Integrand size = 28, antiderivative size = 434 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {11 b f k n}{225 e x^{5/2}}+\frac {5 b f^2 k n}{72 e^2 x^2}-\frac {b f^3 k n}{9 e^3 x^{3/2}}+\frac {2 b f^4 k n}{9 e^4 x}-\frac {7 b f^5 k n}{9 e^5 \sqrt {x}}+\frac {b f^6 k n \log \left (e+f \sqrt {x}\right )}{9 e^6}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{9 x^3}-\frac {2 b f^6 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{3 e^6}-\frac {b f^6 k n \log (x)}{18 e^6}+\frac {b f^6 k n \log ^2(x)}{12 e^6}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{15 e x^{5/2}}+\frac {f^2 k \left (a+b \log \left (c x^n\right )\right )}{12 e^2 x^2}-\frac {f^3 k \left (a+b \log \left (c x^n\right )\right )}{9 e^3 x^{3/2}}+\frac {f^4 k \left (a+b \log \left (c x^n\right )\right )}{6 e^4 x}-\frac {f^5 k \left (a+b \log \left (c x^n\right )\right )}{3 e^5 \sqrt {x}}+\frac {f^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^6}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {f^6 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{6 e^6}-\frac {2 b f^6 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{3 e^6} \]
-11/225*b*f*k*n/e/x^(5/2)+5/72*b*f^2*k*n/e^2/x^2-1/9*b*f^3*k*n/e^3/x^(3/2) +2/9*b*f^4*k*n/e^4/x-1/18*b*f^6*k*n*ln(x)/e^6+1/12*b*f^6*k*n*ln(x)^2/e^6-1 /15*f*k*(a+b*ln(c*x^n))/e/x^(5/2)+1/12*f^2*k*(a+b*ln(c*x^n))/e^2/x^2-1/9*f ^3*k*(a+b*ln(c*x^n))/e^3/x^(3/2)+1/6*f^4*k*(a+b*ln(c*x^n))/e^4/x-1/6*f^6*k *ln(x)*(a+b*ln(c*x^n))/e^6+1/9*b*f^6*k*n*ln(e+f*x^(1/2))/e^6+1/3*f^6*k*(a+ b*ln(c*x^n))*ln(e+f*x^(1/2))/e^6-2/3*b*f^6*k*n*ln(-f*x^(1/2)/e)*ln(e+f*x^( 1/2))/e^6-1/9*b*n*ln(d*(e+f*x^(1/2))^k)/x^3-1/3*(a+b*ln(c*x^n))*ln(d*(e+f* x^(1/2))^k)/x^3-2/3*b*f^6*k*n*polylog(2,1+f*x^(1/2)/e)/e^6-7/9*b*f^5*k*n/e ^5/x^(1/2)-1/3*f^5*k*(a+b*ln(c*x^n))/e^5/x^(1/2)
Time = 0.36 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.05 \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {120 a e^5 f k \sqrt {x}+88 b e^5 f k n \sqrt {x}-150 a e^4 f^2 k x-125 b e^4 f^2 k n x+200 a e^3 f^3 k x^{3/2}+200 b e^3 f^3 k n x^{3/2}-300 a e^2 f^4 k x^2-400 b e^2 f^4 k n x^2+600 a e f^5 k x^{5/2}+1400 b e f^5 k n x^{5/2}+600 a e^6 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+200 b e^6 n \log \left (d \left (e+f \sqrt {x}\right )^k\right )+300 a f^6 k x^3 \log (x)+100 b f^6 k n x^3 \log (x)-600 b f^6 k n x^3 \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-150 b f^6 k n x^3 \log ^2(x)+120 b e^5 f k \sqrt {x} \log \left (c x^n\right )-150 b e^4 f^2 k x \log \left (c x^n\right )+200 b e^3 f^3 k x^{3/2} \log \left (c x^n\right )-300 b e^2 f^4 k x^2 \log \left (c x^n\right )+600 b e f^5 k x^{5/2} \log \left (c x^n\right )+600 b e^6 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+300 b f^6 k x^3 \log (x) \log \left (c x^n\right )-200 f^6 k x^3 \log \left (e+f \sqrt {x}\right ) \left (3 a+b n-3 b n \log (x)+3 b \log \left (c x^n\right )\right )-1200 b f^6 k n x^3 \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{1800 e^6 x^3} \]
-1/1800*(120*a*e^5*f*k*Sqrt[x] + 88*b*e^5*f*k*n*Sqrt[x] - 150*a*e^4*f^2*k* x - 125*b*e^4*f^2*k*n*x + 200*a*e^3*f^3*k*x^(3/2) + 200*b*e^3*f^3*k*n*x^(3 /2) - 300*a*e^2*f^4*k*x^2 - 400*b*e^2*f^4*k*n*x^2 + 600*a*e*f^5*k*x^(5/2) + 1400*b*e*f^5*k*n*x^(5/2) + 600*a*e^6*Log[d*(e + f*Sqrt[x])^k] + 200*b*e^ 6*n*Log[d*(e + f*Sqrt[x])^k] + 300*a*f^6*k*x^3*Log[x] + 100*b*f^6*k*n*x^3* Log[x] - 600*b*f^6*k*n*x^3*Log[1 + (f*Sqrt[x])/e]*Log[x] - 150*b*f^6*k*n*x ^3*Log[x]^2 + 120*b*e^5*f*k*Sqrt[x]*Log[c*x^n] - 150*b*e^4*f^2*k*x*Log[c*x ^n] + 200*b*e^3*f^3*k*x^(3/2)*Log[c*x^n] - 300*b*e^2*f^4*k*x^2*Log[c*x^n] + 600*b*e*f^5*k*x^(5/2)*Log[c*x^n] + 600*b*e^6*Log[d*(e + f*Sqrt[x])^k]*Lo g[c*x^n] + 300*b*f^6*k*x^3*Log[x]*Log[c*x^n] - 200*f^6*k*x^3*Log[e + f*Sqr t[x]]*(3*a + b*n - 3*b*n*Log[x] + 3*b*Log[c*x^n]) - 1200*b*f^6*k*n*x^3*Pol yLog[2, -((f*Sqrt[x])/e)])/(e^6*x^3)
