\(\int \log (d (e+f \sqrt {x})^k) (a+b \log (c x^n)) \, dx\) [123]

Optimal result

Integrand size = 25, antiderivative size = 209 \[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {3 b e k n \sqrt {x}}{f}+b k n x+\frac {b e^2 k n \log \left (e+f \sqrt {x}\right )}{f^2}-b n x \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 b e^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{f^2}+\frac {e k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{f}-\frac {1}{2} k x \left (a+b \log \left (c x^n\right )\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+x \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 b e^2 k n \operatorname {PolyLog}\left (2,1+\frac {f \sqrt {x}}{e}\right )}{f^2} \] Output:

-3*b*e*k*n*x^(1/2)/f+b*k*n*x+b*e^2*k*n*ln(e+f*x^(1/2))/f^2-b*n*x*ln(d*(e+f 
*x^(1/2))^k)+2*b*e^2*k*n*ln(e+f*x^(1/2))*ln(-f*x^(1/2)/e)/f^2+e*k*x^(1/2)* 
(a+b*ln(c*x^n))/f-1/2*k*x*(a+b*ln(c*x^n))-e^2*k*ln(e+f*x^(1/2))*(a+b*ln(c* 
x^n))/f^2+x*ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n))+2*b*e^2*k*n*polylog(2,1+ 
f*x^(1/2)/e)/f^2
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.04 \[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a e k \sqrt {x}}{f}-\frac {3 b e k n \sqrt {x}}{f}-\frac {a k x}{2}+b k n x+a x \log \left (d \left (e+f \sqrt {x}\right )^k\right )-b n x \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {b e^2 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)}{f^2}+\frac {b e k \sqrt {x} \log \left (c x^n\right )}{f}-\frac {1}{2} b k x \log \left (c x^n\right )+b x \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a-b n-b n \log (x)+b \log \left (c x^n\right )\right )}{f^2}-\frac {2 b e^2 k n \operatorname {PolyLog}\left (2,-\frac {f \sqrt {x}}{e}\right )}{f^2} \] Input:

Integrate[Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]
 

Output:

(a*e*k*Sqrt[x])/f - (3*b*e*k*n*Sqrt[x])/f - (a*k*x)/2 + b*k*n*x + a*x*Log[ 
d*(e + f*Sqrt[x])^k] - b*n*x*Log[d*(e + f*Sqrt[x])^k] - (b*e^2*k*n*Log[1 + 
 (f*Sqrt[x])/e]*Log[x])/f^2 + (b*e*k*Sqrt[x]*Log[c*x^n])/f - (b*k*x*Log[c* 
x^n])/2 + b*x*Log[d*(e + f*Sqrt[x])^k]*Log[c*x^n] - (e^2*k*Log[e + f*Sqrt[ 
x]]*(a - b*n - b*n*Log[x] + b*Log[c*x^n]))/f^2 - (2*b*e^2*k*n*PolyLog[2, - 
((f*Sqrt[x])/e)])/f^2
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2817, 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.

\(\displaystyle \int \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \, dx\)

\(\Big \downarrow \) 2817

\(\displaystyle -b n \int \left (-\frac {k \log \left (e+f \sqrt {x}\right ) e^2}{f^2 x}+\frac {k e}{f \sqrt {x}}-\frac {k}{2}+\log \left (d \left (e+f \sqrt {x}\right )^k\right )\right )dx+x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+\frac {e k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{f}-\frac {1}{2} k x \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+\frac {e k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{f}-\frac {1}{2} k x \left (a+b \log \left (c x^n\right )\right )-b n \left (x \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {2 e^2 k \operatorname {PolyLog}\left (2,\frac {\sqrt {x} f}{e}+1\right )}{f^2}-\frac {e^2 k \log \left (e+f \sqrt {x}\right )}{f^2}-\frac {2 e^2 k \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{f^2}+\frac {3 e k \sqrt {x}}{f}-k x\right )\)

Input:

Int[Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]
 

Output:

(e*k*Sqrt[x]*(a + b*Log[c*x^n]))/f - (k*x*(a + b*Log[c*x^n]))/2 - (e^2*k*L 
og[e + f*Sqrt[x]]*(a + b*Log[c*x^n]))/f^2 + x*Log[d*(e + f*Sqrt[x])^k]*(a 
+ b*Log[c*x^n]) - b*n*((3*e*k*Sqrt[x])/f - k*x - (e^2*k*Log[e + f*Sqrt[x]] 
)/f^2 + x*Log[d*(e + f*Sqrt[x])^k] - (2*e^2*k*Log[e + f*Sqrt[x]]*Log[-((f* 
Sqrt[x])/e)])/f^2 - (2*e^2*k*PolyLog[2, 1 + (f*Sqrt[x])/e])/f^2)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2817

