Integrand size = 30, antiderivative size = 367 \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {24 b e^4 k n \sqrt {x}}{25 f^4}-\frac {7 b e^3 k n x}{25 f^3}+\frac {32 b e^2 k n x^{3/2}}{225 f^2}-\frac {9 b e k n x^2}{100 f}+\frac {8}{125} b k n x^{5/2}-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right )}{25 f^5}-\frac {4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac {2}{5} x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b e^5 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{5 f^5} \] Output:
24/25*b*e^4*k*n*x^(1/2)/f^4-7/25*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+8/125*b*k*n*x^(5/2)-4/25*b*e^5*k*n*ln(e+f*x^(1/2)) /f^5-4/25*b*n*x^(5/2)*ln(d*(e+f*x^(1/2))^k)-4/5*b*e^5*k*n*ln(e+f*x^(1/2))* ln(-f*x^(1/2)/e)/f^5-2/5*e^4*k*x^(1/2)*(a+b*ln(c*x^n))/f^4+1/5*e^3*k*x*(a+ b*ln(c*x^n))/f^3-2/15*e^2*k*x^(3/2)*(a+b*ln(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))+2/5*e^5*k*ln(e+f*x^(1/2))*(a+b *ln(c*x^n))/f^5+2/5*x^(5/2)*ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n))-4/5*b*e^ 5*k*n*polylog(2,1+f*x^(1/2)/e)/f^5
Time = 0.52 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.07 \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-1800 a e^4 f k \sqrt {x}+4320 b e^4 f k n \sqrt {x}+900 a e^3 f^2 k x-1260 b e^3 f^2 k n x-600 a e^2 f^3 k x^{3/2}+640 b e^2 f^3 k n x^{3/2}+450 a e f^4 k x^2-405 b e f^4 k n x^2-360 a f^5 k x^{5/2}+288 b f^5 k n x^{5/2}+1800 a f^5 x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )-720 b f^5 n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )+1800 b e^5 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-1800 b e^4 f k \sqrt {x} \log \left (c x^n\right )+900 b e^3 f^2 k x \log \left (c x^n\right )-600 b e^2 f^3 k x^{3/2} \log \left (c x^n\right )+450 b e f^4 k x^2 \log \left (c x^n\right )-360 b f^5 k x^{5/2} \log \left (c x^n\right )+1800 b f^5 x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+360 e^5 k \log \left (e+f \sqrt {x}\right ) \left (5 a-2 b n-5 b n \log (x)+5 b \log \left (c x^n\right )\right )+3600 b e^5 k n \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{4500 f^5} \] Input:
Integrate[x^(3/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]
Output:
(-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 = 0.61 (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 |
| \(\displaystyle \frac {2}{5} x^{5/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {4}{25} x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {4 e^5 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{5 f^5}+\frac {4 e^5 k \log \left (e+f \sqrt {x}\right )}{25 f^5}+\frac {4 e^5 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 f^5}-\frac {24 e^4 k \sqrt {x}}{25 f^4}+\frac {7 e^3 k x}{25 f^3}-\frac {32 e^2 k x^{3/2}}{225 f^2}+\frac {9 e k x^2}{100 f}-\frac {8}{125} k x^{5/2}\right )\) |
Input:
Int[x^(3/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]
Output:
(-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))
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]
\[\int x^{\frac {3}{2}} \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right ) \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Input:
int(x^(3/2)*ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n)),x)
Output:
int(x^(3/2)*ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n)),x)
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:
integrate(x^(3/2)*log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="fr icas")
Output:
integral((b*x^(3/2)*log(c*x^n) + a*x^(3/2))*log((f*sqrt(x) + e)^k*d), x)
Timed out. \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x**(3/2)*ln(d*(e+f*x**(1/2))**k)*(a+b*ln(c*x**n)),x)
Output:
Timed out
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:
integrate(x^(3/2)*log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="ma xima")
Output:
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)
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac {3}{2}} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:
integrate(x^(3/2)*log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="gi ac")
Output:
integrate((b*log(c*x^n) + a)*x^(3/2)*log((f*sqrt(x) + e)^k*d), x)
Timed out. \[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{3/2}\,\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:
int(x^(3/2)*log(d*(e + f*x^(1/2))^k)*(a + b*log(c*x^n)),x)
Output:
int(x^(3/2)*log(d*(e + f*x^(1/2))^k)*(a + b*log(c*x^n)), x)
\[ \int x^{3/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1800 \sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{5} k \,x^{2}+1800 \sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) a \,f^{5} k \,x^{2}-720 \sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) b \,f^{5} k n \,x^{2}-1800 \sqrt {x}\, \mathrm {log}\left (x^{n} c \right ) b \,e^{4} f \,k^{2}-600 \sqrt {x}\, \mathrm {log}\left (x^{n} c \right ) b \,e^{2} f^{3} k^{2} x -360 \sqrt {x}\, \mathrm {log}\left (x^{n} c \right ) b \,f^{5} k^{2} x^{2}-1800 \sqrt {x}\, a \,e^{4} f \,k^{2}-600 \sqrt {x}\, a \,e^{2} f^{3} k^{2} x -360 \sqrt {x}\, a \,f^{5} k^{2} x^{2}+4320 \sqrt {x}\, b \,e^{4} f \,k^{2} n +640 \sqrt {x}\, b \,e^{2} f^{3} k^{2} n x +288 \sqrt {x}\, b \,f^{5} k^{2} n \,x^{2}-1800 \left (\int \frac {\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )}{-f^{2} x^{2}+e^{2} x}d x \right ) b \,e^{7} k n +1800 \left (\int \frac {\sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )}{-f^{2} x^{2}+e^{2} x}d x \right ) b \,e^{6} f k n +1800 \,\mathrm {log}\left (\sqrt {x}\, f +e \right ) a \,e^{5} k^{2}-720 \,\mathrm {log}\left (\sqrt {x}\, f +e \right ) b \,e^{5} k^{2} n -1800 \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )^{2} b \,e^{5} n +1800 \,\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{5} k +900 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} f^{2} k^{2} x +450 \,\mathrm {log}\left (x^{n} c \right ) b e \,f^{4} k^{2} x^{2}+900 a \,e^{3} f^{2} k^{2} x +450 a e \,f^{4} k^{2} x^{2}-1260 b \,e^{3} f^{2} k^{2} n x -405 b e \,f^{4} k^{2} n \,x^{2}}{4500 f^{5} k} \] Input:
int(x^(3/2)*log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x)
Output:
(1800*sqrt(x)*log((sqrt(x)*f + e)**k*d)*log(x**n*c)*b*f**5*k*x**2 + 1800*s qrt(x)*log((sqrt(x)*f + e)**k*d)*a*f**5*k*x**2 - 720*sqrt(x)*log((sqrt(x)* f + e)**k*d)*b*f**5*k*n*x**2 - 1800*sqrt(x)*log(x**n*c)*b*e**4*f*k**2 - 60 0*sqrt(x)*log(x**n*c)*b*e**2*f**3*k**2*x - 360*sqrt(x)*log(x**n*c)*b*f**5* k**2*x**2 - 1800*sqrt(x)*a*e**4*f*k**2 - 600*sqrt(x)*a*e**2*f**3*k**2*x - 360*sqrt(x)*a*f**5*k**2*x**2 + 4320*sqrt(x)*b*e**4*f*k**2*n + 640*sqrt(x)* b*e**2*f**3*k**2*n*x + 288*sqrt(x)*b*f**5*k**2*n*x**2 - 1800*int(log((sqrt (x)*f + e)**k*d)/(e**2*x - f**2*x**2),x)*b*e**7*k*n + 1800*int((sqrt(x)*lo g((sqrt(x)*f + e)**k*d))/(e**2*x - f**2*x**2),x)*b*e**6*f*k*n + 1800*log(s qrt(x)*f + e)*a*e**5*k**2 - 720*log(sqrt(x)*f + e)*b*e**5*k**2*n - 1800*lo g((sqrt(x)*f + e)**k*d)**2*b*e**5*n + 1800*log((sqrt(x)*f + e)**k*d)*log(x **n*c)*b*e**5*k + 900*log(x**n*c)*b*e**3*f**2*k**2*x + 450*log(x**n*c)*b*e *f**4*k**2*x**2 + 900*a*e**3*f**2*k**2*x + 450*a*e*f**4*k**2*x**2 - 1260*b *e**3*f**2*k**2*n*x - 405*b*e*f**4*k**2*n*x**2)/(4500*f**5*k)