\(\int x^2 (a+b \log (c x^n)) \log (d (e+f x^2)^m) \, dx\) [101]

Optimal result

Integrand size = 26, antiderivative size = 251 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {8 b e m n x}{9 f}+\frac {4}{27} b m n x^3+\frac {2 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}-\frac {1}{9} b n x^3 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {i b e^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}} \] Output:

-8/9*b*e*m*n*x/f+4/27*b*m*n*x^3+2/9*b*e^(3/2)*m*n*arctan(f^(1/2)*x/e^(1/2) 
)/f^(3/2)+2/3*e*m*x*(a+b*ln(c*x^n))/f-2/9*m*x^3*(a+b*ln(c*x^n))-2/3*e^(3/2 
)*m*arctan(f^(1/2)*x/e^(1/2))*(a+b*ln(c*x^n))/f^(3/2)-1/9*b*n*x^3*ln(d*(f* 
x^2+e)^m)+1/3*x^3*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)+1/3*I*b*e^(3/2)*m*n*po 
lylog(2,-I*f^(1/2)*x/e^(1/2))/f^(3/2)-1/3*I*b*e^(3/2)*m*n*polylog(2,I*f^(1 
/2)*x/e^(1/2))/f^(3/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.55 \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {18 a e \sqrt {f} m x-24 b e \sqrt {f} m n x-6 a f^{3/2} m x^3+4 b f^{3/2} m n x^3-18 a e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+6 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+18 b e^{3/2} m n \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)+18 b e \sqrt {f} m x \log \left (c x^n\right )-6 b f^{3/2} m x^3 \log \left (c x^n\right )-18 b e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-9 i b e^{3/2} m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+9 i b e^{3/2} m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+9 a f^{3/2} x^3 \log \left (d \left (e+f x^2\right )^m\right )-3 b f^{3/2} n x^3 \log \left (d \left (e+f x^2\right )^m\right )+9 b f^{3/2} x^3 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+9 i b e^{3/2} m n \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-9 i b e^{3/2} m n \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{27 f^{3/2}} \] Input:

Integrate[x^2*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m],x]
 

Output:

(18*a*e*Sqrt[f]*m*x - 24*b*e*Sqrt[f]*m*n*x - 6*a*f^(3/2)*m*x^3 + 4*b*f^(3/ 
2)*m*n*x^3 - 18*a*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 6*b*e^(3/2)*m*n* 
ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 18*b*e^(3/2)*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]] 
*Log[x] + 18*b*e*Sqrt[f]*m*x*Log[c*x^n] - 6*b*f^(3/2)*m*x^3*Log[c*x^n] - 1 
8*b*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (9*I)*b*e^(3/2)*m*n 
*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (9*I)*b*e^(3/2)*m*n*Log[x]*Log[1 
+ (I*Sqrt[f]*x)/Sqrt[e]] + 9*a*f^(3/2)*x^3*Log[d*(e + f*x^2)^m] - 3*b*f^(3 
/2)*n*x^3*Log[d*(e + f*x^2)^m] + 9*b*f^(3/2)*x^3*Log[c*x^n]*Log[d*(e + f*x 
^2)^m] + (9*I)*b*e^(3/2)*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] - (9*I)* 
b*e^(3/2)*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(27*f^(3/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

\(\displaystyle \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx\)

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (-\frac {2 m x^2}{9}+\frac {1}{3} \log \left (d \left (f x^2+e\right )^m\right ) x^2+\frac {2 e m}{3 f}-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2} x}\right )dx-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {2 e^{3/2} m \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}+\frac {1}{9} x^3 \log \left (d \left (e+f x^2\right )^m\right )-\frac {i e^{3/2} m \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {i e^{3/2} m \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {8 e m x}{9 f}-\frac {4 m x^3}{27}\right )\)

Input:

Int[x^2*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m],x]
 

Output:

(2*e*m*x*(a + b*Log[c*x^n]))/(3*f) - (2*m*x^3*(a + b*Log[c*x^n]))/9 - (2*e 
^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(3*f^(3/2)) + (x^ 
3*(a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/3 - b*n*((8*e*m*x)/(9*f) - (4*m 
*x^3)/27 - (2*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(9*f^(3/2)) + (x^3*Lo 
g[d*(e + f*x^2)^m])/9 - ((I/3)*e^(3/2)*m*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[ 
e]])/f^(3/2) + ((I/3)*e^(3/2)*m*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/f^(3/2) 
)
 

Defintions of rubi rules used

rule 2009

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

rule 2823

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 102.77 (sec) , antiderivative size = 1082, normalized size of antiderivative = 4.31

method	result	size
risch	\(\text {Expression too large to display}\)	\(1082\)

