\(\int \frac {e^{20-19 x} (-132+(4864-2508 x) \log (\frac {1}{32} (64-33 x)))}{(-64+33 x) \log ^2(\frac {1}{32} (64-33 x))} \, dx\) [5471]

Optimal result

Integrand size = 45, antiderivative size = 21 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=\frac {4 e^{-20 (-1+x)+x}}{\log \left (2-\frac {33 x}{32}\right )} \]

[Out]

4/exp(ln(ln(-33/32*x+2))+19*x-20)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2326} \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=\frac {4 e^{20-19 x}}{\log \left (\frac {1}{32} (64-33 x)\right )} \]

[In]

Int[(E^(20 - 19*x)*(-132 + (4864 - 2508*x)*Log[(64 - 33*x)/32]))/((-64 + 33*x)*Log[(64 - 33*x)/32]^2),x]

[Out]

(4*E^(20 - 19*x))/Log[(64 - 33*x)/32]

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{align*} \text {integral}& = \frac {4 e^{20-19 x}}{\log \left (\frac {1}{32} (64-33 x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=\frac {4 e^{20-19 x}}{\log \left (2-\frac {33 x}{32}\right )} \]

[In]

Integrate[(E^(20 - 19*x)*(-132 + (4864 - 2508*x)*Log[(64 - 33*x)/32]))/((-64 + 33*x)*Log[(64 - 33*x)/32]^2),x]

[Out]

(4*E^(20 - 19*x))/Log[2 - (33*x)/32]

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{20-19 x}}{\ln \left (-\frac {33 x}{32}+2\right )}\)	\(17\)
norman	\(\frac {4 \,{\mathrm e}^{20-19 x}}{\ln \left (-\frac {33 x}{32}+2\right )}\)	\(18\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{20-19 x}}{\ln \left (-\frac {33 x}{32}+2\right )}\)	\(18\)

[In]

int(((-2508*x+4864)*ln(-33/32*x+2)-132)/(33*x-64)/ln(-33/32*x+2)/exp(ln(ln(-33/32*x+2))+19*x-20),x,method=_RET
URNVERBOSE)

[Out]

4/ln(-33/32*x+2)*exp(20-19*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=4 \, e^{\left (-19 \, x - \log \left (\log \left (-\frac {33}{32} \, x + 2\right )\right ) + 20\right )} \]

[In]

integrate(((-2508*x+4864)*log(-33/32*x+2)-132)/(33*x-64)/log(-33/32*x+2)/exp(log(log(-33/32*x+2))+19*x-20),x,
algorithm="fricas")

[Out]

4*e^(-19*x - log(log(-33/32*x + 2)) + 20)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=\frac {4 e^{20 - 19 x}}{\log {\left (2 - \frac {33 x}{32} \right )}} \]

[In]

integrate(((-2508*x+4864)*ln(-33/32*x+2)-132)/(33*x-64)/ln(-33/32*x+2)/exp(ln(ln(-33/32*x+2))+19*x-20),x)

[Out]

4*exp(20 - 19*x)/log(2 - 33*x/32)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=-\frac {4 \, e^{20}}{5 \, e^{\left (19 \, x\right )} \log \left (2\right ) - e^{\left (19 \, x\right )} \log \left (-33 \, x + 64\right )} \]

[In]

integrate(((-2508*x+4864)*log(-33/32*x+2)-132)/(33*x-64)/log(-33/32*x+2)/exp(log(log(-33/32*x+2))+19*x-20),x,
algorithm="maxima")

[Out]

-4*e^20/(5*e^(19*x)*log(2) - e^(19*x)*log(-33*x + 64))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=-\frac {4 \, e^{\left (-19 \, x + 20\right )}}{5 \, \log \left (2\right ) - \log \left (-33 \, x + 64\right )} \]

[In]

integrate(((-2508*x+4864)*log(-33/32*x+2)-132)/(33*x-64)/log(-33/32*x+2)/exp(log(log(-33/32*x+2))+19*x-20),x,
algorithm="giac")

[Out]

-4*e^(-19*x + 20)/(5*log(2) - log(-33*x + 64))

Mupad [B] (verification not implemented)

Time = 11.95 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{20-19 x} \left (-132+(4864-2508 x) \log \left (\frac {1}{32} (64-33 x)\right )\right )}{(-64+33 x) \log ^2\left (\frac {1}{32} (64-33 x)\right )} \, dx=\frac {4\,{\mathrm {e}}^{-19\,x}\,{\mathrm {e}}^{20}}{\ln \left (2-\frac {33\,x}{32}\right )} \]

[In]

int(-(exp(20 - log(log(2 - (33*x)/32)) - 19*x)*(log(2 - (33*x)/32)*(2508*x - 4864) + 132))/(log(2 - (33*x)/32)
*(33*x - 64)),x)

[Out]

(4*exp(-19*x)*exp(20))/log(2 - (33*x)/32)