3.622 \(\int \frac {\log ^2(f x^p) (a+b \log (c (d+e x^m)^n))}{x} \, dx\)

Optimal. Leaf size=132 \[ \frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}+\frac {2 b n p \log \left (f x^p\right ) \text {Li}_3\left (-\frac {e x^m}{d}\right )}{m^2}-\frac {b n \log ^2\left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}-\frac {b n \log ^3\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{3 p}-\frac {2 b n p^2 \text {Li}_4\left (-\frac {e x^m}{d}\right )}{m^3} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2481, 2337, 2374, 2383, 6589} \[ \frac {2 b n p \log \left (f x^p\right ) \text {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{m^2}-\frac {b n \log ^2\left (f x^p\right ) \text {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{m}-\frac {2 b n p^2 \text {PolyLog}\left (4,-\frac {e x^m}{d}\right )}{m^3}+\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {b n \log ^3\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{3 p} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[k + 1, e*x^q]*(a + b*Log[c*x^n])^p)/q, x] - Dist[(b*n*p)/q, Int[(PolyLog[k + 1, e*x^q]*(a + b*Log[c*x^n])^(
p - 1))/x, x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

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 ^2\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{x} \, dx &=\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {(b e m n) \int \frac {x^{-1+m} \log ^3\left (f x^p\right )}{d+e x^m} \, dx}{3 p}\\ &=\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {b n \log ^3\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{3 p}+(b n) \int \frac {\log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{x} \, dx\\ &=\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {b n \log ^3\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{3 p}-\frac {b n \log ^2\left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}+\frac {(2 b n p) \int \frac {\log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{x} \, dx}{m}\\ &=\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {b n \log ^3\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{3 p}-\frac {b n \log ^2\left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}+\frac {2 b n p \log \left (f x^p\right ) \text {Li}_3\left (-\frac {e x^m}{d}\right )}{m^2}-\frac {\left (2 b n p^2\right ) \int \frac {\text {Li}_3\left (-\frac {e x^m}{d}\right )}{x} \, dx}{m^2}\\ &=\frac {\log ^3\left (f x^p\right ) \left (a+b \log \left (c \left (d+e x^m\right )^n\right )\right )}{3 p}-\frac {b n \log ^3\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{3 p}-\frac {b n \log ^2\left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{m}+\frac {2 b n p \log \left (f x^p\right ) \text {Li}_3\left (-\frac {e x^m}{d}\right )}{m^2}-\frac {2 b n p^2 \text {Li}_4\left (-\frac {e x^m}{d}\right )}{m^3}\\ \end {align*}

