3.90 \(\int \frac{x^5 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac{\left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^3 c^6}-\frac{2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^6}-\frac{a+b \sinh ^{-1}(c x)}{\pi c^6 \sqrt{\pi c^2 x^2+\pi }}-\frac{b x^3}{9 \pi ^{3/2} c^3}+\frac{5 b x}{3 \pi ^{3/2} c^5}+\frac{b \tan ^{-1}(c x)}{\pi ^{3/2} c^6} \]

[Out]

(5*b*x)/(3*c^5*Pi^(3/2)) - (b*x^3)/(9*c^3*Pi^(3/2)) - (a + b*ArcSinh[c*x])/(c^6*Pi*Sqrt[Pi + c^2*Pi*x^2]) - (2
*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(c^6*Pi^2) + ((Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^
6*Pi^3) + (b*ArcTan[c*x])/(c^6*Pi^(3/2))

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Rubi [A]  time = 0.172302, antiderivative size = 140, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {266, 43, 5732, 1153, 205} \[ \frac{\left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{3/2} c^6}-\frac{2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} c^6}-\frac{a+b \sinh ^{-1}(c x)}{\pi ^{3/2} c^6 \sqrt{c^2 x^2+1}}-\frac{b x^3}{9 \pi ^{3/2} c^3}+\frac{5 b x}{3 \pi ^{3/2} c^5}+\frac{b \tan ^{-1}(c x)}{\pi ^{3/2} c^6} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

(5*b*x)/(3*c^5*Pi^(3/2)) - (b*x^3)/(9*c^3*Pi^(3/2)) - (a + b*ArcSinh[c*x])/(c^6*Pi^(3/2)*Sqrt[1 + c^2*x^2]) -
(2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(c^6*Pi^(3/2)) + ((1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/(3*c^6*
Pi^(3/2)) + (b*ArcTan[c*x])/(c^6*Pi^(3/2))

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])

Rule 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 205

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

Rubi steps

\begin{align*} \int \frac{x^5 \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^6 \pi ^{3/2} \sqrt{1+c^2 x^2}}-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{3/2}}+\frac{\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 \pi ^{3/2}}-\frac{(b c) \int \frac{-8-4 c^2 x^2+c^4 x^4}{3 c^6+3 c^8 x^2} \, dx}{\pi ^{3/2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{c^6 \pi ^{3/2} \sqrt{1+c^2 x^2}}-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{3/2}}+\frac{\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 \pi ^{3/2}}-\frac{(b c) \int \left (-\frac{5}{3 c^6}+\frac{x^2}{3 c^4}-\frac{3}{3 c^6+3 c^8 x^2}\right ) \, dx}{\pi ^{3/2}}\\ &=\frac{5 b x}{3 c^5 \pi ^{3/2}}-\frac{b x^3}{9 c^3 \pi ^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{c^6 \pi ^{3/2} \sqrt{1+c^2 x^2}}-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{3/2}}+\frac{\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 \pi ^{3/2}}+\frac{(3 b c) \int \frac{1}{3 c^6+3 c^8 x^2} \, dx}{\pi ^{3/2}}\\ &=\frac{5 b x}{3 c^5 \pi ^{3/2}}-\frac{b x^3}{9 c^3 \pi ^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{c^6 \pi ^{3/2} \sqrt{1+c^2 x^2}}-\frac{2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{3/2}}+\frac{\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 \pi ^{3/2}}+\frac{b \tan ^{-1}(c x)}{c^6 \pi ^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.19999, size = 131, normalized size = 0.96 \[ \frac{3 a c^4 x^4-12 a c^2 x^2-24 a-b c^3 x^3 \sqrt{c^2 x^2+1}+15 b c x \sqrt{c^2 x^2+1}+9 b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)+3 b \left (c^4 x^4-4 c^2 x^2-8\right ) \sinh ^{-1}(c x)}{9 \pi ^{3/2} c^6 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

(-24*a - 12*a*c^2*x^2 + 3*a*c^4*x^4 + 15*b*c*x*Sqrt[1 + c^2*x^2] - b*c^3*x^3*Sqrt[1 + c^2*x^2] + 3*b*(-8 - 4*c
^2*x^2 + c^4*x^4)*ArcSinh[c*x] + 9*b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(9*c^6*Pi^(3/2)*Sqrt[1 + c^2*x^2])

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Maple [C]  time = 0.237, size = 224, normalized size = 1.6 \begin{align*}{\frac{a{x}^{4}}{3\,\pi \,{c}^{2}}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{4\,a{x}^{2}}{3\,{c}^{4}\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{8\,a}{3\,{c}^{6}\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{b{x}^{3}}{9\,{c}^{3}{\pi }^{3/2}}}+{\frac{5\,bx}{3\,{c}^{5}{\pi }^{3/2}}}-{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{6}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}-i \right ) }-{\frac{5\,b{\it Arcsinh} \left ( cx \right ) }{3\,{\pi }^{3/2}{c}^{6}}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{6}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ) }-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}{c}^{6}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{3\,{\pi }^{3/2}{c}^{4}}\sqrt{{c}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

