3.363 \(\int \frac{(f x)^{-1+m} (a+b \log (c x^n))^2}{d+e x^m} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 b n x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{e x^m}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e m^2}-\frac{2 b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (3,-\frac{e x^m}{d}\right )}{e m^3}+\frac{x^{1-m} (f x)^{m-1} \log \left (\frac{e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m} \]

[Out]

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

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Rubi [A]  time = 0.301131, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2339, 2337, 2374, 6589} \[ \frac{2 b n x^{1-m} (f x)^{m-1} \text{PolyLog}\left (2,-\frac{e x^m}{d}\right ) \left (a+b \log \left (c x^n\right )\right )}{e m^2}-\frac{2 b^2 n^2 x^{1-m} (f x)^{m-1} \text{PolyLog}\left (3,-\frac{e x^m}{d}\right )}{e m^3}+\frac{x^{1-m} (f x)^{m-1} \log \left (\frac{e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{e m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2339

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

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

Mathematica [B]  time = 0.250461, size = 502, normalized size = 3.89 \[ \frac{x^{-m} (f x)^m \left (-6 b m n \text{PolyLog}\left (2,\frac{e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-6 b^2 n^2 \text{PolyLog}\left (3,-\frac{d x^{-m}}{e}\right )-6 b^2 m n^2 \log (x) \text{PolyLog}\left (2,-\frac{d x^{-m}}{e}\right )+3 a^2 m^2 \log \left (d-d x^m\right )+3 a^2 m^3 \log (x)+6 a b m^2 \log \left (c x^n\right ) \log \left (d-d x^m\right )+6 a b m^3 \log (x) \log \left (c x^n\right )+6 a b m^2 n \log (x) \log \left (d+e x^m\right )-6 a b m n \log \left (-\frac{e x^m}{d}\right ) \log \left (d+e x^m\right )-6 a b m^2 n \log (x) \log \left (d-d x^m\right )-6 a b m^3 n \log ^2(x)+6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d+e x^m\right )-6 b^2 m n \log \left (c x^n\right ) \log \left (-\frac{e x^m}{d}\right ) \log \left (d+e x^m\right )+3 b^2 m^2 \log ^2\left (c x^n\right ) \log \left (d-d x^m\right )-6 b^2 m^2 n \log (x) \log \left (c x^n\right ) \log \left (d-d x^m\right )+3 b^2 m^3 \log (x) \log ^2\left (c x^n\right )-6 b^2 m^3 n \log ^2(x) \log \left (c x^n\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (\frac{d x^{-m}}{e}+1\right )-6 b^2 m^2 n^2 \log ^2(x) \log \left (d+e x^m\right )+6 b^2 m n^2 \log (x) \log \left (-\frac{e x^m}{d}\right ) \log \left (d+e x^m\right )+3 b^2 m^2 n^2 \log ^2(x) \log \left (d-d x^m\right )+4 b^2 m^3 n^2 \log ^3(x)\right )}{3 e f m^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 1.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{-1+m} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{2}}{d+e{x}^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} f^{m - 1} \log \left (\frac{e x^{m} + d}{e}\right )}{e m} + \int \frac{b^{2} f^{m} x^{m} \log \left (x^{n}\right )^{2} + 2 \,{\left (b^{2} f^{m} \log \left (c\right ) + a b f^{m}\right )} x^{m} \log \left (x^{n}\right ) +{\left (b^{2} f^{m} \log \left (c\right )^{2} + 2 \, a b f^{m} \log \left (c\right )\right )} x^{m}}{e f x x^{m} + d f x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.39031, size = 419, normalized size = 3.25 \begin{align*} -\frac{2 \, b^{2} f^{m - 1} n^{2}{\rm polylog}\left (3, -\frac{e x^{m}}{d}\right ) - 2 \,{\left (b^{2} m n^{2} \log \left (x\right ) + b^{2} m n \log \left (c\right ) + a b m n\right )} f^{m - 1}{\rm Li}_2\left (-\frac{e x^{m} + d}{d} + 1\right ) -{\left (b^{2} m^{2} \log \left (c\right )^{2} + 2 \, a b m^{2} \log \left (c\right ) + a^{2} m^{2}\right )} f^{m - 1} \log \left (e x^{m} + d\right ) -{\left (b^{2} m^{2} n^{2} \log \left (x\right )^{2} + 2 \,{\left (b^{2} m^{2} n \log \left (c\right ) + a b m^{2} n\right )} \log \left (x\right )\right )} f^{m - 1} \log \left (\frac{e x^{m} + d}{d}\right )}{e m^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )}^{2} \left (f x\right )^{m - 1}}{e x^{m} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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