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Mathematica [B]  time = 0.25, size = 456, normalized size = 3.45 \[ \frac {a \log ^3\left (f x^p\right )}{3 p}-b p \log ^2(x) \log \left (f x^p\right ) \log \left (c \left (d+e x^m\right )^n\right )+b \log (x) \log ^2\left (f x^p\right ) \log \left (c \left (d+e x^m\right )^n\right )+\frac {1}{3} b p^2 \log ^3(x) \log \left (c \left (d+e x^m\right )^n\right )+\frac {2 b n p \log \left (f x^p\right ) \text {Li}_3\left (-\frac {d x^{-m}}{e}\right )}{m^2}-\frac {b n p \log (x) \left (p \log (x)-2 \log \left (f x^p\right )\right ) \text {Li}_2\left (-\frac {d x^{-m}}{e}\right )}{m}+\frac {b n \left (\log \left (f x^p\right )-p \log (x)\right )^2 \text {Li}_2\left (\frac {e x^m}{d}+1\right )}{m}-b n p \log ^2(x) \log \left (f x^p\right ) \log \left (\frac {d x^{-m}}{e}+1\right )+2 b n p \log ^2(x) \log \left (f x^p\right ) \log \left (d+e x^m\right )-b n \log (x) \log ^2\left (f x^p\right ) \log \left (d+e x^m\right )+\frac {b n \log ^2\left (f x^p\right ) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )}{m}-\frac {2 b n p \log (x) \log \left (f x^p\right ) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )}{m}+\frac {2 b n p^2 \text {Li}_4\left (-\frac {d x^{-m}}{e}\right )}{m^3}+\frac {2}{3} b n p^2 \log ^3(x) \log \left (\frac {d x^{-m}}{e}+1\right )-b n p^2 \log ^3(x) \log \left (d+e x^m\right )+\frac {b n p^2 \log ^2(x) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )}{m}-\frac {1}{3} b m n p \log ^3(x) \log \left (f x^p\right )+\frac {1}{4} b m n p^2 \log ^4(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.72, size = 281, normalized size = 2.13 \[ -\frac {6 \, b n p^{2} {\rm polylog}\left (4, -\frac {e x^{m}}{d}\right ) - 3 \, {\left (b m^{3} \log \relax (c) + a m^{3}\right )} \log \relax (f)^{2} \log \relax (x) - 3 \, {\left (b m^{3} p \log \relax (c) + a m^{3} p\right )} \log \relax (f) \log \relax (x)^{2} - {\left (b m^{3} p^{2} \log \relax (c) + a m^{3} p^{2}\right )} \log \relax (x)^{3} + 3 \, {\left (b m^{2} n p^{2} \log \relax (x)^{2} + 2 \, b m^{2} n p \log \relax (f) \log \relax (x) + b m^{2} n \log \relax (f)^{2}\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) - {\left (b m^{3} n p^{2} \log \relax (x)^{3} + 3 \, b m^{3} n p \log \relax (f) \log \relax (x)^{2} + 3 \, b m^{3} n \log \relax (f)^{2} \log \relax (x)\right )} \log \left (e x^{m} + d\right ) + {\left (b m^{3} n p^{2} \log \relax (x)^{3} + 3 \, b m^{3} n p \log \relax (f) \log \relax (x)^{2} + 3 \, b m^{3} n \log \relax (f)^{2} \log \relax (x)\right )} \log \left (\frac {e x^{m} + d}{d}\right ) - 6 \, {\left (b m n p^{2} \log \relax (x) + b m n p \log \relax (f)\right )} {\rm polylog}\left (3, -\frac {e x^{m}}{d}\right )}{3 \, m^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*b*n*p^2*polylog(4, -e*x^m/d) - 3*(b*m^3*log(c) + a*m^3)*log(f)^2*log(x) - 3*(b*m^3*p*log(c) + a*m^3*p)
*log(f)*log(x)^2 - (b*m^3*p^2*log(c) + a*m^3*p^2)*log(x)^3 + 3*(b*m^2*n*p^2*log(x)^2 + 2*b*m^2*n*p*log(f)*log(
x) + b*m^2*n*log(f)^2)*dilog(-(e*x^m + d)/d + 1) - (b*m^3*n*p^2*log(x)^3 + 3*b*m^3*n*p*log(f)*log(x)^2 + 3*b*m
^3*n*log(f)^2*log(x))*log(e*x^m + d) + (b*m^3*n*p^2*log(x)^3 + 3*b*m^3*n*p*log(f)*log(x)^2 + 3*b*m^3*n*log(f)^
2*log(x))*log((e*x^m + d)/d) - 6*(b*m*n*p^2*log(x) + b*m*n*p*log(f))*polylog(3, -e*x^m/d))/m^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(f*x^p)^2*(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)^2/x, x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e \,x^{m}+d \right )^{n}\right )+a \right ) \ln \left (f \,x^{p}\right )^{2}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (b p^{2} \log \relax (x)^{3} - 3 \, b p \log \relax (f) \log \relax (x)^{2} + 3 \, b \log \relax (f)^{2} \log \relax (x) + 3 \, b \log \relax (x) \log \left (x^{p}\right )^{2} - 3 \, {\left (b p \log \relax (x)^{2} - 2 \, b \log \relax (f) \log \relax (x)\right )} \log \left (x^{p}\right )\right )} \log \left ({\left (e x^{m} + d\right )}^{n}\right ) - \int -\frac {3 \, b d \log \relax (c) \log \relax (f)^{2} + 3 \, a d \log \relax (f)^{2} + 3 \, {\left (b d \log \relax (c) + a d - {\left (b e m n \log \relax (x) - b e \log \relax (c) - a e\right )} x^{m}\right )} \log \left (x^{p}\right )^{2} - {\left (b e m n p^{2} \log \relax (x)^{3} - 3 \, b e m n p \log \relax (f) \log \relax (x)^{2} + 3 \, b e m n \log \relax (f)^{2} \log \relax (x) - 3 \, b e \log \relax (c) \log \relax (f)^{2} - 3 \, a e \log \relax (f)^{2}\right )} x^{m} + 3 \, {\left (2 \, b d \log \relax (c) \log \relax (f) + 2 \, a d \log \relax (f) + {\left (b e m n p \log \relax (x)^{2} - 2 \, b e m n \log \relax (f) \log \relax (x) + 2 \, b e \log \relax (c) \log \relax (f) + 2 \, a e \log \relax (f)\right )} x^{m}\right )} \log \left (x^{p}\right )}{3 \, {\left (e x x^{m} + d x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\ln \left (f\,x^p\right )}^2\,\left (a+b\,\ln \left (c\,{\left (d+e\,x^m\right )}^n\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*log(c*(d + e*x**m)**n))*log(f*x**p)**2/x, x)

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