3.58.76 \(\int \frac {e^{-4+x} (-4+4 x)+(-36 e^{-3+x}+4 x) \log ^2(e^{3-x} (-9 e^{-3+x}+x))}{(9 e^{-7+2 x}-e^{-4+x} x) \log ^2(e^{3-x} (-9 e^{-3+x}+x))} \, dx\)

Optimal. Leaf size=26 \[ 4 e^{4-x}-\frac {4}{\log \left (-9+e^{3-x} x\right )} \]

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Rubi [A]  time = 1.45, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6741, 6688, 12, 6742, 2194, 6686} \begin {gather*} 4 e^{4-x}-\frac {4}{\log \left (e^{3-x} x-9\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 + x)*(-4 + 4*x) + (-36*E^(-3 + x) + 4*x)*Log[E^(3 - x)*(-9*E^(-3 + x) + x)]^2)/((9*E^(-7 + 2*x) - E
^(-4 + x)*x)*Log[E^(3 - x)*(-9*E^(-3 + x) + x)]^2),x]

[Out]

4*E^(4 - x) - 4/Log[-9 + E^(3 - x)*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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{7-x} \left (e^{-4+x} (-4+4 x)+\left (-36 e^{-3+x}+4 x\right ) \log ^2\left (e^{3-x} \left (-9 e^{-3+x}+x\right )\right )\right )}{\left (9 e^x-e^3 x\right ) \log ^2\left (e^{3-x} \left (-9 e^{-3+x}+x\right )\right )} \, dx\\ &=\int \frac {4 e^3 \left (-1+x+\left (-9 e+e^{4-x} x\right ) \log ^2\left (-9+e^{3-x} x\right )\right )}{\left (9 e^x-e^3 x\right ) \log ^2\left (-9+e^{3-x} x\right )} \, dx\\ &=\left (4 e^3\right ) \int \frac {-1+x+\left (-9 e+e^{4-x} x\right ) \log ^2\left (-9+e^{3-x} x\right )}{\left (9 e^x-e^3 x\right ) \log ^2\left (-9+e^{3-x} x\right )} \, dx\\ &=\left (4 e^3\right ) \int \left (-e^{1-x}-\frac {-1+x}{\left (-9 e^x+e^3 x\right ) \log ^2\left (-9+e^{3-x} x\right )}\right ) \, dx\\ &=-\left (\left (4 e^3\right ) \int e^{1-x} \, dx\right )-\left (4 e^3\right ) \int \frac {-1+x}{\left (-9 e^x+e^3 x\right ) \log ^2\left (-9+e^{3-x} x\right )} \, dx\\ &=4 e^{4-x}-\frac {4}{\log \left (-9+e^{3-x} x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 32, normalized size = 1.23 \begin {gather*} 4 e^3 \left (e^{1-x}-\frac {1}{e^3 \log \left (-9+e^{3-x} x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + x)*(-4 + 4*x) + (-36*E^(-3 + x) + 4*x)*Log[E^(3 - x)*(-9*E^(-3 + x) + x)]^2)/((9*E^(-7 + 2*
x) - E^(-4 + x)*x)*Log[E^(3 - x)*(-9*E^(-3 + x) + x)]^2),x]

[Out]

4*E^3*(E^(1 - x) - 1/(E^3*Log[-9 + E^(3 - x)*x]))

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fricas [B]  time = 0.86, size = 52, normalized size = 2.00 \begin {gather*} \frac {4 \, {\left (e \log \left ({\left (x - 9 \, e^{\left (x - 3\right )}\right )} e^{\left (-x + 3\right )}\right ) - e^{\left (x - 3\right )}\right )} e^{\left (-x + 3\right )}}{\log \left ({\left (x - 9 \, e^{\left (x - 3\right )}\right )} e^{\left (-x + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-36*exp(x-3)+4*x)*log((-9*exp(x-3)+x)/exp(x-3))^2+(4*x-4)*exp(x-4))/(9*exp(x-4)*exp(x-3)-x*exp(x-4
))/log((-9*exp(x-3)+x)/exp(x-3))^2,x, algorithm="fricas")

[Out]

4*(e*log((x - 9*e^(x - 3))*e^(-x + 3)) - e^(x - 3))*e^(-x + 3)/log((x - 9*e^(x - 3))*e^(-x + 3))

