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

Optimal. Leaf size=208 \[ \frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 \sqrt{\pi c^2 x^2+\pi }}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi x \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{3 \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{b c^3}{6 \pi ^{5/2} \left (c^2 x^2+1\right )}-\frac{4 b c^3 \log \left (c^2 x^2+1\right )}{3 \pi ^{5/2}}-\frac{8 b c^3 \log (x)}{3 \pi ^{5/2}}-\frac{b c}{6 \pi ^{5/2} x^2} \]

[Out]

-(b*c)/(6*Pi^(5/2)*x^2) + (b*c^3)/(6*Pi^(5/2)*(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])/(3*Pi*x^3*(Pi + c^2*Pi*x^2
)^(3/2)) + (2*c^2*(a + b*ArcSinh[c*x]))/(Pi*x*(Pi + c^2*Pi*x^2)^(3/2)) + (8*c^4*x*(a + b*ArcSinh[c*x]))/(3*Pi*
(Pi + c^2*Pi*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSinh[c*x]))/(3*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) - (8*b*c^3*Log[x])/(
3*Pi^(5/2)) - (4*b*c^3*Log[1 + c^2*x^2])/(3*Pi^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.239877, antiderivative size = 212, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {271, 192, 191, 5732, 12, 1799, 1620} \[ \frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{c^2 x^2+1}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (c^2 x^2+1\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (c^2 x^2+1\right )^{3/2}}-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (c^2 x^2+1\right )^{3/2}}+\frac{b c^3}{6 \pi ^{5/2} \left (c^2 x^2+1\right )}-\frac{4 b c^3 \log \left (c^2 x^2+1\right )}{3 \pi ^{5/2}}-\frac{8 b c^3 \log (x)}{3 \pi ^{5/2}}-\frac{b c}{6 \pi ^{5/2} x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(6*Pi^(5/2)*x^2) + (b*c^3)/(6*Pi^(5/2)*(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])/(3*Pi^(5/2)*x^3*(1 + c^2*x
^2)^(3/2)) + (2*c^2*(a + b*ArcSinh[c*x]))/(Pi^(5/2)*x*(1 + c^2*x^2)^(3/2)) + (8*c^4*x*(a + b*ArcSinh[c*x]))/(3
*Pi^(5/2)*(1 + c^2*x^2)^(3/2)) + (16*c^4*x*(a + b*ArcSinh[c*x]))/(3*Pi^(5/2)*Sqrt[1 + c^2*x^2]) - (8*b*c^3*Log
[x])/(3*Pi^(5/2)) - (4*b*c^3*Log[1 + c^2*x^2])/(3*Pi^(5/2))

Rule 271

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

Rule 192

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{x^4 \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (1+c^2 x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{1+c^2 x^2}}-\frac{(b c) \int \frac{-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6}{3 x^3 \left (1+c^2 x^2\right )^2} \, dx}{\pi ^{5/2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (1+c^2 x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{1+c^2 x^2}}-\frac{(b c) \int \frac{-1+6 c^2 x^2+24 c^4 x^4+16 c^6 x^6}{x^3 \left (1+c^2 x^2\right )^2} \, dx}{3 \pi ^{5/2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (1+c^2 x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{1+c^2 x^2}}-\frac{(b c) \operatorname{Subst}\left (\int \frac{-1+6 c^2 x+24 c^4 x^2+16 c^6 x^3}{x^2 \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 \pi ^{5/2}}\\ &=-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (1+c^2 x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{1+c^2 x^2}}-\frac{(b c) \operatorname{Subst}\left (\int \left (-\frac{1}{x^2}+\frac{8 c^2}{x}+\frac{c^4}{\left (1+c^2 x\right )^2}+\frac{8 c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 \pi ^{5/2}}\\ &=-\frac{b c}{6 \pi ^{5/2} x^2}+\frac{b c^3}{6 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} x^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 c^2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} x \left (1+c^2 x^2\right )^{3/2}}+\frac{8 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac{16 c^4 x \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^{5/2} \sqrt{1+c^2 x^2}}-\frac{8 b c^3 \log (x)}{3 \pi ^{5/2}}-\frac{4 b c^3 \log \left (1+c^2 x^2\right )}{3 \pi ^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.237217, size = 142, normalized size = 0.68 \[ \frac{2 a \left (16 c^6 x^6+24 c^4 x^4+6 c^2 x^2-1\right )-b c x \sqrt{c^2 x^2+1}+2 b \left (16 c^6 x^6+24 c^4 x^4+6 c^2 x^2-1\right ) \sinh ^{-1}(c x)}{6 \pi ^{5/2} x^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{8 \left (\frac{1}{2} b c^3 \log \left (c^2 x^2+1\right )+b c^3 \log (x)\right )}{3 \pi ^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*c*x*Sqrt[1 + c^2*x^2]) + 2*a*(-1 + 6*c^2*x^2 + 24*c^4*x^4 + 16*c^6*x^6) + 2*b*(-1 + 6*c^2*x^2 + 24*c^4*x^
4 + 16*c^6*x^6)*ArcSinh[c*x])/(6*Pi^(5/2)*x^3*(1 + c^2*x^2)^(3/2)) - (8*(b*c^3*Log[x] + (b*c^3*Log[1 + c^2*x^2
])/2))/(3*Pi^(5/2))

