3.82.8 \(\int \frac {e^{-2 x} (-45 x+45 x^2-9 x^3) \log (3)}{-1000+600 x-120 x^2+8 x^3} \, dx\) [8108]

Optimal. Leaf size=21 \[ 3+\frac {9 e^{-2 x} x^2 \log (3)}{16 (-5+x)^2} \]

[Out]

9/16*x^2/exp(x)^2/(-5+x)^2*ln(3)+3

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 1608, 2326} \begin {gather*} \frac {9 e^{-2 x} x \left (5 x-x^2\right ) \log (3)}{16 \left (-x^3+15 x^2-75 x+125\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-45*x + 45*x^2 - 9*x^3)*Log[3])/(E^(2*x)*(-1000 + 600*x - 120*x^2 + 8*x^3)),x]

[Out]

(9*x*(5*x - x^2)*Log[3])/(16*E^(2*x)*(125 - 75*x + 15*x^2 - x^3))

Rule 12

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

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \int \frac {e^{-2 x} \left (-45 x+45 x^2-9 x^3\right )}{-1000+600 x-120 x^2+8 x^3} \, dx\\ &=\log (3) \int \frac {e^{-2 x} x \left (-45+45 x-9 x^2\right )}{-1000+600 x-120 x^2+8 x^3} \, dx\\ &=\frac {9 e^{-2 x} x \left (5 x-x^2\right ) \log (3)}{16 \left (125-75 x+15 x^2-x^3\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 19, normalized size = 0.90 \begin {gather*} \frac {9 e^{-2 x} x^2 \log (3)}{16 (-5+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-45*x + 45*x^2 - 9*x^3)*Log[3])/(E^(2*x)*(-1000 + 600*x - 120*x^2 + 8*x^3)),x]

[Out]

(9*x^2*Log[3])/(16*E^(2*x)*(-5 + x)^2)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(74\) vs. \(2(18)=36\).
time = 0.42, size = 75, normalized size = 3.57

method	result	size
norman	\(\frac {9 x^{2} {\mathrm e}^{-2 x} \ln \left (3\right )}{16 \left (x -5\right )^{2}}\)	\(17\)
risch	\(\frac {9 x^{2} {\mathrm e}^{-2 x} \ln \left (3\right )}{16 \left (x -5\right )^{2}}\)	\(17\)
gosper	\(\frac {9 x^{2} \ln \left (3\right ) {\mathrm e}^{-2 x}}{16 \left (x^{2}-10 x +25\right )}\)	\(22\)
default	\(\frac {9 \ln \left (3\right ) \left (-\frac {5 \,{\mathrm e}^{-2 x} \left (8 x -45\right )}{2 \left (x^{2}-10 x +25\right )}+\frac {25 \,{\mathrm e}^{-2 x} \left (6 x -35\right )}{2 \left (x^{2}-10 x +25\right )}+\frac {{\mathrm e}^{-2 x}}{2}-\frac {25 \,{\mathrm e}^{-2 x} \left (4 x -25\right )}{2 \left (x^{2}-10 x +25\right )}\right )}{8}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*x^3+45*x^2-45*x)*ln(3)/(8*x^3-120*x^2+600*x-1000)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

9/8*ln(3)*(-5/2*exp(-2*x)*(8*x-45)/(x^2-10*x+25)+25/2*exp(-2*x)*(6*x-35)/(x^2-10*x+25)+1/2*exp(-2*x)-25/2*exp(
-2*x)*(4*x-25)/(x^2-10*x+25))

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (3\right )}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3+45*x^2-45*x)*log(3)/(8*x^3-120*x^2+600*x-1000)/exp(x)^2,x, algorithm="maxima")

[Out]

9/16*x^2*e^(-2*x)*log(3)/(x^2 - 10*x + 25)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (3\right )}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3+45*x^2-45*x)*log(3)/(8*x^3-120*x^2+600*x-1000)/exp(x)^2,x, algorithm="fricas")

[Out]

9/16*x^2*e^(-2*x)*log(3)/(x^2 - 10*x + 25)

________________________________________________________________________________________

Sympy [A]
time = 0.06, size = 22, normalized size = 1.05 \begin {gather*} \frac {9 x^{2} e^{- 2 x} \log {\left (3 \right )}}{16 x^{2} - 160 x + 400} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x**3+45*x**2-45*x)*ln(3)/(8*x**3-120*x**2+600*x-1000)/exp(x)**2,x)

[Out]

9*x**2*exp(-2*x)*log(3)/(16*x**2 - 160*x + 400)

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 21, normalized size = 1.00 \begin {gather*} \frac {9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (3\right )}{16 \, {\left (x^{2} - 10 \, x + 25\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*x^3+45*x^2-45*x)*log(3)/(8*x^3-120*x^2+600*x-1000)/exp(x)^2,x, algorithm="giac")

[Out]

9/16*x^2*e^(-2*x)*log(3)/(x^2 - 10*x + 25)

________________________________________________________________________________________

Mupad [B]
time = 0.16, size = 16, normalized size = 0.76 \begin {gather*} \frac {9\,x^2\,{\mathrm {e}}^{-2\,x}\,\ln \left (3\right )}{16\,{\left (x-5\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*log(3)*(45*x - 45*x^2 + 9*x^3))/(600*x - 120*x^2 + 8*x^3 - 1000),x)

[Out]

(9*x^2*exp(-2*x)*log(3))/(16*(x - 5)^2)

________________________________________________________________________________________