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

Optimal. Leaf size=141 \[ \frac{x^2 e^{-\frac{2 d}{e m}} \left (f x^m\right )^{-2/m} \left (a+b \log \left (c x^n\right )\right ) \text{Ei}\left (\frac{2 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e m}-\frac{b n x^2 e^{-\frac{2 d}{e m}} \left (f x^m\right )^{-2/m} \left (d+e \log \left (f x^m\right )\right ) \text{Ei}\left (\frac{2 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e^2 m^2}+\frac{b n x^2}{2 e m} \]

[Out]

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

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Rubi [A]  time = 0.153998, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2310, 2178, 2366, 12, 15, 6482} \[ \frac{x^2 e^{-\frac{2 d}{e m}} \left (f x^m\right )^{-2/m} \left (a+b \log \left (c x^n\right )\right ) \text{Ei}\left (\frac{2 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e m}-\frac{b n x^2 e^{-\frac{2 d}{e m}} \left (f x^m\right )^{-2/m} \left (d+e \log \left (f x^m\right )\right ) \text{Ei}\left (\frac{2 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e^2 m^2}+\frac{b n x^2}{2 e m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 6482

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((a + b*x)*ExpIntegralEi[a + b*x])/b, x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.145354, size = 93, normalized size = 0.66 \[ \frac{x^2 \left (2 e^{-\frac{2 d}{e m}} \left (f x^m\right )^{-2/m} \text{Ei}\left (\frac{2 \left (d+e \log \left (f x^m\right )\right )}{e m}\right ) \left (a e m+b e m \log \left (c x^n\right )-b d n-b e n \log \left (f x^m\right )\right )+b e m n\right )}{2 e^2 m^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.092, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }{d+e\ln \left ( f{x}^{m} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.888993, size = 242, normalized size = 1.72 \begin{align*} \frac{{\left (b e m n x^{2} e^{\left (\frac{2 \,{\left (e \log \left (f\right ) + d\right )}}{e m}\right )} + 2 \,{\left (b e m \log \left (c\right ) - b e n \log \left (f\right ) + a e m - b d n\right )} \logintegral \left (x^{2} e^{\left (\frac{2 \,{\left (e \log \left (f\right ) + d\right )}}{e m}\right )}\right )\right )} e^{\left (-\frac{2 \,{\left (e \log \left (f\right ) + d\right )}}{e m}\right )}}{2 \, e^{2} m^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.35574, size = 262, normalized size = 1.86 \begin{align*} -\frac{b d n{\rm Ei}\left (\frac{2 \, d e^{\left (-1\right )}}{m} + \frac{2 \, \log \left (f\right )}{m} + 2 \, \log \left (x\right )\right ) e^{\left (-\frac{2 \, d e^{\left (-1\right )}}{m} - 2\right )}}{f^{\frac{2}{m}} m^{2}} + \frac{b{\rm Ei}\left (\frac{2 \, d e^{\left (-1\right )}}{m} + \frac{2 \, \log \left (f\right )}{m} + 2 \, \log \left (x\right )\right ) e^{\left (-\frac{2 \, d e^{\left (-1\right )}}{m} - 1\right )} \log \left (c\right )}{f^{\frac{2}{m}} m} - \frac{b n{\rm Ei}\left (\frac{2 \, d e^{\left (-1\right )}}{m} + \frac{2 \, \log \left (f\right )}{m} + 2 \, \log \left (x\right )\right ) e^{\left (-\frac{2 \, d e^{\left (-1\right )}}{m} - 1\right )} \log \left (f\right )}{f^{\frac{2}{m}} m^{2}} + \frac{a{\rm Ei}\left (\frac{2 \, d e^{\left (-1\right )}}{m} + \frac{2 \, \log \left (f\right )}{m} + 2 \, \log \left (x\right )\right ) e^{\left (-\frac{2 \, d e^{\left (-1\right )}}{m} - 1\right )}}{f^{\frac{2}{m}} m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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