3.68.30 \(\int \frac {15+60 e^{1-4 x}}{-100+75 e^2-75 x+(-40+30 e^2-30 x) \log (5)+(-4+3 e^2-3 x) \log ^2(5)+e^{1-4 x} (75+30 \log (5)+3 \log ^2(5))+(-40+30 e^2-30 x+(-8+6 e^2-6 x) \log (5)+e^{1-4 x} (30+6 \log (5))) \log (-4+3 e^2+3 e^{1-4 x}-3 x)+(-4+3 e^2+3 e^{1-4 x}-3 x) \log ^2(-4+3 e^2+3 e^{1-4 x}-3 x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {5}{5+\log (5)+\log \left (-4+3 \left (e^2+e^{\frac {x-4 x^2}{x}}-x\right )\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 28, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, integrand size = 172, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 6686} \begin {gather*} \frac {5}{\log \left (-3 x+3 e^{1-4 x}+3 e^2-4\right )+5+\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 60*E^(1 - 4*x))/(-100 + 75*E^2 - 75*x + (-40 + 30*E^2 - 30*x)*Log[5] + (-4 + 3*E^2 - 3*x)*Log[5]^2 +
E^(1 - 4*x)*(75 + 30*Log[5] + 3*Log[5]^2) + (-40 + 30*E^2 - 30*x + (-8 + 6*E^2 - 6*x)*Log[5] + E^(1 - 4*x)*(3
0 + 6*Log[5]))*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x] + (-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x)*Log[-4 + 3*E^2 + 3*E
^(1 - 4*x) - 3*x]^2),x]

[Out]

5/(5 + Log[5] + Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*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 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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 \left (4 e+e^{4 x}\right )}{\left (3 e+3 e^{2+4 x}-e^{4 x} (4+3 x)\right ) \left (5 \left (1+\frac {\log (5)}{5}\right )+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )\right )^2} \, dx\\ &=15 \int \frac {4 e+e^{4 x}}{\left (3 e+3 e^{2+4 x}-e^{4 x} (4+3 x)\right ) \left (5 \left (1+\frac {\log (5)}{5}\right )+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )\right )^2} \, dx\\ &=\frac {5}{5+\log (5)+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 28, normalized size = 0.85 \begin {gather*} \frac {5}{5+\log (5)+\log \left (-4+3 e^2+3 e^{1-4 x}-3 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 60*E^(1 - 4*x))/(-100 + 75*E^2 - 75*x + (-40 + 30*E^2 - 30*x)*Log[5] + (-4 + 3*E^2 - 3*x)*Log[
5]^2 + E^(1 - 4*x)*(75 + 30*Log[5] + 3*Log[5]^2) + (-40 + 30*E^2 - 30*x + (-8 + 6*E^2 - 6*x)*Log[5] + E^(1 - 4
*x)*(30 + 6*Log[5]))*Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x] + (-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x)*Log[-4 + 3*E^2
+ 3*E^(1 - 4*x) - 3*x]^2),x]

[Out]

5/(5 + Log[5] + Log[-4 + 3*E^2 + 3*E^(1 - 4*x) - 3*x])

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 26, normalized size = 0.79 \begin {gather*} \frac {5}{\log \relax (5) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+3
0)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5)+30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log
(5)+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75*exp(2)-75*x-100),x, algorithm="fri
cas")

[Out]

5/(log(5) + log(-3*x + 3*e^2 + 3*e^(-4*x + 1) - 4) + 5)

________________________________________________________________________________________

giac [A]  time = 1.02, size = 26, normalized size = 0.79 \begin {gather*} \frac {5}{\log \relax (5) + \log \left (-3 \, x + 3 \, e^{2} + 3 \, e^{\left (-4 \, x + 1\right )} - 4\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+3
0)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5)+30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log
(5)+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75*exp(2)-75*x-100),x, algorithm="gia
c")

[Out]