Time = 0.66 (sec) , antiderivative size = 417, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x^4} \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {k \log \left (e+f \sqrt {x}\right ) f^6}{3 e^6 x}-\frac {k \log (x) f^6}{6 e^6 x}-\frac {k f^5}{3 e^5 x^{3/2}}+\frac {k f^4}{6 e^4 x^2}-\frac {k f^3}{9 e^3 x^{5/2}}+\frac {k f^2}{12 e^2 x^3}-\frac {k f}{15 e x^{7/2}}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{3 x^4}\right )dx-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{3 x^3}+\frac {f^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^6}-\frac {f^6 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{6 e^6}-\frac {f^5 k \left (a+b \log \left (c x^n\right )\right )}{3 e^5 \sqrt {x}}+\frac {f^4 k \left (a+b \log \left (c x^n\right )\right )}{6 e^4 x}-\frac {f^3 k \left (a+b \log \left (c x^n\right )\right )}{9 e^3 x^{3/2}}+\frac {f^2 k \left (a+b \log \left (c x^n\right )\right )}{12 e^2 x^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{15 e x^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{3 x^3}+\frac {f^6 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^6}-\frac {f^6 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{6 e^6}-\frac {f^5 k \left (a+b \log \left (c x^n\right )\right )}{3 e^5 \sqrt {x}}+\frac {f^4 k \left (a+b \log \left (c x^n\right )\right )}{6 e^4 x}-\frac {f^3 k \left (a+b \log \left (c x^n\right )\right )}{9 e^3 x^{3/2}}+\frac {f^2 k \left (a+b \log \left (c x^n\right )\right )}{12 e^2 x^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{15 e x^{5/2}}-b n \left (\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{9 x^3}+\frac {2 f^6 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{3 e^6}-\frac {f^6 k \log ^2(x)}{12 e^6}-\frac {f^6 k \log \left (e+f \sqrt {x}\right )}{9 e^6}+\frac {2 f^6 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{3 e^6}+\frac {f^6 k \log (x)}{18 e^6}+\frac {7 f^5 k}{9 e^5 \sqrt {x}}-\frac {2 f^4 k}{9 e^4 x}+\frac {f^3 k}{9 e^3 x^{3/2}}-\frac {5 f^2 k}{72 e^2 x^2}+\frac {11 f k}{225 e x^{5/2}}\right )\) |
-1/15*(f*k*(a + b*Log[c*x^n]))/(e*x^(5/2)) + (f^2*k*(a + b*Log[c*x^n]))/(1 2*e^2*x^2) - (f^3*k*(a + b*Log[c*x^n]))/(9*e^3*x^(3/2)) + (f^4*k*(a + b*Lo g[c*x^n]))/(6*e^4*x) - (f^5*k*(a + b*Log[c*x^n]))/(3*e^5*Sqrt[x]) + (f^6*k *Log[e + f*Sqrt[x]]*(a + b*Log[c*x^n]))/(3*e^6) - (Log[d*(e + f*Sqrt[x])^k ]*(a + b*Log[c*x^n]))/(3*x^3) - (f^6*k*Log[x]*(a + b*Log[c*x^n]))/(6*e^6) - b*n*((11*f*k)/(225*e*x^(5/2)) - (5*f^2*k)/(72*e^2*x^2) + (f^3*k)/(9*e^3* x^(3/2)) - (2*f^4*k)/(9*e^4*x) + (7*f^5*k)/(9*e^5*Sqrt[x]) - (f^6*k*Log[e + f*Sqrt[x]])/(9*e^6) + Log[d*(e + f*Sqrt[x])^k]/(9*x^3) + (2*f^6*k*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(3*e^6) + (f^6*k*Log[x])/(18*e^6) - (f ^6*k*Log[x]^2)/(12*e^6) + (2*f^6*k*PolyLog[2, 1 + (f*Sqrt[x])/e])/(3*e^6))
3.2.21.3.1 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]
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )}{x^{4}}d x\]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\text {Timed out} \]
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{4}} \,d x } \]
-1/225*(75*b*e*log(d)*log(x^n) + 75*a*e*log(d) + 25*(e*n*log(d) + 3*e*log( c)*log(d))*b + 25*(3*b*e*log(x^n) + (e*n + 3*e*log(c))*b + 3*a*e)*log((f*s qrt(x) + e)^k) + (15*b*f*k*x*log(x^n) + (15*a*f*k + (11*f*k*n + 15*f*k*log (c))*b)*x)/sqrt(x))/(e*x^3) - integrate(1/18*(3*b*f^2*k*log(x^n) + 3*a*f^2 *k + (f^2*k*n + 3*f^2*k*log(c))*b)/(e*f*x^(7/2) + e^2*x^3), x)
\[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\int \frac {\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \]