Integrand size = 28, antiderivative size = 117 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}-\frac {k \log \left (1+\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}-2 k \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )+4 b k n \operatorname {PolyLog}\left (3,-\frac {f \sqrt {x}}{e}\right ) \]
1/2*(a+b*ln(c*x^n))^2*ln(d*(e+f*x^(1/2))^k)/b/n-1/2*k*(a+b*ln(c*x^n))^2*ln (1+f*x^(1/2)/e)/b/n-2*k*(a+b*ln(c*x^n))*polylog(2,-f*x^(1/2)/e)+4*b*k*n*po lylog(3,-f*x^(1/2)/e)
Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.59 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {1}{2} \left (4 a \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )-b n \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log ^2(x)+b k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log ^2(x)+2 b \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log (x) \log \left (c x^n\right )-2 b k \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x) \log \left (c x^n\right )+4 a k \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )-4 b k \log \left (c x^n\right ) \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )+8 b k n \operatorname {PolyLog}\left (3,-\frac {f \sqrt {x}}{e}\right )\right ) \]
(4*a*Log[d*(e + f*Sqrt[x])^k]*Log[-((f*Sqrt[x])/e)] - b*n*Log[d*(e + f*Sqr t[x])^k]*Log[x]^2 + b*k*n*Log[1 + (f*Sqrt[x])/e]*Log[x]^2 + 2*b*Log[d*(e + f*Sqrt[x])^k]*Log[x]*Log[c*x^n] - 2*b*k*Log[1 + (f*Sqrt[x])/e]*Log[x]*Log [c*x^n] + 4*a*k*PolyLog[2, 1 + (f*Sqrt[x])/e] - 4*b*k*Log[c*x^n]*PolyLog[2 , -((f*Sqrt[x])/e)] + 8*b*k*n*PolyLog[3, -((f*Sqrt[x])/e)])/2
Time = 0.49 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2822, 2775, 2821, 7143}
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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x} \, dx\) |
\(\Big \downarrow \) 2822 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{2 b n}-\frac {f k \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{\left (e+f \sqrt {x}\right ) \sqrt {x}}dx}{4 b n}\) |
\(\Big \downarrow \) 2775 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{2 b n}-\frac {f k \left (\frac {2 \log \left (\frac {f \sqrt {x}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{f}-\frac {4 b n \int \frac {\log \left (\frac {\sqrt {x} f}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{x}dx}{f}\right )}{4 b n}\) |
\(\Big \downarrow \) 2821 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{2 b n}-\frac {f k \left (\frac {2 \log \left (\frac {f \sqrt {x}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{f}-\frac {4 b n \left (2 b n \int \frac {\operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{x}dx-2 \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )\right )}{f}\right )}{4 b n}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{2 b n}-\frac {f k \left (\frac {2 \log \left (\frac {f \sqrt {x}}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{f}-\frac {4 b n \left (4 b n \operatorname {PolyLog}\left (3,-\frac {f \sqrt {x}}{e}\right )-2 \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right ) \left (a+b \log \left (c x^n\right )\right )\right )}{f}\right )}{4 b n}\) |
(Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n])^2)/(2*b*n) - (f*k*((2*Log[1 + (f*Sqrt[x])/e]*(a + b*Log[c*x^n])^2)/f - (4*b*n*(-2*(a + b*Log[c*x^n])*Po lyLog[2, -((f*Sqrt[x])/e)] + 4*b*n*PolyLog[3, -((f*Sqrt[x])/e)]))/f))/(4*b *n)
3.2.18.3.1 Defintions of rubi rules used
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Simp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c* x^n])^p/(e*r)), x] - Simp[b*f^m*n*(p/(e*r)) Int[Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] & & EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b _.))^(p_.))/(x_), x_Symbol] :> Simp[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c *x^n])^p/m), x] + Simp[b*n*(p/m) Int[PolyLog[2, (-d)*f*x^m]*((a + b*Log[c *x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]
Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_ .)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp[Log[d*(e + f*x^m)^r]*((a + b*Log[ c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[f*m*(r/(b*n*(p + 1))) Int[x^(m - 1)*((a + b*Log[c*x^n])^(p + 1)/(e + f*x^m)), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && NeQ[d*e, 1]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )}{x}d x\]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x} \,d x } \]
-1/2*(b*e*n*log(d)*log(x)^2 - 2*b*e*log(d)*log(x)*log(x^n) + (b*e*n*log(x) ^2 - 2*b*e*log(x)*log(x^n) - 2*(b*e*log(c) + a*e)*log(x))*log((f*sqrt(x) + e)^k) - 2*(b*e*log(c)*log(d) + a*e*log(d))*log(x) - (b*f*k*n*x*log(x)^2 - 2*(b*f*k*log(c) + a*f*k)*x*log(x) + 4*(a*f*k - (2*f*k*n - f*k*log(c))*b)* x - 2*(b*f*k*x*log(x) - 2*b*f*k*x)*log(x^n))/sqrt(x))/e + integrate(-1/4*( b*f^2*k*n*log(x)^2 - 2*b*f^2*k*log(x)*log(x^n) - 2*(b*f^2*k*log(c) + a*f^2 *k)*log(x))/(e*f*sqrt(x) + e^2), x)
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]