Input:

int(x^2*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m),x,method=_RETURNVERBOSE)
 

Output:

-2/3*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*ln(c)+2/9*m/f*e^2/(e*f) 
^(1/2)*arctan(x*f/(e*f)^(1/2))*n*b-2/3*m/f*b*e^2/(e*f)^(1/2)*arctan(x*f/(e 
*f)^(1/2))*ln(x^n)-1/3*m/f*b*n*e^2/(-e*f)^(1/2)*dilog((-f*x+(-e*f)^(1/2))/ 
(-e*f)^(1/2))-2/9*x^3*ln(c)*b*m+2/3*m/f*b*e^2/(e*f)^(1/2)*arctan(x*f/(e*f) 
^(1/2))*n*ln(x)-1/3*m/f*b*n*e^2*ln(x)/(-e*f)^(1/2)*ln((-f*x+(-e*f)^(1/2))/ 
(-e*f)^(1/2))+1/3*m/f*b*n*e^2*ln(x)/(-e*f)^(1/2)*ln((f*x+(-e*f)^(1/2))/(-e 
*f)^(1/2))-1/3*I*m/f*e*x*b*Pi*csgn(I*c*x^n)^3+1/9*I*m*x^3*b*Pi*csgn(I*x^n) 
*csgn(I*c*x^n)*csgn(I*c)+(1/3*b*x^3*ln(x^n)+1/18*x^3*(3*I*b*Pi*csgn(I*x^n) 
*csgn(I*c*x^n)^2-3*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3*I*b*Pi*csg 
n(I*c*x^n)^3+3*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+6*b*ln(c)-2*n*b+6*a))*ln(( 
f*x^2+e)^m)+(-1/4*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m) 
+1/4*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2+1/4*I*Pi*csgn(I*(f*x^2+e)^m)*c 
sgn(I*d*(f*x^2+e)^m)^2-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3+1/2*ln(d))*(1/3*(I 
*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I* 
c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2*a)* 
x^3+2/3*b*x^3*ln(x^n)-2/9*b*n*x^3)-2/9*m*b*ln(x^n)*x^3-2/3*m/f*e^2/(e*f)^( 
1/2)*arctan(x*f/(e*f)^(1/2))*a+1/9*I*m*x^3*b*Pi*csgn(I*c*x^n)^3+1/3*m/f*b* 
n*e^2/(-e*f)^(1/2)*dilog((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/9*I*m*x^3*b*Pi 
*csgn(I*x^n)*csgn(I*c*x^n)^2-1/9*I*m*x^3*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)+1/ 
3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I...
 

Fricas [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \] Input:

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

Output:

integral((b*x^2*log(c*x^n) + a*x^2)*log((f*x^2 + e)^m*d), x)
 

Sympy [F(-1)]

Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ) \,d x } \] Input:

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

Output:

integrate((b*log(c*x^n) + a)*x^2*log((f*x^2 + e)^m*d), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\int x^2\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {-18 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) a e m +6 \sqrt {f}\, \sqrt {e}\, \mathit {atan} \left (\frac {f x}{\sqrt {f}\, \sqrt {e}}\right ) b e m n -18 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{f \,x^{2}+e}d x \right ) b \,e^{2} f m +9 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) \mathrm {log}\left (x^{n} c \right ) b \,f^{2} x^{3}+9 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) a \,f^{2} x^{3}-3 \,\mathrm {log}\left (\left (f \,x^{2}+e \right )^{m} d \right ) b \,f^{2} n \,x^{3}+18 \,\mathrm {log}\left (x^{n} c \right ) b e f m x -6 \,\mathrm {log}\left (x^{n} c \right ) b \,f^{2} m \,x^{3}+18 a e f m x -6 a \,f^{2} m \,x^{3}-24 b e f m n x +4 b \,f^{2} m n \,x^{3}}{27 f^{2}} \] Input:

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

Output:

( - 18*sqrt(f)*sqrt(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*a*e*m + 6*sqrt(f)*sqr 
t(e)*atan((f*x)/(sqrt(f)*sqrt(e)))*b*e*m*n - 18*int(log(x**n*c)/(e + f*x** 
2),x)*b*e**2*f*m + 9*log((e + f*x**2)**m*d)*log(x**n*c)*b*f**2*x**3 + 9*lo 
g((e + f*x**2)**m*d)*a*f**2*x**3 - 3*log((e + f*x**2)**m*d)*b*f**2*n*x**3 
+ 18*log(x**n*c)*b*e*f*m*x - 6*log(x**n*c)*b*f**2*m*x**3 + 18*a*e*f*m*x - 
6*a*f**2*m*x**3 - 24*b*e*f*m*n*x + 4*b*f**2*m*n*x**3)/(27*f**2)