\(\int \frac {e^{-2 x} ((-6+12 x) \log (2)+e^{2 x} (e^4+6 x \log (2)+\log (2) \log (5)))}{\log (2)} \, dx\) [7153]

Optimal result

Integrand size = 39, antiderivative size = 23 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=x \left (-6 e^{-2 x}+3 x+\frac {e^4}{\log (2)}+\log (5)\right ) \]

[Out]

x*(3*x+ln(5)+exp(4)/ln(2)-6/exp(2*x))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 6874, 2207, 2225} \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=3 x^2-3 e^{-2 x}+3 e^{-2 x} (1-2 x)+\frac {x \left (e^4+\log (2) \log (5)\right )}{\log (2)} \]

[In]

Int[((-6 + 12*x)*Log[2] + E^(2*x)*(E^4 + 6*x*Log[2] + Log[2]*Log[5]))/(E^(2*x)*Log[2]),x]

[Out]

-3/E^(2*x) + (3*(1 - 2*x))/E^(2*x) + 3*x^2 + (x*(E^4 + Log[2]*Log[5]))/Log[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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right ) \, dx}{\log (2)} \\ & = \frac {\int \left (e^4+6 x \log (2)+6 e^{-2 x} (-1+2 x) \log (2)+\log (2) \log (5)\right ) \, dx}{\log (2)} \\ & = 3 x^2+\frac {x \left (e^4+\log (2) \log (5)\right )}{\log (2)}+6 \int e^{-2 x} (-1+2 x) \, dx \\ & = 3 e^{-2 x} (1-2 x)+3 x^2+\frac {x \left (e^4+\log (2) \log (5)\right )}{\log (2)}+6 \int e^{-2 x} \, dx \\ & = -3 e^{-2 x}+3 e^{-2 x} (1-2 x)+3 x^2+\frac {x \left (e^4+\log (2) \log (5)\right )}{\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=-6 e^{-2 x} x+3 x^2+\frac {e^4 x}{\log (2)}+x \log (5) \]

[In]

Integrate[((-6 + 12*x)*Log[2] + E^(2*x)*(E^4 + 6*x*Log[2] + Log[2]*Log[5]))/(E^(2*x)*Log[2]),x]

[Out]

(-6*x)/E^(2*x) + 3*x^2 + (E^4*x)/Log[2] + x*Log[5]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13

method	result	size
risch	\(x \ln \left (5\right )+3 x^{2}+\frac {x \,{\mathrm e}^{4}}{\ln \left (2\right )}-6 \,{\mathrm e}^{-2 x} x\)	\(26\)
parts	\(x \ln \left (5\right )+3 x^{2}+\frac {x \,{\mathrm e}^{4}}{\ln \left (2\right )}-6 \,{\mathrm e}^{-2 x} x\)	\(28\)
norman	\(\left (\frac {\left (\ln \left (2\right ) \ln \left (5\right )+{\mathrm e}^{4}\right ) x \,{\mathrm e}^{2 x}}{\ln \left (2\right )}-6 x +3 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-2 x}\)	\(39\)
parallelrisch	\(\frac {\left (\ln \left (2\right ) \ln \left (5\right ) x \,{\mathrm e}^{2 x}+3 x^{2} \ln \left (2\right ) {\mathrm e}^{2 x}+{\mathrm e}^{2 x} {\mathrm e}^{4} x -6 x \ln \left (2\right )\right ) {\mathrm e}^{-2 x}}{\ln \left (2\right )}\)	\(47\)
default	\(\frac {x \ln \left (2\right ) \ln \left (5\right )+3 x^{2} \ln \left (2\right )+3 \ln \left (2\right ) {\mathrm e}^{-2 x}+3 \ln \left (2\right ) \left (-2 \,{\mathrm e}^{-2 x} x -{\mathrm e}^{-2 x}\right )+x \,{\mathrm e}^{4}}{\ln \left (2\right )}\)	\(56\)
derivativedivides	\(\frac {2 x \ln \left (2\right ) \ln \left (5\right )+6 x^{2} \ln \left (2\right )+6 \ln \left (2\right ) {\mathrm e}^{-2 x}+6 \ln \left (2\right ) \left (-2 \,{\mathrm e}^{-2 x} x -{\mathrm e}^{-2 x}\right )+2 x \,{\mathrm e}^{4}}{2 \ln \left (2\right )}\)	\(59\)

[In]

int(((ln(2)*ln(5)+6*x*ln(2)+exp(4))*exp(2*x)+(12*x-6)*ln(2))/ln(2)/exp(2*x),x,method=_RETURNVERBOSE)

[Out]

x*ln(5)+3*x^2+1/ln(2)*x*exp(4)-6*exp(-2*x)*x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=\frac {{\left ({\left (3 \, x^{2} \log \left (2\right ) + x \log \left (5\right ) \log \left (2\right ) + x e^{4}\right )} e^{\left (2 \, x\right )} - 6 \, x \log \left (2\right )\right )} e^{\left (-2 \, x\right )}}{\log \left (2\right )} \]

[In]

integrate(((log(2)*log(5)+6*x*log(2)+exp(4))*exp(2*x)+(12*x-6)*log(2))/log(2)/exp(2*x),x, algorithm="fricas")

[Out]

((3*x^2*log(2) + x*log(5)*log(2) + x*e^4)*e^(2*x) - 6*x*log(2))*e^(-2*x)/log(2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=3 x^{2} + \frac {x \left (\log {\left (2 \right )} \log {\left (5 \right )} + e^{4}\right )}{\log {\left (2 \right )}} - 6 x e^{- 2 x} \]

[In]

integrate(((ln(2)*ln(5)+6*x*ln(2)+exp(4))*exp(2*x)+(12*x-6)*ln(2))/ln(2)/exp(2*x),x)

[Out]

3*x**2 + x*(log(2)*log(5) + exp(4))/log(2) - 6*x*exp(-2*x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (21) = 42\).

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=\frac {3 \, x^{2} \log \left (2\right ) - 3 \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} \log \left (2\right ) + x \log \left (5\right ) \log \left (2\right ) + x e^{4} + 3 \, e^{\left (-2 \, x\right )} \log \left (2\right )}{\log \left (2\right )} \]

[In]

integrate(((log(2)*log(5)+6*x*log(2)+exp(4))*exp(2*x)+(12*x-6)*log(2))/log(2)/exp(2*x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=\frac {3 \, x^{2} \log \left (2\right ) - 6 \, x e^{\left (-2 \, x\right )} \log \left (2\right ) + x \log \left (5\right ) \log \left (2\right ) + x e^{4}}{\log \left (2\right )} \]

[In]

integrate(((log(2)*log(5)+6*x*log(2)+exp(4))*exp(2*x)+(12*x-6)*log(2))/log(2)/exp(2*x),x, algorithm="giac")

[Out]

(3*x^2*log(2) - 6*x*e^(-2*x)*log(2) + x*log(5)*log(2) + x*e^4)/log(2)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-2 x} \left ((-6+12 x) \log (2)+e^{2 x} \left (e^4+6 x \log (2)+\log (2) \log (5)\right )\right )}{\log (2)} \, dx=x\,\ln \left (5\right )-6\,x\,{\mathrm {e}}^{-2\,x}+3\,x^2+\frac {x\,{\mathrm {e}}^4}{\ln \left (2\right )} \]

[In]

int((exp(-2*x)*(log(2)*(12*x - 6) + exp(2*x)*(exp(4) + log(2)*log(5) + 6*x*log(2))))/log(2),x)

[Out]

x*log(5) - 6*x*exp(-2*x) + 3*x^2 + (x*exp(4))/log(2)