\(\int (3 e^{4+3 x} (5+4 e^2+2 e^4)+e^{2 x} (1+2 x)) \, dx\) [105]

Optimal result

Integrand size = 33, antiderivative size = 25 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=e^{2 x} \left (e^{4+x} \left (3+2 \left (1+e^2\right )^2\right )+x\right ) \]

[Out]

(exp(ln(3+2*(exp(2)+1)^2)+x+4)+x)*exp(x)^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2225, 2207} \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=\frac {1}{2} e^{2 x} (2 x+1)-\frac {e^{2 x}}{2}+\left (5+4 e^2+2 e^4\right ) e^{3 x+4} \]

[In]

Int[3*E^(4 + 3*x)*(5 + 4*E^2 + 2*E^4) + E^(2*x)*(1 + 2*x),x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} x \]

[In]

Integrate[3*E^(4 + 3*x)*(5 + 4*E^2 + 2*E^4) + E^(2*x)*(1 + 2*x),x]

[Out]

E^(4 + 3*x)*(5 + 4*E^2 + 2*E^4) + E^(2*x)*x

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(x \,{\mathrm e}^{2 x}+{\mathrm e}^{2 x} {\mathrm e}^{\ln \left (2 \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2}+5\right )+4+x}\)	\(30\)
norman	\(x \,{\mathrm e}^{2 x}+\left (2 \left ({\mathrm e}^{4}\right )^{2}+4 \,{\mathrm e}^{2} {\mathrm e}^{4}+5 \,{\mathrm e}^{4}\right ) {\mathrm e}^{3 x}\)	\(32\)
risch	\(5 \,{\mathrm e}^{4+3 x}+2 \,{\mathrm e}^{8+3 x}+4 \,{\mathrm e}^{6+3 x}+x \,{\mathrm e}^{2 x}\)	\(32\)
default	\(x \,{\mathrm e}^{2 x}+2 \,{\mathrm e}^{3 x} \left ({\mathrm e}^{4}\right )^{2}+4 \,{\mathrm e}^{3 x} {\mathrm e}^{4} {\mathrm e}^{2}+5 \,{\mathrm e}^{3 x} {\mathrm e}^{4}\)	\(38\)
meijerg	\(-{\mathrm e}^{\ln \left (2 \,{\mathrm e}^{4}+4 \,{\mathrm e}^{2}+5\right )+4} \left (1-{\mathrm e}^{3 x}\right )+\frac {{\mathrm e}^{2 x}}{2}-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{4}\)	\(43\)

[In]

int(3*exp(x)^2*exp(ln(2*exp(2)^2+4*exp(2)+5)+4+x)+(1+2*x)*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

x*exp(x)^2+exp(x)^2*exp(ln(2*exp(2)^2+4*exp(2)+5)+4+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx={\left (2 \, e^{8} + 4 \, e^{6} + 5 \, e^{4}\right )} e^{\left (3 \, x\right )} + x e^{\left (2 \, x\right )} \]

[In]

integrate(3*exp(x)^2*exp(log(2*exp(2)^2+4*exp(2)+5)+4+x)+(1+2*x)*exp(x)^2,x, algorithm="fricas")

[Out]

(2*e^8 + 4*e^6 + 5*e^4)*e^(3*x) + x*e^(2*x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=x e^{2 x} + \left (5 e^{4} + 4 e^{6} + 2 e^{8}\right ) \left (e^{2 x}\right )^{\frac {3}{2}} \]

[In]

integrate(3*exp(x)**2*exp(ln(2*exp(2)**2+4*exp(2)+5)+4+x)+(1+2*x)*exp(x)**2,x)

[Out]

x*exp(2*x) + (5*exp(4) + 4*exp(6) + 2*exp(8))*exp(2*x)**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=\frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + {\left (2 \, e^{4} + 4 \, e^{2} + 5\right )} e^{\left (3 \, x + 4\right )} + \frac {1}{2} \, e^{\left (2 \, x\right )} \]

[In]

integrate(3*exp(x)^2*exp(log(2*exp(2)^2+4*exp(2)+5)+4+x)+(1+2*x)*exp(x)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=x e^{\left (2 \, x\right )} + e^{\left (3 \, x + \log \left (2 \, e^{4} + 4 \, e^{2} + 5\right ) + 4\right )} \]

[In]

integrate(3*exp(x)^2*exp(log(2*exp(2)^2+4*exp(2)+5)+4+x)+(1+2*x)*exp(x)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \left (3 e^{4+3 x} \left (5+4 e^2+2 e^4\right )+e^{2 x} (1+2 x)\right ) \, dx=x\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x+4}\,\left (4\,{\mathrm {e}}^2+2\,{\mathrm {e}}^4+5\right ) \]

[In]

int(3*exp(2*x)*exp(x + log(4*exp(2) + 2*exp(4) + 5) + 4) + exp(2*x)*(2*x + 1),x)

[Out]

x*exp(2*x) + exp(3*x + 4)*(4*exp(2) + 2*exp(4) + 5)