### 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) - 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

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Rubi [A]  time = 0.0970317, antiderivative size = 60, normalized size of antiderivative = 1., 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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.108683, 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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Maple [B]  time = 0.195, size = 210, normalized size = 3.5 \begin{align*} -{\frac{1}{2\,x}\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{{c}^{-2}}+\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}\sqrt{{c}^{-2}}{x}^{2}{c}^{2}-2\,\sqrt{{c}^{-2}}\sqrt{-{\frac{ \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }{{c}^{4}}}}{x}^{2}{c}^{2}-\ln \left ( 2\,{\frac{1}{x{c}^{2}} \left ( \sqrt{{c}^{-2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{c}^{2}+1 \right ) } \right ){x}^{2} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}{\frac{1}{\sqrt{{c}^{-2}}}}}-{\frac{1}{2\,c{x}^{2}}}-c\ln \left ( x \right ) +{\frac{c\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2}} \end{align*}

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)*(-(x*c^2+(-c^2)^(1/2))*(-x*c^2+(-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., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, c \log \left (c^{2} x^{2} + 1\right ) - c \log \left (x\right ) - \frac{1}{2 \, c x^{2}} + \int \frac{\sqrt{c^{2} x^{2} + 1}}{c^{3} x^{5} + c x^{3}}\,{d x} \end{align*}

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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Fricas [B]  time = 2.61346, size = 293, normalized size = 4.88 \begin{align*} \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 \left (x\right ) - c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1}{2 \, c x^{2}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Giac [B]  time = 1.14435, size = 154, normalized size = 2.57 \begin{align*} \frac{1}{2} \, c \log \left (c^{2} x^{2} + 1\right ) + \frac{1}{4} \,{\left ({\left | c \right |} \mathrm{sgn}\left (x\right ) - 2 \, c\right )} \log \left (\sqrt{c^{2} x^{2} + 1} + 1\right ) - \frac{1}{4} \,{\left ({\left | c \right |} \mathrm{sgn}\left (x\right ) + 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}\left (x\right ) + c}{2 \,{\left (\sqrt{c^{2} x^{2} + 1} + 1\right )}{\left (\sqrt{c^{2} x^{2} + 1} - 1\right )}} \end{align*}

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))