\(\int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx\) [6556]

Optimal result

Integrand size = 63, antiderivative size = 20 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \log \left (e^{12+2 x}-\log (\log (2+10 x))\right ) \]

[Out]

4*ln(exp(6+x)^2-ln(ln(10*x+2)))

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6873, 6816} \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \log \left (e^{2 x+12}-\log (\log (10 x+2))\right ) \]

[In]

Int[(20 + E^(12 + 2*x)*(-8 - 40*x)*Log[2 + 10*x])/(E^(12 + 2*x)*(-1 - 5*x)*Log[2 + 10*x] + (1 + 5*x)*Log[2 + 1
0*x]*Log[Log[2 + 10*x]]),x]

[Out]

4*Log[E^(12 + 2*x) - Log[Log[2 + 10*x]]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-20-e^{12+2 x} (-8-40 x) \log (2+10 x)}{(1+5 x) \log (2+10 x) \left (e^{12+2 x}-\log (\log (2+10 x))\right )} \, dx \\ & = 4 \log \left (e^{12+2 x}-\log (\log (2+10 x))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \log \left (e^{2 (6+x)}-\log (\log (2+10 x))\right ) \]

[In]

Integrate[(20 + E^(12 + 2*x)*(-8 - 40*x)*Log[2 + 10*x])/(E^(12 + 2*x)*(-1 - 5*x)*Log[2 + 10*x] + (1 + 5*x)*Log
[2 + 10*x]*Log[Log[2 + 10*x]]),x]

[Out]

4*Log[E^(2*(6 + x)) - Log[Log[2 + 10*x]]]

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
risch	\(4 \ln \left (-{\mathrm e}^{2 x +12}+\ln \left (\ln \left (10 x +2\right )\right )\right )\)	\(20\)
parallelrisch	\(4 \ln \left ({\mathrm e}^{2 x +12}-\ln \left (\ln \left (10 x +2\right )\right )\right )\)	\(20\)

[In]

int(((-40*x-8)*exp(6+x)^2*ln(10*x+2)+20)/((1+5*x)*ln(10*x+2)*ln(ln(10*x+2))+(-5*x-1)*exp(6+x)^2*ln(10*x+2)),x,
method=_RETURNVERBOSE)

[Out]

4*ln(-exp(2*x+12)+ln(ln(10*x+2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \, \log \left (-e^{\left (2 \, x + 12\right )} + \log \left (\log \left (10 \, x + 2\right )\right )\right ) \]

[In]

integrate(((-40*x-8)*exp(6+x)^2*log(10*x+2)+20)/((1+5*x)*log(10*x+2)*log(log(10*x+2))+(-5*x-1)*exp(6+x)^2*log(
10*x+2)),x, algorithm="fricas")

[Out]

4*log(-e^(2*x + 12) + log(log(10*x + 2)))

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \log {\left (e^{2 x + 12} - \log {\left (\log {\left (10 x + 2 \right )} \right )} \right )} \]

[In]

integrate(((-40*x-8)*exp(6+x)**2*ln(10*x+2)+20)/((1+5*x)*ln(10*x+2)*ln(ln(10*x+2))+(-5*x-1)*exp(6+x)**2*ln(10*
x+2)),x)

[Out]

4*log(exp(2*x + 12) - log(log(10*x + 2)))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \, \log \left (-e^{\left (2 \, x + 12\right )} + \log \left (\log \left (2\right ) + \log \left (5 \, x + 1\right )\right )\right ) \]

[In]

integrate(((-40*x-8)*exp(6+x)^2*log(10*x+2)+20)/((1+5*x)*log(10*x+2)*log(log(10*x+2))+(-5*x-1)*exp(6+x)^2*log(
10*x+2)),x, algorithm="maxima")

[Out]

4*log(-e^(2*x + 12) + log(log(2) + log(5*x + 1)))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4 \, \log \left (-e^{\left (2 \, x + 12\right )} + \log \left (\log \left (10 \, x + 2\right )\right )\right ) \]

[In]

integrate(((-40*x-8)*exp(6+x)^2*log(10*x+2)+20)/((1+5*x)*log(10*x+2)*log(log(10*x+2))+(-5*x-1)*exp(6+x)^2*log(
10*x+2)),x, algorithm="giac")

[Out]

4*log(-e^(2*x + 12) + log(log(10*x + 2)))

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {20+e^{12+2 x} (-8-40 x) \log (2+10 x)}{e^{12+2 x} (-1-5 x) \log (2+10 x)+(1+5 x) \log (2+10 x) \log (\log (2+10 x))} \, dx=4\,\ln \left (\ln \left (\ln \left (10\,x+2\right )\right )-{\mathrm {e}}^{2\,x+12}\right ) \]

[In]

int(-(exp(2*x + 12)*log(10*x + 2)*(40*x + 8) - 20)/(log(log(10*x + 2))*log(10*x + 2)*(5*x + 1) - exp(2*x + 12)
*log(10*x + 2)*(5*x + 1)),x)

[Out]

4*log(log(log(10*x + 2)) - exp(2*x + 12))