\(\int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} (-2 e^2 x-32 x^2) \log (4)}{e^2 \log (4)} \, dx\) [10208]

Optimal result

Integrand size = 44, antiderivative size = 24 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)} \]

[Out]

x-exp(16*x/exp(1)^2)^2*x^2-1/2*x^2/ln(2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 1607, 2227, 2207, 2225} \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}+x \]

[In]

Int[(-2*E^2*x + E^2*Log[4] + E^((32*x)/E^2)*(-2*E^2*x - 32*x^2)*Log[4])/(E^2*Log[4]),x]

[Out]

x - E^((32*x)/E^2)*x^2 - x^2/Log[4]

Rule 12

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

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2227

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)\right ) \, dx}{e^2 \log (4)} \\ & = x-\frac {x^2}{\log (4)}+\frac {\int e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \, dx}{e^2} \\ & = x-\frac {x^2}{\log (4)}+\frac {\int e^{\frac {32 x}{e^2}} \left (-2 e^2-32 x\right ) x \, dx}{e^2} \\ & = x-\frac {x^2}{\log (4)}+\frac {\int \left (-2 e^{2+\frac {32 x}{e^2}} x-32 e^{\frac {32 x}{e^2}} x^2\right ) \, dx}{e^2} \\ & = x-\frac {x^2}{\log (4)}-\frac {2 \int e^{2+\frac {32 x}{e^2}} x \, dx}{e^2}-\frac {32 \int e^{\frac {32 x}{e^2}} x^2 \, dx}{e^2} \\ & = x-\frac {1}{16} e^{2+\frac {32 x}{e^2}} x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}+\frac {1}{16} \int e^{2+\frac {32 x}{e^2}} \, dx+2 \int e^{\frac {32 x}{e^2}} x \, dx \\ & = \frac {1}{512} e^{4+\frac {32 x}{e^2}}+x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)}-\frac {1}{16} e^2 \int e^{\frac {32 x}{e^2}} \, dx \\ & = x-e^{\frac {32 x}{e^2}} x^2-\frac {x^2}{\log (4)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=x \left (1-e^{\frac {32 x}{e^2}} x-\frac {x}{\log (4)}\right ) \]

[In]

Integrate[(-2*E^2*x + E^2*Log[4] + E^((32*x)/E^2)*(-2*E^2*x - 32*x^2)*Log[4])/(E^2*Log[4]),x]

[Out]

x*(1 - E^((32*x)/E^2)*x - x/Log[4])

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(x -\frac {x^{2}}{2 \ln \left (2\right )}-{\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}\)	\(23\)
derivativedivides	\(\frac {256 x \ln \left (2\right )-128 x^{2}-256 \ln \left (2\right ) {\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}}{256 \ln \left (2\right )}\)	\(35\)
norman	\(\left (x \,{\mathrm e}-x^{2} {\mathrm e} \,{\mathrm e}^{32 x \,{\mathrm e}^{-2}}-\frac {{\mathrm e} x^{2}}{2 \ln \left (2\right )}\right ) {\mathrm e}^{-1}\)	\(39\)
default	\(\frac {{\mathrm e}^{-2} \left (x \,{\mathrm e}^{2} \ln \left (2\right )-{\mathrm e}^{2} \ln \left (2\right ) {\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}-\frac {x^{2} {\mathrm e}^{2}}{2}\right )}{\ln \left (2\right )}\)	\(49\)
parallelrisch	\(\frac {{\mathrm e}^{-2} \left (-x^{2} {\mathrm e}^{2}-2 \,{\mathrm e}^{2} \ln \left (2\right ) {\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2}+2 x \,{\mathrm e}^{2} \ln \left (2\right )\right )}{2 \ln \left (2\right )}\)	\(51\)
parts	\(x -\frac {x^{2}}{2 \ln \left (2\right )}-\frac {{\mathrm e}^{2} \left ({\mathrm e}^{2} \left (8 x \,{\mathrm e}^{-2} {\mathrm e}^{32 x \,{\mathrm e}^{-2}}-\frac {{\mathrm e}^{32 x \,{\mathrm e}^{-2}}}{4}\right )+{\mathrm e}^{2} \left (128 \,{\mathrm e}^{32 x \,{\mathrm e}^{-2}} x^{2} {\mathrm e}^{-4}-8 x \,{\mathrm e}^{-2} {\mathrm e}^{32 x \,{\mathrm e}^{-2}}+\frac {{\mathrm e}^{32 x \,{\mathrm e}^{-2}}}{4}\right )\right )}{128}\)	\(108\)

