\(\int x^2 \log (f x^m) (a+b \log (c (d+e x)^n)) \, dx\) [359]

Optimal result

Integrand size = 24, antiderivative size = 195 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {4 b d^2 m n x}{9 e^2}-\frac {5 b d m n x^2}{36 e}+\frac {2}{27} b m n x^3-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^3}+\frac {b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^3} \]

[Out]

4/9*b*d^2*m*n*x/e^2-5/36*b*d*m*n*x^2/e+2/27*b*m*n*x^3-1/3*b*d^2*n*x*ln(f*x^m)/e^2+1/6*b*d*n*x^2*ln(f*x^m)/e-1/
9*b*n*x^3*ln(f*x^m)-1/9*b*d^3*m*n*ln(e*x+d)/e^3-1/9*(m*x^3-3*x^3*ln(f*x^m))*(a+b*ln(c*(e*x+d)^n))+1/3*b*d^3*n*
ln(f*x^m)*ln(1+e*x/d)/e^3+1/3*b*d^3*m*n*polylog(2,-e*x/d)/e^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2473, 45, 2393, 2332, 2341, 2354, 2438} \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{3 e^3}+\frac {b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{3 e^3}-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {4 b d^2 m n x}{9 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {5 b d m n x^2}{36 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )+\frac {2}{27} b m n x^3 \]

[In]

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

[Out]

(4*b*d^2*m*n*x)/(9*e^2) - (5*b*d*m*n*x^2)/(36*e) + (2*b*m*n*x^3)/27 - (b*d^2*n*x*Log[f*x^m])/(3*e^2) + (b*d*n*
x^2*Log[f*x^m])/(6*e) - (b*n*x^3*Log[f*x^m])/9 - (b*d^3*m*n*Log[d + e*x])/(9*e^3) - ((m*x^3 - 3*x^3*Log[f*x^m]
)*(a + b*Log[c*(d + e*x)^n]))/9 + (b*d^3*n*Log[f*x^m]*Log[1 + (e*x)/d])/(3*e^3) + (b*d^3*m*n*PolyLog[2, -((e*x
)/d)])/(3*e^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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 2473

Int[Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :
> Simp[(-(g*(q + 1))^(-1))*(m*((g*x)^(q + 1)/(q + 1)) - (g*x)^(q + 1)*Log[f*x^m])*(a + b*Log[c*(d + e*x)^n]),
x] + (-Dist[b*e*(n/(g*(q + 1))), Int[(g*x)^(q + 1)*(Log[f*x^m]/(d + e*x)), x], x] + Dist[b*e*m*(n/(g*(q + 1)^2
)), Int[(g*x)^(q + 1)/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && NeQ[q, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} (b e n) \int \frac {x^3 \log \left (f x^m\right )}{d+e x} \, dx+\frac {1}{9} (b e m n) \int \frac {x^3}{d+e x} \, dx \\ & = -\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} (b e n) \int \left (\frac {d^2 \log \left (f x^m\right )}{e^3}-\frac {d x \log \left (f x^m\right )}{e^2}+\frac {x^2 \log \left (f x^m\right )}{e}-\frac {d^3 \log \left (f x^m\right )}{e^3 (d+e x)}\right ) \, dx+\frac {1}{9} (b e m n) \int \left (\frac {d^2}{e^3}-\frac {d x}{e^2}+\frac {x^2}{e}-\frac {d^3}{e^3 (d+e x)}\right ) \, dx \\ & = \frac {b d^2 m n x}{9 e^2}-\frac {b d m n x^2}{18 e}+\frac {1}{27} b m n x^3-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{3} (b n) \int x^2 \log \left (f x^m\right ) \, dx-\frac {\left (b d^2 n\right ) \int \log \left (f x^m\right ) \, dx}{3 e^2}+\frac {\left (b d^3 n\right ) \int \frac {\log \left (f x^m\right )}{d+e x} \, dx}{3 e^2}+\frac {(b d n) \int x \log \left (f x^m\right ) \, dx}{3 e} \\ & = \frac {4 b d^2 m n x}{9 e^2}-\frac {5 b d m n x^2}{36 e}+\frac {2}{27} b m n x^3-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^3}-\frac {\left (b d^3 m n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{3 e^3} \\ & = \frac {4 b d^2 m n x}{9 e^2}-\frac {5 b d m n x^2}{36 e}+\frac {2}{27} b m n x^3-\frac {b d^2 n x \log \left (f x^m\right )}{3 e^2}+\frac {b d n x^2 \log \left (f x^m\right )}{6 e}-\frac {1}{9} b n x^3 \log \left (f x^m\right )-\frac {b d^3 m n \log (d+e x)}{9 e^3}-\frac {1}{9} \left (m x^3-3 x^3 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b d^3 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{3 e^3}+\frac {b d^3 m n \text {Li}_2\left (-\frac {e x}{d}\right )}{3 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.01 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {6 \log \left (f x^m\right ) \left (6 a e^3 x^3+b e n x \left (-6 d^2+3 d e x-2 e^2 x^2\right )+6 b d^3 n \log (d+e x)+6 b e^3 x^3 \log \left (c (d+e x)^n\right )\right )+m \left (48 b d^2 e n x-15 b d e^2 n x^2-12 a e^3 x^3+8 b e^3 n x^3-12 b d^3 n (1+3 \log (x)) \log (d+e x)-12 b e^3 x^3 \log \left (c (d+e x)^n\right )+36 b d^3 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )+36 b d^3 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{108 e^3} \]

