3.21 \(\int \frac{(a+b \text{csch}^{-1}(c x))^2}{x^3} \, dx\)

Optimal. Leaf size=87 \[ \frac{b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )}{2 x}-\frac{1}{2} a b c^2 \text{csch}^{-1}(c x)-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{4} b^2 c^2 \text{csch}^{-1}(c x)^2-\frac{b^2}{4 x^2} \]

[Out]

-b^2/(4*x^2) - (a*b*c^2*ArcCsch[c*x])/2 - (b^2*c^2*ArcCsch[c*x]^2)/4 + (b*c*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCs
ch[c*x]))/(2*x) - (a + b*ArcCsch[c*x])^2/(2*x^2)

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Rubi [A]  time = 0.0824147, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6286, 5446, 3310} \[ \frac{b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )}{2 x}-\frac{1}{2} a b c^2 \text{csch}^{-1}(c x)-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{4} b^2 c^2 \text{csch}^{-1}(c x)^2-\frac{b^2}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])^2/x^3,x]

[Out]

-b^2/(4*x^2) - (a*b*c^2*ArcCsch[c*x])/2 - (b^2*c^2*ArcCsch[c*x]^2)/4 + (b*c*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCs
ch[c*x]))/(2*x) - (a + b*ArcCsch[c*x])^2/(2*x^2)

Rule 6286

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Csch[x]^(m + 1)*Coth[x], x], x, ArcCsch[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rule 5446

Int[Cosh[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c
+ d*x)^m*Sinh[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sinh[a + b*x]^
(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{x^3} \, dx &=-\left (c^2 \operatorname{Subst}\left (\int (a+b x)^2 \cosh (x) \sinh (x) \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{2 x^2}+\left (b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=-\frac{b^2}{4 x^2}+\frac{b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )}{2 x}-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{2 x^2}-\frac{1}{2} \left (b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=-\frac{b^2}{4 x^2}-\frac{1}{2} a b c^2 \text{csch}^{-1}(c x)-\frac{1}{4} b^2 c^2 \text{csch}^{-1}(c x)^2+\frac{b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )}{2 x}-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^2}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.136994, size = 100, normalized size = 1.15 \[ -\frac{2 a^2-2 a b c x \sqrt{\frac{1}{c^2 x^2}+1}+2 a b c^2 x^2 \sinh ^{-1}\left (\frac{1}{c x}\right )-2 b \text{csch}^{-1}(c x) \left (b c x \sqrt{\frac{1}{c^2 x^2}+1}-2 a\right )+b^2 \left (c^2 x^2+2\right ) \text{csch}^{-1}(c x)^2+b^2}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*a^2 + b^2 - 2*a*b*c*Sqrt[1 + 1/(c^2*x^2)]*x - 2*b*(-2*a + b*c*Sqrt[1 + 1/(c^2*x^2)]*x)*ArcCsch[c*x] + b^2*
(2 + c^2*x^2)*ArcCsch[c*x]^2 + 2*a*b*c^2*x^2*ArcSinh[1/(c*x)])/(4*x^2)

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Maple [F]  time = 0.187, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\rm arccsch} \left (cx\right ) \right ) ^{2}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a b{\left (\frac{\frac{2 \, c^{4} x \sqrt{\frac{1}{c^{2} x^{2}} + 1}}{c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} + 1\right )} - 1} - c^{3} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} + 1} + 1\right ) + c^{3} \log \left (c x \sqrt{\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c} - \frac{4 \, \operatorname{arcsch}\left (c x\right )}{x^{2}}\right )} - \frac{1}{2} \, b^{2}{\left (\frac{\log \left (\sqrt{c^{2} x^{2} + 1} + 1\right )^{2}}{x^{2}} + 2 \, \int -\frac{c^{2} x^{2} \log \left (c\right )^{2} +{\left (c^{2} x^{2} + 1\right )} \log \left (x\right )^{2} + \log \left (c\right )^{2} + 2 \,{\left (c^{2} x^{2} \log \left (c\right ) + \log \left (c\right )\right )} \log \left (x\right ) -{\left (2 \, c^{2} x^{2} \log \left (c\right ) + 2 \,{\left (c^{2} x^{2} + 1\right )} \log \left (x\right ) +{\left (c^{2} x^{2}{\left (2 \, \log \left (c\right ) - 1\right )} + 2 \,{\left (c^{2} x^{2} + 1\right )} \log \left (x\right ) + 2 \, \log \left (c\right )\right )} \sqrt{c^{2} x^{2} + 1} + 2 \, \log \left (c\right )\right )} \log \left (\sqrt{c^{2} x^{2} + 1} + 1\right ) +{\left (c^{2} x^{2} \log \left (c\right )^{2} +{\left (c^{2} x^{2} + 1\right )} \log \left (x\right )^{2} + \log \left (c\right )^{2} + 2 \,{\left (c^{2} x^{2} \log \left (c\right ) + \log \left (c\right )\right )} \log \left (x\right )\right )} \sqrt{c^{2} x^{2} + 1}}{c^{2} x^{5} + x^{3} +{\left (c^{2} x^{5} + x^{3}\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x}\right )} - \frac{a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*b*((2*c^4*x*sqrt(1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) + 1) - 1) - c^3*log(c*x*sqrt(1/(c^2*x^2) + 1) +
1) + c^3*log(c*x*sqrt(1/(c^2*x^2) + 1) - 1))/c - 4*arccsch(c*x)/x^2) - 1/2*b^2*(log(sqrt(c^2*x^2 + 1) + 1)^2/x
^2 + 2*integrate(-(c^2*x^2*log(c)^2 + (c^2*x^2 + 1)*log(x)^2 + log(c)^2 + 2*(c^2*x^2*log(c) + log(c))*log(x) -
(2*c^2*x^2*log(c) + 2*(c^2*x^2 + 1)*log(x) + (c^2*x^2*(2*log(c) - 1) + 2*(c^2*x^2 + 1)*log(x) + 2*log(c))*sqr
t(c^2*x^2 + 1) + 2*log(c))*log(sqrt(c^2*x^2 + 1) + 1) + (c^2*x^2*log(c)^2 + (c^2*x^2 + 1)*log(x)^2 + log(c)^2
+ 2*(c^2*x^2*log(c) + log(c))*log(x))*sqrt(c^2*x^2 + 1))/(c^2*x^5 + x^3 + (c^2*x^5 + x^3)*sqrt(c^2*x^2 + 1)),
x)) - 1/2*a^2/x^2

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Fricas [B]  time = 2.11415, size = 350, normalized size = 4.02 \begin{align*} \frac{2 \, a b c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} -{\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - 2 \, a^{2} - b^{2} - 2 \,{\left (a b c^{2} x^{2} - b^{2} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, a b\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))**2/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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