### 3.49 $$\int e^{2 \text{csch}^{-1}(a x)} x^4 \, dx$$

Optimal. Leaf size=85 $\frac{x^4 \sqrt{\frac{1}{a^2 x^2}+1}}{2 a}+\frac{2 x^3}{3 a^2}+\frac{x^2 \sqrt{\frac{1}{a^2 x^2}+1}}{4 a^3}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{4 a^5}+\frac{x^5}{5}$

[Out]

(Sqrt[1 + 1/(a^2*x^2)]*x^2)/(4*a^3) + (2*x^3)/(3*a^2) + (Sqrt[1 + 1/(a^2*x^2)]*x^4)/(2*a) + x^5/5 - ArcTanh[Sq
rt[1 + 1/(a^2*x^2)]]/(4*a^5)

________________________________________________________________________________________

Rubi [A]  time = 0.254138, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.583, Rules used = {6338, 6742, 266, 47, 51, 63, 208} $\frac{x^4 \sqrt{\frac{1}{a^2 x^2}+1}}{2 a}+\frac{2 x^3}{3 a^2}+\frac{x^2 \sqrt{\frac{1}{a^2 x^2}+1}}{4 a^3}-\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{4 a^5}+\frac{x^5}{5}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*ArcCsch[a*x])*x^4,x]

[Out]

(Sqrt[1 + 1/(a^2*x^2)]*x^2)/(4*a^3) + (2*x^3)/(3*a^2) + (Sqrt[1 + 1/(a^2*x^2)]*x^4)/(2*a) + x^5/5 - ArcTanh[Sq
rt[1 + 1/(a^2*x^2)]]/(4*a^5)

Rule 6338

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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 47

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

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]

Rubi steps

\begin{align*} \int e^{2 \text{csch}^{-1}(a x)} x^4 \, dx &=\int \left (\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )^2 x^4 \, dx\\ &=\int \left (\frac{2 x^2}{a^2}+\frac{2 \sqrt{1+\frac{1}{a^2 x^2}} x^3}{a}+x^4\right ) \, dx\\ &=\frac{2 x^3}{3 a^2}+\frac{x^5}{5}+\frac{2 \int \sqrt{1+\frac{1}{a^2 x^2}} x^3 \, dx}{a}\\ &=\frac{2 x^3}{3 a^2}+\frac{x^5}{5}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{x}{a^2}}}{x^3} \, dx,x,\frac{1}{x^2}\right )}{a}\\ &=\frac{2 x^3}{3 a^2}+\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^4}{2 a}+\frac{x^5}{5}-\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 a^3}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{4 a^3}+\frac{2 x^3}{3 a^2}+\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^4}{2 a}+\frac{x^5}{5}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{8 a^5}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{4 a^3}+\frac{2 x^3}{3 a^2}+\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^4}{2 a}+\frac{x^5}{5}+\frac{\operatorname{Subst}\left (\int \frac{1}{-a^2+a^2 x^2} \, dx,x,\sqrt{1+\frac{1}{a^2 x^2}}\right )}{4 a^3}\\ &=\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^2}{4 a^3}+\frac{2 x^3}{3 a^2}+\frac{\sqrt{1+\frac{1}{a^2 x^2}} x^4}{2 a}+\frac{x^5}{5}-\frac{\tanh ^{-1}\left (\sqrt{1+\frac{1}{a^2 x^2}}\right )}{4 a^5}\\ \end{align*}

Mathematica [A]  time = 0.0630435, size = 84, normalized size = 0.99 $\frac{a^2 x^2 \left (12 a^3 x^3+30 a^2 x^2 \sqrt{\frac{1}{a^2 x^2}+1}+15 \sqrt{\frac{1}{a^2 x^2}+1}+40 a x\right )-15 \log \left (x \left (\sqrt{\frac{1}{a^2 x^2}+1}+1\right )\right )}{60 a^5}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCsch[a*x])*x^4,x]

[Out]

(a^2*x^2*(15*Sqrt[1 + 1/(a^2*x^2)] + 40*a*x + 30*a^2*Sqrt[1 + 1/(a^2*x^2)]*x^2 + 12*a^3*x^3) - 15*Log[(1 + Sqr
t[1 + 1/(a^2*x^2)])*x])/(60*a^5)

________________________________________________________________________________________

Maple [A]  time = 0.174, size = 117, normalized size = 1.4 \begin{align*}{\frac{{x}^{5}}{5}}+{\frac{2\,{x}^{3}}{3\,{a}^{2}}}+{\frac{x}{4\,{a}^{5}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( 2\,x \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{3/2}{a}^{4}-x\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}{a}^{2}-\ln \left ( x+\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x)

[Out]

1/5*x^5+2/3*x^3/a^2+1/4/a^5*((a^2*x^2+1)/a^2/x^2)^(1/2)*x*(2*x*((a^2*x^2+1)/a^2)^(3/2)*a^4-x*((a^2*x^2+1)/a^2)
^(1/2)*a^2-ln(x+((a^2*x^2+1)/a^2)^(1/2)))/((a^2*x^2+1)/a^2)^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.03488, size = 158, normalized size = 1.86 \begin{align*} \frac{1}{5} \, x^{5} + \frac{2 \, x^{3}}{3 \, a^{2}} + \frac{\frac{2 \,{\left ({\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + \sqrt{\frac{1}{a^{2} x^{2}} + 1}\right )}}{a^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )} + a^{4}} - \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right )}{a^{4}} + \frac{\log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right )}{a^{4}}}{8 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="maxima")

[Out]

1/5*x^5 + 2/3*x^3/a^2 + 1/8*(2*((1/(a^2*x^2) + 1)^(3/2) + sqrt(1/(a^2*x^2) + 1))/(a^4*(1/(a^2*x^2) + 1)^2 - 2*
a^4*(1/(a^2*x^2) + 1) + a^4) - log(sqrt(1/(a^2*x^2) + 1) + 1)/a^4 + log(sqrt(1/(a^2*x^2) + 1) - 1)/a^4)/a

________________________________________________________________________________________

Fricas [A]  time = 2.56241, size = 193, normalized size = 2.27 \begin{align*} \frac{12 \, a^{5} x^{5} + 40 \, a^{3} x^{3} + 15 \,{\left (2 \, a^{4} x^{4} + a^{2} x^{2}\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} + 15 \, \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{60 \, a^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="fricas")

[Out]

1/60*(12*a^5*x^5 + 40*a^3*x^3 + 15*(2*a^4*x^4 + a^2*x^2)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 15*log(a*x*sqrt((a^2*
x^2 + 1)/(a^2*x^2)) - a*x))/a^5

________________________________________________________________________________________

Sympy [A]  time = 6.68562, size = 82, normalized size = 0.96 \begin{align*} \frac{x^{5}}{5} + \frac{x^{5}}{2 \sqrt{a^{2} x^{2} + 1}} + \frac{2 x^{3}}{3 a^{2}} + \frac{3 x^{3}}{4 a^{2} \sqrt{a^{2} x^{2} + 1}} + \frac{x}{4 a^{4} \sqrt{a^{2} x^{2} + 1}} - \frac{\operatorname{asinh}{\left (a x \right )}}{4 a^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))**2*x**4,x)

[Out]

x**5/5 + x**5/(2*sqrt(a**2*x**2 + 1)) + 2*x**3/(3*a**2) + 3*x**3/(4*a**2*sqrt(a**2*x**2 + 1)) + x/(4*a**4*sqrt
(a**2*x**2 + 1)) - asinh(a*x)/(4*a**5)

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2*x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError