3.1.95 \(\int x^2 (a+b \log (c x^n)) \log (d (e+f x^2)^m) \, dx\) [95]

Optimal. Leaf size=251 \[ -\frac {8 b e m n x}{9 f}+\frac {4}{27} b m n x^3+\frac {2 b e^{3/2} m n \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}+\frac {2 e m x \left (a+b \log \left (c x^n\right )\right )}{3 f}-\frac {2}{9} m x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {2 e^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 f^{3/2}}-\frac {1}{9} b n x^3 \log \left (d \left (e+f x^2\right )^m\right )+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac {i b e^{3/2} m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2505, 308, 211, 2423, 4940, 2438} \begin {gather*} \frac {i b e^{3/2} m n \text {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \text {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {2 e^{3/2} m \text {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 )+\frac {2 b e^{3/2} m n \text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 f^{3/2}}-\frac {1}{9} b n x^3 \log \left (d \left (e+f x^2\right )^m\right )-\frac {8 b e m n x}{9 f}+\frac {4}{27} b m n x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2423

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], 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]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \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 {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 \tan ^{-1}\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 )-(b n) \int \left (\frac {2 e m}{3 f}-\frac {2 m x^2}{9}-\frac {2 e^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2} x}+\frac {1}{3} x^2 \log \left (d \left (e+f x^2\right )^m\right )\right ) \, dx\\ &=-\frac {2 b e m n x}{3 f}+\frac {2}{27} b m n x^3+\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 \tan ^{-1}\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 {1}{3} (b n) \int x^2 \log \left (d \left (e+f x^2\right )^m\right ) \, dx+\frac {\left (2 b e^{3/2} m n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 f^{3/2}}\\ &=-\frac {2 b e m n x}{3 f}+\frac {2}{27} b m n x^3+\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 \tan ^{-1}\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 {\left (i b e^{3/2} m n\right ) \int \frac {\log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 f^{3/2}}-\frac {\left (i b e^{3/2} m n\right ) \int \frac {\log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 f^{3/2}}+\frac {1}{9} (2 b f m n) \int \frac {x^4}{e+f x^2} \, dx\\ &=-\frac {2 b e m n x}{3 f}+\frac {2}{27} b m n x^3+\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 \tan ^{-1}\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 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {1}{9} (2 b f m n) \int \left (-\frac {e}{f^2}+\frac {x^2}{f}+\frac {e^2}{f^2 \left (e+f x^2\right )}\right ) \, dx\\ &=-\frac {8 b e m n x}{9 f}+\frac {4}{27} b m n x^3+\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 \tan ^{-1}\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 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}+\frac {\left (2 b e^2 m n\right ) \int \frac {1}{e+f x^2} \, dx}{9 f}\\ &=-\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 \tan ^{-1}\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 \tan ^{-1}\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 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}-\frac {i b e^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 f^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 389, normalized size = 1.55 \begin {gather*} \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 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+6 b e^{3/2} m n \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+18 b e^{3/2} m n \tan ^{-1}\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 \tan ^{-1}\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 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-9 i b e^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{27 f^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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] - 18*b*e^(3/2)*m*ArcTan[(Sqrt[f]*x)/S
qrt[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]*Lo
g[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))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.19, size = 2321, normalized size = 9.25

