3.4.55 \(\int \frac {(f x)^{-1+m} (a+b \log (c x^n))}{d+e x^m} \, dx\) [355]

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

[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 = 0.13, 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 = {2377, 2375, 2438} \begin {gather*} \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} \end {gather*}

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 2375

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 2377

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 2438

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 = 0.09, size = 141, normalized size = 1.83 \begin {gather*} \frac {x^{-m} (f x)^m \left (-b m^2 n \log ^2(x)+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 \log \left (-\frac {e x^m}{d}\right ) \log \left (d+e x^m\right )+m \log (x) \left (a m+b m \log \left (c x^n\right )-b n \log \left (d-d x^m\right )+b n \log \left (d+e x^m\right )\right )-b n \text {Li}_2\left (1+\frac {e x^m}{d}\right )\right )}{e f m^2} \end {gather*}

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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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (f x \right )^{-1+m} \left (a +b \ln \left (c \,x^{n}\right )\right )}{d +e \,x^{m}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [A]
time = 0.36, size = 79, normalized size = 1.03 \begin {gather*} \frac {{\left (b f^{m - 1} m n \log \left (x\right ) \log \left (\frac {x^{m} e + d}{d}\right ) + b f^{m - 1} n {\rm Li}_2\left (-\frac {x^{m} e + d}{d} + 1\right ) + {\left (b m \log \left (c\right ) + a m\right )} f^{m - 1} \log \left (x^{m} e + d\right )\right )} e^{\left (-1\right )}}{m^{2}} \end {gather*}

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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x^m*e + d), x)

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

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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