3.11 \(\int \frac {a+b \text {csch}^{-1}(c x)}{x^4} \, dx\)

Optimal. Leaf size=58 \[ -\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}+\frac {1}{9} b c^3 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}-\frac {1}{3} b c^3 \sqrt {\frac {1}{c^2 x^2}+1} \]

[Out]

1/9*b*c^3*(1+1/c^2/x^2)^(3/2)+1/3*(-a-b*arccsch(c*x))/x^3-1/3*b*c^3*(1+1/c^2/x^2)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 58, 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} \[ -\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}+\frac {1}{9} b c^3 \left (\frac {1}{c^2 x^2}+1\right )^{3/2}-\frac {1}{3} b c^3 \sqrt {\frac {1}{c^2 x^2}+1} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x^4,x]

[Out]

-(b*c^3*Sqrt[1 + 1/(c^2*x^2)])/3 + (b*c^3*(1 + 1/(c^2*x^2))^(3/2))/9 - (a + b*ArcCsch[c*x])/(3*x^3)

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^4} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^5} \, dx}{3 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}+\frac {b \operatorname {Subst}\left (\int \frac {x}{\sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}+\frac {b \operatorname {Subst}\left (\int \left (-\frac {c^2}{\sqrt {1+\frac {x}{c^2}}}+c^2 \sqrt {1+\frac {x}{c^2}}\right ) \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=-\frac {1}{3} b c^3 \sqrt {1+\frac {1}{c^2 x^2}}+\frac {1}{9} b c^3 \left (1+\frac {1}{c^2 x^2}\right )^{3/2}-\frac {a+b \text {csch}^{-1}(c x)}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 59, normalized size = 1.02 \[ -\frac {a}{3 x^3}+b \left (\frac {c}{9 x^2}-\frac {2 c^3}{9}\right ) \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x^4,x]

[Out]

-1/3*a/x^3 + b*((-2*c^3)/9 + c/(9*x^2))*Sqrt[(1 + c^2*x^2)/(c^2*x^2)] - (b*ArcCsch[c*x])/(3*x^3)

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fricas [A]  time = 1.00, size = 77, normalized size = 1.33 \[ -\frac {3 \, b \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (2 \, b c^{3} x^{3} - b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 3 \, a}{9 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(3*b*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (2*b*c^3*x^3 - b*c*x)*sqrt((c^2*x^2 + 1)/(c^2*x
^2)) + 3*a)/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)/x^4, x)

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maple [A]  time = 0.05, size = 75, normalized size = 1.29 \[ c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\mathrm {arccsch}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\left (c^{2} x^{2}+1\right ) \left (2 c^{2} x^{2}-1\right )}{9 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))/x^4,x)

[Out]

c^3*(-1/3/c^3/x^3*a+b*(-1/3/c^3/x^3*arccsch(c*x)-1/9*(c^2*x^2+1)*(2*c^2*x^2-1)/((c^2*x^2+1)/c^2/x^2)^(1/2)/c^4
/x^4))

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maxima [A]  time = 0.37, size = 56, normalized size = 0.97 \[ \frac {1}{9} \, b {\left (\frac {c^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {3 \, \operatorname {arcsch}\left (c x\right )}{x^{3}}\right )} - \frac {a}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/x^4,x, algorithm="maxima")

[Out]

1/9*b*((c^4*(1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(1/(c^2*x^2) + 1))/c - 3*arccsch(c*x)/x^3) - 1/3*a/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x^4,x)

[Out]

int((a + b*asinh(1/(c*x)))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))/x**4,x)

[Out]

Integral((a + b*acsch(c*x))/x**4, x)

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