3.91 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\pi c^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{3 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 c^4}-\frac{3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{3/2} b c^5}-\frac{b x^2}{4 \pi ^{3/2} c^3}-\frac{b \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2} c^5} \]

[Out]

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

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Rubi [A]  time = 0.257467, antiderivative size = 181, normalized size of antiderivative = 1.38, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5751, 5758, 5675, 30, 266, 43} \[ -\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\pi c^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{3 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^2 c^4}-\frac{3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{3/2} b c^5}-\frac{b x^2 \sqrt{c^2 x^2+1}}{4 \pi c^3 \sqrt{\pi c^2 x^2+\pi }}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{2 \pi c^5 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

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

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]

Rule 5758

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

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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])

Rubi steps

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

Mathematica [A]  time = 0.347651, size = 147, normalized size = 1.12 \[ \frac{\sinh ^{-1}(c x) \left (-12 a \sqrt{c^2 x^2+1}+9 b c x+b \sinh \left (3 \sinh ^{-1}(c x)\right )\right )+4 a c^3 x^3+12 a c x-4 b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )-6 b \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)^2-b \sqrt{c^2 x^2+1} \cosh \left (2 \sinh ^{-1}(c x)\right )}{8 \pi ^{3/2} c^5 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]

[Out]

(12*a*c*x + 4*a*c^3*x^3 - 6*b*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - b*Sqrt[1 + c^2*x^2]*Cosh[2*ArcSinh[c*x]] - 4*
b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + ArcSinh[c*x]*(9*b*c*x - 12*a*Sqrt[1 + c^2*x^2] + b*Sinh[3*ArcSinh[c*x]]
))/(8*c^5*Pi^(3/2)*Sqrt[1 + c^2*x^2])

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Maple [B]  time = 0.178, size = 269, normalized size = 2.1 \begin{align*}{\frac{a{x}^{3}}{2\,\pi \,{c}^{2}}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}+{\frac{3\,ax}{2\,{c}^{4}\pi }{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-{\frac{3\,a}{2\,{c}^{4}\pi }\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}-{\frac{3\,b \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{4\,{\pi }^{3/2}{c}^{5}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) x}{2\,{\pi }^{3/2}{c}^{4}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{x}^{2}}{4\,{c}^{3}{\pi }^{3/2}}}+2\,{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{3/2}{c}^{5}}}-{\frac{b}{8\,{\pi }^{3/2}{c}^{5}}}-{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{{c}^{3}{\pi }^{{\frac{3}{2}}} \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ) x}{{\pi }^{{\frac{3}{2}}}{c}^{4}}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}{c}^{5} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b}{{\pi }^{{\frac{3}{2}}}{c}^{5}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

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

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

[Out]

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

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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 x^{4} \operatorname{arsinh}\left (c x\right ) + a x^{4}\right )}}{\pi ^{2} c^{4} x^{4} + 2 \, \pi ^{2} c^{2} x^{2} + \pi ^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(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^4*arcsinh(c*x) + a*x^4)/(pi^2*c^4*x^4 + 2*pi^2*c^2*x^2 + pi^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{4}}{c^{2} x^{2} \sqrt{c^{2} x^{2} + 1} + \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b x^{4} \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**4*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a*x**4/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) + Integral(b*x**4*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^{4}}{{\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^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^4/(pi + pi*c^2*x^2)^(3/2), x)