### 3.62 $$\int \frac{e^{\text{csch}^{-1}(c x)} x^3}{1+c^2 x^2} \, dx$$

Optimal. Leaf size=59 $\frac{x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{2 c^2}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{2 c^4}+\frac{x}{c^3}-\frac{\tan ^{-1}(c x)}{c^4}$

[Out]

x/c^3 + (Sqrt[1 + 1/(c^2*x^2)]*x^2)/(2*c^2) - ArcTan[c*x]/c^4 - ArcTanh[Sqrt[1 + 1/(c^2*x^2)]]/(2*c^4)

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Rubi [A]  time = 0.0897483, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {6342, 266, 51, 63, 208, 321, 203} $\frac{x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{2 c^2}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{2 c^4}+\frac{x}{c^3}-\frac{\tan ^{-1}(c x)}{c^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/c^3 + (Sqrt[1 + 1/(c^2*x^2)]*x^2)/(2*c^2) - ArcTan[c*x]/c^4 - ArcTanh[Sqrt[1 + 1/(c^2*x^2)]]/(2*c^4)

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/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 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}{1+c^2 x^2} \, dx &=\frac{\int \frac{x}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{c^2}+\frac{\int \frac{x^2}{1+c^2 x^2} \, dx}{c}\\ &=\frac{x}{c^3}-\frac{\int \frac{1}{1+c^2 x^2} \, dx}{c^3}-\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 c^2}\\ &=\frac{x}{c^3}+\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c^2}-\frac{\tan ^{-1}(c x)}{c^4}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 c^4}\\ &=\frac{x}{c^3}+\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c^2}-\frac{\tan ^{-1}(c x)}{c^4}+\frac{\operatorname{Subst}\left (\int \frac{1}{-c^2+c^2 x^2} \, dx,x,\sqrt{1+\frac{1}{c^2 x^2}}\right )}{2 c^2}\\ &=\frac{x}{c^3}+\frac{\sqrt{1+\frac{1}{c^2 x^2}} x^2}{2 c^2}-\frac{\tan ^{-1}(c x)}{c^4}-\frac{\tanh ^{-1}\left (\sqrt{1+\frac{1}{c^2 x^2}}\right )}{2 c^4}\\ \end{align*}

Mathematica [A]  time = 0.131322, size = 54, normalized size = 0.92 $-\frac{-c x \left (c x \sqrt{\frac{1}{c^2 x^2}+1}+2\right )+\log \left (x \left (\sqrt{\frac{1}{c^2 x^2}+1}+1\right )\right )+2 \tan ^{-1}(c x)}{2 c^4}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-(c*x*(2 + c*Sqrt[1 + 1/(c^2*x^2)]*x)) + 2*ArcTan[c*x] + Log[(1 + Sqrt[1 + 1/(c^2*x^2)])*x])/(2*c^4)

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Maple [B]  time = 0.15, size = 133, normalized size = 2.3 \begin{align*}{\frac{x}{2\,{c}^{4}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}} \left ( x\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{c}^{2}+\ln \left ( x+\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}} \right ) -2\,\ln \left ( x+\sqrt{-{\frac{ \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }{{c}^{4}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}+{\frac{x}{{c}^{3}}}-{\frac{\arctan \left ( cx \right ) }{{c}^{4}}} \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]

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

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Maxima [B]  time = 1.49774, size = 142, normalized size = 2.41 \begin{align*} \frac{x}{c^{3}} + \frac{\frac{2 \, \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}} - 1} - \log \left (\sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1\right ) + \log \left (\sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1\right )}{4 \, c^{4}} - \frac{\arctan \left (c x\right )}{c^{4}} \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]

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

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Fricas [A]  time = 2.61218, size = 162, normalized size = 2.75 \begin{align*} \frac{c^{2} x^{2} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, c x - 2 \, \arctan \left (c x\right ) + \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{2 \, c^{4}} \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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac{c x^{3} \sqrt{1 + \frac{1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, 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**3/(c**2*x**2+1),x)

[Out]

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

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Giac [A]  time = 1.13793, size = 82, normalized size = 1.39 \begin{align*} \frac{\sqrt{c^{2} x^{2} + 1} x{\left | c \right |} \mathrm{sgn}\left (x\right )}{2 \, c^{4}} + \frac{x}{c^{3}} + \frac{\log \left (-x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1}\right ) \mathrm{sgn}\left (x\right )}{2 \, c^{4}} - \frac{\arctan \left (c x\right )}{c^{4}} \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]

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