[In]

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

[Out]

(6*Log[f*x^m]*(6*a*e^3*x^3 + b*e*n*x*(-6*d^2 + 3*d*e*x - 2*e^2*x^2) + 6*b*d^3*n*Log[d + e*x] + 6*b*e^3*x^3*Log
[c*(d + e*x)^n]) + m*(48*b*d^2*e*n*x - 15*b*d*e^2*n*x^2 - 12*a*e^3*x^3 + 8*b*e^3*n*x^3 - 12*b*d^3*n*(1 + 3*Log
[x])*Log[d + e*x] - 12*b*e^3*x^3*Log[c*(d + e*x)^n] + 36*b*d^3*n*Log[x]*Log[1 + (e*x)/d]) + 36*b*d^3*m*n*PolyL
og[2, -((e*x)/d)])/(108*e^3)

Maple [C] (warning: unable to verify)

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

Time = 32.94 (sec) , antiderivative size = 1012, normalized size of antiderivative = 5.19

method	result	size
risch	\(\text {Expression too large to display}\)	\(1012\)

[In]

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

[Out]

-1/12*I/e*n*b*d*x^2*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/18*I*n*b*x^3*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/18*I*
n*b*x^3*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/6*I/e^2*n*b*d^2*x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/6*I/e^3*n*
b*d^3*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/3*m/e^3*b*d^3*n*dilog(-e*x/d)-1/12*I/e*n*b*d*x^2*Pi*c
sgn(I*f*x^m)^3+1/6*I/e^2*n*b*d^2*x*Pi*csgn(I*f*x^m)^3+1/18*I*n*b*x^3*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/
3/e^2*n*b*ln(x^m)*x*d^2+49/108/e^3*n*b*m*d^3+(1/3*b*x^3*ln(x^m)+1/18*b*x^3*(-3*I*Pi*csgn(I*f)*csgn(I*x^m)*csgn
(I*f*x^m)+3*I*Pi*csgn(I*f)*csgn(I*f*x^m)^2+3*I*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-3*I*Pi*csgn(I*f*x^m)^3+6*ln(f)-2
*m))*ln((e*x+d)^n)-1/3*m/e^3*b*d^3*n*ln(e*x+d)*ln(-e*x/d)-1/6*I/e^3*n*b*d^3*ln(e*x+d)*Pi*csgn(I*f*x^m)^3+2/27*
b*m*n*x^3+1/6*I/e^3*n*b*d^3*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/6*I/e^3*n*b*d^3*ln(e*x+d)*Pi*csgn(I*x^m)*
csgn(I*f*x^m)^2+(-1/4*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*
x+d)^n)^2+1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3+1/2*b*ln(c)+1/2*
a)*(1/3*(-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*csgn(I*f)*csgn(I*f*x^m)^2+I*Pi*csgn(I*x^m)*csgn(I*f*x^
m)^2-I*Pi*csgn(I*f*x^m)^3+2*ln(f))*x^3+2/3*x^3*ln(x^m)-2/9*m*x^3)+1/3/e^3*n*b*d^3*ln(e*x+d)*ln(f)+1/18*I*n*b*x
^3*Pi*csgn(I*f*x^m)^3+1/3/e^3*n*b*ln(x^m)*d^3*ln(e*x+d)+1/6/e*n*b*ln(x^m)*d*x^2+1/6/e*n*b*d*x^2*ln(f)-1/3/e^2*
n*b*d^2*x*ln(f)-1/9*n*b*x^3*ln(f)+1/12*I/e*n*b*d*x^2*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/12*I/e*n*b*d*x^2*Pi*csgn(I
*x^m)*csgn(I*f*x^m)^2-1/6*I/e^2*n*b*d^2*x*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/6*I/e^2*n*b*d^2*x*Pi*csgn(I*x^m)*csgn
(I*f*x^m)^2-1/9*n*b*ln(x^m)*x^3+4/9*b*d^2*m*n*x/e^2-5/36*b*d*m*n*x^2/e-1/9*b*d^3*m*n*ln(e*x+d)/e^3

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.06 \[ \int x^2 \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {1}{108} \, {\left (\frac {36 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{3} n}{e^{3}} + \frac {12 \, b e^{3} x^{3} \log \left ({\left (e x + d\right )}^{n}\right ) + 15 \, b d e^{2} n x^{2} - 48 \, b d^{2} e n x + 12 \, b d^{3} n \log \left (e x + d\right ) + 4 \, {\left (3 \, a e^{3} - {\left (2 \, e^{3} n - 3 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3}}{e^{3}}\right )} m + \frac {1}{18} \, {\left (6 \, b x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + 6 \, a x^{3} + b e n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )}\right )} \log \left (f x^{m}\right ) \]

[In]

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

[Out]

-1/108*(36*(log(e*x + d)*log(-(e*x + d)/d + 1) + dilog((e*x + d)/d))*b*d^3*n/e^3 + (12*b*e^3*x^3*log((e*x + d)
^n) + 15*b*d*e^2*n*x^2 - 48*b*d^2*e*n*x + 12*b*d^3*n*log(e*x + d) + 4*(3*a*e^3 - (2*e^3*n - 3*e^3*log(c))*b)*x
^3)/e^3)*m + 1/18*(6*b*x^3*log((e*x + d)^n*c) + 6*a*x^3 + b*e*n*(6*d^3*log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x
^2 + 6*d^2*x)/e^3))*log(f*x^m)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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