3.620 \(\int \frac {x^{-1+m} \log ^2(f x^p)}{d+e x^m} \, dx\)

Optimal. Leaf size=75 \[ \frac {2 p \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2}+\frac {\log ^2\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{e m}-\frac {2 p^2 \text {Li}_3\left (-\frac {e x^m}{d}\right )}{e m^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2337, 2374, 6589} \[ \frac {2 p \log \left (f x^p\right ) \text {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2}-\frac {2 p^2 \text {PolyLog}\left (3,-\frac {e x^m}{d}\right )}{e m^3}+\frac {\log ^2\left (f x^p\right ) \log \left (\frac {e x^m}{d}+1\right )}{e m} \]

Antiderivative was successfully verified.

[In]

Int[(x^(-1 + m)*Log[f*x^p]^2)/(d + e*x^m),x]

[Out]

(Log[f*x^p]^2*Log[1 + (e*x^m)/d])/(e*m) + (2*p*Log[f*x^p]*PolyLog[2, -((e*x^m)/d)])/(e*m^2) - (2*p^2*PolyLog[3
, -((e*x^m)/d)])/(e*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 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 {x^{-1+m} \log ^2\left (f x^p\right )}{d+e x^m} \, dx &=\frac {\log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}-\frac {(2 p) \int \frac {\log \left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{x} \, dx}{e m}\\ &=\frac {\log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {2 p \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2}-\frac {\left (2 p^2\right ) \int \frac {\text {Li}_2\left (-\frac {e x^m}{d}\right )}{x} \, dx}{e m^2}\\ &=\frac {\log ^2\left (f x^p\right ) \log \left (1+\frac {e x^m}{d}\right )}{e m}+\frac {2 p \log \left (f x^p\right ) \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2}-\frac {2 p^2 \text {Li}_3\left (-\frac {e x^m}{d}\right )}{e m^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.14, size = 210, normalized size = 2.80 \[ \frac {3 m^2 \log ^2\left (f x^p\right ) \log \left (d+e x^m\right )+6 m p \left (p \log (x)-\log \left (f x^p\right )\right ) \text {Li}_2\left (\frac {e x^m}{d}+1\right )-6 m p \log \left (f x^p\right ) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+3 m^2 p^2 \log ^2(x) \log \left (\frac {d x^{-m}}{e}+1\right )-3 m^2 p^2 \log ^2(x) \log \left (d+e x^m\right )-6 p^2 \text {Li}_3\left (-\frac {d x^{-m}}{e}\right )-6 m p^2 \log (x) \text {Li}_2\left (-\frac {d x^{-m}}{e}\right )+6 m p^2 \log (x) \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+m^3 p^2 \log ^3(x)}{3 e m^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(-1 + m)*Log[f*x^p]^2)/(d + e*x^m),x]

[Out]

(m^3*p^2*Log[x]^3 + 3*m^2*p^2*Log[x]^2*Log[1 + d/(e*x^m)] - 3*m^2*p^2*Log[x]^2*Log[d + e*x^m] + 6*m*p^2*Log[x]
*Log[-((e*x^m)/d)]*Log[d + e*x^m] - 6*m*p*Log[-((e*x^m)/d)]*Log[f*x^p]*Log[d + e*x^m] + 3*m^2*Log[f*x^p]^2*Log
[d + e*x^m] - 6*m*p^2*Log[x]*PolyLog[2, -(d/(e*x^m))] + 6*m*p*(p*Log[x] - Log[f*x^p])*PolyLog[2, 1 + (e*x^m)/d
] - 6*p^2*PolyLog[3, -(d/(e*x^m))])/(3*e*m^3)

________________________________________________________________________________________

fricas [C]  time = 0.82, size = 105, normalized size = 1.40 \[ \frac {m^{2} \log \left (e x^{m} + d\right ) \log \relax (f)^{2} - 2 \, p^{2} {\rm polylog}\left (3, -\frac {e x^{m}}{d}\right ) + 2 \, {\left (m p^{2} \log \relax (x) + m p \log \relax (f)\right )} {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) + {\left (m^{2} p^{2} \log \relax (x)^{2} + 2 \, m^{2} p \log \relax (f) \log \relax (x)\right )} \log \left (\frac {e x^{m} + d}{d}\right )}{e m^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+m)*log(f*x^p)^2/(d+e*x^m),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+m)*log(f*x^p)^2/(d+e*x^m),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.48, size = 1373, normalized size = 18.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m-1)*ln(f*x^p)^2/(e*x^m+d),x)

[Out]