________________________________________________________________________________________

Maple [B]  time = 0.158, size = 1153, normalized size = 5.5 \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))/x^4/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

-128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^4*c^7+1/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^
2*x^2+1)^(3/2)/x^3*arcsinh(c*x)+128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^10*c^13+128*b/Pi^(5/2
)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^8*c^11+128*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^6*c^9-
2*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)*x^2*c^5+128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2
+1)*x^4*c^7+1/6*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)/x^2*c-1/3*a/Pi/x^3/(Pi*c^2*x^2+Pi)^(3/2)-8/3*
b*c^3/Pi^(5/2)*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)+32/3*b*c^3/Pi^(5/2)*arcsinh(c*x)+16/3*b/Pi^(5/2)/(12*c^4*x^4+12
*c^2*x^2-1)/(c^2*x^2+1)^2*arcsinh(c*x)*c^3-128/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^12*c^15-
512/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^10*c^13-256*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c
^2*x^2+1)^2*x^8*c^11-512/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^6*c^9-2*b/Pi^(5/2)/(12*c^4*x^4
+12*c^2*x^2-1)/(c^2*x^2+1)*c^3-6*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/x*arcsinh(c*x)*c^2+64*
b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^7*arcsinh(c*x)*c^10+160*b/Pi^(5/2)/(12*c^4*x^4+12*c^2
*x^2-1)/(c^2*x^2+1)^(3/2)*x^5*arcsinh(c*x)*c^8+344/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x^
3*arcsinh(c*x)*c^6+12*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*x*arcsinh(c*x)*c^4-64*b/Pi^(5/2)/
(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^8*arcsinh(c*x)*c^11-560/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*
x^2+1)^2*x^4*arcsinh(c*x)*c^7-160/3*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^2*arcsinh(c*x)*c^5-19
2*b/Pi^(5/2)/(12*c^4*x^4+12*c^2*x^2-1)/(c^2*x^2+1)^2*x^6*arcsinh(c*x)*c^9+8/3*a*c^4/Pi*x/(Pi*c^2*x^2+Pi)^(3/2)
+16/3*a*c^4/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)+2*a*c^2/Pi/x/(Pi*c^2*x^2+Pi)^(3/2)

________________________________________________________________________________________

Maxima [A]  time = 1.21905, size = 319, normalized size = 1.53 \begin{align*} -\frac{1}{6} \, b c{\left (\frac{8 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac{5}{2}}} + \frac{16 \, c^{2} \log \left (x\right )}{\pi ^{\frac{5}{2}}} + \frac{1}{\pi ^{\frac{5}{2}} c^{2} x^{4} + \pi ^{\frac{5}{2}} x^{2}}\right )} + \frac{1}{3} \,{\left (\frac{8 \, c^{4} x}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}}} + \frac{16 \, c^{4} x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}}} + \frac{6 \, c^{2}}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x} - \frac{1}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x^{3}}\right )} b \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \,{\left (\frac{8 \, c^{4} x}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}}} + \frac{16 \, c^{4} x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}}} + \frac{6 \, c^{2}}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x} - \frac{1}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} x^{3}}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*c*(8*c^2*log(c^2*x^2 + 1)/pi^(5/2) + 16*c^2*log(x)/pi^(5/2) + 1/(pi^(5/2)*c^2*x^4 + pi^(5/2)*x^2)) + 1/
3*(8*c^4*x/(pi*(pi + pi*c^2*x^2)^(3/2)) + 16*c^4*x/(pi^2*sqrt(pi + pi*c^2*x^2)) + 6*c^2/(pi*(pi + pi*c^2*x^2)^
(3/2)*x) - 1/(pi*(pi + pi*c^2*x^2)^(3/2)*x^3))*b*arcsinh(c*x) + 1/3*(8*c^4*x/(pi*(pi + pi*c^2*x^2)^(3/2)) + 16
*c^4*x/(pi^2*sqrt(pi + pi*c^2*x^2)) + 6*c^2/(pi*(pi + pi*c^2*x^2)^(3/2)*x) - 1/(pi*(pi + pi*c^2*x^2)^(3/2)*x^3
))*a

________________________________________________________________________________________

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^{10} + 3 \, \pi ^{3} c^{4} x^{8} + 3 \, \pi ^{3} c^{2} x^{6} + \pi ^{3} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^4/(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^10 + 3*pi^3*c^4*x^8 + 3*pi^3*c^2*x^6 + pi^3*x^
4), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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}} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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