^(5/2)*
c^3/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^4+17/3*b/Pi^(5/2)*c^2/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(c
*x)*x^3+4*b/Pi^(5/2)*c^3/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^4-3/2*b/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)*x^2-22/3*b
/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c*x)*x^2+4*b/Pi^(5/2)/(3*c^2*x^2+4)/(c^2*x^2+1)^(3/2)*arcsinh(
c*x)*x+8/3*b/Pi^(5/2)*c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*x^2-8/3*b/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2*arcsinh(c
*x)+2/3*b/Pi^(5/2)/c/(3*c^2*x^2+4)/(c^2*x^2+1)^2-2/3*b/c/Pi^(5/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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Maxima [A]  time = 1.12087, size = 170, normalized size = 1.57 \begin{align*} \frac{1}{6} \, b c{\left (\frac{1}{\pi ^{\frac{5}{2}} c^{4} x^{2} + \pi ^{\frac{5}{2}} c^{2}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac{5}{2}} c^{2}}\right )} + \frac{1}{3} \, b{\left (\frac{x}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}}} + \frac{2 \, x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{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}}} + \frac{2 \, x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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^4*x^2 + pi^(5/2)*c^2) - 2*log(c^2*x^2 + 1)/(pi^(5/2)*c^2)) + 1/3*b*(x/(pi*(pi + pi*c^2*
x^2)^(3/2)) + 2*x/(pi^2*sqrt(pi + pi*c^2*x^2)))*arcsinh(c*x) + 1/3*a*(x/(pi*(pi + pi*c^2*x^2)^(3/2)) + 2*x/(pi
^2*sqrt(pi + pi*c^2*x^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 \operatorname{arsinh}\left (c x\right ) + a\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((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*arcsinh(c*x) + a)/(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}{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 \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((a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(5/2),x)

[Out]

(Integral(a/(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) + Inte
gral(b*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{b \operatorname{arsinh}\left (c x\right ) + a}{{\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((a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

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