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

Optimal. Leaf size=97 \[ \frac{2 c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x}-\frac{\sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x^3}-\frac{2 b c^3 \log (x)}{3 \sqrt{\pi }}-\frac{b c}{6 \sqrt{\pi } x^2} \]

[Out]

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

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Rubi [A]  time = 0.181917, antiderivative size = 141, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {5747, 5723, 29, 30} \[ \frac{2 c^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x}-\frac{\sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{3 \pi x^3}-\frac{b c \sqrt{c^2 x^2+1}}{6 x^2 \sqrt{\pi c^2 x^2+\pi }}-\frac{2 b c^3 \sqrt{c^2 x^2+1} \log (x)}{3 \sqrt{\pi c^2 x^2+\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.15839, size = 99, normalized size = 1.02 \[ \frac{2 a \sqrt{c^2 x^2+1} \left (2 c^2 x^2-1\right )+b c x \left (6 c^2 x^2-1\right )+2 b \sqrt{c^2 x^2+1} \left (2 c^2 x^2-1\right ) \sinh ^{-1}(c x)}{6 \sqrt{\pi } x^3}-\frac{2 b c^3 \log (x)}{3 \sqrt{\pi }} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[1 + c^2*x^2]*(-1 + 2*c^2*x^2) + b*c*x*(-1 + 6*c^2*x^2) + 2*b*Sqrt[1 + c^2*x^2]*(-1 + 2*c^2*x^2)*ArcS
inh[c*x])/(6*Sqrt[Pi]*x^3) - (2*b*c^3*Log[x])/(3*Sqrt[Pi])

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Maple [B]  time = 0.09, size = 372, normalized size = 3.8 \begin{align*} -{\frac{a}{3\,\pi \,{x}^{3}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{2\,a{c}^{2}}{3\,\pi \,x}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}+{\frac{4\,b{c}^{3}{\it Arcsinh} \left ( cx \right ) }{3\,\sqrt{\pi }}}-{\frac{2\,b{x}^{4}{c}^{7}}{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}+{\frac{2\,b{x}^{2} \left ({c}^{2}{x}^{2}+1 \right ){c}^{5}}{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}-2\,{\frac{b{x}^{2}{\it Arcsinh} \left ( cx \right ){c}^{5}}{\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}+2\,{\frac{bx{\it Arcsinh} \left ( cx \right ) \sqrt{{c}^{2}{x}^{2}+1}{c}^{4}}{\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}-{\frac{2\,b \left ({c}^{2}{x}^{2}+1 \right ){c}^{3}}{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}+{\frac{2\,b{c}^{3}{\it Arcsinh} \left ( cx \right ) }{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) }}-{\frac{5\,b{\it Arcsinh} \left ( cx \right ){c}^{2}}{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ) x}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{b \left ({c}^{2}{x}^{2}+1 \right ) c}{6\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ){x}^{2}}}+{\frac{b{\it Arcsinh} \left ( cx \right ) }{3\,\sqrt{\pi } \left ( 3\,{c}^{2}{x}^{2}-1 \right ){x}^{3}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{2\,b{c}^{3}}{3\,\sqrt{\pi }}\ln \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ) } \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)^(1/2),x)

[Out]

-1/3*a/Pi/x^3*(Pi*c^2*x^2+Pi)^(1/2)+2/3*a/Pi*c^2/x*(Pi*c^2*x^2+Pi)^(1/2)+4/3*b*c^3/Pi^(1/2)*arcsinh(c*x)-2/3*b
/Pi^(1/2)/(3*c^2*x^2-1)*x^4*c^7+2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*x^2*(c^2*x^2+1)*c^5-2*b/Pi^(1/2)/(3*c^2*x^2-1)*x^
2*arcsinh(c*x)*c^5+2*b/Pi^(1/2)/(3*c^2*x^2-1)*x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^4-2/3*b/Pi^(1/2)/(3*c^2*x^2-1
)*(c^2*x^2+1)*c^3+2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*arcsinh(c*x)*c^3-5/3*b/Pi^(1/2)/(3*c^2*x^2-1)/x*arcsinh(c*x)*(c
^2*x^2+1)^(1/2)*c^2+1/6*b/Pi^(1/2)/(3*c^2*x^2-1)/x^2*(c^2*x^2+1)*c+1/3*b/Pi^(1/2)/(3*c^2*x^2-1)/x^3*arcsinh(c*
x)*(c^2*x^2+1)^(1/2)-2/3*b*c^3/Pi^(1/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)

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Maxima [A]  time = 1.19238, size = 163, normalized size = 1.68 \begin{align*} -\frac{1}{6} \,{\left (\frac{4 \, c^{2} \log \left (x\right )}{\sqrt{\pi }} + \frac{1}{\sqrt{\pi } x^{2}}\right )} b c + \frac{1}{3} \, b{\left (\frac{2 \, \sqrt{\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac{\sqrt{\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{2 \, \sqrt{\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac{\sqrt{\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.03319, size = 495, normalized size = 5.1 \begin{align*} \frac{2 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (2 \, b c^{4} x^{4} + b c^{2} x^{2} - b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \, \sqrt{\pi }{\left (b c^{5} x^{5} + b c^{3} x^{3}\right )} \log \left (\frac{\pi + \pi c^{2} x^{6} + \pi c^{2} x^{2} + \pi x^{4} - \sqrt{\pi } \sqrt{\pi + \pi c^{2} x^{2}} \sqrt{c^{2} x^{2} + 1}{\left (x^{4} - 1\right )}}{c^{2} x^{4} + x^{2}}\right ) + \sqrt{\pi + \pi c^{2} x^{2}}{\left (4 \, a c^{4} x^{4} + 2 \, a c^{2} x^{2} +{\left (b c x^{3} - b c x\right )} \sqrt{c^{2} x^{2} + 1} - 2 \, a\right )}}{6 \,{\left (\pi c^{2} x^{5} + \pi x^{3}\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)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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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}{\sqrt{\pi + \pi c^{2} x^{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)^(1/2),x, algorithm="giac")

[Out]

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