\(\int \frac {-6 e^x+e^x (-11+3 x) \log (121-66 x+9 x^2)}{e^{25} (-11+3 x) \log ^2(121-66 x+9 x^2)} \, dx\) [7792]

Optimal result

Integrand size = 50, antiderivative size = 20 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{-25+x}}{\log \left ((2+3 (3-x))^2\right )} \]

[Out]

exp(x)/ln((11-3*x)^2)/exp(25)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {12, 6820, 2326} \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{x-25}}{\log \left ((11-3 x)^2\right )} \]

[In]

Int[(-6*E^x + E^x*(-11 + 3*x)*Log[121 - 66*x + 9*x^2])/(E^25*(-11 + 3*x)*Log[121 - 66*x + 9*x^2]^2),x]

[Out]

E^(-25 + x)/Log[(11 - 3*x)^2]

Rule 12

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

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]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{(-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx}{e^{25}} \\ & = \frac {\int \frac {e^x \left (\frac {6}{11-3 x}+\log \left ((11-3 x)^2\right )\right )}{\log ^2\left ((11-3 x)^2\right )} \, dx}{e^{25}} \\ & = \frac {e^{-25+x}}{\log \left ((11-3 x)^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{-25+x}}{\log \left ((11-3 x)^2\right )} \]

[In]

Integrate[(-6*E^x + E^x*(-11 + 3*x)*Log[121 - 66*x + 9*x^2])/(E^25*(-11 + 3*x)*Log[121 - 66*x + 9*x^2]^2),x]

[Out]

E^(-25 + x)/Log[(11 - 3*x)^2]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {{\mathrm e}^{-25} {\mathrm e}^{x}}{\ln \left (\left (-11+3 x \right )^{2}\right )}\)	\(18\)
norman	\(\frac {{\mathrm e}^{-25} {\mathrm e}^{x}}{\ln \left (9 x^{2}-66 x +121\right )}\)	\(21\)
parallelrisch	\(\frac {{\mathrm e}^{-25} {\mathrm e}^{x}}{\ln \left (9 x^{2}-66 x +121\right )}\)	\(21\)
risch	\(\frac {2 i {\mathrm e}^{x -25}}{\pi \operatorname {csgn}\left (i \left (x -\frac {11}{3}\right )\right )^{2} \operatorname {csgn}\left (i \left (x -\frac {11}{3}\right )^{2}\right )-2 \pi \,\operatorname {csgn}\left (i \left (x -\frac {11}{3}\right )\right ) \operatorname {csgn}\left (i \left (x -\frac {11}{3}\right )^{2}\right )^{2}+\pi \operatorname {csgn}\left (i \left (x -\frac {11}{3}\right )^{2}\right )^{3}+4 i \ln \left (x -\frac {11}{3}\right )}\)	\(72\)

[In]

int(((-11+3*x)*exp(x)*ln(9*x^2-66*x+121)-6*exp(x))/(-11+3*x)/exp(25)/ln(9*x^2-66*x+121)^2,x,method=_RETURNVERB
OSE)

[Out]

1/exp(25)*exp(x)/ln((-11+3*x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{\left (x - 25\right )}}{\log \left (9 \, x^{2} - 66 \, x + 121\right )} \]

[In]

integrate(((-11+3*x)*exp(x)*log(9*x^2-66*x+121)-6*exp(x))/(-11+3*x)/exp(25)/log(9*x^2-66*x+121)^2,x, algorithm
="fricas")

[Out]

e^(x - 25)/log(9*x^2 - 66*x + 121)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{x}}{e^{25} \log {\left (9 x^{2} - 66 x + 121 \right )}} \]

[In]

integrate(((-11+3*x)*exp(x)*ln(9*x**2-66*x+121)-6*exp(x))/(-11+3*x)/exp(25)/ln(9*x**2-66*x+121)**2,x)

[Out]

exp(-25)*exp(x)/log(9*x**2 - 66*x + 121)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{\left (x - 25\right )}}{2 \, \log \left (3 \, x - 11\right )} \]

[In]

integrate(((-11+3*x)*exp(x)*log(9*x^2-66*x+121)-6*exp(x))/(-11+3*x)/exp(25)/log(9*x^2-66*x+121)^2,x, algorithm
="maxima")

[Out]

1/2*e^(x - 25)/log(3*x - 11)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {e^{\left (x - 25\right )}}{\log \left (9 \, x^{2} - 66 \, x + 121\right )} \]

[In]

integrate(((-11+3*x)*exp(x)*log(9*x^2-66*x+121)-6*exp(x))/(-11+3*x)/exp(25)/log(9*x^2-66*x+121)^2,x, algorithm
="giac")

[Out]

e^(x - 25)/log(9*x^2 - 66*x + 121)

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-6 e^x+e^x (-11+3 x) \log \left (121-66 x+9 x^2\right )}{e^{25} (-11+3 x) \log ^2\left (121-66 x+9 x^2\right )} \, dx=\frac {{\mathrm {e}}^{-25}\,{\mathrm {e}}^x}{\ln \left (9\,x^2-66\,x+121\right )} \]

[In]

int(-(exp(-25)*(6*exp(x) - exp(x)*log(9*x^2 - 66*x + 121)*(3*x - 11)))/(log(9*x^2 - 66*x + 121)^2*(3*x - 11)),
x)

[Out]

(exp(-25)*exp(x))/log(9*x^2 - 66*x + 121)