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

Optimal. Leaf size=50 \[ -\frac {a+b \text {csch}^{-1}(c x)}{2 x^2}+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{4 x}-\frac {1}{4} b c^2 \text {csch}^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6284, 335, 321, 215} \[ -\frac {a+b \text {csch}^{-1}(c x)}{2 x^2}+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{4 x}-\frac {1}{4} b c^2 \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x^3,x]

[Out]

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
 x] && GtQ[a, 0] && PosQ[b]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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

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Mathematica [A]  time = 0.04, size = 66, normalized size = 1.32 \[ -\frac {a}{2 x^2}+\frac {b c \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 \sinh ^{-1}\left (\frac {1}{c x}\right )-\frac {b \text {csch}^{-1}(c x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x^3,x]

[Out]

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

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fricas [A]  time = 0.98, size = 76, normalized size = 1.52 \[ \frac {b c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - {\left (b c^{2} x^{2} + 2 \, b\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \, a}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 100, normalized size = 2.00 \[ c^{2} \left (-\frac {a}{2 c^{2} x^{2}}+b \left (-\frac {\mathrm {arccsch}\left (c x \right )}{2 c^{2} x^{2}}-\frac {\sqrt {c^{2} x^{2}+1}\, \left (\arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\right )}{4 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{3} x^{3}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 105, normalized size = 2.10 \[ \frac {1}{8} \, b {\left (\frac {\frac {2 \, c^{4} x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} - 1} - c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right ) + c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c} - \frac {4 \, \operatorname {arcsch}\left (c x\right )}{x^{2}}\right )} - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*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*a/x^2

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mupad [B]  time = 2.27, size = 51, normalized size = 1.02 \[ \frac {b\,c\,\sqrt {\frac {1}{c^2\,x^2}+1}}{4\,x}-\frac {b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\,\left (\frac {c^2\,x}{4}+\frac {1}{2\,x}\right )}{x}-\frac {a}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x^3,x)

[Out]

(b*c*(1/(c^2*x^2) + 1)^(1/2))/(4*x) - (b*asinh(1/(c*x))*((c^2*x)/4 + 1/(2*x)))/x - a/(2*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))/x**3, x)

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