3.623 \(\int \frac {\log (f x^p) (a+b \log (c (d+e x^m)^n))}{x} \, dx\)

Optimal. Leaf size=102 \[ \frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {b n \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}-\frac {b n \log ^2\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{2 p}+\frac {b n p \text {Li}_3\left (-\frac {e x^m}{d}\right )}{m^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2481, 2337, 2374, 6589} \[ -\frac {b n \log \left (f x^p\right ) \text {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{m}+\frac {b n p \text {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{m^2}+\frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {b n \log ^2\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{2 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2481

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

Rule 6589

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

Rubi steps

\begin {align*} \int \frac {\log \left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{x} \, dx &=\frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {(b e m n) \int \frac {x^{-1+m} \log ^2\left (f x^p\right )}{d+e x^m} \, dx}{2 p}\\ &=\frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {b n \log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{2 p}+(b n) \int \frac {\log \left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{x} \, dx\\ &=\frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {b n \log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{2 p}-\frac {b n \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}+\frac {(b n p) \int \frac {\text {Li}_2\left (-\frac {e x^m}{d}\right )}{x} \, dx}{m}\\ &=\frac {\log ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{2 p}-\frac {b n \log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{2 p}-\frac {b n \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}+\frac {b n p \text {Li}_3\left (-\frac {e x^m}{d}\right )}{m^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.19, size = 265, normalized size = 2.60 \[ \frac {a \log ^2\left (f x^p\right )}{2 p}+b \log (x) \log \left (f x^p\right ) \log \left (c \left (d+e x^m\right )^n\right )-\frac {1}{2} b p \log ^2(x) \log \left (c \left (d+e x^m\right )^n\right )-\frac {b n \left (p \log (x)-\log \left (f x^p\right )\right ) \text {Li}_2\left (\frac {e x^m}{d}+1\right )}{m}-b n \log (x) \log \left (f x^p\right ) \log \left (d+e x^m\right )+\frac {b n \log \left (f x^p\right ) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )}{m}+\frac {b n p \text {Li}_3\left (-\frac {d x^{-m}}{e}\right )}{m^2}+\frac {b n p \log (x) \text {Li}_2\left (-\frac {d x^{-m}}{e}\right )}{m}-\frac {1}{2} b n p \log ^2(x) \log \left (\frac {d x^{-m}}{e}+1\right )+b n p \log ^2(x) \log \left (d+e x^m\right )-\frac {b n p \log (x) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )}{m}-\frac {1}{6} b m n p \log ^3(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*m*n*p*Log[x]^3) + (a*Log[f*x^p]^2)/(2*p) - (b*n*p*Log[x]^2*Log[1 + d/(e*x^m)])/2 + b*n*p*Log[x]^2*Log[
d + e*x^m] - (b*n*p*Log[x]*Log[-((e*x^m)/d)]*Log[d + e*x^m])/m - b*n*Log[x]*Log[f*x^p]*Log[d + e*x^m] + (b*n*L
og[-((e*x^m)/d)]*Log[f*x^p]*Log[d + e*x^m])/m - (b*p*Log[x]^2*Log[c*(d + e*x^m)^n])/2 + b*Log[x]*Log[f*x^p]*Lo
g[c*(d + e*x^m)^n] + (b*n*p*Log[x]*PolyLog[2, -(d/(e*x^m))])/m - (b*n*(p*Log[x] - Log[f*x^p])*PolyLog[2, 1 + (
e*x^m)/d])/m + (b*n*p*PolyLog[3, -(d/(e*x^m))])/m^2

________________________________________________________________________________________

fricas [C]  time = 0.74, size = 161, normalized size = 1.58 \[ \frac {2 \, b n p {\rm polylog}\left (3, -\frac {e x^{m}}{d}\right ) + 2 \, {\left (b m^{2} \log \relax (c) + a m^{2}\right )} \log \relax (f) \log \relax (x) + {\left (b m^{2} p \log \relax (c) + a m^{2} p\right )} \log \relax (x)^{2} - 2 \, {\left (b m n p \log \relax (x) + b m n \log \relax (f)\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) + {\left (b m^{2} n p \log \relax (x)^{2} + 2 \, b m^{2} n \log \relax (f) \log \relax (x)\right )} \log \left (e x^{m} + d\right ) - {\left (b m^{2} n p \log \relax (x)^{2} + 2 \, b m^{2} n \log \relax (f) \log \relax (x)\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{2 \, m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left ({\left (e x^{m} + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{p}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e \,x^{m}+d \right )^{n}\right )+a \right ) \ln \left (f \,x^{p}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (b p \log \relax (x)^{2} - 2 \, b \log \relax (f) \log \relax (x) - 2 \, b \log \relax (x) \log \left (x^{p}\right )\right )} \log \left ({\left (e x^{m} + d\right )}^{n}\right ) - \int -\frac {2 \, b d \log \relax (c) \log \relax (f) + 2 \, a d \log \relax (f) + {\left (b e m n p \log \relax (x)^{2} - 2 \, b e m n \log \relax (f) \log \relax (x) + 2 \, b e \log \relax (c) \log \relax (f) + 2 \, a e \log \relax (f)\right )} x^{m} + 2 \, {\left (b d \log \relax (c) + a d - {\left (b e m n \log \relax (x) - b e \log \relax (c) - a e\right )} x^{m}\right )} \log \left (x^{p}\right )}{2 \, {\left (e x x^{m} + d x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (f\,x^p\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x^m\right )}^n\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________