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))^(p_.), x_Symbol] :> With[{u = IntHide[Log[d*(e + f*x^m)^r], x]}, 
Simp[(a + b*Log[c*x^n])^p   u, x] - Simp[b*n*p   Int[(a + b*Log[c*x^n])^(p 
- 1)/x   u, x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] 
&& RationalQ[m] && (EqQ[p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 
 1] && EqQ[m, 1] && EqQ[d*e, 1]))
 

Maple [F]

\[\int \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right ) \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]

Input:

int(ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n)),x)
 

Output:

int(ln(d*(e+f*x^(1/2))^k)*(a+b*ln(c*x^n)),x)
 

Fricas [F]

\[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:

integrate(log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="fricas")
 

Output:

integral((b*log(c*x^n) + a)*log((f*sqrt(x) + e)^k*d), x)
 

Sympy [F(-1)]

Timed out. \[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \] Input:

integrate(ln(d*(e+f*x**(1/2))**k)*(a+b*ln(c*x**n)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:

integrate(log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="maxima")
 

Output:

1/9*(9*b*e*x*log(d)*log(x^n) + 9*(a*e*log(d) - (e*n*log(d) - e*log(c)*log( 
d))*b)*x + 9*(b*e*x*log(x^n) - ((e*n - e*log(c))*b - a*e)*x)*log((f*sqrt(x 
) + e)^k) - (3*b*f*k*x^2*log(x^n) + (3*a*f*k - (5*f*k*n - 3*f*k*log(c))*b) 
*x^2)/sqrt(x))/e + integrate(1/2*(b*f^2*k*x*log(x^n) + (a*f^2*k - (f^2*k*n 
 - f^2*k*log(c))*b)*x)/(e*f*sqrt(x) + e^2), x)
 

Giac [F]

\[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right ) \,d x } \] Input:

integrate(log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*log((f*sqrt(x) + e)^k*d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:

int(log(d*(e + f*x^(1/2))^k)*(a + b*log(c*x^n)),x)
 

Output:

int(log(d*(e + f*x^(1/2))^k)*(a + b*log(c*x^n)), x)
 

Reduce [F]

\[ \int \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 \sqrt {x}\, \mathrm {log}\left (x^{n} c \right ) b e f \,k^{2}+2 \sqrt {x}\, a e f \,k^{2}-6 \sqrt {x}\, b e f \,k^{2} n +2 \left (\int \frac {\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )}{-f^{2} x^{2}+e^{2} x}d x \right ) b \,e^{4} k n -2 \left (\int \frac {\sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )}{-f^{2} x^{2}+e^{2} x}d x \right ) b \,e^{3} f k n -2 \,\mathrm {log}\left (\sqrt {x}\, f +e \right ) a \,e^{2} k^{2}+2 \,\mathrm {log}\left (\sqrt {x}\, f +e \right ) b \,e^{2} k^{2} n +2 \mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right )^{2} b \,e^{2} n -2 \,\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,e^{2} k +2 \,\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{2} k x +2 \,\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) a \,f^{2} k x -2 \,\mathrm {log}\left (\left (\sqrt {x}\, f +e \right )^{k} d \right ) b \,f^{2} k n x -\mathrm {log}\left (x^{n} c \right ) b \,f^{2} k^{2} x -a \,f^{2} k^{2} x +2 b \,f^{2} k^{2} n x}{2 f^{2} k} \] Input:

int(log(d*(e+f*x^(1/2))^k)*(a+b*log(c*x^n)),x)
 

Output:

(2*sqrt(x)*log(x**n*c)*b*e*f*k**2 + 2*sqrt(x)*a*e*f*k**2 - 6*sqrt(x)*b*e*f 
*k**2*n + 2*int(log((sqrt(x)*f + e)**k*d)/(e**2*x - f**2*x**2),x)*b*e**4*k 
*n - 2*int((sqrt(x)*log((sqrt(x)*f + e)**k*d))/(e**2*x - f**2*x**2),x)*b*e 
**3*f*k*n - 2*log(sqrt(x)*f + e)*a*e**2*k**2 + 2*log(sqrt(x)*f + e)*b*e**2 
*k**2*n + 2*log((sqrt(x)*f + e)**k*d)**2*b*e**2*n - 2*log((sqrt(x)*f + e)* 
*k*d)*log(x**n*c)*b*e**2*k + 2*log((sqrt(x)*f + e)**k*d)*log(x**n*c)*b*f** 
2*k*x + 2*log((sqrt(x)*f + e)**k*d)*a*f**2*k*x - 2*log((sqrt(x)*f + e)**k* 
d)*b*f**2*k*n*x - log(x**n*c)*b*f**2*k**2*x - a*f**2*k**2*x + 2*b*f**2*k** 
2*n*x)/(2*f**2*k)