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

Optimal. Leaf size=123 \[ -\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 x^2}+\frac{3 b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}+\frac{3 b^3 c \sqrt{\frac{1}{c^2 x^2}+1}}{8 x}-\frac{3}{8} b^3 c^2 \text{csch}^{-1}(c x) \]

[Out]

(3*b^3*c*Sqrt[1 + 1/(c^2*x^2)])/(8*x) - (3*b^3*c^2*ArcCsch[c*x])/8 - (3*b^2*(a + b*ArcCsch[c*x]))/(4*x^2) + (3
*b*c*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(4*x) - (c^2*(a + b*ArcCsch[c*x])^3)/4 - (a + b*ArcCsch[c*x
])^3/(2*x^2)

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Rubi [A]  time = 0.106324, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6286, 5446, 3311, 32, 2635, 8} \[ -\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 x^2}+\frac{3 b c \sqrt{\frac{1}{c^2 x^2}+1} \left (a+b \text{csch}^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}+\frac{3 b^3 c \sqrt{\frac{1}{c^2 x^2}+1}}{8 x}-\frac{3}{8} b^3 c^2 \text{csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])^3/x^3,x]

[Out]

(3*b^3*c*Sqrt[1 + 1/(c^2*x^2)])/(8*x) - (3*b^3*c^2*ArcCsch[c*x])/8 - (3*b^2*(a + b*ArcCsch[c*x]))/(4*x^2) + (3
*b*c*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x])^2)/(4*x) - (c^2*(a + b*ArcCsch[c*x])^3)/4 - (a + b*ArcCsch[c*x
])^3/(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 3311

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{x^3} \, dx &=-\left (c^2 \operatorname{Subst}\left (\int (a+b x)^3 \cosh (x) \sinh (x) \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}+\frac{1}{2} \left (3 b c^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{4 x}-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}-\frac{1}{4} \left (3 b c^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \, dx,x,\text{csch}^{-1}(c x)\right )+\frac{1}{4} \left (3 b^3 c^2\right ) \operatorname{Subst}\left (\int \sinh ^2(x) \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{3 b^3 c \sqrt{1+\frac{1}{c^2 x^2}}}{8 x}-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}-\frac{1}{8} \left (3 b^3 c^2\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{3 b^3 c \sqrt{1+\frac{1}{c^2 x^2}}}{8 x}-\frac{3}{8} b^3 c^2 \text{csch}^{-1}(c x)-\frac{3 b^2 \left (a+b \text{csch}^{-1}(c x)\right )}{4 x^2}+\frac{3 b c \sqrt{1+\frac{1}{c^2 x^2}} \left (a+b \text{csch}^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \text{csch}^{-1}(c x)\right )^3-\frac{\left (a+b \text{csch}^{-1}(c x)\right )^3}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.310988, size = 182, normalized size = 1.48 \[ -\frac{3 b c^2 x^2 \left (2 a^2+b^2\right ) \sinh ^{-1}\left (\frac{1}{c x}\right )+6 b \text{csch}^{-1}(c x) \left (2 a^2-2 a b c x \sqrt{\frac{1}{c^2 x^2}+1}+b^2\right )-6 a^2 b c x \sqrt{\frac{1}{c^2 x^2}+1}+4 a^3+6 b^2 \text{csch}^{-1}(c x)^2 \left (a \left (c^2 x^2+2\right )-b c x \sqrt{\frac{1}{c^2 x^2}+1}\right )+6 a b^2-3 b^3 c x \sqrt{\frac{1}{c^2 x^2}+1}+2 b^3 \left (c^2 x^2+2\right ) \text{csch}^{-1}(c x)^3}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*a^3 + 6*a*b^2 - 6*a^2*b*c*Sqrt[1 + 1/(c^2*x^2)]*x - 3*b^3*c*Sqrt[1 + 1/(c^2*x^2)]*x + 6*b*(2*a^2 + b^2 - 2
*a*b*c*Sqrt[1 + 1/(c^2*x^2)]*x)*ArcCsch[c*x] + 6*b^2*(-(b*c*Sqrt[1 + 1/(c^2*x^2)]*x) + a*(2 + c^2*x^2))*ArcCsc
h[c*x]^2 + 2*b^3*(2 + c^2*x^2)*ArcCsch[c*x]^3 + 3*b*(2*a^2 + b^2)*c^2*x^2*ArcSinh[1/(c*x)])/(8*x^2)

