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

Optimal. Leaf size=86 \[ \frac{\sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^4}+\frac{a+b \sinh ^{-1}(c x)}{\pi c^4 \sqrt{\pi c^2 x^2+\pi }}-\frac{b x}{\pi ^{3/2} c^3}-\frac{b \tan ^{-1}(c x)}{\pi ^{3/2} c^4} \]

[Out]

-((b*x)/(c^3*Pi^(3/2))) + (a + b*ArcSinh[c*x])/(c^4*Pi*Sqrt[Pi + c^2*Pi*x^2]) + (Sqrt[Pi + c^2*Pi*x^2]*(a + b*
ArcSinh[c*x]))/(c^4*Pi^2) - (b*ArcTan[c*x])/(c^4*Pi^(3/2))

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Rubi [A]  time = 0.142452, antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {266, 43, 5732, 388, 205} \[ \frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} c^4}+\frac{a+b \sinh ^{-1}(c x)}{\pi ^{3/2} c^4 \sqrt{c^2 x^2+1}}-\frac{b x}{\pi ^{3/2} c^3}-\frac{b \tan ^{-1}(c x)}{\pi ^{3/2} c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x)/(c^3*Pi^(3/2))) + (a + b*ArcSinh[c*x])/(c^4*Pi^(3/2)*Sqrt[1 + c^2*x^2]) + (Sqrt[1 + c^2*x^2]*(a + b*Ar
cSinh[c*x]))/(c^4*Pi^(3/2)) - (b*ArcTan[c*x])/(c^4*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 388

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

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

Mathematica [A]  time = 0.159627, size = 87, normalized size = 1.01 \[ \frac{a c^2 x^2+2 a-b c x \sqrt{c^2 x^2+1}-b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)+b \left (c^2 x^2+2\right ) \sinh ^{-1}(c x)}{\pi ^{3/2} c^4 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.181, size = 158, normalized size = 1.8 \begin{align*}{\frac{a{x}^{2}}{\pi \,{c}^{2}}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}+2\,{\frac{a}{\pi \,{c}^{4}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}{c}^{4}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{bx}{{c}^{3}{\pi }^{{\frac{3}{2}}}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}{c}^{4}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{4}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}-i \right ) }-{\frac{ib}{{\pi }^{{\frac{3}{2}}}{c}^{4}}\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^3*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.7572, size = 161, normalized size = 1.87 \begin{align*} -b c{\left (\frac{x}{\pi ^{\frac{3}{2}} c^{4}} + \frac{\arctan \left (c x\right )}{\pi ^{\frac{3}{2}} c^{5}}\right )} + b{\left (\frac{x^{2}}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{2}} + \frac{2}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{4}}\right )} \operatorname{arsinh}\left (c x\right ) + a{\left (\frac{x^{2}}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{2}} + \frac{2}{\pi \sqrt{\pi + \pi c^{2} x^{2}} c^{4}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.98134, size = 382, normalized size = 4.44 \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}}{\left (b c^{2} x^{2} + 2 \, b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (a c^{2} x^{2} - \sqrt{c^{2} x^{2} + 1} b c x + 2 \, a\right )}}{2 \,{\left (\pi ^{2} c^{6} x^{2} + \pi ^{2} c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

(Integral(a*x**3/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b*x**3*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^{3}}{{\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^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

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