3.1.100 \(\int x (a+b \log (c x^n))^2 \log (d (e+f x^2)^m) \, dx\) [100]

3.1.100.1 Optimal result
3.1.100.2 Mathematica [C] (verified)
3.1.100.3 Rubi [A] (verified)
3.1.100.4 Maple [C] (warning: unable to verify)
3.1.100.5 Fricas [F]
3.1.100.6 Sympy [F(-1)]
3.1.100.7 Maxima [F]
3.1.100.8 Giac [F]
3.1.100.9 Mupad [F(-1)]

3.1.100.1 Optimal result

Integrand size = 26, antiderivative size = 310 \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=-\frac {3}{4} b^2 m n^2 x^2+b m n x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )^2+\frac {b^2 e m n^2 \log \left (e+f x^2\right )}{4 f}+\frac {1}{4} b^2 n^2 x^2 \log \left (d \left (e+f x^2\right )^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )-\frac {b e m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x^2}{e}\right )}{2 f}+\frac {e m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x^2}{e}\right )}{2 f}-\frac {b^2 e m n^2 \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{4 f}+\frac {b e m n \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {f x^2}{e}\right )}{2 f}-\frac {b^2 e m n^2 \operatorname {PolyLog}\left (3,-\frac {f x^2}{e}\right )}{4 f} \]

output
-3/4*b^2*m*n^2*x^2+b*m*n*x^2*(a+b*ln(c*x^n))-1/2*m*x^2*(a+b*ln(c*x^n))^2+1 
/4*b^2*e*m*n^2*ln(f*x^2+e)/f+1/4*b^2*n^2*x^2*ln(d*(f*x^2+e)^m)-1/2*b*n*x^2 
*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)+1/2*x^2*(a+b*ln(c*x^n))^2*ln(d*(f*x^2+e 
)^m)-1/2*b*e*m*n*(a+b*ln(c*x^n))*ln(1+f*x^2/e)/f+1/2*e*m*(a+b*ln(c*x^n))^2 
*ln(1+f*x^2/e)/f-1/4*b^2*e*m*n^2*polylog(2,-f*x^2/e)/f+1/2*b*e*m*n*(a+b*ln 
(c*x^n))*polylog(2,-f*x^2/e)/f-1/4*b^2*e*m*n^2*polylog(3,-f*x^2/e)/f
 
3.1.100.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 814, normalized size of antiderivative = 2.63 \[ \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right ) \, dx=\frac {-2 a^2 f m x^2+4 a b f m n x^2-3 b^2 f m n^2 x^2-4 a b f m x^2 \log \left (c x^n\right )+4 b^2 f m n x^2 \log \left (c x^n\right )-2 b^2 f m x^2 \log ^2\left (c x^n\right )+4 a b e m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b^2 e m n^2 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b^2 e m n^2 \log ^2(x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b^2 e m n \log (x) \log \left (c x^n\right ) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 a b e m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b^2 e m n^2 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b^2 e m n^2 \log ^2(x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+4 b^2 e m n \log (x) \log \left (c x^n\right ) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 a^2 e m \log \left (e+f x^2\right )-2 a b e m n \log \left (e+f x^2\right )+b^2 e m n^2 \log \left (e+f x^2\right )-4 a b e m n \log (x) \log \left (e+f x^2\right )+2 b^2 e m n^2 \log (x) \log \left (e+f x^2\right )+2 b^2 e m n^2 \log ^2(x) \log \left (e+f x^2\right )+4 a b e m \log \left (c x^n\right ) \log \left (e+f x^2\right )-2 b^2 e m n \log \left (c x^n\right ) \log \left (e+f x^2\right )-4 b^2 e m n \log (x) \log \left (c x^n\right ) \log \left (e+f x^2\right )+2 b^2 e m \log ^2\left (c x^n\right ) \log \left (e+f x^2\right )+2 a^2 f x^2 \log \left (d \left (e+f x^2\right )^m\right )-2 a b f n x^2 \log \left (d \left (e+f x^2\right )^m\right )+b^2 f n^2 x^2 \log \left (d \left (e+f x^2\right )^m\right )+4 a b f x^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-2 b^2 f n x^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b^2 f x^2 \log ^2\left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b e m n \left (2 a-b n+2 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b e m n \left (2 a-b n+2 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b^2 e m n^2 \operatorname {PolyLog}\left (3,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-4 b^2 e m n^2 \operatorname {PolyLog}\left (3,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{4 f} \]

input
Integrate[x*(a + b*Log[c*x^n])^2*Log[d*(e + f*x^2)^m],x]
 