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Maple [F]  time = 0.204, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\rm arccsch} \left (cx\right ) \right ) ^{3}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*a^2*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^3*log(sqrt(c^2*x^2 + 1) + 1)^3/
x^2 - 1/2*a^3/x^2 - integrate(1/2*(2*b^3*log(c)^3 - 6*a*b^2*log(c)^2 + 2*(b^3*c^2*x^2 + b^3)*log(x)^3 + 2*(b^3
*c^2*log(c)^3 - 3*a*b^2*c^2*log(c)^2)*x^2 + 6*(b^3*log(c) - a*b^2 + (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*log(x)^2
+ 3*(2*b^3*log(c) - 2*a*b^2 + 2*(b^3*c^2*log(c) - a*b^2*c^2)*x^2 + 2*(b^3*c^2*x^2 + b^3)*log(x) + sqrt(c^2*x^
2 + 1)*(2*b^3*log(c) - 2*a*b^2 + (b^3*c^2*(2*log(c) - 1) - 2*a*b^2*c^2)*x^2 + 2*(b^3*c^2*x^2 + b^3)*log(x)))*l
og(sqrt(c^2*x^2 + 1) + 1)^2 + 6*(b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2)*
log(x) - 6*(b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2 + (b^3*c^2*x^2 + b^3)*
log(x)^2 + 2*(b^3*log(c) - a*b^2 + (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*log(x) + (b^3*log(c)^2 - 2*a*b^2*log(c) +
(b^3*c^2*log(c)^2 - 2*a*b^2*c^2*log(c))*x^2 + (b^3*c^2*x^2 + b^3)*log(x)^2 + 2*(b^3*log(c) - a*b^2 + (b^3*c^2
*log(c) - a*b^2*c^2)*x^2)*log(x))*sqrt(c^2*x^2 + 1))*log(sqrt(c^2*x^2 + 1) + 1) + 2*(b^3*log(c)^3 - 3*a*b^2*lo
g(c)^2 + (b^3*c^2*x^2 + b^3)*log(x)^3 + (b^3*c^2*log(c)^3 - 3*a*b^2*c^2*log(c)^2)*x^2 + 3*(b^3*log(c) - a*b^2
+ (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*log(x)^2 + 3*(b^3*log(c)^2 - 2*a*b^2*log(c) + (b^3*c^2*log(c)^2 - 2*a*b^2*
c^2*log(c))*x^2)*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)

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Fricas [B]  time = 2.46831, size = 576, normalized size = 4.68 \begin{align*} -\frac{2 \,{\left (b^{3} c^{2} x^{2} + 2 \, b^{3}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{3} - 3 \,{\left (2 \, a^{2} b + b^{3}\right )} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 4 \, a^{3} + 6 \, a b^{2} + 6 \,{\left (a b^{2} c^{2} x^{2} - b^{3} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, a b^{2}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - 3 \,{\left (4 \, a b^{2} c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} -{\left (2 \, a^{2} b + b^{3}\right )} c^{2} x^{2} - 4 \, a^{2} b - 2 \, b^{3}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(b^3*c^2*x^2 + 2*b^3)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x))^3 - 3*(2*a^2*b + b^3)*c*x*sqr
t((c^2*x^2 + 1)/(c^2*x^2)) + 4*a^3 + 6*a*b^2 + 6*(a*b^2*c^2*x^2 - b^3*c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 2*a*
b^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x))^2 - 3*(4*a*b^2*c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - (2*
a^2*b + b^3)*c^2*x^2 - 4*a^2*b - 2*b^3)*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 )^{3}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))**3/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 )}^{3}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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