3.13 \(\int \frac {a+b \text {csch}^{-1}(c x)}{x^6} \, dx\)

Optimal. Leaf size=79 \[ -\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}+\frac {1}{25} b c^5 \left (\frac {1}{c^2 x^2}+1\right )^{5/2}-\frac {2}{15} b c^5 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}+\frac {1}{5} b c^5 \sqrt {\frac {1}{c^2 x^2}+1} \]

[Out]

-2/15*b*c^5*(1+1/c^2/x^2)^(3/2)+1/25*b*c^5*(1+1/c^2/x^2)^(5/2)+1/5*(-a-b*arccsch(c*x))/x^5+1/5*b*c^5*(1+1/c^2/
x^2)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6284, 266, 43} \[ -\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}+\frac {1}{25} b c^5 \left (\frac {1}{c^2 x^2}+1\right )^{5/2}-\frac {2}{15} b c^5 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}+\frac {1}{5} b c^5 \sqrt {\frac {1}{c^2 x^2}+1} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x^6,x]

[Out]

(b*c^5*Sqrt[1 + 1/(c^2*x^2)])/5 - (2*b*c^5*(1 + 1/(c^2*x^2))^(3/2))/15 + (b*c^5*(1 + 1/(c^2*x^2))^(5/2))/25 -
(a + b*ArcCsch[c*x])/(5*x^5)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6284

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCsch[c*
x]))/(d*(m + 1)), x] + Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 + 1/(c^2*x^2)], x], x] /; FreeQ[{a, b,
c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \text {csch}^{-1}(c x)}{x^6} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^7} \, dx}{5 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}+\frac {b \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{10 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}+\frac {b \operatorname {Subst}\left (\int \left (\frac {c^4}{\sqrt {1+\frac {x}{c^2}}}-2 c^4 \sqrt {1+\frac {x}{c^2}}+c^4 \left (1+\frac {x}{c^2}\right )^{3/2}\right ) \, dx,x,\frac {1}{x^2}\right )}{10 c}\\ &=\frac {1}{5} b c^5 \sqrt {1+\frac {1}{c^2 x^2}}-\frac {2}{15} b c^5 \left (1+\frac {1}{c^2 x^2}\right )^{3/2}+\frac {1}{25} b c^5 \left (1+\frac {1}{c^2 x^2}\right )^{5/2}-\frac {a+b \text {csch}^{-1}(c x)}{5 x^5}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 69, normalized size = 0.87 \[ -\frac {a}{5 x^5}+b \left (\frac {8 c^5}{75}-\frac {4 c^3}{75 x^2}+\frac {c}{25 x^4}\right ) \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x^6,x]

[Out]

-1/5*a/x^5 + b*((8*c^5)/75 + c/(25*x^4) - (4*c^3)/(75*x^2))*Sqrt[(1 + c^2*x^2)/(c^2*x^2)] - (b*ArcCsch[c*x])/(
5*x^5)

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fricas [A]  time = 0.48, size = 87, normalized size = 1.10 \[ -\frac {15 \, b \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (8 \, b c^{5} x^{5} - 4 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 15 \, a}{75 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/75*(15*b*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (8*b*c^5*x^5 - 4*b*c^3*x^3 + 3*b*c*x)*sqrt((c
^2*x^2 + 1)/(c^2*x^2)) + 15*a)/x^5

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^6,x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)/x^6, x)

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maple [A]  time = 0.05, size = 83, normalized size = 1.05 \[ c^{5} \left (-\frac {a}{5 c^{5} x^{5}}+b \left (-\frac {\mathrm {arccsch}\left (c x \right )}{5 c^{5} x^{5}}+\frac {\left (c^{2} x^{2}+1\right ) \left (8 c^{4} x^{4}-4 c^{2} x^{2}+3\right )}{75 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{6} x^{6}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))/x^6,x)

[Out]

c^5*(-1/5*a/c^5/x^5+b*(-1/5/c^5/x^5*arccsch(c*x)+1/75*(c^2*x^2+1)*(8*c^4*x^4-4*c^2*x^2+3)/((c^2*x^2+1)/c^2/x^2
)^(1/2)/c^6/x^6))

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maxima [A]  time = 0.31, size = 73, normalized size = 0.92 \[ \frac {1}{75} \, b {\left (\frac {3 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{6} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, c^{6} \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {15 \, \operatorname {arcsch}\left (c x\right )}{x^{5}}\right )} - \frac {a}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^6,x, algorithm="maxima")

[Out]

1/75*b*((3*c^6*(1/(c^2*x^2) + 1)^(5/2) - 10*c^6*(1/(c^2*x^2) + 1)^(3/2) + 15*c^6*sqrt(1/(c^2*x^2) + 1))/c - 15
*arccsch(c*x)/x^5) - 1/5*a/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x^6,x)

[Out]

int((a + b*asinh(1/(c*x)))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))/x**6,x)

[Out]

Integral((a + b*acsch(c*x))/x**6, x)

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