[In]

int(1/2*(2*(-2*x*exp(1)^2-32*x^2)*ln(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*ln(2)-2*x*exp(1)^2)/exp(1)^2/ln(2),x,m
ethod=_RETURNVERBOSE)

[Out]

x-1/2*x^2/ln(2)-exp(32*x*exp(-2))*x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=-\frac {2 \, x^{2} e^{\left (32 \, x e^{\left (-2\right )}\right )} \log \left (2\right ) + x^{2} - 2 \, x \log \left (2\right )}{2 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="fricas")

[Out]

-1/2*(2*x^2*e^(32*x*e^(-2))*log(2) + x^2 - 2*x*log(2))/log(2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=- x^{2} e^{\frac {32 x}{e^{2}}} - \frac {x^{2}}{2 \log {\left (2 \right )}} + x \]

[In]

integrate(1/2*(2*(-2*x*exp(1)**2-32*x**2)*ln(2)*exp(16*x/exp(1)**2)**2+2*exp(1)**2*ln(2)-2*x*exp(1)**2)/exp(1)
**2/ln(2),x)

[Out]

-x**2*exp(32*x*exp(-2)) - x**2/(2*log(2)) + x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=-\frac {{\left (2 \, x^{2} e^{\left (32 \, x e^{\left (-2\right )} + 2\right )} \log \left (2\right ) + x^{2} e^{2} - 2 \, x e^{2} \log \left (2\right )\right )} e^{\left (-2\right )}}{2 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="maxima")

[Out]

-1/2*(2*x^2*e^(32*x*e^(-2) + 2)*log(2) + x^2*e^2 - 2*x*e^2*log(2))*e^(-2)/log(2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (22) = 44\).

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=-\frac {{\left (256 \, x^{2} e^{2} - 512 \, x e^{2} \log \left (2\right ) + {\left ({\left (32 \, x e^{2} - e^{4}\right )} e^{\left (2 \, {\left (16 \, x + e^{2}\right )} e^{\left (-2\right )}\right )} + {\left (512 \, x^{2} e^{2} - 32 \, x e^{4} + e^{6}\right )} e^{\left (32 \, x e^{\left (-2\right )}\right )}\right )} \log \left (2\right )\right )} e^{\left (-2\right )}}{512 \, \log \left (2\right )} \]

[In]

integrate(1/2*(2*(-2*x*exp(1)^2-32*x^2)*log(2)*exp(16*x/exp(1)^2)^2+2*exp(1)^2*log(2)-2*x*exp(1)^2)/exp(1)^2/l
og(2),x, algorithm="giac")

[Out]

-1/512*(256*x^2*e^2 - 512*x*e^2*log(2) + ((32*x*e^2 - e^4)*e^(2*(16*x + e^2)*e^(-2)) + (512*x^2*e^2 - 32*x*e^4
+ e^6)*e^(32*x*e^(-2)))*log(2))*e^(-2)/log(2)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 e^2 x+e^2 \log (4)+e^{\frac {32 x}{e^2}} \left (-2 e^2 x-32 x^2\right ) \log (4)}{e^2 \log (4)} \, dx=x-\frac {x^2}{2\,\ln \left (2\right )}-x^2\,{\mathrm {e}}^{32\,x\,{\mathrm {e}}^{-2}} \]

[In]

int(-(exp(-2)*(x*exp(2) - exp(2)*log(2) + exp(32*x*exp(-2))*log(2)*(2*x*exp(2) + 32*x^2)))/log(2),x)

[Out]

x - x^2/(2*log(2)) - x^2*exp(32*x*exp(-2))