Integrand size = 26, antiderivative size = 71 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=-\frac {e r \left (a+b \log \left (c x^n\right )\right )^{2+p}}{b^2 n^2 (1+p) (2+p)}+\frac {\left (a+b \log \left (c x^n\right )\right )^{1+p} \left (d+e \log \left (f x^r\right )\right )}{b n (1+p)} \]
-e*r*(a+b*ln(c*x^n))^(2+p)/b^2/n^2/(p+1)/(2+p)+(a+b*ln(c*x^n))^(p+1)*(d+e* ln(f*x^r))/b/n/(p+1)
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=\frac {\left (a+b \log \left (c x^n\right )\right )^{1+p} \left (2 b d n+b d n p-a e r-b e r \log \left (c x^n\right )+b e n (2+p) \log \left (f x^r\right )\right )}{b^2 n^2 (1+p) (2+p)} \]
((a + b*Log[c*x^n])^(1 + p)*(2*b*d*n + b*d*n*p - a*e*r - b*e*r*Log[c*x^n] + b*e*n*(2 + p)*Log[f*x^r]))/(b^2*n^2*(1 + p)*(2 + p))
Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2813, 27, 2739, 15}
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.
\(\displaystyle \int \frac {\left (d+e \log \left (f x^r\right )\right ) \left (a+b \log \left (c x^n\right )\right )^p}{x} \, dx\) |
\(\Big \downarrow \) 2813 |
\(\displaystyle \frac {\left (d+e \log \left (f x^r\right )\right ) \left (a+b \log \left (c x^n\right )\right )^{p+1}}{b n (p+1)}-e r \int \frac {\left (a+b \log \left (c x^n\right )\right )^{p+1}}{b n (p+1) x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (d+e \log \left (f x^r\right )\right ) \left (a+b \log \left (c x^n\right )\right )^{p+1}}{b n (p+1)}-\frac {e r \int \frac {\left (a+b \log \left (c x^n\right )\right )^{p+1}}{x}dx}{b n (p+1)}\) |
\(\Big \downarrow \) 2739 |
\(\displaystyle \frac {\left (d+e \log \left (f x^r\right )\right ) \left (a+b \log \left (c x^n\right )\right )^{p+1}}{b n (p+1)}-\frac {e r \int \left (a+b \log \left (c x^n\right )\right )^{p+1}d\left (a+b \log \left (c x^n\right )\right )}{b^2 n^2 (p+1)}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\left (d+e \log \left (f x^r\right )\right ) \left (a+b \log \left (c x^n\right )\right )^{p+1}}{b n (p+1)}-\frac {e r \left (a+b \log \left (c x^n\right )\right )^{p+2}}{b^2 n^2 (p+1) (p+2)}\) |
-((e*r*(a + b*Log[c*x^n])^(2 + p))/(b^2*n^2*(1 + p)*(2 + p))) + ((a + b*Lo g[c*x^n])^(1 + p)*(d + e*Log[f*x^r]))/(b*n*(1 + p))
3.2.82.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[1/( b*n) Subst[Int[x^p, x], x, a + b*Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p} , x]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ .)]*(e_.))*((g_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Simp[(d + e*Log[f*x^r]) u, x] - Simp[e*r Int[Simp lifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] && !(EqQ[p, 1] && EqQ[a, 0] && NeQ[d, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(71)=142\).
Time = 24.20 (sec) , antiderivative size = 323, normalized size of antiderivative = 4.55
method | result | size |
parallelrisch | \(\frac {2 \ln \left (c \,x^{n}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} b^{5} d \,n^{3}+2 {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a \,b^{4} d \,n^{3}+2 \ln \left (f \,x^{r}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a \,b^{4} e \,n^{3}+{\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a \,b^{4} d \,n^{3} p +\ln \left (f \,x^{r}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a \,b^{4} e \,n^{3} p -2 \ln \left (c \,x^{n}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a \,b^{4} e \,n^{2} r +\ln \left (c \,x^{n}\right ) \ln \left (f \,x^{r}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} b^{5} e \,n^{3} p -{\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} a^{2} b^{3} e \,n^{2} r -\ln \left (c \,x^{n}\right )^{2} {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} b^{5} e \,n^{2} r +2 \ln \left (c \,x^{n}\right ) \ln \left (f \,x^{r}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} b^{5} e \,n^{3}+\ln \left (c \,x^{n}\right ) {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p} b^{5} d \,n^{3} p}{\left (p^{2}+3 p +2\right ) b^{5} n^{4}}\) | \(323\) |
risch | \(\text {Expression too large to display}\) | \(854\) |
