Integrand size = 20, antiderivative size = 169 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=\frac {e^{-\frac {a}{b m n}} (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right )}{2 b^3 f m^3 n^3}-\frac {e+f x}{2 b f m n \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2}-\frac {e+f x}{2 b^2 f m^2 n^2 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )} \]
1/2*(f*x+e)*Ei((a+b*ln(c*(d*(f*x+e)^m)^n))/b/m/n)/b^3/exp(a/b/m/n)/f/m^3/n ^3/((c*(d*(f*x+e)^m)^n)^(1/m/n))+1/2*(-f*x-e)/b/f/m/n/(a+b*ln(c*(d*(f*x+e) ^m)^n))^2+1/2*(-f*x-e)/b^2/f/m^2/n^2/(a+b*ln(c*(d*(f*x+e)^m)^n))
Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=-\frac {e^{-\frac {a}{b m n}} (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \left (-\operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right ) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2+b e^{\frac {a}{b m n}} m n \left (c \left (d (e+f x)^m\right )^n\right )^{\frac {1}{m n}} \left (a+b m n+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )\right )}{2 b^3 f m^3 n^3 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^2} \]
-1/2*((e + f*x)*(-(ExpIntegralEi[(a + b*Log[c*(d*(e + f*x)^m)^n])/(b*m*n)] *(a + b*Log[c*(d*(e + f*x)^m)^n])^2) + b*E^(a/(b*m*n))*m*n*(c*(d*(e + f*x) ^m)^n)^(1/(m*n))*(a + b*m*n + b*Log[c*(d*(e + f*x)^m)^n])))/(b^3*E^(a/(b*m *n))*f*m^3*n^3*(c*(d*(e + f*x)^m)^n)^(1/(m*n))*(a + b*Log[c*(d*(e + f*x)^m )^n])^2)
Time = 0.61 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2895, 2836, 2734, 2734, 2737, 2609}
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.
| \(\displaystyle \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx\) |
| \(\Big \downarrow \) 2895 |
| \(\displaystyle \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3}dx\) |
| \(\Big \downarrow \) 2836 |
| \(\displaystyle \frac {\int \frac {1}{\left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^3}d(e+f x)}{f}\) |
| \(\Big \downarrow \) 2734 |
| \(\displaystyle \frac {\frac {\int \frac {1}{\left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2}d(e+f x)}{2 b m n}-\frac {e+f x}{2 b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2}}{f}\) |
| \(\Big \downarrow \) 2734 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {1}{a+b \log \left (c d^n (e+f x)^{m n}\right )}d(e+f x)}{b m n}-\frac {e+f x}{b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )}}{2 b m n}-\frac {e+f x}{2 b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2}}{f}\) |
| \(\Big \downarrow \) 2737 |
| \(\displaystyle \frac {\frac {\frac {(e+f x) \left (c d^n (e+f x)^{m n}\right )^{-\frac {1}{m n}} \int \frac {\left (c d^n (e+f x)^{m n}\right )^{\frac {1}{m n}}}{a+b \log \left (c d^n (e+f x)^{m n}\right )}d\log \left (c d^n (e+f x)^{m n}\right )}{b m^2 n^2}-\frac {e+f x}{b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )}}{2 b m n}-\frac {e+f x}{2 b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2}}{f}\) |
| \(\Big \downarrow \) 2609 |
| \(\displaystyle \frac {\frac {\frac {(e+f x) e^{-\frac {a}{b m n}} \left (c d^n (e+f x)^{m n}\right )^{-\frac {1}{m n}} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c d^n (e+f x)^{m n}\right )}{b m n}\right )}{b^2 m^2 n^2}-\frac {e+f x}{b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )}}{2 b m n}-\frac {e+f x}{2 b m n \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^2}}{f}\) |
(-1/2*(e + f*x)/(b*m*n*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^2) + (((e + f*x) *ExpIntegralEi[(a + b*Log[c*d^n*(e + f*x)^(m*n)])/(b*m*n)])/(b^2*E^(a/(b*m *n))*m^2*n^2*(c*d^n*(e + f*x)^(m*n))^(1/(m*n))) - (e + f*x)/(b*m*n*(a + b* Log[c*d^n*(e + f*x)^(m*n)])))/(2*b*m*n))/f
3.5.10.3.1 Defintions of rubi rules used
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b *Log[c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[1/(b*n*(p + 1)) Int[(a + b *Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] && Int egerQ[2*p]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x/(n*(c*x ^n)^(1/n)) Subst[Int[E^(x/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ [{a, b, c, n, p}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[1/e Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ a, b, c, d, e, n, p}, x]
Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_. )*(u_.), x_Symbol] :> Subst[Int[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[n] && !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[ IntHide[u*(a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x]]
\[\int \frac {1}{{\left (a +b \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )\right )}^{3}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (163) = 326\).
