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

Optimal. Leaf size=133 \[ \frac {e^{\frac {d}{e m}} \left (f x^m\right )^{\frac {1}{m}} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (-\frac {d+e \log \left (f x^m\right )}{e m}\right )}{e m x}-\frac {b n e^{\frac {d}{e m}} \left (f x^m\right )^{\frac {1}{m}} \left (d+e \log \left (f x^m\right )\right ) \text {Ei}\left (-\frac {d+e \log \left (f x^m\right )}{e m}\right )}{e^2 m^2 x}-\frac {b n}{e m x} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2310, 2178, 2366, 12, 15, 6482} \[ \frac {e^{\frac {d}{e m}} \left (f x^m\right )^{\frac {1}{m}} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (-\frac {d+e \log \left (f x^m\right )}{e m}\right )}{e m x}-\frac {b n e^{\frac {d}{e m}} \left (f x^m\right )^{\frac {1}{m}} \left (d+e \log \left (f x^m\right )\right ) \text {Ei}\left (-\frac {d+e \log \left (f x^m\right )}{e m}\right )}{e^2 m^2 x}-\frac {b n}{e m x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 87, normalized size = 0.65 \[ \frac {e^{\frac {d}{e m}} \left (f x^m\right )^{\frac {1}{m}} \text {Ei}\left (-\frac {d+e \log \left (f x^m\right )}{e m}\right ) \left (a e m+b e m \log \left (c x^n\right )-b d n-b e n \log \left (f x^m\right )\right )-b e m n}{e^2 m^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*e*m*n) + E^(d/(e*m))*(f*x^m)^m^(-1)*ExpIntegralEi[-((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*m^2*x)

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fricas [A]  time = 0.61, size = 81, normalized size = 0.61 \[ -\frac {b e m n - {\left (b e m x \log \relax (c) - b e n x \log \relax (f) + {\left (a e m - b d n\right )} x\right )} e^{\left (\frac {e \log \relax (f) + d}{e m}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {e \log \relax (f) + d}{e m}\right )}}{x}\right )}{e^{2} m^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.27, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\left (e \ln \left (f \,x^{m}\right )+d \right ) x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*log(c*x**n))/(x**2*(d + e*log(f*x**m))), x)

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