3.103 \(\int \frac{x^3 (a+b \sinh ^{-1}(c x))}{(\pi +c^2 \pi x^2)^{5/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{a+b \sinh ^{-1}(c x)}{\pi ^2 c^4 \sqrt{\pi c^2 x^2+\pi }}+\frac{a+b \sinh ^{-1}(c x)}{3 \pi c^4 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac{b x}{6 \pi ^{5/2} c^3 \left (c^2 x^2+1\right )}+\frac{5 b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^4} \]

[Out]

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

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Rubi [A]  time = 0.146583, antiderivative size = 107, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {266, 43, 5732, 12, 385, 203} \[ -\frac{a+b \sinh ^{-1}(c x)}{\pi ^{5/2} c^4 \sqrt{c^2 x^2+1}}+\frac{a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} c^4 \left (c^2 x^2+1\right )^{3/2}}-\frac{b x}{6 \pi ^{5/2} c^3 \left (c^2 x^2+1\right )}+\frac{5 b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

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

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 12

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

Rule 385

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.184065, size = 93, normalized size = 0.89 \[ \frac{-6 a c^2 x^2-4 a-b c x \sqrt{c^2 x^2+1}+5 b \left (c^2 x^2+1\right )^{3/2} \tan ^{-1}(c x)-2 b \left (3 c^2 x^2+2\right ) \sinh ^{-1}(c x)}{6 \pi ^{5/2} c^4 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(5/2),x]

[Out]

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

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Maple [C]  time = 0.221, size = 175, normalized size = 1.7 \begin{align*} -{\frac{a{x}^{2}}{\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,a}{3\,\pi \,{c}^{4}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{-{\frac{3}{2}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}}{{\pi }^{{\frac{5}{2}}}{c}^{2}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{bx}{6\,{c}^{3}{\pi }^{5/2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{2\,b{\it Arcsinh} \left ( cx \right ) }{3\,{\pi }^{5/2}{c}^{4}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{{\frac{5\,i}{6}}b}{{\pi }^{{\frac{5}{2}}}{c}^{4}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ) }-{\frac{{\frac{5\,i}{6}}b}{{\pi }^{{\frac{5}{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)^(5/2),x)

[Out]

-a*x^2/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)-2/3*a/Pi/c^4/(Pi*c^2*x^2+Pi)^(3/2)-b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^2*arcsin
h(c*x)*x^2-1/6*b*x/c^3/Pi^(5/2)/(c^2*x^2+1)-2/3*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^4*arcsinh(c*x)+5/6*I*b/c^4/Pi^(
5/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)-5/6*I*b/c^4/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)-I)

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Maxima [A]  time = 1.8499, size = 186, normalized size = 1.77 \begin{align*} -\frac{1}{6} \, b c{\left (\frac{x}{\pi ^{\frac{5}{2}} c^{6} x^{2} + \pi ^{\frac{5}{2}} c^{4}} - \frac{5 \, \arctan \left (c x\right )}{\pi ^{\frac{5}{2}} c^{5}}\right )} - \frac{1}{3} \, b{\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 )} \operatorname{arsinh}\left (c x\right ) - \frac{1}{3} \, a{\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 )} \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)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(x/(pi^(5/2)*c^6*x^2 + pi^(5/2)*c^4) - 5*arctan(c*x)/(pi^(5/2)*c^5)) - 1/3*b*(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))*arcsinh(c*x) - 1/3*a*(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))

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Fricas [B]  time = 3.11247, size = 435, normalized size = 4.14 \begin{align*} -\frac{5 \, \sqrt{\pi }{\left (b c^{4} x^{4} + 2 \, 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 ) + 4 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (3 \, 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 (6 \, a c^{2} x^{2} + \sqrt{c^{2} x^{2} + 1} b c x + 4 \, a\right )}}{12 \,{\left (\pi ^{3} c^{8} x^{4} + 2 \, \pi ^{3} c^{6} x^{2} + \pi ^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(5*sqrt(pi)*(b*c^4*x^4 + 2*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)) + 4*sqrt(pi + pi*c^2*x^2)*(3*b*c^2*x^2 + 2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*sqrt(pi + p
i*c^2*x^2)*(6*a*c^2*x^2 + sqrt(c^2*x^2 + 1)*b*c*x + 4*a))/(pi^3*c^8*x^4 + 2*pi^3*c^6*x^2 + pi^3*c^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{3}}{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^{3} \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**3*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(5/2),x)

[Out]

(Integral(a*x**3/(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**3*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^{3}}{{\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^3*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^3/(pi + pi*c^2*x^2)^(5/2), x)