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

Optimal. Leaf size=141 \[ \frac {x^3 e^{-\frac {3 d}{e m}} \left (f x^m\right )^{-3/m} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e m}-\frac {b n x^3 e^{-\frac {3 d}{e m}} \left (f x^m\right )^{-3/m} \left (d+e \log \left (f x^m\right )\right ) \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e^2 m^2}+\frac {b n x^3}{3 e m} \]

[Out]

1/3*b*n*x^3/e/m-b*n*x^3*Ei(3*(d+e*ln(f*x^m))/e/m)*(d+e*ln(f*x^m))/e^2/exp(3*d/e/m)/m^2/((f*x^m)^(3/m))+x^3*Ei(
3*(d+e*ln(f*x^m))/e/m)*(a+b*ln(c*x^n))/e/exp(3*d/e/m)/m/((f*x^m)^(3/m))

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Rubi [A]  time = 0.18, antiderivative size = 141, 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} \[ \frac {x^3 e^{-\frac {3 d}{e m}} \left (f x^m\right )^{-3/m} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e m}-\frac {b n x^3 e^{-\frac {3 d}{e m}} \left (f x^m\right )^{-3/m} \left (d+e \log \left (f x^m\right )\right ) \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e^2 m^2}+\frac {b n x^3}{3 e m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {x^2 \left (a+b \log \left (c x^n\right )\right )}{d+e \log \left (f x^m\right )} \, dx &=\frac {e^{-\frac {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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 {3 d}{e m}} x^2 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{e m} \, dx\\ &=\frac {e^{-\frac {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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 {3 d}{e m}} n\right ) \int x^2 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right ) \, dx}{e m}\\ &=\frac {e^{-\frac {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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 {3 d}{e m}} n x^3 \left (f x^m\right )^{-3/m}\right ) \int \frac {\text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right )}{x} \, dx}{e m}\\ &=\frac {e^{-\frac {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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 {3 d}{e m}} n x^3 \left (f x^m\right )^{-3/m}\right ) \operatorname {Subst}\left (\int \text {Ei}\left (\frac {3 (d+e x)}{e m}\right ) \, dx,x,\log \left (f x^m\right )\right )}{e m^2}\\ &=\frac {e^{-\frac {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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 {3 d}{e m}} n x^3 \left (f x^m\right )^{-3/m}\right ) \operatorname {Subst}\left (\int \text {Ei}(x) \, dx,x,\frac {3 d}{e m}+\frac {3 \log \left (f x^m\right )}{m}\right )}{3 e m}\\ &=\frac {b n x^3}{3 e m}-\frac {b e^{-\frac {3 d}{e m}} n x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 d}{e m}+\frac {3 \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 {3 d}{e m}} x^3 \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \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*}

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Mathematica [A]  time = 0.17, size = 93, normalized size = 0.66 \[ \frac {x^3 \left (3 e^{-\frac {3 d}{e m}} \left (f x^m\right )^{-3/m} \text {Ei}\left (\frac {3 \left (d+e \log \left (f x^m\right )\right )}{e m}\right ) \left (a e m+b e m \log \left (c x^n\right )-b d n-b e n \log \left (f x^m\right )\right )+b e m n\right )}{3 e^2 m^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(b*e*m*n + (3*ExpIntegralEi[(3*(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^((3*d)/(e*m))*(f*x^m)^(3/m))))/(3*e^2*m^2)

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fricas [A]  time = 0.79, size = 92, normalized size = 0.65 \[ \frac {{\left (b e m n x^{3} e^{\left (\frac {3 \, {\left (e \log \relax (f) + d\right )}}{e m}\right )} + 3 \, {\left (b e m \log \relax (c) - b e n \log \relax (f) + a e m - b d n\right )} \operatorname {log\_integral}\left (x^{3} e^{\left (\frac {3 \, {\left (e \log \relax (f) + d\right )}}{e m}\right )}\right )\right )} e^{\left (-\frac {3 \, {\left (e \log \relax (f) + d\right )}}{e m}\right )}}{3 \, e^{2} m^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*e*m*n*x^3*e^(3*(e*log(f) + d)/(e*m)) + 3*(b*e*m*log(c) - b*e*n*log(f) + a*e*m - b*d*n)*log_integral(x^3
*e^(3*(e*log(f) + d)/(e*m))))*e^(-3*(e*log(f) + d)/(e*m))/(e^2*m^2)

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giac [A]  time = 0.43, size = 206, normalized size = 1.46 \[ \frac {b n x^{3} e^{\left (-1\right )}}{3 \, m} - \frac {b d n {\rm Ei}\left (\frac {3 \, d e^{\left (-1\right )}}{m} + \frac {3 \, \log \relax (f)}{m} + 3 \, \log \relax (x)\right ) e^{\left (-\frac {3 \, d e^{\left (-1\right )}}{m} - 2\right )}}{f^{\frac {3}{m}} m^{2}} + \frac {b {\rm Ei}\left (\frac {3 \, d e^{\left (-1\right )}}{m} + \frac {3 \, \log \relax (f)}{m} + 3 \, \log \relax (x)\right ) e^{\left (-\frac {3 \, d e^{\left (-1\right )}}{m} - 1\right )} \log \relax (c)}{f^{\frac {3}{m}} m} - \frac {b n {\rm Ei}\left (\frac {3 \, d e^{\left (-1\right )}}{m} + \frac {3 \, \log \relax (f)}{m} + 3 \, \log \relax (x)\right ) e^{\left (-\frac {3 \, d e^{\left (-1\right )}}{m} - 1\right )} \log \relax (f)}{f^{\frac {3}{m}} m^{2}} + \frac {a {\rm Ei}\left (\frac {3 \, d e^{\left (-1\right )}}{m} + \frac {3 \, \log \relax (f)}{m} + 3 \, \log \relax (x)\right ) e^{\left (-\frac {3 \, d e^{\left (-1\right )}}{m} - 1\right )}}{f^{\frac {3}{m}} m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*n*x^3*e^(-1)/m - b*d*n*Ei(3*d*e^(-1)/m + 3*log(f)/m + 3*log(x))*e^(-3*d*e^(-1)/m - 2)/(f^(3/m)*m^2) + b*
Ei(3*d*e^(-1)/m + 3*log(f)/m + 3*log(x))*e^(-3*d*e^(-1)/m - 1)*log(c)/(f^(3/m)*m) - b*n*Ei(3*d*e^(-1)/m + 3*lo
g(f)/m + 3*log(x))*e^(-3*d*e^(-1)/m - 1)*log(f)/(f^(3/m)*m^2) + a*Ei(3*d*e^(-1)/m + 3*log(f)/m + 3*log(x))*e^(
-3*d*e^(-1)/m - 1)/(f^(3/m)*m)

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maple [F]  time = 1.42, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2}}{e \ln \left (f \,x^{m}\right )+d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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