\(\int x \log (f x^m) (a+b \log (c (d+e x)^n)) \, dx\) [360]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2473, 45, 2393, 2332, 2341, 2354, 2438} \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (\frac {e x}{d}+1\right ) \log \left (f x^m\right )}{2 e^2}-\frac {b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{2 e^2}+\frac {b d^2 m n \log (d+e x)}{4 e^2}+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {3 b d m n x}{4 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {1}{4} b m n x^2 \]

[In]

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

[Out]

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

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}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b e n) \int \frac {x^2 \log \left (f x^m\right )}{d+e x} \, dx+\frac {1}{4} (b e m n) \int \frac {x^2}{d+e x} \, dx \\ & = -\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b e n) \int \left (-\frac {d \log \left (f x^m\right )}{e^2}+\frac {x \log \left (f x^m\right )}{e}+\frac {d^2 \log \left (f x^m\right )}{e^2 (d+e x)}\right ) \, dx+\frac {1}{4} (b e m n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx \\ & = -\frac {b d m n x}{4 e}+\frac {1}{8} b m n x^2+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b n) \int x \log \left (f x^m\right ) \, dx+\frac {(b d n) \int \log \left (f x^m\right ) \, dx}{2 e}-\frac {\left (b d^2 n\right ) \int \frac {\log \left (f x^m\right )}{d+e x} \, dx}{2 e} \\ & = -\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}+\frac {\left (b d^2 m n\right ) \int \frac {\log \left (1+\frac {e x}{d}\right )}{x} \, dx}{2 e^2} \\ & = -\frac {3 b d m n x}{4 e}+\frac {1}{4} b m n x^2+\frac {b d n x \log \left (f x^m\right )}{2 e}-\frac {1}{4} b n x^2 \log \left (f x^m\right )+\frac {b d^2 m n \log (d+e x)}{4 e^2}-\frac {1}{4} \left (m x^2-2 x^2 \log \left (f x^m\right )\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b d^2 n \log \left (f x^m\right ) \log \left (1+\frac {e x}{d}\right )}{2 e^2}-\frac {b d^2 m n \text {Li}_2\left (-\frac {e x}{d}\right )}{2 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {\log \left (f x^m\right ) \left (-2 b d^2 n \log (d+e x)+e x \left (2 b d n+2 a e x-b e n x+2 b e x \log \left (c (d+e x)^n\right )\right )\right )+m \left (-3 b d e n x-a e^2 x^2+b e^2 n x^2+b d^2 n (1+2 \log (x)) \log (d+e x)-b e^2 x^2 \log \left (c (d+e x)^n\right )-2 b d^2 n \log (x) \log \left (1+\frac {e x}{d}\right )\right )-2 b d^2 m n \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 e^2} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 13.04 (sec) , antiderivative size = 843, normalized size of antiderivative = 5.34

method	result	size
risch	\(\text {Expression too large to display}\)	\(843\)

[In]

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

[Out]

-1/4*I/e*n*b*d*x*Pi*csgn(I*f*x^m)^3+1/4*I/e^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*f*x^m)^3+1/8*I*n*b*x^2*Pi*csgn(I*f)*
csgn(I*x^m)*csgn(I*f*x^m)-1/2/e^2*n*b*ln(x^m)*d^2*ln(e*x+d)+1/2/e^2*n*b*m*d^2*dilog(-e*x/d)-1/2/e^2*n*b*d^2*ln
(e*x+d)*ln(f)+1/8*I*n*b*x^2*Pi*csgn(I*f*x^m)^3-1/4*I/e^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/
e^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/8*I*n*b*x^2*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*I/e*n*b*
d*x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I/e^2*n*b*d^2*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+
1/4*I/e*n*b*d*x*Pi*csgn(I*f)*csgn(I*f*x^m)^2+1/4*I/e*n*b*d*x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/2/e^2*n*b*m*d^2*
ln(e*x+d)*ln(-e*x/d)+1/4*b*m*n*x^2+(1/2*b*x^2*ln(x^m)+1/4*b*x^2*(-I*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*P
i*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)-m))*ln((e*x+d)^n)-1/
8*I*n*b*x^2*Pi*csgn(I*f)*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)*(1/2*(-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*P
i*csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi*csgn(I*f*x^m)^3+2*ln(f))*x^2+x^2*ln(x^m)-1/2*m*x^2)-1/4*n*b*x^2*ln(f)-1/4*n
*b*ln(x^m)*x^2-5/8/e^2*n*b*m*d^2+1/2/e*n*b*d*x*ln(f)+1/2/e*n*b*ln(x^m)*d*x-3/4*b*d*m*n*x/e+1/4*b*d^2*m*n*ln(e*
x+d)/e^2

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.13 \[ \int x \log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (e x + d\right ) \log \left (-\frac {e x + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x + d}{d}\right )\right )} b d^{2} n}{e^{2}} - \frac {b e^{2} x^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + 3 \, b d e n x - b d^{2} n \log \left (e x + d\right ) + {\left (a e^{2} - {\left (e^{2} n - e^{2} \log \left (c\right )\right )} b\right )} x^{2}}{e^{2}}\right )} m - \frac {1}{4} \, {\left (b e n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} - 2 \, b x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) - 2 \, a x^{2}\right )} \log \left (f x^{m}\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

integrate(x*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*log(f*x^m), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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