Integrand size = 30, antiderivative size = 367 \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {24 b e^4 k n \sqrt {x}}{25 f^4}-\frac {7 b e^3 k n x}{25 f^3}+\frac {32 b e^2 k n x^{3/2}}{225 f^2}-\frac {9 b e k n x^2}{100 f}+\frac {8}{125} b k n x^{5/2}-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right )}{25 f^5}-\frac {4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac {2}{5} x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b e^5 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{5 f^5} \]
-7/25*b*e^3*k*n*x/f^3+32/225*b*e^2*k*n*x^(3/2)/f^2-9/100*b*e*k*n*x^2/f+8/1 25*b*k*n*x^(5/2)+1/5*e^3*k*x*(a+b*ln(c*x^n))/f^3-2/15*e^2*k*x^(3/2)*(a+b*l n(c*x^n))/f^2+1/10*e*k*x^2*(a+b*ln(c*x^n))/f-2/25*k*x^(5/2)*(a+b*ln(c*x^n) )-4/25*b*e^5*k*n*ln(e+f*x^(1/2))/f^5+2/5*e^5*k*(a+b*ln(c*x^n))*ln(e+f*x^(1 /2))/f^5-4/5*b*e^5*k*n*ln(-f*x^(1/2)/e)*ln(e+f*x^(1/2))/f^5-4/25*b*n*x^(5/ 2)*ln(d*(e+f*x^(1/2))^k)+2/5*x^(5/2)*(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k) -4/5*b*e^5*k*n*polylog(2,1+f*x^(1/2)/e)/f^5+24/25*b*e^4*k*n*x^(1/2)/f^4-2/ 5*e^4*k*(a+b*ln(c*x^n))*x^(1/2)/f^4
Time = 0.29 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.07 \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-1800 a e^4 f k \sqrt {x}+4320 b e^4 f k n \sqrt {x}+900 a e^3 f^2 k x-1260 b e^3 f^2 k n x-600 a e^2 f^3 k x^{3/2}+640 b e^2 f^3 k n x^{3/2}+450 a e f^4 k x^2-405 b e f^4 k n x^2-360 a f^5 k x^{5/2}+288 b f^5 k n x^{5/2}+1800 a f^5 x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )-720 b f^5 n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )+1800 b e^5 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-1800 b e^4 f k \sqrt {x} \log \left (c x^n\right )+900 b e^3 f^2 k x \log \left (c x^n\right )-600 b e^2 f^3 k x^{3/2} \log \left (c x^n\right )+450 b e f^4 k x^2 \log \left (c x^n\right )-360 b f^5 k x^{5/2} \log \left (c x^n\right )+1800 b f^5 x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+360 e^5 k \log \left (e+f \sqrt {x}\right ) \left (5 a-2 b n-5 b n \log (x)+5 b \log \left (c x^n\right )\right )+3600 b e^5 k n \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{4500 f^5} \]
(-1800*a*e^4*f*k*Sqrt[x] + 4320*b*e^4*f*k*n*Sqrt[x] + 900*a*e^3*f^2*k*x - 1260*b*e^3*f^2*k*n*x - 600*a*e^2*f^3*k*x^(3/2) + 640*b*e^2*f^3*k*n*x^(3/2) + 450*a*e*f^4*k*x^2 - 405*b*e*f^4*k*n*x^2 - 360*a*f^5*k*x^(5/2) + 288*b*f ^5*k*n*x^(5/2) + 1800*a*f^5*x^(5/2)*Log[d*(e + f*Sqrt[x])^k] - 720*b*f^5*n *x^(5/2)*Log[d*(e + f*Sqrt[x])^k] + 1800*b*e^5*k*n*Log[1 + (f*Sqrt[x])/e]* Log[x] - 1800*b*e^4*f*k*Sqrt[x]*Log[c*x^n] + 900*b*e^3*f^2*k*x*Log[c*x^n] - 600*b*e^2*f^3*k*x^(3/2)*Log[c*x^n] + 450*b*e*f^4*k*x^2*Log[c*x^n] - 360* b*f^5*k*x^(5/2)*Log[c*x^n] + 1800*b*f^5*x^(5/2)*Log[d*(e + f*Sqrt[x])^k]*L og[c*x^n] + 360*e^5*k*Log[e + f*Sqrt[x]]*(5*a - 2*b*n - 5*b*n*Log[x] + 5*b *Log[c*x^n]) + 3600*b*e^5*k*n*PolyLog[2, -((f*Sqrt[x])/e)])/(4500*f^5)
