3.1.23 \(\int \frac {(a+b \text {csch}^{-1}(c x))^2}{x^5} \, dx\) [23]

Optimal. Leaf size=132 \[ -\frac {b^2}{32 x^4}+\frac {3 b^2 c^2}{32 x^2}+\frac {3}{16} a b c^4 \text {csch}^{-1}(c x)+\frac {3}{32} b^2 c^4 \text {csch}^{-1}(c x)^2+\frac {b c \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{8 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4} \]

[Out]

-1/32*b^2/x^4+3/32*b^2*c^2/x^2+3/16*a*b*c^4*arccsch(c*x)+3/32*b^2*c^4*arccsch(c*x)^2-1/4*(a+b*arccsch(c*x))^2/
x^4+1/8*b*c*(a+b*arccsch(c*x))*(1+1/c^2/x^2)^(1/2)/x^3-3/16*b*c^3*(a+b*arccsch(c*x))*(1+1/c^2/x^2)^(1/2)/x

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Rubi [A]
time = 0.08, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6421, 5554, 3391} \begin {gather*} \frac {3}{16} a b c^4 \text {csch}^{-1}(c x)+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1} \left (a+b \text {csch}^{-1}(c x)\right )}{8 x^3}-\frac {3 b c^3 \sqrt {\frac {1}{c^2 x^2}+1} \left (a+b \text {csch}^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4}+\frac {3}{32} b^2 c^4 \text {csch}^{-1}(c x)^2+\frac {3 b^2 c^2}{32 x^2}-\frac {b^2}{32 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])^2/x^5,x]

[Out]

-1/32*b^2/x^4 + (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcCsch[c*x])/16 + (3*b^2*c^4*ArcCsch[c*x]^2)/32 + (b*c*Sqrt
[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x]))/(8*x^3) - (3*b*c^3*Sqrt[1 + 1/(c^2*x^2)]*(a + b*ArcCsch[c*x]))/(16*x)
- (a + b*ArcCsch[c*x])^2/(4*x^4)

Rule 3391

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]

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 )^2}{x^5} \, dx &=-\left (c^4 \text {Subst}\left (\int (a+b x)^2 \cosh (x) \sinh ^3(x) \, dx,x,\text {csch}^{-1}(c x)\right )\right )\\ &=-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{2} \left (b c^4\right ) \text {Subst}\left (\int (a+b x) \sinh ^4(x) \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}+\frac {b c \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{8 x^3}-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4}-\frac {1}{8} \left (3 b c^4\right ) \text {Subst}\left (\int (a+b x) \sinh ^2(x) \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}+\frac {3 b^2 c^2}{32 x^2}+\frac {b c \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{8 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{16} \left (3 b c^4\right ) \text {Subst}\left (\int (a+b x) \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}+\frac {3 b^2 c^2}{32 x^2}+\frac {3}{16} a b c^4 \text {csch}^{-1}(c x)+\frac {3}{32} b^2 c^4 \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 )}{8 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}} \left (a+b \text {csch}^{-1}(c x)\right )}{16 x}-\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 147, normalized size = 1.11 \begin {gather*} \frac {-8 a^2-b^2+4 a b c \sqrt {1+\frac {1}{c^2 x^2}} x+3 b^2 c^2 x^2-6 a b c^3 \sqrt {1+\frac {1}{c^2 x^2}} x^3-2 b \left (8 a+b c \sqrt {1+\frac {1}{c^2 x^2}} x \left (-2+3 c^2 x^2\right )\right ) \text {csch}^{-1}(c x)+b^2 \left (-8+3 c^4 x^4\right ) \text {csch}^{-1}(c x)^2+6 a b c^4 x^4 \sinh ^{-1}\left (\frac {1}{c x}\right )}{32 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])^2/x^5,x]

[Out]

(-8*a^2 - b^2 + 4*a*b*c*Sqrt[1 + 1/(c^2*x^2)]*x + 3*b^2*c^2*x^2 - 6*a*b*c^3*Sqrt[1 + 1/(c^2*x^2)]*x^3 - 2*b*(8
*a + b*c*Sqrt[1 + 1/(c^2*x^2)]*x*(-2 + 3*c^2*x^2))*ArcCsch[c*x] + b^2*(-8 + 3*c^4*x^4)*ArcCsch[c*x]^2 + 6*a*b*
c^4*x^4*ArcSinh[1/(c*x)])/(32*x^4)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,\mathrm {arccsch}\left (c x \right )\right )^{2}}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))^2/x^5,x)

[Out]

int((a+b*arccsch(c*x))^2/x^5,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))^2/x^5,x, algorithm="maxima")

[Out]

1/32*a*b*((3*c^5*log(c*x*sqrt(1/(c^2*x^2) + 1) + 1) - 3*c^5*log(c*x*sqrt(1/(c^2*x^2) + 1) - 1) - 2*(3*c^8*x^3*
(1/(c^2*x^2) + 1)^(3/2) - 5*c^6*x*sqrt(1/(c^2*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) + 1)^2 - 2*c^2*x^2*(1/(c^2*x^2)
+ 1) + 1))/c - 16*arccsch(c*x)/x^4) - 1/4*b^2*(log(sqrt(c^2*x^2 + 1) + 1)^2/x^4 + 4*integrate(-1/2*(2*c^2*x^2
*log(c)^2 + 2*(c^2*x^2 + 1)*log(x)^2 + 2*log(c)^2 + 4*(c^2*x^2*log(c) + log(c))*log(x) - (4*c^2*x^2*log(c) + 4
*(c^2*x^2 + 1)*log(x) + (c^2*x^2*(4*log(c) - 1) + 4*(c^2*x^2 + 1)*log(x) + 4*log(c))*sqrt(c^2*x^2 + 1) + 4*log
(c))*log(sqrt(c^2*x^2 + 1) + 1) + 2*(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^7 + x^5 + (c^2*x^7 + x^5)*sqrt(c^2*x^2 + 1)), x)) - 1/4*a^2/x^4

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Fricas [A]
time = 0.45, size = 202, normalized size = 1.53 \begin {gather*} \frac {3 \, b^{2} c^{2} x^{2} + {\left (3 \, b^{2} c^{4} x^{4} - 8 \, b^{2}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - 8 \, a^{2} - b^{2} + 2 \, {\left (3 \, a b c^{4} x^{4} - 8 \, a b - {\left (3 \, b^{2} c^{3} x^{3} - 2 \, b^{2} c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \, {\left (3 \, a b c^{3} x^{3} - 2 \, a b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{32 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))^2/x^5,x, algorithm="fricas")

[Out]

1/32*(3*b^2*c^2*x^2 + (3*b^2*c^4*x^4 - 8*b^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x))^2 - 8*a^2 - b
^2 + 2*(3*a*b*c^4*x^4 - 8*a*b - (3*b^2*c^3*x^3 - 2*b^2*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*log((c*x*sqrt((c^2*
x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(3*a*b*c^3*x^3 - 2*a*b*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/x^4

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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 )^{2}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))**2/x**5,x)

[Out]

Integral((a + b*acsch(c*x))**2/x**5, 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))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)^2/x^5, 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 )}^2}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))^2/x^5,x)

[Out]

int((a + b*asinh(1/(c*x)))^2/x^5, x)

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