3.1.30 \(\int \frac {(a+b \text {csch}^{-1}(c x))^3}{x^3} \, dx\) [30]

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

[Out]

-3/8*b^3*c^2*arccsch(c*x)-3/4*b^2*(a+b*arccsch(c*x))/x^2-1/4*c^2*(a+b*arccsch(c*x))^3-1/2*(a+b*arccsch(c*x))^3
/x^2+3/8*b^3*c*(1+1/c^2/x^2)^(1/2)/x+3/4*b*c*(a+b*arccsch(c*x))^2*(1+1/c^2/x^2)^(1/2)/x

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Rubi [A]
time = 0.08, antiderivative size = 123, normalized size of antiderivative = 1.00, 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 = {6421, 5554, 3392, 32, 2715, 8} \begin {gather*} -\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) \end {gather*}

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 8

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

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 2715

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] && Integ
erQ[2*n]

Rule 3392

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 5554

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 6421

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])

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 182, normalized size = 1.48 \begin {gather*} -\frac {4 a^3+6 a b^2-6 a^2 b c \sqrt {1+\frac {1}{c^2 x^2}} x-3 b^3 c \sqrt {1+\frac {1}{c^2 x^2}} x+6 b \left (2 a^2+b^2-2 a b c \sqrt {1+\frac {1}{c^2 x^2}} x\right ) \text {csch}^{-1}(c x)+6 b^2 \left (-b c \sqrt {1+\frac {1}{c^2 x^2}} x+a \left (2+c^2 x^2\right )\right ) \text {csch}^{-1}(c x)^2+2 b^3 \left (2+c^2 x^2\right ) \text {csch}^{-1}(c x)^3+3 b \left (2 a^2+b^2\right ) c^2 x^2 \sinh ^{-1}\left (\frac {1}{c x}\right )}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(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))*Ar
cCsch[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)])/x^2

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {arccsch}\left (c x \right )\right )^{3}}{x^{3}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (107) = 214\).
time = 0.39, size = 267, normalized size = 2.17 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {acsch}{\left (c x \right )}\right )^{3}}{x^{3}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}^3}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))^3/x^3,x)

[Out]

int((a + b*asinh(1/(c*x)))^3/x^3, x)

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