Time = 0.55 (sec) , antiderivative size = 354, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
| \(\displaystyle \int x^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \, dx\) |
| \(\Big \downarrow \) 2823 |
| \(\displaystyle -b n \int \left (\frac {2 k \log \left (e+f \sqrt {x}\right ) e^5}{5 f^5 x}-\frac {2 k e^4}{5 f^4 \sqrt {x}}+\frac {k e^3}{5 f^3}-\frac {2 k \sqrt {x} e^2}{15 f^2}+\frac {k x e}{10 f}-\frac {2}{25} k x^{3/2}+\frac {2}{5} x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )\right )dx+\frac {2}{5} x^{5/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{5} x^{5/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {4}{25} x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {4 e^5 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{5 f^5}+\frac {4 e^5 k \log \left (e+f \sqrt {x}\right )}{25 f^5}+\frac {4 e^5 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 f^5}-\frac {24 e^4 k \sqrt {x}}{25 f^4}+\frac {7 e^3 k x}{25 f^3}-\frac {32 e^2 k x^{3/2}}{225 f^2}+\frac {9 e k x^2}{100 f}-\frac {8}{125} k x^{5/2}\right )\) |
(-2*e^4*k*Sqrt[x]*(a + b*Log[c*x^n]))/(5*f^4) + (e^3*k*x*(a + b*Log[c*x^n] ))/(5*f^3) - (2*e^2*k*x^(3/2)*(a + b*Log[c*x^n]))/(15*f^2) + (e*k*x^2*(a + b*Log[c*x^n]))/(10*f) - (2*k*x^(5/2)*(a + b*Log[c*x^n]))/25 + (2*e^5*k*Lo g[e + f*Sqrt[x]]*(a + b*Log[c*x^n]))/(5*f^5) + (2*x^(5/2)*Log[d*(e + f*Sqr t[x])^k]*(a + b*Log[c*x^n]))/5 - b*n*((-24*e^4*k*Sqrt[x])/(25*f^4) + (7*e^ 3*k*x)/(25*f^3) - (32*e^2*k*x^(3/2))/(225*f^2) + (9*e*k*x^2)/(100*f) - (8* k*x^(5/2))/125 + (4*e^5*k*Log[e + f*Sqrt[x]])/(25*f^5) + (4*x^(5/2)*Log[d* (e + f*Sqrt[x])^k])/25 + (4*e^5*k*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)] )/(5*f^5) + (4*e^5*k*PolyLog[2, 1 + (f*Sqrt[x])/e])/(5*f^5))
3.2.33.3.1 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]
\[\int x^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )d x\]
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
1/500*(50*b*e*k*x^2*log(x^n) + 40*(5*b*f*x*log(x^n) - ((2*f*n - 5*f*log(c) )*b - 5*a*f)*x)*x^(3/2)*log((f*sqrt(x) + e)^k) + 5*(10*a*e*k - (9*e*k*n - 10*e*k*log(c))*b)*x^2 + 40*(5*b*f*x*log(d)*log(x^n) + (5*a*f*log(d) - (2*f *n*log(d) - 5*f*log(c)*log(d))*b)*x)*x^(3/2) - 8*(5*b*f*k*x^2*log(x^n) + ( 5*a*f*k - (4*f*k*n - 5*f*k*log(c))*b)*x^2)*sqrt(x))/f - integrate(1/25*(5* b*e^2*k*x*log(x^n) + (5*a*e^2*k - (2*e^2*k*n - 5*e^2*k*log(c))*b)*x)/(f^2* sqrt(x) + e*f), x)
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \]
Timed out. \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{3/2}\,\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]