3.100 \(\int \frac{x^6 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=192 \[ -\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c^4 \sqrt{\pi c^2 x^2+\pi }}+\frac{5 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^3 c^6}-\frac{5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{5/2} b c^7}-\frac{b x^2}{4 \pi ^{5/2} c^5}-\frac{b}{6 \pi ^{5/2} c^7 \left (c^2 x^2+1\right )}-\frac{7 b \log \left (c^2 x^2+1\right )}{6 \pi ^{5/2} c^7} \]

[Out]

-(b*x^2)/(4*c^5*Pi^(5/2)) - b/(6*c^7*Pi^(5/2)*(1 + c^2*x^2)) - (x^5*(a + b*ArcSinh[c*x]))/(3*c^2*Pi*(Pi + c^2*
Pi*x^2)^(3/2)) - (5*x^3*(a + b*ArcSinh[c*x]))/(3*c^4*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (5*x*Sqrt[Pi + c^2*Pi*x^2]*
(a + b*ArcSinh[c*x]))/(2*c^6*Pi^3) - (5*(a + b*ArcSinh[c*x])^2)/(4*b*c^7*Pi^(5/2)) - (7*b*Log[1 + c^2*x^2])/(6
*c^7*Pi^(5/2))

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Rubi [A]  time = 0.425205, antiderivative size = 256, normalized size of antiderivative = 1.33, number of steps used = 11, 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} \[ -\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi ^2 c^4 \sqrt{\pi c^2 x^2+\pi }}+\frac{5 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{2 \pi ^3 c^6}-\frac{5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 \pi ^{5/2} b c^7}-\frac{b x^2 \sqrt{c^2 x^2+1}}{4 \pi ^2 c^5 \sqrt{\pi c^2 x^2+\pi }}-\frac{b}{6 \pi ^2 c^7 \sqrt{c^2 x^2+1} \sqrt{\pi c^2 x^2+\pi }}-\frac{7 b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 \pi ^2 c^7 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

-b/(6*c^7*Pi^2*Sqrt[1 + c^2*x^2]*Sqrt[Pi + c^2*Pi*x^2]) - (b*x^2*Sqrt[1 + c^2*x^2])/(4*c^5*Pi^2*Sqrt[Pi + c^2*
Pi*x^2]) - (x^5*(a + b*ArcSinh[c*x]))/(3*c^2*Pi*(Pi + c^2*Pi*x^2)^(3/2)) - (5*x^3*(a + b*ArcSinh[c*x]))/(3*c^4
*Pi^2*Sqrt[Pi + c^2*Pi*x^2]) + (5*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(2*c^6*Pi^3) - (5*(a + b*ArcSi
nh[c*x])^2)/(4*b*c^7*Pi^(5/2)) - (7*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^7*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 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^6 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=-\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac{5 \int \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx}{3 c^2 \pi }+\frac{\left (b \sqrt{1+c^2 x^2}\right ) \int \frac{x^5}{\left (1+c^2 x^2\right )^2} \, dx}{3 c \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{5 \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{c^4 \pi ^2}+\frac{\left (5 b \sqrt{1+c^2 x^2}\right ) \int \frac{x^3}{1+c^2 x^2} \, dx}{3 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^2}{\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^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{5 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 \pi ^3}-\frac{5 \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{\pi +c^2 \pi x^2}} \, dx}{2 c^6 \pi ^2}-\frac{\left (5 b \sqrt{1+c^2 x^2}\right ) \int x \, dx}{2 c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (5 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )}{6 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 \left (\frac{1}{c^4}+\frac{1}{c^4 \left (1+c^2 x\right )^2}-\frac{2}{c^4 \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^7 \pi ^2 \sqrt{1+c^2 x^2} \sqrt{\pi +c^2 \pi x^2}}-\frac{13 b x^2 \sqrt{1+c^2 x^2}}{12 c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}-\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{5 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 \pi ^3}-\frac{5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^7 \pi ^{5/2}}-\frac{b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{3 c^7 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{\left (5 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 )}{6 c^3 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ &=-\frac{b}{6 c^7 \pi ^2 \sqrt{1+c^2 x^2} \sqrt{\pi +c^2 \pi x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2}}{4 c^5 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}-\frac{x^5 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^2 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}-\frac{5 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}+\frac{5 x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^6 \pi ^3}-\frac{5 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c^7 \pi ^{5/2}}-\frac{7 b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^7 \pi ^2 \sqrt{\pi +c^2 \pi x^2}}\\ \end{align*}

