3.102 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^4 \sqrt{\pi c^2 x^2+\pi }}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 \pi ^{5/2} b c^5}+\frac{b}{6 \pi ^{5/2} c^5 \left (c^2 x^2+1\right )}+\frac{2 b \log \left (c^2 x^2+1\right )}{3 \pi ^{5/2} c^5} \]

[Out]

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

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Rubi [A]  time = 0.282824, antiderivative size = 178, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {5751, 5675, 260, 266, 43} \[ -\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^4 \sqrt{\pi c^2 x^2+\pi }}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 \pi ^{5/2} b c^5}+\frac{b}{6 \pi ^2 c^5 \sqrt{c^2 x^2+1} \sqrt{\pi c^2 x^2+\pi }}+\frac{2 b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{3 \pi ^2 c^5 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

b/(6*c^5*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*Pi*x^2]) - (x^3*(a + b*ArcSinh[c*x]))/(3*c^2*Pi*(Pi + c^2*Pi*x^2
)^(3/2)) - (x*(a + b*ArcSinh[c*x]))/(c^4*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (a + b*ArcSinh[c*x])^2/(2*b*c^5*Pi^(5/2
)) + (2*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(3*c^5*Pi^2*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 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 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 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 )^{5/2}} \, dx &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac{\int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx}{c^2 \pi }+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{x^3}{\left (1+c^2 x^2\right )^2} \, dx}{3 c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{c^4 \pi ^2}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{x}{1+c^2 x^2} \, dx}{c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^5 \pi ^{5/2}}+\frac{b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{2 c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2 \left (1+c^2 x\right )^2}+\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=\frac{b}{6 c^5 \pi ^2 \sqrt{1+c^2 x^2} \sqrt{\pi +c^2 \pi x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c^5 \pi ^{5/2}}+\frac{2 b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ \end{align*}

Mathematica [A]  time = 0.350127, size = 166, normalized size = 1.19 \[ \frac{2 \sinh ^{-1}(c x) \left (3 a \left (c^2 x^2+1\right )^2-b c x \sqrt{c^2 x^2+1} \left (4 c^2 x^2+3\right )\right )-8 a c^3 x^3 \sqrt{c^2 x^2+1}-6 a c x \sqrt{c^2 x^2+1}+b c^2 x^2+4 b \left (c^2 x^2+1\right )^2 \log \left (c^2 x^2+1\right )+3 b \left (c^2 x^2+1\right )^2 \sinh ^{-1}(c x)^2+b}{6 \pi ^{5/2} c^5 \left (c^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

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

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Maple [B]  time = 0.203, size = 897, normalized size = 6.5 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

-1/3*a*x^3/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)-a/Pi^2/c^4*x/(Pi*c^2*x^2+Pi)^(1/2)+a/Pi^2/c^4*ln(Pi*x*c^2/(Pi*c^2)^(1/
2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/2*b/c^5/Pi^(5/2)*arcsinh(c*x)^2-8/3*b/c^5/Pi^(5/2)*arcsinh(c*x)+32*
b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2*c^3*arcsinh(c*x)*x^8-32*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+
16)/(c^2*x^2+1)^(3/2)*c^2*arcsinh(c*x)*x^7+8/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2*c^3*x^8-8/3
*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)*c*x^6+116*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)
^2*c*arcsinh(c*x)*x^6-76*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^5+32/3*b/Pi^(5
/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2*c*x^6-4*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)/c*x^4+4
72/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2/c*arcsinh(c*x)*x^4-181/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^
2*x^2+16)/(c^2*x^2+1)^(3/2)/c^2*arcsinh(c*x)*x^3+16*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2/c*x^4-
3/2*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)/c^3*x^2+284/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2
*x^2+1)^2/c^3*arcsinh(c*x)*x^2-16*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^(3/2)/c^4*arcsinh(c*x)*x+3
2/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2/c^3*x^2+64/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^
2*x^2+1)^2/c^5*arcsinh(c*x)+8/3*b/Pi^(5/2)/(24*c^4*x^4+39*c^2*x^2+16)/(c^2*x^2+1)^2/c^5+4/3*b/c^5/Pi^(5/2)*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}{3} \,{\left (x{\left (\frac{3 \, x^{2}}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2}} + \frac{2}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{4}}\right )} + \frac{x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} c^{4}} - \frac{3 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\pi ^{2} \sqrt{\pi c^{2}} c^{4}}\right )} a + 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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

(Integral(a*x**4/(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) +
Integral(b*x**4*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{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{4}}{{\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(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^4/(pi + pi*c^2*x^2)^(5/2), x)