output
(-2*a^2*f*m*x^2 + 4*a*b*f*m*n*x^2 - 3*b^2*f*m*n^2*x^2 - 4*a*b*f*m*x^2*Log[ 
c*x^n] + 4*b^2*f*m*n*x^2*Log[c*x^n] - 2*b^2*f*m*x^2*Log[c*x^n]^2 + 4*a*b*e 
*m*n*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*e*m*n^2*Log[x]*Log[1 - 
(I*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*e*m*n^2*Log[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt 
[e]] + 4*b^2*e*m*n*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*a* 
b*e*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*e*m*n^2*Log[x]*Log[1 
 + (I*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*e*m*n^2*Log[x]^2*Log[1 + (I*Sqrt[f]*x)/S 
qrt[e]] + 4*b^2*e*m*n*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 2 
*a^2*e*m*Log[e + f*x^2] - 2*a*b*e*m*n*Log[e + f*x^2] + b^2*e*m*n^2*Log[e + 
 f*x^2] - 4*a*b*e*m*n*Log[x]*Log[e + f*x^2] + 2*b^2*e*m*n^2*Log[x]*Log[e + 
 f*x^2] + 2*b^2*e*m*n^2*Log[x]^2*Log[e + f*x^2] + 4*a*b*e*m*Log[c*x^n]*Log 
[e + f*x^2] - 2*b^2*e*m*n*Log[c*x^n]*Log[e + f*x^2] - 4*b^2*e*m*n*Log[x]*L 
og[c*x^n]*Log[e + f*x^2] + 2*b^2*e*m*Log[c*x^n]^2*Log[e + f*x^2] + 2*a^2*f 
*x^2*Log[d*(e + f*x^2)^m] - 2*a*b*f*n*x^2*Log[d*(e + f*x^2)^m] + b^2*f*n^2 
*x^2*Log[d*(e + f*x^2)^m] + 4*a*b*f*x^2*Log[c*x^n]*Log[d*(e + f*x^2)^m] - 
2*b^2*f*n*x^2*Log[c*x^n]*Log[d*(e + f*x^2)^m] + 2*b^2*f*x^2*Log[c*x^n]^2*L 
og[d*(e + f*x^2)^m] + 2*b*e*m*n*(2*a - b*n + 2*b*Log[c*x^n])*PolyLog[2, (( 
-I)*Sqrt[f]*x)/Sqrt[e]] + 2*b*e*m*n*(2*a - b*n + 2*b*Log[c*x^n])*PolyLog[2 
, (I*Sqrt[f]*x)/Sqrt[e]] - 4*b^2*e*m*n^2*PolyLog[3, ((-I)*Sqrt[f]*x)/Sqrt[ 
e]] - 4*b^2*e*m*n^2*PolyLog[3, (I*Sqrt[f]*x)/Sqrt[e]])/(4*f)
 
3.1.100.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2825, 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.

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

\(\Big \downarrow \) 2825

\(\displaystyle -2 f m \int \left (\frac {\left (a+b \log \left (c x^n\right )\right )^2 x^3}{2 \left (f x^2+e\right )}-\frac {b n \left (a+b \log \left (c x^n\right )\right ) x^3}{2 \left (f x^2+e\right )}+\frac {b^2 n^2 x^3}{4 \left (f x^2+e\right )}\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{4} b^2 n^2 x^2 \log \left (d \left (e+f x^2\right )^m\right )\)

\(\Big \downarrow \) 2009

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

input
Int[x*(a + b*Log[c*x^n])^2*Log[d*(e + f*x^2)^m],x]
 
output
(b^2*n^2*x^2*Log[d*(e + f*x^2)^m])/4 - (b*n*x^2*(a + b*Log[c*x^n])*Log[d*( 
e + f*x^2)^m])/2 + (x^2*(a + b*Log[c*x^n])^2*Log[d*(e + f*x^2)^m])/2 - 2*f 
*m*((3*b^2*n^2*x^2)/(8*f) - (b*n*x^2*(a + b*Log[c*x^n]))/(2*f) + (x^2*(a + 
 b*Log[c*x^n])^2)/(4*f) - (b^2*e*n^2*Log[e + f*x^2])/(8*f^2) + (b*e*n*(a + 
 b*Log[c*x^n])*Log[1 + (f*x^2)/e])/(4*f^2) - (e*(a + b*Log[c*x^n])^2*Log[1 
 + (f*x^2)/e])/(4*f^2) + (b^2*e*n^2*PolyLog[2, -((f*x^2)/e)])/(8*f^2) - (b 
*e*n*(a + b*Log[c*x^n])*PolyLog[2, -((f*x^2)/e)])/(4*f^2) + (b^2*e*n^2*Pol 
yLog[3, -((f*x^2)/e)])/(8*f^2))
 

3.1.100.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2825
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q* 
(a + b*Log[c*x^n])^p, x]}, Simp[Log[d*(e + f*x^m)^r]   u, x] - Simp[f*m*r 
 Int[x^(m - 1)/(e + f*x^m)   u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m 
, n, q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q]
 
