Integrand size = 28, antiderivative size = 248 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {3 b f k n}{e \sqrt {x}}+\frac {b f^2 k n \log \left (e+f \sqrt {x}\right )}{e^2}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}-\frac {b f^2 k n \log (x)}{2 e^2}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 b f^2 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{e^2} \]
-1/2*b*f^2*k*n*ln(x)/e^2+1/4*b*f^2*k*n*ln(x)^2/e^2-1/2*f^2*k*ln(x)*(a+b*ln (c*x^n))/e^2+b*f^2*k*n*ln(e+f*x^(1/2))/e^2+f^2*k*(a+b*ln(c*x^n))*ln(e+f*x^ (1/2))/e^2-2*b*f^2*k*n*ln(-f*x^(1/2)/e)*ln(e+f*x^(1/2))/e^2-b*n*ln(d*(e+f* x^(1/2))^k)/x-(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k)/x-2*b*f^2*k*n*polylog( 2,1+f*x^(1/2)/e)/e^2-3*b*f*k*n/e/x^(1/2)-f*k*(a+b*ln(c*x^n))/e/x^(1/2)
Time = 0.24 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {4 a e f k \sqrt {x}+12 b e f k n \sqrt {x}+4 a e^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+4 b e^2 n \log \left (d \left (e+f \sqrt {x}\right )^k\right )+2 a f^2 k x \log (x)+2 b f^2 k n x \log (x)-4 b f^2 k n x \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-b f^2 k n x \log ^2(x)+4 b e f k \sqrt {x} \log \left (c x^n\right )+4 b e^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+2 b f^2 k x \log (x) \log \left (c x^n\right )-4 f^2 k x \log \left (e+f \sqrt {x}\right ) \left (a+b n-b n \log (x)+b \log \left (c x^n\right )\right )-8 b f^2 k n x \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{4 e^2 x} \]
-1/4*(4*a*e*f*k*Sqrt[x] + 12*b*e*f*k*n*Sqrt[x] + 4*a*e^2*Log[d*(e + f*Sqrt [x])^k] + 4*b*e^2*n*Log[d*(e + f*Sqrt[x])^k] + 2*a*f^2*k*x*Log[x] + 2*b*f^ 2*k*n*x*Log[x] - 4*b*f^2*k*n*x*Log[1 + (f*Sqrt[x])/e]*Log[x] - b*f^2*k*n*x *Log[x]^2 + 4*b*e*f*k*Sqrt[x]*Log[c*x^n] + 4*b*e^2*Log[d*(e + f*Sqrt[x])^k ]*Log[c*x^n] + 2*b*f^2*k*x*Log[x]*Log[c*x^n] - 4*f^2*k*x*Log[e + f*Sqrt[x] ]*(a + b*n - b*n*Log[x] + b*Log[c*x^n]) - 8*b*f^2*k*n*x*PolyLog[2, -((f*Sq rt[x])/e)])/(e^2*x)
Time = 0.45 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2823, 2009}
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.
\(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2823 |
\(\displaystyle -b n \int \left (\frac {k \log \left (e+f \sqrt {x}\right ) f^2}{e^2 x}-\frac {k \log (x) f^2}{2 e^2 x}-\frac {k f}{e x^{3/2}}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x^2}\right )dx-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}-b n \left (\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}+\frac {2 f^2 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{e^2}-\frac {f^2 k \log ^2(x)}{4 e^2}-\frac {f^2 k \log \left (e+f \sqrt {x}\right )}{e^2}+\frac {2 f^2 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}+\frac {f^2 k \log (x)}{2 e^2}+\frac {3 f k}{e \sqrt {x}}\right )\) |
-((f*k*(a + b*Log[c*x^n]))/(e*Sqrt[x])) + (f^2*k*Log[e + f*Sqrt[x]]*(a + b *Log[c*x^n]))/e^2 - (Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/x - (f^2 *k*Log[x]*(a + b*Log[c*x^n]))/(2*e^2) - b*n*((3*f*k)/(e*Sqrt[x]) - (f^2*k* Log[e + f*Sqrt[x]])/e^2 + Log[d*(e + f*Sqrt[x])^k]/x + (2*f^2*k*Log[e + f* Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/e^2 + (f^2*k*Log[x])/(2*e^2) - (f^2*k*Log[ x]^2)/(4*e^2) + (2*f^2*k*PolyLog[2, 1 + (f*Sqrt[x])/e])/e^2)
3.2.19.3.1 Defintions of rubi rules used
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. )]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* (e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )}{x^{2}}d x\]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{2}} \,d x } \]
-(b*e*log(d)*log(x^n) + a*e*log(d) + (e*n*log(d) + e*log(c)*log(d))*b + (b *e*log(x^n) + (e*n + e*log(c))*b + a*e)*log((f*sqrt(x) + e)^k) + (b*f*k*x* log(x^n) + (a*f*k + (3*f*k*n + f*k*log(c))*b)*x)/sqrt(x))/(e*x) - integrat e(1/2*(b*f^2*k*log(x^n) + a*f^2*k + (f^2*k*n + f^2*k*log(c))*b)/(e*f*x^(3/ 2) + e^2*x), x)
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^2} \,d x \]