Integrand size = 10, antiderivative size = 59 \[ \int \frac {1}{x^4 W(a x)} \, dx=-\frac {1}{12 x^3}-\frac {9}{8} a^3 \operatorname {ExpIntegralEi}(-3 W(a x))-\frac {1}{4 x^3 W(a x)}+\frac {W(a x)}{8 x^3}-\frac {3 W(a x)^2}{8 x^3} \] Output:
-1/12/x^3-9/8*a^3*Ei(-3*LambertW(a*x))-1/4/x^3/LambertW(a*x)+1/8*LambertW( a*x)/x^3-3/8*LambertW(a*x)^2/x^3
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 W(a x)} \, dx=-\frac {1}{12 x^3}-\frac {9}{8} a^3 \operatorname {ExpIntegralEi}(-3 W(a x))-\frac {1}{4 x^3 W(a x)}+\frac {W(a x)}{8 x^3}-\frac {3 W(a x)^2}{8 x^3} \] Input:
Integrate[1/(x^4*ProductLog[a*x]),x]
Output:
-1/12*1/x^3 - (9*a^3*ExpIntegralEi[-3*ProductLog[a*x]])/8 - 1/(4*x^3*Produ ctLog[a*x]) + ProductLog[a*x]/(8*x^3) - (3*ProductLog[a*x]^2)/(8*x^3)
Time = 0.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {7173, 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.
\(\displaystyle \int \frac {1}{x^4 W(a x)} \, dx\) |
\(\Big \downarrow \) 7173 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^4 (W(a x)+1)}dx-\frac {1}{4 x^3 W(a x)}\) |
\(\Big \downarrow \) 7196 |
\(\displaystyle \frac {1}{4} \left (-\int \frac {W(a x)}{x^4 (W(a x)+1)}dx-\frac {1}{3 x^3}\right )-\frac {1}{4 x^3 W(a x)}\) |
\(\Big \downarrow \) 7206 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{2} \int \frac {W(a x)^2}{x^4 (W(a x)+1)}dx+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\right )-\frac {1}{4 x^3 W(a x)}\) |
\(\Big \downarrow \) 7206 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{2} \left (-3 \int \frac {W(a x)^3}{x^4 (W(a x)+1)}dx-\frac {W(a x)^2}{x^3}\right )+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\right )-\frac {1}{4 x^3 W(a x)}\) |
\(\Big \downarrow \) 7202 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{2} \left (-3 a^3 \operatorname {ExpIntegralEi}(-3 W(a x))-\frac {W(a x)^2}{x^3}\right )+\frac {W(a x)}{2 x^3}-\frac {1}{3 x^3}\right )-\frac {1}{4 x^3 W(a x)}\) |
Input:
Int[1/(x^4*ProductLog[a*x]),x]
Output:
-1/4*1/(x^3*ProductLog[a*x]) + (-1/3*1/x^3 + ProductLog[a*x]/(2*x^3) + (3* (-3*a^3*ExpIntegralEi[-3*ProductLog[a*x]] - ProductLog[a*x]^2/x^3))/2)/4
Int[(x_)^(m_.)*((c_.)*ProductLog[(a_.)*(x_)^(n_.)])^(p_.), x_Symbol] :> Sim p[x^(m + 1)*((c*ProductLog[a*x^n])^p/(m + n*p + 1)), x] + Simp[n*(p/(c*(m + n*p + 1))) Int[x^m*((c*ProductLog[a*x^n])^(p + 1)/(1 + ProductLog[a*x^n] )), x], x] /; FreeQ[{a, c, m, n, p}, x] && (EqQ[m, -1] || (IntegerQ[p - 1/2 ] && ILtQ[Simplify[p + (m + 1)/n] - 1/2, 0]) || ( !IntegerQ[p - 1/2] && ILt Q[Simplify[p + (m + 1)/n], 0]))
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 = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08
method | result | size |
derivativedivides | \(a^{3} \left (-\frac {1}{4 \operatorname {LambertW}\left (x a \right ) x^{3} a^{3}}-\frac {1}{12 x^{3} a^{3}}+\frac {\operatorname {LambertW}\left (x a \right )}{8 x^{3} a^{3}}-\frac {3 \operatorname {LambertW}\left (x a \right )^{2}}{8 x^{3} a^{3}}+\frac {9 \,\operatorname {expIntegral}_{1}\left (3 \operatorname {LambertW}\left (x a \right )\right )}{8}\right )\) | \(64\) |
default | \(a^{3} \left (-\frac {1}{4 \operatorname {LambertW}\left (x a \right ) x^{3} a^{3}}-\frac {1}{12 x^{3} a^{3}}+\frac {\operatorname {LambertW}\left (x a \right )}{8 x^{3} a^{3}}-\frac {3 \operatorname {LambertW}\left (x a \right )^{2}}{8 x^{3} a^{3}}+\frac {9 \,\operatorname {expIntegral}_{1}\left (3 \operatorname {LambertW}\left (x a \right )\right )}{8}\right )\) | \(64\) |
Input:
int(1/x^4/LambertW(x*a),x,method=_RETURNVERBOSE)
Output:
a^3*(-1/4/LambertW(x*a)/x^3/a^3-1/12/x^3/a^3+1/8*LambertW(x*a)/x^3/a^3-3/8 *LambertW(x*a)^2/x^3/a^3+9/8*Ei(1,3*LambertW(x*a)))
\[ \int \frac {1}{x^4 W(a x)} \, dx=\int { \frac {1}{x^{4} \operatorname {W}({a x})} \,d x } \] Input:
integrate(1/x^4/lambert_w(a*x),x, algorithm="fricas")
Output:
integral(1/(x^4*lambert_w(a*x)), x)
\[ \int \frac {1}{x^4 W(a x)} \, dx=\int \frac {1}{x^{4} W\left (a x\right )}\, dx \] Input:
integrate(1/x**4/LambertW(a*x),x)
Output:
Integral(1/(x**4*LambertW(a*x)), x)
\[ \int \frac {1}{x^4 W(a x)} \, dx=\int { \frac {1}{x^{4} \operatorname {W}({a x})} \,d x } \] Input:
integrate(1/x^4/lambert_w(a*x),x, algorithm="maxima")
Output:
integrate(1/(x^4*lambert_w(a*x)), x)
\[ \int \frac {1}{x^4 W(a x)} \, dx=\int { \frac {1}{x^{4} \operatorname {W}({a x})} \,d x } \] Input:
integrate(1/x^4/lambert_w(a*x),x, algorithm="giac")
Output:
integrate(1/(x^4*lambert_w(a*x)), x)
Timed out. \[ \int \frac {1}{x^4 W(a x)} \, dx=\int \frac {1}{x^4\,\mathrm {LambertW}\left (a\,x\right )} \,d x \] Input:
int(1/(x^4*LambertW(a*x)),x)
Output:
int(1/(x^4*LambertW(a*x)), x)
\[ \int \frac {1}{x^4 W(a x)} \, dx=\frac {-3 \left (\int \frac {1}{x^{4}}d x \right ) x^{3}+6 \left (\int \frac {1}{\textit {lambert\_w} \left (a x \right ) x^{4}}d x \right ) x^{3}-1}{6 x^{3}} \] Input:
int(1/x^4/Lambert_W(a*x),x)
Output:
( - 3*int(1/x**4,x)*x**3 + 6*int(1/(lambert_w(a*x)*x**4),x)*x**3 - 1)/(6*x **3)