3.12 \(\int \frac {a+b \text {csch}^{-1}(c x)}{x^5} \, dx\)

Optimal. Leaf size=74 \[ -\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}+\frac {3}{32} b c^4 \text {csch}^{-1}(c x)+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{16 x^3}-\frac {3 b c^3 \sqrt {\frac {1}{c^2 x^2}+1}}{32 x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6284, 335, 321, 215} \[ -\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}-\frac {3 b c^3 \sqrt {\frac {1}{c^2 x^2}+1}}{32 x}+\frac {b c \sqrt {\frac {1}{c^2 x^2}+1}}{16 x^3}+\frac {3}{32} b c^4 \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x^5,x]

[Out]

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

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^5} \, dx &=-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}-\frac {b \int \frac {1}{\sqrt {1+\frac {1}{c^2 x^2}} x^6} \, dx}{4 c}\\ &=-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}+\frac {b \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{4 c}\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{16 x^3}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}-\frac {1}{16} (3 b c) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c \sqrt {1+\frac {1}{c^2 x^2}}}{16 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{32 x}-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}+\frac {1}{32} \left (3 b c^3\right ) \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}}}{16 x^3}-\frac {3 b c^3 \sqrt {1+\frac {1}{c^2 x^2}}}{32 x}+\frac {3}{32} b c^4 \text {csch}^{-1}(c x)-\frac {a+b \text {csch}^{-1}(c x)}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 78, normalized size = 1.05 \[ -\frac {a}{4 x^4}+\frac {3}{32} b c^4 \sinh ^{-1}\left (\frac {1}{c x}\right )+b \left (\frac {c}{16 x^3}-\frac {3 c^3}{32 x}\right ) \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}-\frac {b \text {csch}^{-1}(c x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x^5,x]

[Out]

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

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fricas [A]  time = 1.02, size = 89, normalized size = 1.20 \[ \frac {{\left (3 \, b c^{4} x^{4} - 8 \, b\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (3 \, b c^{3} x^{3} - 2 \, b c x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - 8 \, a}{32 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsch}\left (c x\right ) + a}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 120, normalized size = 1.62 \[ c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\mathrm {arccsch}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {c^{2} x^{2}+1}\, \left (3 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right ) c^{4} x^{4}-3 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 \sqrt {c^{2} x^{2}+1}\right )}{32 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c^{5} x^{5}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^4*(-1/4*a/c^4/x^4+b*(-1/4/c^4/x^4*arccsch(c*x)+1/32*(c^2*x^2+1)^(1/2)*(3*arctanh(1/(c^2*x^2+1)^(1/2))*c^4*x^
4-3*c^2*x^2*(c^2*x^2+1)^(1/2)+2*(c^2*x^2+1)^(1/2))/((c^2*x^2+1)/c^2/x^2)^(1/2)/c^5/x^5))

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maxima [B]  time = 0.33, size = 147, normalized size = 1.99 \[ \frac {1}{64} \, b {\left (\frac {3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right ) - 3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (3 \, c^{8} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 5 \, c^{6} x \sqrt {\frac {1}{c^{2} x^{2}} + 1}\right )}}{c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{2} - 2 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} + 1}}{c} - \frac {16 \, \operatorname {arcsch}\left (c x\right )}{x^{4}}\right )} - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*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*a/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x^5,x)

[Out]

int((a + b*asinh(1/(c*x)))/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))/x**5, x)

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