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

Optimal. Leaf size=93 \[ -\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{\pi \sqrt{\pi c^2 x^2+\pi }}-\frac{a+b \sinh ^{-1}(c x)}{\pi x \sqrt{\pi c^2 x^2+\pi }}+\frac{b c \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2}}+\frac{b c \log (x)}{\pi ^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.138361, antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {271, 191, 5732, 446, 72} \[ -\frac{2 c^2 x \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} \sqrt{c^2 x^2+1}}-\frac{a+b \sinh ^{-1}(c x)}{\pi ^{3/2} x \sqrt{c^2 x^2+1}}+\frac{b c \log \left (c^2 x^2+1\right )}{2 \pi ^{3/2}}+\frac{b c \log (x)}{\pi ^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.160248, size = 69, normalized size = 0.74 \[ \frac{b \left (\frac{1}{2} c \log \left (c^2 x^2+1\right )+c \log (x)\right )}{\pi ^{3/2}}-\frac{\left (2 c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2} x \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.097, size = 180, normalized size = 1.9 \begin{align*} -{\frac{a}{\pi \,x}{\frac{1}{\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}}-2\,{\frac{a{c}^{2}x}{\pi \,\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}}-4\,{\frac{bc{\it Arcsinh} \left ( cx \right ) }{{\pi }^{3/2}}}+2\,{\frac{b{\it Arcsinh} \left ( cx \right ){x}^{2}{c}^{3}}{{\pi }^{3/2} \left ({c}^{2}{x}^{2}+1 \right ) }}-2\,{\frac{b{\it Arcsinh} \left ( cx \right ) x{c}^{2}}{{\pi }^{3/2}\sqrt{{c}^{2}{x}^{2}+1}}}+2\,{\frac{bc{\it Arcsinh} \left ( cx \right ) }{{\pi }^{3/2} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{\pi }^{{\frac{3}{2}}}x}{\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}+{\frac{bc}{{\pi }^{{\frac{3}{2}}}}\ln \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{4}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x^2/(Pi*c^2*x^2+Pi)^(3/2),x)

[Out]

-a/Pi/x/(Pi*c^2*x^2+Pi)^(1/2)-2*a/Pi*c^2*x/(Pi*c^2*x^2+Pi)^(1/2)-4*b*c/Pi^(3/2)*arcsinh(c*x)+2*b/Pi^(3/2)*arcs
inh(c*x)*x^2/(c^2*x^2+1)*c^3-2*b/Pi^(3/2)*arcsinh(c*x)*x/(c^2*x^2+1)^(1/2)*c^2+2*b/Pi^(3/2)*arcsinh(c*x)/(c^2*
x^2+1)*c-b/Pi^(3/2)*arcsinh(c*x)/x/(c^2*x^2+1)^(1/2)+b*c/Pi^(3/2)*ln((c*x+(c^2*x^2+1)^(1/2))^4-1)

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Maxima [A]  time = 1.21045, size = 161, normalized size = 1.73 \begin{align*} \frac{1}{2} \, b c{\left (\frac{\log \left (c^{2} x^{2} + 1\right )}{\pi ^{\frac{3}{2}}} + \frac{2 \, \log \left (x\right )}{\pi ^{\frac{3}{2}}}\right )} -{\left (\frac{2 \, c^{2} x}{\pi \sqrt{\pi + \pi c^{2} x^{2}}} + \frac{1}{\pi \sqrt{\pi + \pi c^{2} x^{2}} x}\right )} b \operatorname{arsinh}\left (c x\right ) -{\left (\frac{2 \, c^{2} x}{\pi \sqrt{\pi + \pi c^{2} x^{2}}} + \frac{1}{\pi \sqrt{\pi + \pi c^{2} x^{2}} x}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxima")

[Out]

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

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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 \operatorname{arsinh}\left (c x\right ) + a\right )}}{\pi ^{2} c^{4} x^{6} + 2 \, \pi ^{2} c^{2} x^{4} + \pi ^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^6 + 2*pi^2*c^2*x^4 + pi^2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{2} x^{4} \sqrt{c^{2} x^{2} + 1} + x^{2} \sqrt{c^{2} x^{2} + 1}}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{4} \sqrt{c^{2} x^{2} + 1} + x^{2} \sqrt{c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x**2/(pi*c**2*x**2+pi)**(3/2),x)

[Out]

(Integral(a/(c**2*x**4*sqrt(c**2*x**2 + 1) + x**2*sqrt(c**2*x**2 + 1)), x) + Integral(b*asinh(c*x)/(c**2*x**4*
sqrt(c**2*x**2 + 1) + x**2*sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

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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{3}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/x^2/(pi*c^2*x^2+pi)^(3/2),x, algorithm="giac")

[Out]

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