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

Optimal. Leaf size=77 \[ \frac {x^{1-m} (f x)^{m-1} \log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e m}+\frac {b n x^{1-m} (f x)^{m-1} \text {Li}_2\left (-\frac {e x^m}{d}\right )}{e m^2} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2339, 2337, 2391} \[ \frac {b n x^{1-m} (f x)^{m-1} \text {PolyLog}\left (2,-\frac {e x^m}{d}\right )}{e m^2}+\frac {x^{1-m} (f x)^{m-1} \log \left (\frac {e x^m}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{e m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 - m)*(f*x)^(-1 + m)*(a + b*Log[c*x^n])*Log[1 + (e*x^m)/d])/(e*m) + (b*n*x^(1 - m)*(f*x)^(-1 + m)*PolyLog
[2, -((e*x^m)/d)])/(e*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 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 2391

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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 141, normalized size = 1.83 \[ \frac {x^{-m} (f x)^m \left (m \log (x) \left (a m+b m \log \left (c x^n\right )+b n \log \left (d+e x^m\right )-b n \log \left (d-d x^m\right )\right )+a m \log \left (d-d x^m\right )+b m \log \left (c x^n\right ) \log \left (d-d x^m\right )-b n \text {Li}_2\left (\frac {e x^m}{d}+1\right )-b n \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )-b m^2 n \log ^2(x)\right )}{e f m^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 77, normalized size = 1.00 \[ \frac {b f^{m - 1} m n \log \relax (x) \log \left (\frac {e x^{m} + d}{d}\right ) + b f^{m - 1} n {\rm Li}_2\left (-\frac {e x^{m} + d}{d} + 1\right ) + {\left (b m \log \relax (c) + a m\right )} f^{m - 1} \log \left (e x^{m} + d\right )}{e m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \left (f x \right )^{m -1}}{e \,x^{m}+d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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