Time = 0.31 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.63 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=-\frac {{\left ({\left (b^{2} e m^{2} n^{2} + a b e m n + {\left (b^{2} f m^{2} n^{2} + a b f m n\right )} x + {\left (b^{2} f m^{2} n^{2} x + b^{2} e m^{2} n^{2}\right )} \log \left (f x + e\right ) + {\left (b^{2} f m n x + b^{2} e m n\right )} \log \left (c\right ) + {\left (b^{2} f m n^{2} x + b^{2} e m n^{2}\right )} \log \left (d\right )\right )} e^{\left (\frac {b n \log \left (d\right ) + b \log \left (c\right ) + a}{b m n}\right )} - {\left (b^{2} m^{2} n^{2} \log \left (f x + e\right )^{2} + b^{2} n^{2} \log \left (d\right )^{2} + b^{2} \log \left (c\right )^{2} + 2 \, a b \log \left (c\right ) + a^{2} + 2 \, {\left (b^{2} m n^{2} \log \left (d\right ) + b^{2} m n \log \left (c\right ) + a b m n\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{2} n \log \left (c\right ) + a b n\right )} \log \left (d\right )\right )} \operatorname {log\_integral}\left ({\left (f x + e\right )} e^{\left (\frac {b n \log \left (d\right ) + b \log \left (c\right ) + a}{b m n}\right )}\right )\right )} e^{\left (-\frac {b n \log \left (d\right ) + b \log \left (c\right ) + a}{b m n}\right )}}{2 \, {\left (b^{5} f m^{5} n^{5} \log \left (f x + e\right )^{2} + b^{5} f m^{3} n^{5} \log \left (d\right )^{2} + b^{5} f m^{3} n^{3} \log \left (c\right )^{2} + 2 \, a b^{4} f m^{3} n^{3} \log \left (c\right ) + a^{2} b^{3} f m^{3} n^{3} + 2 \, {\left (b^{5} f m^{4} n^{5} \log \left (d\right ) + b^{5} f m^{4} n^{4} \log \left (c\right ) + a b^{4} f m^{4} n^{4}\right )} \log \left (f x + e\right ) + 2 \, {\left (b^{5} f m^{3} n^{4} \log \left (c\right ) + a b^{4} f m^{3} n^{4}\right )} \log \left (d\right )\right )}} \]
-1/2*((b^2*e*m^2*n^2 + a*b*e*m*n + (b^2*f*m^2*n^2 + a*b*f*m*n)*x + (b^2*f* m^2*n^2*x + b^2*e*m^2*n^2)*log(f*x + e) + (b^2*f*m*n*x + b^2*e*m*n)*log(c) + (b^2*f*m*n^2*x + b^2*e*m*n^2)*log(d))*e^((b*n*log(d) + b*log(c) + a)/(b *m*n)) - (b^2*m^2*n^2*log(f*x + e)^2 + b^2*n^2*log(d)^2 + b^2*log(c)^2 + 2 *a*b*log(c) + a^2 + 2*(b^2*m*n^2*log(d) + b^2*m*n*log(c) + a*b*m*n)*log(f* x + e) + 2*(b^2*n*log(c) + a*b*n)*log(d))*log_integral((f*x + e)*e^((b*n*l og(d) + b*log(c) + a)/(b*m*n))))*e^(-(b*n*log(d) + b*log(c) + a)/(b*m*n))/ (b^5*f*m^5*n^5*log(f*x + e)^2 + b^5*f*m^3*n^5*log(d)^2 + b^5*f*m^3*n^3*log (c)^2 + 2*a*b^4*f*m^3*n^3*log(c) + a^2*b^3*f*m^3*n^3 + 2*(b^5*f*m^4*n^5*lo g(d) + b^5*f*m^4*n^4*log(c) + a*b^4*f*m^4*n^4)*log(f*x + e) + 2*(b^5*f*m^3 *n^4*log(c) + a*b^4*f*m^3*n^4)*log(d))
\[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=\int \frac {1}{\left (a + b \log {\left (c \left (d \left (e + f x\right )^{m}\right )^{n} \right )}\right )^{3}}\, dx \]
\[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=\int { \frac {1}{{\left (b \log \left (\left ({\left (f x + e\right )}^{m} d\right )^{n} c\right ) + a\right )}^{3}} \,d x } \]
-1/2*((e*m*n + e*n*log(d) + e*log(c))*b + a*e + ((f*m*n + f*n*log(d) + f*l og(c))*b + a*f)*x + (b*f*x + b*e)*log(((f*x + e)^m)^n))/(b^4*f*m^2*n^2*log (((f*x + e)^m)^n)^2 + a^2*b^2*f*m^2*n^2 + 2*(f*m^2*n^3*log(d) + f*m^2*n^2* log(c))*a*b^3 + (f*m^2*n^4*log(d)^2 + 2*f*m^2*n^3*log(c)*log(d) + f*m^2*n^ 2*log(c)^2)*b^4 + 2*(a*b^3*f*m^2*n^2 + (f*m^2*n^3*log(d) + f*m^2*n^2*log(c ))*b^4)*log(((f*x + e)^m)^n)) + integrate(1/2/(b^3*m^2*n^2*log(((f*x + e)^ m)^n) + a*b^2*m^2*n^2 + (m^2*n^3*log(d) + m^2*n^2*log(c))*b^3), x)
Leaf count of result is larger than twice the leaf count of optimal. 3401 vs. \(2 (163) = 326\).
