3.67 \(\int \frac {e^{\text {csch}^{-1}(c x)}}{x^2 (1+c^2 x^2)} \, dx\)

Optimal. Leaf size=60 \[ -\frac {\sqrt {\frac {1}{c^2 x^2}+1}}{2 x}+\frac {1}{2} c \log \left (c^2 x^2+1\right )-\frac {1}{2 c x^2}-c \log (x)+\frac {1}{2} c \text {csch}^{-1}(c x) \]

[Out]

-1/2/c/x^2+1/2*c*arccsch(c*x)-c*ln(x)+1/2*c*ln(c^2*x^2+1)-1/2*(1+1/c^2/x^2)^(1/2)/x

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Rubi [A]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6342, 335, 321, 215, 266, 44} \[ -\frac {\sqrt {\frac {1}{c^2 x^2}+1}}{2 x}+\frac {1}{2} c \log \left (c^2 x^2+1\right )-\frac {1}{2 c x^2}-c \log (x)+\frac {1}{2} c \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCsch[c*x]/(x^2*(1 + c^2*x^2)),x]

[Out]

-1/(2*c*x^2) - Sqrt[1 + 1/(c^2*x^2)]/(2*x) + (c*ArcCsch[c*x])/2 - c*Log[x] + (c*Log[1 + c^2*x^2])/2

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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 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 6342

Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d^2/(a*c^2), Int[(d*x)
^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d,
m}, x] && EqQ[b - a*c^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 58, normalized size = 0.97 \[ \frac {1}{2} \left (-\frac {\sqrt {\frac {1}{c^2 x^2}+1}}{x}+c \log \left (c^2 x^2+1\right )-\frac {1}{c x^2}-2 c \log (x)+c \sinh ^{-1}\left (\frac {1}{c x}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[c*x]/(x^2*(1 + c^2*x^2)),x]

[Out]

(-(1/(c*x^2)) - Sqrt[1 + 1/(c^2*x^2)]/x + c*ArcSinh[1/(c*x)] - 2*c*Log[x] + c*Log[1 + c^2*x^2])/2

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fricas [B]  time = 1.34, size = 130, normalized size = 2.17 \[ \frac {c^{2} x^{2} \log \left (c^{2} x^{2} + 1\right ) + c^{2} x^{2} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - c^{2} x^{2} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) - 2 \, c^{2} x^{2} \log \relax (x) - c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1}{2 \, c x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^2/(c^2*x^2+1),x, algorithm="fricas")

[Out]

1/2*(c^2*x^2*log(c^2*x^2 + 1) + c^2*x^2*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x + 1) - c^2*x^2*log(c*x*sqr
t((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) - 2*c^2*x^2*log(x) - c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 1)/(c*x^2)

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giac [B]  time = 0.15, size = 114, normalized size = 1.90 \[ \frac {1}{2} \, c \log \left (c^{2} x^{2} + 1\right ) + \frac {1}{4} \, {\left ({\left | c \right |} \mathrm {sgn}\relax (x) - 2 \, c\right )} \log \left (\sqrt {c^{2} x^{2} + 1} + 1\right ) - \frac {1}{4} \, {\left ({\left | c \right |} \mathrm {sgn}\relax (x) + 2 \, c\right )} \log \left (\sqrt {c^{2} x^{2} + 1} - 1\right ) - \frac {\sqrt {c^{2} x^{2} + 1} {\left | c \right |} \mathrm {sgn}\relax (x) + c}{2 \, {\left (\sqrt {c^{2} x^{2} + 1} + 1\right )} {\left (\sqrt {c^{2} x^{2} + 1} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^2/(c^2*x^2+1),x, algorithm="giac")

[Out]

1/2*c*log(c^2*x^2 + 1) + 1/4*(abs(c)*sgn(x) - 2*c)*log(sqrt(c^2*x^2 + 1) + 1) - 1/4*(abs(c)*sgn(x) + 2*c)*log(
sqrt(c^2*x^2 + 1) - 1) - 1/2*(sqrt(c^2*x^2 + 1)*abs(c)*sgn(x) + c)/((sqrt(c^2*x^2 + 1) + 1)*(sqrt(c^2*x^2 + 1)
- 1))