5/(log(5) + log(-3*x + 3*e^2 + 3*e^(-4*x + 1) - 4) + 5)

________________________________________________________________________________________

maple [A]  time = 0.34, size = 27, normalized size = 0.82

method	result	size
norman	\(\frac {5}{\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+\ln \relax (5)+5}\)	\(27\)
risch	\(\frac {5}{\ln \left (3 \,{\mathrm e}^{-4 x +1}+3 \,{\mathrm e}^{2}-3 x -4\right )+\ln \relax (5)+5}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)^2+((6*ln(5)+30)*exp(-
4*x+1)+(6*exp(2)-6*x-8)*ln(5)+30*exp(2)-30*x-40)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*ln(5)^2+30*ln(5)+75)*exp(
-4*x+1)+(3*exp(2)-3*x-4)*ln(5)^2+(30*exp(2)-30*x-40)*ln(5)+75*exp(2)-75*x-100),x,method=_RETURNVERBOSE)

[Out]

5/(ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)+ln(5)+5)

________________________________________________________________________________________

maxima [A]  time = 0.66, size = 36, normalized size = 1.09 \begin {gather*} -\frac {5}{4 \, x - \log \relax (5) - \log \left (-{\left (3 \, x - 3 \, e^{2} + 4\right )} e^{\left (4 \, x\right )} + 3 \, e\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)^2+((6*log(5)+3
0)*exp(-4*x+1)+(6*exp(2)-6*x-8)*log(5)+30*exp(2)-30*x-40)*log(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*log(5)^2+30*log
(5)+75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*log(5)^2+(30*exp(2)-30*x-40)*log(5)+75*exp(2)-75*x-100),x, algorithm="max
ima")

[Out]

-5/(4*x - log(5) - log(-(3*x - 3*e^2 + 4)*e^(4*x) + 3*e) - 5)

________________________________________________________________________________________

mupad [B]  time = 7.94, size = 24, normalized size = 0.73 \begin {gather*} \frac {5}{\ln \left (15\,{\mathrm {e}}^2-15\,x+15\,{\mathrm {e}}^{-4\,x}\,\mathrm {e}-20\right )+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(60*exp(1 - 4*x) + 15)/(75*x - 75*exp(2) + log(3*exp(2) - 3*x + 3*exp(1 - 4*x) - 4)^2*(3*x - 3*exp(2) - 3
*exp(1 - 4*x) + 4) + log(5)^2*(3*x - 3*exp(2) + 4) - exp(1 - 4*x)*(30*log(5) + 3*log(5)^2 + 75) + log(3*exp(2)
- 3*x + 3*exp(1 - 4*x) - 4)*(30*x - 30*exp(2) - exp(1 - 4*x)*(6*log(5) + 30) + log(5)*(6*x - 6*exp(2) + 8) +
40) + log(5)*(30*x - 30*exp(2) + 40) + 100),x)

[Out]

5/(log(15*exp(2) - 15*x + 15*exp(-4*x)*exp(1) - 20) + 5)

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 26, normalized size = 0.79 \begin {gather*} \frac {5}{\log {\left (- 3 x + 3 e^{1 - 4 x} - 4 + 3 e^{2} \right )} + \log {\relax (5 )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((60*exp(-4*x+1)+15)/((3*exp(-4*x+1)+3*exp(2)-3*x-4)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)**2+((6*ln(5)+30
)*exp(-4*x+1)+(6*exp(2)-6*x-8)*ln(5)+30*exp(2)-30*x-40)*ln(3*exp(-4*x+1)+3*exp(2)-3*x-4)+(3*ln(5)**2+30*ln(5)+
75)*exp(-4*x+1)+(3*exp(2)-3*x-4)*ln(5)**2+(30*exp(2)-30*x-40)*ln(5)+75*exp(2)-75*x-100),x)

[Out]

5/(log(-3*x + 3*exp(1 - 4*x) - 4 + 3*exp(2)) + log(5) + 5)

________________________________________________________________________________________