Integrand size = 12, antiderivative size = 46 \[ \int \frac {1}{x^4 (1+W(a x))} \, dx=-\frac {1}{3 x^3}-\frac {9}{2} a^3 \operatorname {ExpIntegralEi}(-3 W(a x))+\frac {W(a x)}{2 x^3}-\frac {3 W(a x)^2}{2 x^3} \] Output:
-1/3/x^3-9/2*a^3*Ei(-3*LambertW(a*x))+1/2*LambertW(a*x)/x^3-3/2*LambertW(a *x)^2/x^3
Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 (1+W(a x))} \, dx=-\frac {1}{3 x^3}-\frac {9}{2} a^3 \operatorname {ExpIntegralEi}(-3 W(a x))+\frac {W(a x)}{2 x^3}-\frac {3 W(a x)^2}{2 x^3} \] Input:
Integrate[1/(x^4*(1 + ProductLog[a*x])),x]
Output:
-1/3*1/x^3 - (9*a^3*ExpIntegralEi[-3*ProductLog[a*x]])/2 + ProductLog[a*x] /(2*x^3) - (3*ProductLog[a*x]^2)/(2*x^3)
Time = 0.47 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7196, 7206, 7206, 7202}
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.
\(\displaystyle \int \frac {1}{x^4 (W(a x)+1)} \, dx\) |
\(\Big \downarrow \) 7196 |
\(\displaystyle -\int \frac {W(a x)}{x^4 (W(a x)+1)}dx-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 7206 |
\(\displaystyle \frac {3}{2} \int \frac {W(a x)^2}{x^4 (W(a x)+1)}dx+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 7206 |
\(\displaystyle \frac {3}{2} \left (-3 \int \frac {W(a x)^3}{x^4 (W(a x)+1)}dx-\frac {W(a x)^2}{x^3}\right )+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\) |
\(\Big \downarrow \) 7202 |
\(\displaystyle \frac {3}{2} \left (-3 a^3 \operatorname {ExpIntegralEi}(-3 W(a x))-\frac {W(a x)^2}{x^3}\right )+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\) |
Input:
Int[1/(x^4*(1 + ProductLog[a*x])),x]
Output:
-1/3*1/x^3 + ProductLog[a*x]/(2*x^3) + (3*(-3*a^3*ExpIntegralEi[-3*Product Log[a*x]] - ProductLog[a*x]^2/x^3))/2
Int[(x_)^(m_.)/((d_) + (d_.)*ProductLog[(a_.)*(x_)]), x_Symbol] :> Simp[x^( m + 1)/(d*(m + 1)), x] - Int[x^m*(ProductLog[a*x]/(d + d*ProductLog[a*x])), x] /; FreeQ[{a, d}, x] && LtQ[m, -1]
Int[((x_)^(m_.)*ProductLog[(a_.)*(x_)^(n_.)]^(p_.))/((d_) + (d_.)*ProductLo g[(a_.)*(x_)^(n_.)]), x_Symbol] :> Simp[a^p*(ExpIntegralEi[(-p)*ProductLog[ a*x^n]]/(d*n)), x] /; FreeQ[{a, d, m, n}, x] && IntegerQ[p] && EqQ[m + n*p, -1]
Int[((x_)^(m_.)*((c_.)*ProductLog[(a_.)*(x_)^(n_.)])^(p_.))/((d_) + (d_.)*P roductLog[(a_.)*(x_)^(n_.)]), x_Symbol] :> Simp[x^(m + 1)*((c*ProductLog[a* x^n])^p/(d*(m + n*p + 1))), x] - Simp[(m + 1)/(c*(m + n*p + 1)) Int[x^m*( (c*ProductLog[a*x^n])^(p + 1)/(d + d*ProductLog[a*x^n])), x], x] /; FreeQ[{ a, c, d, m, n, p}, x] && NeQ[m, -1] && LtQ[Simplify[p + (m + 1)/n], 0]
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(a^{3} \left (-\frac {1}{3 x^{3} a^{3}}+\frac {\operatorname {LambertW}\left (x a \right )}{2 x^{3} a^{3}}-\frac {3 \operatorname {LambertW}\left (x a \right )^{2}}{2 x^{3} a^{3}}+\frac {9 \,\operatorname {expIntegral}_{1}\left (3 \operatorname {LambertW}\left (x a \right )\right )}{2}\right )\) | \(50\) |
default | \(a^{3} \left (-\frac {1}{3 x^{3} a^{3}}+\frac {\operatorname {LambertW}\left (x a \right )}{2 x^{3} a^{3}}-\frac {3 \operatorname {LambertW}\left (x a \right )^{2}}{2 x^{3} a^{3}}+\frac {9 \,\operatorname {expIntegral}_{1}\left (3 \operatorname {LambertW}\left (x a \right )\right )}{2}\right )\) | \(50\) |
Input:
int(1/x^4/(1+LambertW(x*a)),x,method=_RETURNVERBOSE)
Output:
a^3*(-1/3/x^3/a^3+1/2*LambertW(x*a)/x^3/a^3-3/2*LambertW(x*a)^2/x^3/a^3+9/ 2*Ei(1,3*LambertW(x*a)))
\[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\int { \frac {1}{x^{4} {\left (\operatorname {W}({a x}) + 1\right )}} \,d x } \] Input:
integrate(1/x^4/(1+lambert_w(a*x)),x, algorithm="fricas")
Output:
integral(1/(x^4*lambert_w(a*x) + x^4), x)
\[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\int \frac {1}{x^{4} \left (W\left (a x\right ) + 1\right )}\, dx \] Input:
integrate(1/x**4/(1+LambertW(a*x)),x)
Output:
Integral(1/(x**4*(LambertW(a*x) + 1)), x)
\[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\int { \frac {1}{x^{4} {\left (\operatorname {W}({a x}) + 1\right )}} \,d x } \] Input:
integrate(1/x^4/(1+lambert_w(a*x)),x, algorithm="maxima")
Output:
integrate(1/(x^4*(lambert_w(a*x) + 1)), x)
\[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\int { \frac {1}{x^{4} {\left (\operatorname {W}({a x}) + 1\right )}} \,d x } \] Input:
integrate(1/x^4/(1+lambert_w(a*x)),x, algorithm="giac")
Output:
integrate(1/(x^4*(lambert_w(a*x) + 1)), x)
Timed out. \[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\int \frac {1}{x^4\,\left (\mathrm {LambertW}\left (a\,x\right )+1\right )} \,d x \] Input:
int(1/(x^4*(LambertW(a*x) + 1)),x)
Output:
int(1/(x^4*(LambertW(a*x) + 1)), x)
\[ \int \frac {1}{x^4 (1+W(a x))} \, dx=\frac {-3 \left (\int \frac {\textit {lambert\_w} \left (a x \right )}{\textit {lambert\_w} \left (a x \right ) x^{4}+x^{4}}d x \right ) x^{3}+9 \left (\int \frac {1}{\textit {lambert\_w} \left (a x \right ) x^{4}+x^{4}}d x \right ) x^{3}-1}{12 x^{3}} \] Input:
int(1/x^4/(1+Lambert_W(a*x)),x)
Output:
( - 3*int(lambert_w(a*x)/(lambert_w(a*x)*x**4 + x**4),x)*x**3 + 9*int(1/(l ambert_w(a*x)*x**4 + x**4),x)*x**3 - 1)/(12*x**3)