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maple [B]  time = 0.07, size = 210, normalized size = 3.50 \[ -\frac {\sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, \left (c^{2} \left (\frac {c^{2} x^{2}+1}{c^{2}}\right )^{\frac {3}{2}} \sqrt {\frac {1}{c^{2}}}+\sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, \sqrt {\frac {1}{c^{2}}}\, x^{2} c^{2}-2 \sqrt {\frac {1}{c^{2}}}\, \sqrt {-\frac {\left (-c^{2} x +\sqrt {-c^{2}}\right ) \left (c^{2} x +\sqrt {-c^{2}}\right )}{c^{4}}}\, x^{2} c^{2}-\ln \left (\frac {2 \sqrt {\frac {1}{c^{2}}}\, \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, c^{2}+2}{c^{2} x}\right ) x^{2}\right )}{2 x \sqrt {\frac {c^{2} x^{2}+1}{c^{2}}}\, \sqrt {\frac {1}{c^{2}}}}-\frac {1}{2 c \,x^{2}}-c \ln \relax (x )+\frac {c \ln \left (c^{2} x^{2}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))/x^2/(c^2*x^2+1),x)

[Out]

-1/2*((c^2*x^2+1)/c^2/x^2)^(1/2)/x*(c^2*((c^2*x^2+1)/c^2)^(3/2)*(1/c^2)^(1/2)+((c^2*x^2+1)/c^2)^(1/2)*(1/c^2)^
(1/2)*x^2*c^2-2*(1/c^2)^(1/2)*(-(-c^2*x+(-c^2)^(1/2))*(c^2*x+(-c^2)^(1/2))/c^4)^(1/2)*x^2*c^2-ln(2*((1/c^2)^(1
/2)*((c^2*x^2+1)/c^2)^(1/2)*c^2+1)/x/c^2)*x^2)/((c^2*x^2+1)/c^2)^(1/2)/(1/c^2)^(1/2)-1/2/c/x^2-c*ln(x)+1/2*c*l
n(c^2*x^2+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, c \log \left (c^{2} x^{2} + 1\right ) - c \log \relax (x) - \frac {1}{2 \, c x^{2}} + \int \frac {\sqrt {c^{2} x^{2} + 1}}{c^{3} x^{5} + c x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))/x^2/(c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*c*log(c^2*x^2 + 1) - c*log(x) - 1/2/(c*x^2) + integrate(sqrt(c^2*x^2 + 1)/(c^3*x^5 + c*x^3), x)

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mupad [B]  time = 2.42, size = 61, normalized size = 1.02 \[ \frac {\mathrm {asinh}\left (\frac {\sqrt {\frac {1}{c^2}}}{x}\right )}{2\,\sqrt {\frac {1}{c^2}}}+\frac {c\,\ln \left (-c^2\,x^2-1\right )}{2}-c\,\ln \relax (x)-\frac {\sqrt {\frac {1}{c^2\,x^2}+1}}{2\,x}-\frac {1}{2\,c\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/(c^2*x^2) + 1)^(1/2) + 1/(c*x))/(x^2*(c^2*x^2 + 1)),x)

[Out]

asinh((1/c^2)^(1/2)/x)/(2*(1/c^2)^(1/2)) + (c*log(- c^2*x^2 - 1))/2 - c*log(x) - (1/(c^2*x^2) + 1)^(1/2)/(2*x)
- 1/(2*c*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c x \sqrt {1 + \frac {1}{c^{2} x^{2}}}}{c^{2} x^{5} + x^{3}}\, dx + \int \frac {1}{c^{2} x^{5} + x^{3}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))/x**2/(c**2*x**2+1),x)

[Out]

(Integral(c*x*sqrt(1 + 1/(c**2*x**2))/(c**2*x**5 + x**3), x) + Integral(1/(c**2*x**5 + x**3), x))/c

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