1/m*ln(e*x^m+d)/e*ln(f)^2+1/m*(-p*ln(x)+ln(x^p))^2*ln(e*x^m+d)/e+I/m*ln(e*x^m+d)/e*p*ln(x)*Pi*csgn(I*f)*csgn(I
*x^p)*csgn(I*f*x^p)+I/m*ln(e*x^m+d)/e*ln(f)*Pi*csgn(I*f)*csgn(I*f*x^p)^2+I/m*ln(e*x^m+d)/e*ln(x^p)*Pi*csgn(I*f
)*csgn(I*f*x^p)^2+I/m*ln(e*x^m+d)/e*ln(f)*Pi*csgn(I*x^p)*csgn(I*f*x^p)^2-I/m*p*ln(x)*ln((e*x^m+d)/d)/e*Pi*csgn
(I*f*x^p)^3-I/m*ln(e*x^m+d)/e*ln(x^p)*Pi*csgn(I*f*x^p)^3+I/m^2*p*dilog((e*x^m+d)/d)/e*Pi*csgn(I*f)*csgn(I*f*x^
p)^2+I/m*ln(e*x^m+d)/e*p*ln(x)*Pi*csgn(I*f*x^p)^3+2/m*p*ln(x)*ln((e*x^m+d)/d)/e*ln(f)-2/m*ln(e*x^m+d)/e*p*ln(x
)*ln(f)+1/2/m*ln(e*x^m+d)/e*Pi^2*csgn(I*x^p)*csgn(I*f*x^p)^5+I/m^2*p*dilog((e*x^m+d)/d)/e*Pi*csgn(I*x^p)*csgn(
I*f*x^p)^2-1/4/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f)^2*csgn(I*f*x^p)^4-1/4/m*ln(e*x^m+d)/e*Pi^2*csgn(I*x^p)^2*csgn(I*
f*x^p)^4+1/2/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f)*csgn(I*f*x^p)^5+I/m*ln(e*x^m+d)/e*ln(x^p)*Pi*csgn(I*x^p)*csgn(I*f*
x^p)^2-2*p^2*polylog(3,-1/d*e*x^m)/e/m^3+2/m*p*(-p*ln(x)+ln(x^p))*ln(x)*ln((e*x^m+d)/d)/e-I/m*p*ln(x)*ln((e*x^
m+d)/d)/e*Pi*csgn(I*f)*csgn(I*x^p)*csgn(I*f*x^p)+I/m*p*ln(x)*ln((e*x^m+d)/d)/e*Pi*csgn(I*x^p)*csgn(I*f*x^p)^2-
I/m^2*p*dilog((e*x^m+d)/d)/e*Pi*csgn(I*f)*csgn(I*x^p)*csgn(I*f*x^p)-I/m*ln(e*x^m+d)/e*ln(x^p)*Pi*csgn(I*f)*csg
n(I*x^p)*csgn(I*f*x^p)+I/m*p*ln(x)*ln((e*x^m+d)/d)/e*Pi*csgn(I*f)*csgn(I*f*x^p)^2-I/m*ln(e*x^m+d)/e*p*ln(x)*Pi
*csgn(I*x^p)*csgn(I*f*x^p)^2-I/m^2*p*dilog((e*x^m+d)/d)/e*Pi*csgn(I*f*x^p)^3-I/m*ln(e*x^m+d)/e*ln(f)*Pi*csgn(I
*f*x^p)^3+1/2/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f)^2*csgn(I*x^p)*csgn(I*f*x^p)^3-1/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f)*cs
gn(I*x^p)*csgn(I*f*x^p)^4+1/2/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f)*csgn(I*x^p)^2*csgn(I*f*x^p)^3-1/4/m*ln(e*x^m+d)/e
*Pi^2*csgn(I*f)^2*csgn(I*x^p)^2*csgn(I*f*x^p)^2-I/m*ln(e*x^m+d)/e*ln(f)*Pi*csgn(I*f)*csgn(I*x^p)*csgn(I*f*x^p)
-I/m*ln(e*x^m+d)/e*p*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^p)^2+2/m^2*p^2/e*ln(x)*polylog(2,-1/d*e*x^m)+2/m^2*p*(-p*ln
(x)+ln(x^p))*dilog((e*x^m+d)/d)/e+2/m*ln(e*x^m+d)/e*ln(x^p)*ln(f)+2/m^2*p*dilog((e*x^m+d)/d)/e*ln(f)+1/m*p^2/e
*ln(x)^2*ln(1/d*e*x^m+1)-1/4/m*ln(e*x^m+d)/e*Pi^2*csgn(I*f*x^p)^6

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+m)*log(f*x^p)^2/(d+e*x^m),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(m - 1)*log(f*x^p)^2)/(d + e*x^m),x)

[Out]

int((x^(m - 1)*log(f*x^p)^2)/(d + e*x^m), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m - 1} \log {\left (f x^{p} \right )}^{2}}{d + e x^{m}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+m)*ln(f*x**p)**2/(d+e*x**m),x)

[Out]

Integral(x**(m - 1)*log(f*x**p)**2/(d + e*x**m), x)

________________________________________________________________________________________