3.68 \(\int \frac{e^{\text{csch}^{-1}(c x)}}{x^3 (1+c^2 x^2)} \, dx\)

Optimal. Leaf size=61 \[ -\frac{1}{3} c^2 \left (\frac{1}{c^2 x^2}+1\right )^{3/2}+c^2 \sqrt{\frac{1}{c^2 x^2}+1}+c^2 \tan ^{-1}(c x)-\frac{1}{3 c x^3}+\frac{c}{x} \]

[Out]

c^2*Sqrt[1 + 1/(c^2*x^2)] - (c^2*(1 + 1/(c^2*x^2))^(3/2))/3 - 1/(3*c*x^3) + c/x + c^2*ArcTan[c*x]

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Rubi [A]  time = 0.0945817, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6342, 266, 43, 325, 203} \[ -\frac{1}{3} c^2 \left (\frac{1}{c^2 x^2}+1\right )^{3/2}+c^2 \sqrt{\frac{1}{c^2 x^2}+1}+c^2 \tan ^{-1}(c x)-\frac{1}{3 c x^3}+\frac{c}{x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCsch[c*x]/(x^3*(1 + c^2*x^2)),x]

[Out]

c^2*Sqrt[1 + 1/(c^2*x^2)] - (c^2*(1 + 1/(c^2*x^2))^(3/2))/3 - 1/(3*c*x^3) + c/x + c^2*ArcTan[c*x]

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 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 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 325

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

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.147343, size = 54, normalized size = 0.89 \[ \frac{\sqrt{\frac{1}{c^2 x^2}+1} \left (2 c^2 x^2-1\right )}{3 x^2}+c^2 \tan ^{-1}(c x)-\frac{1}{3 c x^3}+\frac{c}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCsch[c*x]/(x^3*(1 + c^2*x^2)),x]

[Out]

-1/(3*c*x^3) + c/x + (Sqrt[1 + 1/(c^2*x^2)]*(-1 + 2*c^2*x^2))/(3*x^2) + c^2*ArcTan[c*x]

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Maple [B]  time = 0.196, size = 193, normalized size = 3.2 \begin{align*}{\frac{{c}^{2}}{3\,{x}^{2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}} \left ( 3\, \left ({\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}} \right ) ^{3/2}{x}^{2}{c}^{2}-3\,\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{x}^{4}{c}^{2}-3\,\ln \left ( x+\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}} \right ){x}^{3}+3\,\ln \left ( x+\sqrt{-{\frac{ \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }{{c}^{4}}}} \right ){x}^{3}- \left ({\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}} \right ) ^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}-{\frac{1}{3\,c{x}^{3}}}+{\frac{c}{x}}+{c}^{2}\arctan \left ( cx \right ) \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^3/(c^2*x^2+1),x)

[Out]

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

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Maxima [A]  time = 1.5894, size = 76, normalized size = 1.25 \begin{align*} c^{2} \arctan \left (c x\right ) + \frac{{\left (2 \, c^{2} x^{2} - 1\right )} \sqrt{c^{2} x^{2} + 1}}{3 \, c x^{3}} + \frac{3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \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^3/(c^2*x^2+1),x, algorithm="maxima")

[Out]

c^2*arctan(c*x) + 1/3*(2*c^2*x^2 - 1)*sqrt(c^2*x^2 + 1)/(c*x^3) + 1/3*(3*c^2*x^2 - 1)/(c*x^3)

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Fricas [A]  time = 2.52815, size = 155, normalized size = 2.54 \begin{align*} \frac{3 \, c^{3} x^{3} \arctan \left (c x\right ) + 2 \, c^{3} x^{3} + 3 \, c^{2} x^{2} +{\left (2 \, c^{3} x^{3} - c x\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1}{3 \, c x^{3}} \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^3/(c^2*x^2+1),x, algorithm="fricas")

[Out]

1/3*(3*c^3*x^3*arctan(c*x) + 2*c^3*x^3 + 3*c^2*x^2 + (2*c^3*x^3 - c*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 1)/(c*x
^3)

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Sympy [A]  time = 5.46128, size = 75, normalized size = 1.23 \begin{align*} - 2 c^{5} \left (\frac{\left (1 + \frac{1}{c^{2} x^{2}}\right )^{\frac{3}{2}}}{6 c^{3}} - \frac{\sqrt{1 + \frac{1}{c^{2} x^{2}}}}{2 c^{3}}\right ) - \frac{c^{3} \operatorname{atan}{\left (\frac{1}{x \sqrt{c^{2}}} \right )}}{\sqrt{c^{2}}} + \frac{c}{x} - \frac{1}{3 c x^{3}} \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**3/(c**2*x**2+1),x)

[Out]

-2*c**5*((1 + 1/(c**2*x**2))**(3/2)/(6*c**3) - sqrt(1 + 1/(c**2*x**2))/(2*c**3)) - c**3*atan(1/(x*sqrt(c**2)))
/sqrt(c**2) + c/x - 1/(3*c*x**3)

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Giac [A]  time = 1.16573, size = 111, normalized size = 1.82 \begin{align*} c^{2} \arctan \left (c x\right ) + \frac{4 \,{\left (3 \,{\left (x{\left | c \right |} - \sqrt{c^{2} x^{2} + 1}\right )}^{2} - 1\right )} c^{2} \mathrm{sgn}\left (x\right )}{3 \,{\left ({\left (x{\left | c \right |} - \sqrt{c^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} + \frac{3 \, c^{2} x^{2} - 1}{3 \, c x^{3}} \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^3/(c^2*x^2+1),x, algorithm="giac")

[Out]

c^2*arctan(c*x) + 4/3*(3*(x*abs(c) - sqrt(c^2*x^2 + 1))^2 - 1)*c^2*sgn(x)/((x*abs(c) - sqrt(c^2*x^2 + 1))^2 -
1)^3 + 1/3*(3*c^2*x^2 - 1)/(c*x^3)