Mathematica [A]  time = 0.453433, size = 202, normalized size = 1.05 \[ \frac{4 \sinh ^{-1}(c x) \left (b c x \left (3 c^4 x^4+20 c^2 x^2+15\right )-15 a \left (c^2 x^2+1\right )^{3/2}\right )+12 a c^5 x^5+80 a c^3 x^3+60 a c x-6 b c^4 x^4 \sqrt{c^2 x^2+1}-9 b c^2 x^2 \sqrt{c^2 x^2+1}-7 b \sqrt{c^2 x^2+1}-28 b \left (c^2 x^2+1\right )^{3/2} \log \left (c^2 x^2+1\right )-30 b \left (c^2 x^2+1\right )^{3/2} \sinh ^{-1}(c x)^2}{24 \pi ^{5/2} c^7 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

(60*a*c*x + 80*a*c^3*x^3 + 12*a*c^5*x^5 - 7*b*Sqrt[1 + c^2*x^2] - 9*b*c^2*x^2*Sqrt[1 + c^2*x^2] - 6*b*c^4*x^4*
Sqrt[1 + c^2*x^2] + 4*(-15*a*(1 + c^2*x^2)^(3/2) + b*c*x*(15 + 20*c^2*x^2 + 3*c^4*x^4))*ArcSinh[c*x] - 30*b*(1
+ c^2*x^2)^(3/2)*ArcSinh[c*x]^2 - 28*b*(1 + c^2*x^2)^(3/2)*Log[1 + c^2*x^2])/(24*c^7*Pi^(5/2)*(1 + c^2*x^2)^(
3/2))

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Maple [B]  time = 0.301, size = 970, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

5/2*a/c^6/Pi^2*x/(Pi*c^2*x^2+Pi)^(1/2)-1/4*b*x^2/c^5/Pi^(5/2)+1/2*b/Pi^(5/2)/c^6*(c^2*x^2+1)^(1/2)*arcsinh(c*x
)*x-49/6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2*c*x^8-98/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49
)/(c^2*x^2+1)^2/c*x^6-49*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^3*x^4-98/3*b/Pi^(5/2)/(63*c^4*
x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^5*x^2-343/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^7*arcsi
nh(c*x)+147*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)*arcsinh(c*x)*x^7+49/6*b/Pi^(5/2)/(63*c^4*
x^4+111*c^2*x^2+49)/(c^2*x^2+1)/c*x^6+14*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)/c^3*x^4+6*b/Pi^(5/
2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)/c^5*x^2-1/8*b/Pi^(5/2)/c^7-5/2*a/c^6/Pi^2*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*a*x^5/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)+5/6*a/c^4*x^3/Pi/(Pi*c^2*x^2+Pi
)^(3/2)-49/6*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^7+14/3*b/c^7/Pi^(5/2)*arcsinh(c*x)-5/4*b/c
^7/Pi^(5/2)*arcsinh(c*x)^2-7/3*b/c^7/Pi^(5/2)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+385*b/Pi^(5/2)/(63*c^4*x^4+111*c
^2*x^2+49)/(c^2*x^2+1)^(3/2)/c^2*arcsinh(c*x)*x^5+1009/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3
/2)/c^4*arcsinh(c*x)*x^3+98*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^(3/2)/c^6*arcsinh(c*x)*x-147*b/
Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2*c*arcsinh(c*x)*x^8-553*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+4
9)/(c^2*x^2+1)^2/c*arcsinh(c*x)*x^6-2338/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^3*arcsinh(c*
x)*x^4-1463/3*b/Pi^(5/2)/(63*c^4*x^4+111*c^2*x^2+49)/(c^2*x^2+1)^2/c^5*arcsinh(c*x)*x^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, a{\left (\frac{3 \, x^{5}}{\pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2}} + \frac{5 \, 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 )}}{c^{2}} + \frac{5 \, x}{\pi ^{2} \sqrt{\pi + \pi c^{2} x^{2}} c^{6}} - \frac{15 \, \operatorname{arsinh}\left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\pi ^{2} \sqrt{\pi c^{2}} c^{6}}\right )} + b \int \frac{x^{6} \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^6*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="maxima")

[Out]

1/6*a*(3*x^5/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2) + 5*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))/c^2 + 5*x/(pi^2*sqrt(pi + pi*c^2*x^2)*c^6) - 15*arcsinh(c^2*x/sqrt(c^2))/(pi^2*sqrt(pi*c^2
)*c^6)) + b*integrate(x^6*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^{6} \operatorname{arsinh}\left (c x\right ) + a x^{6}\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^6*(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^6*arcsinh(c*x) + a*x^6)/(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(-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(x**6*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(5/2),x)

[Out]

Timed out

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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^{6}}{{\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^6*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^6/(pi + pi*c^2*x^2)^(5/2), x)