(2*ln(c*x^n)*(a+b*ln(c*x^n))^p*b^5*d*n^3+2*(a+b*ln(c*x^n))^p*a*b^4*d*n^3+2 *ln(f*x^r)*(a+b*ln(c*x^n))^p*a*b^4*e*n^3+(a+b*ln(c*x^n))^p*a*b^4*d*n^3*p+l n(f*x^r)*(a+b*ln(c*x^n))^p*a*b^4*e*n^3*p-2*ln(c*x^n)*(a+b*ln(c*x^n))^p*a*b ^4*e*n^2*r+ln(c*x^n)*ln(f*x^r)*(a+b*ln(c*x^n))^p*b^5*e*n^3*p-(a+b*ln(c*x^n ))^p*a^2*b^3*e*n^2*r-ln(c*x^n)^2*(a+b*ln(c*x^n))^p*b^5*e*n^2*r+2*ln(c*x^n) *ln(f*x^r)*(a+b*ln(c*x^n))^p*b^5*e*n^3+ln(c*x^n)*(a+b*ln(c*x^n))^p*b^5*d*n ^3*p)/(p^2+3*p+2)/b^5/n^4
Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.13 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=-\frac {{\left (b^{2} e r \log \left (c\right )^{2} - a b d n p - 2 \, a b d n + a^{2} e r - {\left (b^{2} e n^{2} p + b^{2} e n^{2}\right )} r \log \left (x\right )^{2} - {\left (b^{2} d n p + 2 \, b^{2} d n - 2 \, a b e r\right )} \log \left (c\right ) - {\left (a b e n p + 2 \, a b e n + {\left (b^{2} e n p + 2 \, b^{2} e n\right )} \log \left (c\right )\right )} \log \left (f\right ) - {\left (b^{2} e n p r \log \left (c\right ) + b^{2} d n^{2} p + a b e n p r + 2 \, b^{2} d n^{2} + {\left (b^{2} e n^{2} p + 2 \, b^{2} e n^{2}\right )} \log \left (f\right )\right )} \log \left (x\right )\right )} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p}}{b^{2} n^{2} p^{2} + 3 \, b^{2} n^{2} p + 2 \, b^{2} n^{2}} \]
-(b^2*e*r*log(c)^2 - a*b*d*n*p - 2*a*b*d*n + a^2*e*r - (b^2*e*n^2*p + b^2* e*n^2)*r*log(x)^2 - (b^2*d*n*p + 2*b^2*d*n - 2*a*b*e*r)*log(c) - (a*b*e*n* p + 2*a*b*e*n + (b^2*e*n*p + 2*b^2*e*n)*log(c))*log(f) - (b^2*e*n*p*r*log( c) + b^2*d*n^2*p + a*b*e*n*p*r + 2*b^2*d*n^2 + (b^2*e*n^2*p + 2*b^2*e*n^2) *log(f))*log(x))*(b*n*log(x) + b*log(c) + a)^p/(b^2*n^2*p^2 + 3*b^2*n^2*p + 2*b^2*n^2)
\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{p} \left (d + e \log {\left (f x^{r} \right )}\right )}{x}\, dx \]
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=\frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{p + 1} e \log \left (f x^{r}\right )}{b n {\left (p + 1\right )}} + \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{p + 1} d}{b n {\left (p + 1\right )}} - \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{p + 2} e r}{b^{2} n^{2} {\left (p + 2\right )} {\left (p + 1\right )}} \]
(b*log(c*x^n) + a)^(p + 1)*e*log(f*x^r)/(b*n*(p + 1)) + (b*log(c*x^n) + a) ^(p + 1)*d/(b*n*(p + 1)) - (b*log(c*x^n) + a)^(p + 2)*e*r/(b^2*n^2*(p + 2) *(p + 1))
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (71) = 142\).
Time = 0.30 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.44 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=\frac {\frac {{\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p + 1} e \log \left (f\right )}{p + 1} + \frac {{\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p + 1} d}{p + 1} - \frac {{\left ({\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} b p \log \left (c\right ) - {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{2} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} p + {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} a p + 2 \, {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} b \log \left (c\right ) - {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{2} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} + 2 \, {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )} {\left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}^{p} a\right )} e r}{{\left (p^{2} + 3 \, p + 2\right )} b n}}{b n} \]
((b*n*log(x) + b*log(c) + a)^(p + 1)*e*log(f)/(p + 1) + (b*n*log(x) + b*lo g(c) + a)^(p + 1)*d/(p + 1) - ((b*n*log(x) + b*log(c) + a)*(b*n*log(x) + b *log(c) + a)^p*b*p*log(c) - (b*n*log(x) + b*log(c) + a)^2*(b*n*log(x) + b* log(c) + a)^p*p + (b*n*log(x) + b*log(c) + a)*(b*n*log(x) + b*log(c) + a)^ p*a*p + 2*(b*n*log(x) + b*log(c) + a)*(b*n*log(x) + b*log(c) + a)^p*b*log( c) - (b*n*log(x) + b*log(c) + a)^2*(b*n*log(x) + b*log(c) + a)^p + 2*(b*n* log(x) + b*log(c) + a)*(b*n*log(x) + b*log(c) + a)^p*a)*e*r/((p^2 + 3*p + 2)*b*n))/(b*n)
Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^p \left (d+e \log \left (f x^r\right )\right )}{x} \, dx=\int \frac {\left (d+e\,\ln \left (f\,x^r\right )\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^p}{x} \,d x \]