Time = 0.38 (sec) , antiderivative size = 3401, normalized size of antiderivative = 20.12 \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=\text {Too large to display} \]
-1/2*(f*x + e)*b^2*m^2*n^2*log(f*x + e)/(b^5*f*m^5*n^5*log(f*x + e)^2 + 2* b^5*f*m^4*n^5*log(f*x + e)*log(d) + 2*b^5*f*m^4*n^4*log(f*x + e)*log(c) + b^5*f*m^3*n^5*log(d)^2 + 2*a*b^4*f*m^4*n^4*log(f*x + e) + 2*b^5*f*m^3*n^4* log(c)*log(d) + b^5*f*m^3*n^3*log(c)^2 + 2*a*b^4*f*m^3*n^4*log(d) + 2*a*b^ 4*f*m^3*n^3*log(c) + a^2*b^3*f*m^3*n^3) + 1/2*b^2*m^2*n^2*Ei(log(d)/m + lo g(c)/(m*n) + a/(b*m*n) + log(f*x + e))*e^(-a/(b*m*n))*log(f*x + e)^2/((b^5 *f*m^5*n^5*log(f*x + e)^2 + 2*b^5*f*m^4*n^5*log(f*x + e)*log(d) + 2*b^5*f* m^4*n^4*log(f*x + e)*log(c) + b^5*f*m^3*n^5*log(d)^2 + 2*a*b^4*f*m^4*n^4*l og(f*x + e) + 2*b^5*f*m^3*n^4*log(c)*log(d) + b^5*f*m^3*n^3*log(c)^2 + 2*a *b^4*f*m^3*n^4*log(d) + 2*a*b^4*f*m^3*n^3*log(c) + a^2*b^3*f*m^3*n^3)*c^(1 /(m*n))*d^(1/m)) - 1/2*(f*x + e)*b^2*m^2*n^2/(b^5*f*m^5*n^5*log(f*x + e)^2 + 2*b^5*f*m^4*n^5*log(f*x + e)*log(d) + 2*b^5*f*m^4*n^4*log(f*x + e)*log( c) + b^5*f*m^3*n^5*log(d)^2 + 2*a*b^4*f*m^4*n^4*log(f*x + e) + 2*b^5*f*m^3 *n^4*log(c)*log(d) + b^5*f*m^3*n^3*log(c)^2 + 2*a*b^4*f*m^3*n^4*log(d) + 2 *a*b^4*f*m^3*n^3*log(c) + a^2*b^3*f*m^3*n^3) - 1/2*(f*x + e)*b^2*m*n^2*log (d)/(b^5*f*m^5*n^5*log(f*x + e)^2 + 2*b^5*f*m^4*n^5*log(f*x + e)*log(d) + 2*b^5*f*m^4*n^4*log(f*x + e)*log(c) + b^5*f*m^3*n^5*log(d)^2 + 2*a*b^4*f*m ^4*n^4*log(f*x + e) + 2*b^5*f*m^3*n^4*log(c)*log(d) + b^5*f*m^3*n^3*log(c) ^2 + 2*a*b^4*f*m^3*n^4*log(d) + 2*a*b^4*f*m^3*n^3*log(c) + a^2*b^3*f*m^3*n ^3) + b^2*m*n^2*Ei(log(d)/m + log(c)/(m*n) + a/(b*m*n) + log(f*x + e))*...
Timed out. \[ \int \frac {1}{\left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^3} \, dx=\int \frac {1}{{\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\right )}^3} \,d x \]