method	result	size
risch	\(\text {Expression too large to display}\)	\(2321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*I*m*x^3*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^3
*a+1/6*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*ln(c)+1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*Pi
*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+4/27*b*m*n*x^3+1/3*x^3*ln(d)*a-1/6*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*x^3*a-1
/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^3*b*csgn(I*c*x^n)^3-2/3*m/f*b*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^
n)-2/9*x^3*a*m+2/3*a*e*m/f*x-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6
*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b*x^3*ln(x^n)-1/18*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b*x^3*n+(1/3
*x^3*b*ln(x^n)+1/18*x^3*(-3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+3*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+3*I*
b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I*b*Pi*csgn(I*c*x^n)^3+6*b*ln(c)-2*b*n+6*a))*ln((f*x^2+e)^m)+1/6*I*Pi*csgn(
I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*a+1/6*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^3*a-1/6*I*Pi*csgn(I*
d*(f*x^2+e)^m)^3*x^3*b*ln(c)-1/6*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*b*x^3*ln(x^n)-1/9*I*m*x^3*Pi*b*csgn(I*c)*csgn(I*
c*x^n)^2-2/9*m*b*ln(x^n)*x^3+1/3*ln(d)*b*x^3*ln(x^n)+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/(-e*f)^(1/2)*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/3*m/f*b*n*e^2/(-e*f)^(1/2)*ln(x)*ln
((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/6*I*x^3*Pi*ln(d)*b*csgn(I*c*x^n)^3-2/3*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f
)^(1/2))*b*ln(c)+1/3*I*m/f*x*e*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*x^3*ln(d)*ln(c)*b-2/9*x^3*ln(c)*b*m-1/9*ln
(d)*b*n*x^3+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/12*Pi^2*cs
gn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/12*Pi^2*csgn(I*d)*csgn(I
*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/12*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*
csgn(I*d*(f*x^2+e)^m)*x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e
)^m)*b*x^3*ln(x^n)+1/18*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b*x^3*n-1/12*Pi^2*csgn(I*d)*c
sgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/3*I*m/f*x*e*Pi*b*csgn(I*c
)*csgn(I*c*x^n)^2+2/3*m/f*b*ln(x^n)*x*e+1/18*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*b*x^3*n+1/9*I*m*x^3*Pi*b*csgn(I*c*x^
n)^3+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*c*x^n)^3+1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*
(f*x^2+e)^m)^2*x^3*b*csgn(I*c*x^n)^3+1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^3*b*csgn(I*c)*csgn(I*c*x^n)^2+1/6*I*x
^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*c*x^n)^2-8/9*b*e*m*n*x/f+2/9*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*n-1/
3*m/f*b*n*e^2/(-e*f)^(1/2)*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/3*m/f*b*n*e^2/(-e*f)^(1/2)*dilog((f*x+(-e
*f)^(1/2))/(-e*f)^(1/2))-1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*c)*csgn(I*c*x^n)^2
-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csg
n(I*d*(f*x^2+e)^m)^2*x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/3*I*m/f*x*e*Pi*b*csgn(I*c*x^n)^3+1/9*I*m*x^3*Pi*b*csg
n(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^3*b*ln(c)+1/3*
I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*c*x^n)^3+1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^3*b*csg
n(I*x^n)*csgn(I*c*x^n)^2+1/6*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^3*b*ln(c)+1/6*I*Pi*csgn(I*(f*x
^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b*x^3*ln(x^n)-1/18*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b*x^3*n-1
/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/12*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e
)^m)*csgn(I*d*(f*x^2+e)^m)*x^3*b*csgn(I*c*x^n)^3+2/3/f*ln(c)*b*e*m*x+1/6*I*x^3*Pi*ln(d)*b*csgn(I*x^n)*csgn(I*c
*x^n)^2-1/6*I*x^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2/3*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2
))*a-1/3*I*m/f*x*e*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*
Pi*b*csgn(I*c)*csgn(I*c*x^n)^2-1/3*I*m/f*e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*Pi*b*csgn(I*x^n)*csgn(I*c*x^n
)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*b*m*x^3*log(x^n) - ((m*n - 3*m*log(c))*b - 3*a*m)*x^3)*log(f*x^2 + e) + integrate(-1/9*((3*(2*f*m - 3*f
*log(d))*a - (2*f*m*n - 3*(2*f*m - 3*f*log(d))*log(c))*b)*x^4 - 9*(b*log(c)*log(d) + a*log(d))*x^2*e + 3*((2*f
*m - 3*f*log(d))*b*x^4 - 3*b*x^2*e*log(d))*log(x^n))/(f*x^2 + e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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