Integrand size = 10, antiderivative size = 75 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {\sqrt {1+\frac {1}{a^2 x^2}} x^2}{8 a^2}+\frac {x^3}{3 a}+\frac {1}{4} \sqrt {1+\frac {1}{a^2 x^2}} x^4-\frac {\text {arctanh}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right )}{8 a^4} \] Output:
1/8*(1+1/a^2/x^2)^(1/2)*x^2/a^2+1/3*x^3/a+1/4*(1+1/a^2/x^2)^(1/2)*x^4-1/8* arctanh((1+1/a^2/x^2)^(1/2))/a^4
Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {a^2 x^2 \left (3 \sqrt {1+\frac {1}{a^2 x^2}}+8 a x+6 a^2 \sqrt {1+\frac {1}{a^2 x^2}} x^2\right )-3 \log \left (\left (1+\sqrt {1+\frac {1}{a^2 x^2}}\right ) x\right )}{24 a^4} \] Input:
Integrate[E^ArcCsch[a*x]*x^3,x]
Output:
(a^2*x^2*(3*Sqrt[1 + 1/(a^2*x^2)] + 8*a*x + 6*a^2*Sqrt[1 + 1/(a^2*x^2)]*x^ 2) - 3*Log[(1 + Sqrt[1 + 1/(a^2*x^2)])*x])/(24*a^4)
Time = 0.42 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6890, 15, 798, 51, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 e^{\text {csch}^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6890 |
\(\displaystyle \int \sqrt {1+\frac {1}{a^2 x^2}} x^3dx+\frac {\int x^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \int \sqrt {1+\frac {1}{a^2 x^2}} x^3dx+\frac {x^3}{3 a}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {x^3}{3 a}-\frac {1}{2} \int \sqrt {1+\frac {1}{a^2 x^2}} x^6d\frac {1}{x^2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \sqrt {\frac {1}{a^2 x^2}+1}-\frac {\int \frac {x^4}{\sqrt {1+\frac {1}{a^2 x^2}}}d\frac {1}{x^2}}{4 a^2}\right )+\frac {x^3}{3 a}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \sqrt {\frac {1}{a^2 x^2}+1}-\frac {x^2 \left (-\sqrt {\frac {1}{a^2 x^2}+1}\right )-\frac {\int \frac {x^2}{\sqrt {1+\frac {1}{a^2 x^2}}}d\frac {1}{x^2}}{2 a^2}}{4 a^2}\right )+\frac {x^3}{3 a}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \sqrt {\frac {1}{a^2 x^2}+1}-\frac {x^2 \left (-\sqrt {\frac {1}{a^2 x^2}+1}\right )-\int \frac {1}{\frac {a^2}{x^4}-a^2}d\sqrt {1+\frac {1}{a^2 x^2}}}{4 a^2}\right )+\frac {x^3}{3 a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \sqrt {\frac {1}{a^2 x^2}+1}-\frac {\frac {\text {arctanh}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )}{a^2}-x^2 \sqrt {\frac {1}{a^2 x^2}+1}}{4 a^2}\right )+\frac {x^3}{3 a}\) |
Input:
Int[E^ArcCsch[a*x]*x^3,x]
Output:
x^3/(3*a) + ((Sqrt[1 + 1/(a^2*x^2)]*x^4)/2 - (-(Sqrt[1 + 1/(a^2*x^2)]*x^2) + ArcTanh[Sqrt[1 + 1/(a^2*x^2)]]/a^2)/(4*a^2))/2
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Simp[1/a Int[x^( m - p), x], x] + Int[x^m*Sqrt[1 + 1/(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45
method | result | size |
default | \(-\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, x \left (-2 x \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} a^{4}+x \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right )\right )}{8 \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{4}}+\frac {x^{3}}{3 a}\) | \(109\) |
Input:
int((1/a/x+(1+1/a^2/x^2)^(1/2))*x^3,x,method=_RETURNVERBOSE)
Output:
-1/8*((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)/a^4+1/3*x^3/a
