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

Optimal. Leaf size=75 \[ -\frac{a+b \sinh ^{-1}(c x)}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{b x}{6 \pi ^{5/2} c \left (c^2 x^2+1\right )}+\frac{b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^2} \]

[Out]

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

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Rubi [A]  time = 0.0802926, antiderivative size = 114, normalized size of antiderivative = 1.52, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5717, 199, 203} \[ -\frac{a+b \sinh ^{-1}(c x)}{3 \pi c^2 \left (\pi c^2 x^2+\pi \right )^{3/2}}+\frac{b x}{6 \pi ^2 c \sqrt{c^2 x^2+1} \sqrt{\pi c^2 x^2+\pi }}+\frac{b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 \pi ^2 c^2 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 199

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.135417, size = 72, normalized size = 0.96 \[ \frac{-2 a+b c x \sqrt{c^2 x^2+1}+b \left (c^2 x^2+1\right )^{3/2} \tan ^{-1}(c x)-2 b \sinh ^{-1}(c x)}{6 \pi ^{5/2} c^2 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.088, size = 124, normalized size = 1.7 \begin{align*} -{\frac{a}{3\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{-{\frac{3}{2}}}}+{\frac{bx}{6\,c{\pi }^{5/2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{3\,{\pi }^{5/2}{c}^{2}} \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{\frac{{\frac{i}{6}}b}{{\pi }^{{\frac{5}{2}}}{c}^{2}}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ) }-{\frac{{\frac{i}{6}}b}{{\pi }^{{\frac{5}{2}}}{c}^{2}}\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*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{x \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{5}{2}}}\,{d x} - \frac{a}{3 \, \pi{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{3}{2}} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(x*log(c*x + sqrt(c^2*x^2 + 1))/(pi + pi*c^2*x^2)^(5/2), x) - 1/3*a/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2
)

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Fricas [B]  time = 3.11376, size = 389, normalized size = 5.19 \begin{align*} -\frac{\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}} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (\sqrt{c^{2} x^{2} + 1} b c x - 2 \, a\right )}}{12 \,{\left (\pi ^{3} c^{6} x^{4} + 2 \, \pi ^{3} c^{4} x^{2} + \pi ^{3} c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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