3.1.100.4 Maple [C] (warning: unable to verify)

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

Time = 174.33 (sec) , antiderivative size = 4839, normalized size of antiderivative = 15.61

method result size
risch \(\text {Expression too large to display}\) \(4839\)

input
int(x*(a+b*ln(c*x^n))^2*ln(d*(f*x^2+e)^m),x,method=_RETURNVERBOSE)
 
output
-1/2*I*m/f*e*ln(f*x^2+e)*Pi*a*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/2*I* 
m/f*ln(x^n)*e*ln(f*x^2+e)*b^2*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/2*I 
*m/f*n*e*ln(x)*ln(f*x^2+e)*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/2*I*m/f*n*e* 
ln(x)*ln(f*x^2+e)*b^2*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*m/f*e*ln(f*x^2+ 
e)*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/4*I*m/f*e*ln(f*x^2+e 
)*Pi*b^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2*I*m/f*n*e*ln(x)*ln((-f* 
x+(-e*f)^(1/2))/(-e*f)^(1/2))*b^2*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*m/f*n 
*e*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))*b^2*Pi*csgn(I*x^n)*csgn(I*c* 
x^n)^2+1/2*I*m/f*n*e*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))*b^2*Pi*csgn 
(I*c)*csgn(I*c*x^n)^2-m*ln(x^n)*x^2*b^2*ln(c)+m*n*x^2*b^2*ln(c)-m*b*ln(x^n 
)*x^2*a+m*b*n*x^2*a+1/8*m*x^2*Pi^2*b^2*csgn(I*c)^2*csgn(I*c*x^n)^4-1/4*m*x 
^2*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5+1/8*m*x^2*Pi^2*b^2*csgn(I*x^n)^2*csg 
n(I*c*x^n)^4-1/4*m*x^2*Pi^2*b^2*csgn(I*x^n)*csgn(I*c*x^n)^5+(1/2*x^2*b^2*l 
n(x^n)^2+1/2*b*x^2*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csg 
n(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c* 
x^n)^3+2*b*ln(c)-b*n+2*a)*ln(x^n)+1/8*x^2*(4*a^2+2*I*Pi*b^2*n*csgn(I*c)*cs 
gn(I*x^n)*csgn(I*c*x^n)+2*Pi^2*b^2*csgn(I*c)*csgn(I*c*x^n)^5+4*I*ln(c)*Pi* 
b^2*csgn(I*x^n)*csgn(I*c*x^n)^2+4*I*Pi*a*b*csgn(I*x^n)*csgn(I*c*x^n)^2+2*b 
^2*n^2-4*I*ln(c)*Pi*b^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2*I*Pi*b^2*n*c 
sgn(I*c)*csgn(I*c*x^n)^2+8*ln(c)*a*b+4*ln(c)^2*b^2-4*b^2*ln(c)*n-4*a*b*...
 
3.1.100.5 Fricas [F]

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

input
integrate(x*(a+b*log(c*x^n))^2*log(d*(f*x^2+e)^m),x, algorithm="fricas")
 
output
integral((b^2*x*log(c*x^n)^2 + 2*a*b*x*log(c*x^n) + a^2*x)*log((f*x^2 + e) 
^m*d), x)
 
3.1.100.6 Sympy [F(-1)]

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

input
integrate(x*(a+b*ln(c*x**n))**2*ln(d*(f*x**2+e)**m),x)
 
output
Timed out
 
3.1.100.7 Maxima [F]

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

input
integrate(x*(a+b*log(c*x^n))^2*log(d*(f*x^2+e)^m),x, algorithm="maxima")
 
output
1/4*(2*b^2*x^2*log(x^n)^2 - 2*(b^2*(n - 2*log(c)) - 2*a*b)*x^2*log(x^n) + 
((n^2 - 2*n*log(c) + 2*log(c)^2)*b^2 - 2*a*b*(n - 2*log(c)) + 2*a^2)*x^2)* 
log((f*x^2 + e)^m) + integrate(-1/2*((2*(f*m - f*log(d))*a^2 - 2*(f*m*n - 
2*(f*m - f*log(d))*log(c))*a*b + (f*m*n^2 - 2*f*m*n*log(c) + 2*(f*m - f*lo 
g(d))*log(c)^2)*b^2)*x^3 + 2*((f*m - f*log(d))*b^2*x^3 - b^2*e*x*log(d))*l 
og(x^n)^2 - 2*(b^2*e*log(c)^2*log(d) + 2*a*b*e*log(c)*log(d) + a^2*e*log(d 
))*x + 2*((2*(f*m - f*log(d))*a*b - (f*m*n - 2*(f*m - f*log(d))*log(c))*b^ 
2)*x^3 - 2*(b^2*e*log(c)*log(d) + a*b*e*log(d))*x)*log(x^n))/(f*x^2 + e), 
x)
 
3.1.100.8 Giac [F]

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

input
integrate(x*(a+b*log(c*x^n))^2*log(d*(f*x^2+e)^m),x, algorithm="giac")
 
output
integrate((b*log(c*x^n) + a)^2*x*log((f*x^2 + e)^m*d), x)
 
3.1.100.9 Mupad [F(-1)]

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

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