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giac [A]  time = 0.15, size = 44, normalized size = 1.69 \begin {gather*} \frac {4 \, {\left (x e^{4} - e^{4} \log \left (x e^{3} - 9 \, e^{x}\right ) + e^{x}\right )}}{x e^{x} - e^{x} \log \left (x e^{3} - 9 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-36*exp(x-3)+4*x)*log((-9*exp(x-3)+x)/exp(x-3))^2+(4*x-4)*exp(x-4))/(9*exp(x-4)*exp(x-3)-x*exp(x-4
))/log((-9*exp(x-3)+x)/exp(x-3))^2,x, algorithm="giac")

[Out]

4*(x*e^4 - e^4*log(x*e^3 - 9*e^x) + e^x)/(x*e^x - e^x*log(x*e^3 - 9*e^x))

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maple [A]  time = 0.44, size = 30, normalized size = 1.15

method	result	size
default	\(-\frac {4}{\ln \left (\left (-9 \,{\mathrm e}^{x -3}+x \right ) {\mathrm e}^{3-x}\right )}+4 \,{\mathrm e}^{-x +4}\)	\(30\)
norman	\(\frac {\left (-4 \,{\mathrm e}^{x -4}+4 \ln \left (\left (-9 \,{\mathrm e} \,{\mathrm e}^{x -4}+x \right ) {\mathrm e}^{-1} {\mathrm e}^{-x +4}\right )\right ) {\mathrm e}^{-x +4}}{\ln \left (\left (-9 \,{\mathrm e} \,{\mathrm e}^{x -4}+x \right ) {\mathrm e}^{-1} {\mathrm e}^{-x +4}\right )}\)	\(63\)
risch	\(4 \,{\mathrm e}^{-x +4}-\frac {8 i}{2 \pi \mathrm {csgn}\left (i {\mathrm e}^{-x +4} \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x +4}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x +4} \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right )+\pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x +4} \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x +4}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x +4} \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-x +4} \left (-{\mathrm e}^{x -3}+\frac {x}{9}\right )\right )^{3}-2 \pi +4 i \ln \relax (3)+2 i \ln \left ({\mathrm e}^{x -3}-\frac {x}{9}\right )-2 i \ln \left ({\mathrm e}^{x -4}\right )-2 i}\)	\(214\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-36*exp(x-3)+4*x)*ln((-9*exp(x-3)+x)/exp(x-3))^2+(4*x-4)*exp(x-4))/(9*exp(x-4)*exp(x-3)-x*exp(x-4))/ln((
-9*exp(x-3)+x)/exp(x-3))^2,x,method=_RETURNVERBOSE)

[Out]

-4/ln((-9*exp(x-3)+x)/exp(x-3))+4/exp(x-4)

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maxima [A]  time = 0.40, size = 44, normalized size = 1.69 \begin {gather*} \frac {4 \, {\left (x e^{4} - e^{4} \log \left (x e^{3} - 9 \, e^{x}\right ) + e^{x}\right )}}{x e^{x} - e^{x} \log \left (x e^{3} - 9 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-36*exp(x-3)+4*x)*log((-9*exp(x-3)+x)/exp(x-3))^2+(4*x-4)*exp(x-4))/(9*exp(x-4)*exp(x-3)-x*exp(x-4
))/log((-9*exp(x-3)+x)/exp(x-3))^2,x, algorithm="maxima")

[Out]

4*(x*e^4 - e^4*log(x*e^3 - 9*e^x) + e^x)/(x*e^x - e^x*log(x*e^3 - 9*e^x))

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mupad [B]  time = 4.38, size = 24, normalized size = 0.92 \begin {gather*} 4\,{\mathrm {e}}^{4-x}-\frac {4}{\ln \left (x\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3-9\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(3 - x)*(x - 9*exp(x - 3)))^2*(4*x - 36*exp(x - 3)) + exp(x - 4)*(4*x - 4))/(log(exp(3 - x)*(x -
9*exp(x - 3)))^2*(x*exp(x - 4) - 9*exp(x - 3)*exp(x - 4))),x)

[Out]

4*exp(4 - x) - 4/log(x*exp(-x)*exp(3) - 9)

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sympy [A]  time = 0.21, size = 26, normalized size = 1.00 \begin {gather*} 4 e e^{3 - x} - \frac {4}{\log {\left (\left (x - 9 e^{x - 3}\right ) e^{3 - x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-36*exp(x-3)+4*x)*ln((-9*exp(x-3)+x)/exp(x-3))**2+(4*x-4)*exp(x-4))/(9*exp(x-4)*exp(x-3)-x*exp(x-4
))/ln((-9*exp(x-3)+x)/exp(x-3))**2,x)

[Out]

4*E*exp(3 - x) - 4/log((x - 9*exp(x - 3))*exp(3 - x))

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