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {8 \, a^{3} x^{3} + 3 \, {\left (2 \, a^{4} x^{4} + a^{2} x^{2}\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + 3 \, \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right )}{24 \, a^{4}} \] Input:
integrate((1/a/x+(1+1/a^2/x^2)^(1/2))*x^3,x, algorithm="fricas")
Output:
1/24*(8*a^3*x^3 + 3*(2*a^4*x^4 + a^2*x^2)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 3*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x))/a^4
Time = 2.68 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {a x^{5}}{4 \sqrt {a^{2} x^{2} + 1}} + \frac {x^{3}}{3 a} + \frac {3 x^{3}}{8 a \sqrt {a^{2} x^{2} + 1}} + \frac {x}{8 a^{3} \sqrt {a^{2} x^{2} + 1}} - \frac {\operatorname {asinh}{\left (a x \right )}}{8 a^{4}} \] Input:
integrate((1/a/x+(1+1/a**2/x**2)**(1/2))*x**3,x)
Output:
a*x**5/(4*sqrt(a**2*x**2 + 1)) + x**3/(3*a) + 3*x**3/(8*a*sqrt(a**2*x**2 + 1)) + x/(8*a**3*sqrt(a**2*x**2 + 1)) - asinh(a*x)/(8*a**4)
Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.43 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {x^{3}}{3 \, a} + \frac {{\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + \sqrt {\frac {1}{a^{2} x^{2}} + 1}}{8 \, {\left (a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} + a^{4}\right )}} - \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right )}{16 \, a^{4}} + \frac {\log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right )}{16 \, a^{4}} \] Input:
integrate((1/a/x+(1+1/a^2/x^2)^(1/2))*x^3,x, algorithm="maxima")
Output:
1/3*x^3/a + 1/8*((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) - 1/16*log(sqrt(1/(a^2*x ^2) + 1) + 1)/a^4 + 1/16*log(sqrt(1/(a^2*x^2) + 1) - 1)/a^4
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {1}{8} \, \sqrt {a^{2} x^{2} + 1} {\left (\frac {2 \, x^{2} {\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{2}} + \frac {{\left | a \right |} \mathrm {sgn}\left (x\right )}{a^{4}}\right )} x + \frac {x^{3}}{3 \, a} + \frac {\log \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} + 1}\right ) \mathrm {sgn}\left (x\right )}{8 \, a^{4}} \] Input:
integrate((1/a/x+(1+1/a^2/x^2)^(1/2))*x^3,x, algorithm="giac")
Output:
1/8*sqrt(a^2*x^2 + 1)*(2*x^2*abs(a)*sgn(x)/a^2 + abs(a)*sgn(x)/a^4)*x + 1/ 3*x^3/a + 1/8*log(-x*abs(a) + sqrt(a^2*x^2 + 1))*sgn(x)/a^4
Time = 24.77 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {x^4\,\sqrt {\frac {1}{a^2\,x^2}+1}}{4}-\frac {\mathrm {atanh}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\right )}{8\,a^4}+\frac {x^3}{3\,a}+\frac {x^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{8\,a^2} \] Input:
int(x^3*((1/(a^2*x^2) + 1)^(1/2) + 1/(a*x)),x)
Output:
(x^4*(1/(a^2*x^2) + 1)^(1/2))/4 - atanh((1/(a^2*x^2) + 1)^(1/2))/(8*a^4) + x^3/(3*a) + (x^2*(1/(a^2*x^2) + 1)^(1/2))/(8*a^2)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int e^{\text {csch}^{-1}(a x)} x^3 \, dx=\frac {6 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+3 \sqrt {a^{2} x^{2}+1}\, a x -3 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x \right )+8 a^{3} x^{3}}{24 a^{4}} \] Input:
int((1/a/x+(1+1/a^2/x^2)^(1/2))*x^3,x)
Output:
(6*sqrt(a**2*x**2 + 1)*a**3*x**3 + 3*sqrt(a**2*x**2 + 1)*a*x - 3*log(sqrt( a**2*x**2 + 1) + a*x) + 8*a**3*x**3)/(24*a**4)