1/3*a*x^4/Pi/c^2/(Pi*c^2*x^2+Pi)^(1/2)-4/3*a/c^4*x^2/Pi/(Pi*c^2*x^2+Pi)^(1/2)-8/3*a/c^6/Pi/(Pi*c^2*x^2+Pi)^(1/
2)-1/9*b*x^3/c^3/Pi^(3/2)+5/3*b*x/c^5/Pi^(3/2)-I*b/c^6/Pi^(3/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I)-5/3*b/Pi^(3/2)/c^6
*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+I*b/c^6/Pi^(3/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)-b/Pi^(3/2)/(c^2*x^2+1)^(1/2)/c^6*
arcsinh(c*x)+1/3*b/Pi^(3/2)/c^4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a{\left (\frac{x^{4}}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{2}} - \frac{4 \, x^{2}}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{4}} - \frac{8}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{6}}\right )} + \frac{1}{3} \, b{\left (\frac{{\left (\sqrt{\pi } c^{4} x^{4} - 4 \, \sqrt{\pi } c^{2} x^{2} - 8 \, \sqrt{\pi }\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{\pi ^{2} \sqrt{c^{2} x^{2} + 1} c^{6}} - \frac{\frac{1}{3} \, \sqrt{\pi } \sqrt{c^{2} x^{2} + 1} c^{2} x^{2} + 8 \, \sqrt{\pi } \operatorname{arsinh}\left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - \frac{14}{3} \, \sqrt{\pi } \sqrt{c^{2} x^{2} + 1}}{\pi ^{2} c^{6}} + 3 \, \int \frac{\sqrt{\pi } c^{4} x^{4} - 4 \, \sqrt{\pi } c^{2} x^{2} - 8 \, \sqrt{\pi }}{3 \,{\left (\pi ^{2} c^{9} x^{4} + \pi ^{2} c^{7} x^{2} +{\left (\pi ^{2} c^{8} x^{3} + \pi ^{2} c^{6} x\right )} \sqrt{c^{2} x^{2} + 1}\right )}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxima")

[Out]

1/3*a*(x^4/(pi*sqrt(pi + pi*c^2*x^2)*c^2) - 4*x^2/(pi*sqrt(pi + pi*c^2*x^2)*c^4) - 8/(pi*sqrt(pi + pi*c^2*x^2)
*c^6)) + 1/3*b*((sqrt(pi)*c^4*x^4 - 4*sqrt(pi)*c^2*x^2 - 8*sqrt(pi))*log(c*x + sqrt(c^2*x^2 + 1))/(pi^2*sqrt(c
^2*x^2 + 1)*c^6) - integrate((sqrt(pi)*c^4*x^4 - 4*sqrt(pi)*c^2*x^2 - 8*sqrt(pi))/(sqrt(c^2*x^2 + 1)*x), x)/(p
i^2*c^6) + 3*integrate(1/3*(sqrt(pi)*c^4*x^4 - 4*sqrt(pi)*c^2*x^2 - 8*sqrt(pi))/(pi^2*c^9*x^4 + pi^2*c^7*x^2 +
(pi^2*c^8*x^3 + pi^2*c^6*x)*sqrt(c^2*x^2 + 1)), x))

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Fricas [A]  time = 2.96779, size = 454, normalized size = 3.31 \begin{align*} -\frac{9 \, \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 ) - 6 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (b c^{4} x^{4} - 4 \, b c^{2} x^{2} - 8 \, b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (3 \, a c^{4} x^{4} - 12 \, a c^{2} x^{2} -{\left (b c^{3} x^{3} - 15 \, b c x\right )} \sqrt{c^{2} x^{2} + 1} - 24 \, a\right )}}{18 \,{\left (\pi ^{2} c^{8} x^{2} + \pi ^{2} c^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="fricas")

[Out]

-1/18*(9*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)) - 6*sqrt(pi + pi*c^2*x^2)*(b*c^4*x^4 - 4*b*c^2*x^2 - 8*b)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*sqrt(pi + pi*
c^2*x^2)*(3*a*c^4*x^4 - 12*a*c^2*x^2 - (b*c^3*x^3 - 15*b*c*x)*sqrt(c^2*x^2 + 1) - 24*a))/(pi^2*c^8*x^2 + pi^2*
c^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{5}}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b x^{5} \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**5*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a*x**5/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b*x**5*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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{5}}{{\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^5*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^5/(pi + pi*c^2*x^2)^(3/2), x)