\(\int \frac {e^4 (-6 x+2 e^3 x+6 x^2)+e^{7+x} \log (3)}{e^3} \, dx\) [7427]

Optimal result

Integrand size = 32, antiderivative size = 26 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx=e^4 \left (1+x \left (x+\frac {x (-3+2 x)}{e^3}\right )+e^x \log (3)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 6, 2225} \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx=2 e x^3-e \left (3-e^3\right ) x^2+e^{x+4} \log (3) \]

[In]

Int[(E^4*(-6*x + 2*E^3*x + 6*x^2) + E^(7 + x)*Log[3])/E^3,x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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}& = \frac {\int \left (e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)\right ) \, dx}{e^3} \\ & = e \int \left (-6 x+2 e^3 x+6 x^2\right ) \, dx+\frac {\log (3) \int e^{7+x} \, dx}{e^3} \\ & = e^{4+x} \log (3)+e \int \left (\left (-6+2 e^3\right ) x+6 x^2\right ) \, dx \\ & = -e \left (3-e^3\right ) x^2+2 e x^3+e^{4+x} \log (3) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx=e \left (\left (-3+e^3\right ) x^2+2 x^3+e^{3+x} \log (3)\right ) \]

[In]

Integrate[(E^4*(-6*x + 2*E^3*x + 6*x^2) + E^(7 + x)*Log[3])/E^3,x]

[Out]

E*((-3 + E^3)*x^2 + 2*x^3 + E^(3 + x)*Log[3])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12

method	result	size
risch	\(x^{2} {\mathrm e}^{4}+2 x^{3} {\mathrm e}-3 x^{2} {\mathrm e}+\ln \left (3\right ) {\mathrm e}^{4+x}\)	\(29\)
parts	\({\mathrm e}^{4} \ln \left (3\right ) {\mathrm e}^{x}+2 \,{\mathrm e}^{-3} {\mathrm e}^{4} \left (x^{3}+\frac {\left ({\mathrm e}^{3}-3\right ) x^{2}}{2}\right )\)	\(30\)
default	\({\mathrm e}^{-3} \left (2 \,{\mathrm e}^{4} \left (x^{3}+\frac {\left ({\mathrm e}^{3}-3\right ) x^{2}}{2}\right )+{\mathrm e}^{3} {\mathrm e}^{4} \ln \left (3\right ) {\mathrm e}^{x}\right )\)	\(33\)
norman	\({\mathrm e}^{4} \ln \left (3\right ) {\mathrm e}^{x}+{\mathrm e}^{4} \left ({\mathrm e}^{3}-3\right ) {\mathrm e}^{-3} x^{2}+2 \,{\mathrm e}^{-3} {\mathrm e}^{4} x^{3}\)	\(34\)
parallelrisch	\({\mathrm e}^{-3} \left ({\mathrm e}^{3} {\mathrm e}^{4} \ln \left (3\right ) {\mathrm e}^{x}+x^{2} {\mathrm e}^{3} {\mathrm e}^{4}+2 x^{3} {\mathrm e}^{4}-3 x^{2} {\mathrm e}^{4}\right )\)	\(38\)

[In]

int((exp(3)*exp(4)*ln(3)*exp(x)+(2*x*exp(3)+6*x^2-6*x)*exp(4))/exp(3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx={\left (x^{2} e^{7} + {\left (2 \, x^{3} - 3 \, x^{2}\right )} e^{4} + e^{\left (x + 7\right )} \log \left (3\right )\right )} e^{\left (-3\right )} \]

[In]

integrate((exp(3)*exp(4)*log(3)*exp(x)+(2*x*exp(3)+6*x^2-6*x)*exp(4))/exp(3),x, algorithm="fricas")

[Out]

(x^2*e^7 + (2*x^3 - 3*x^2)*e^4 + e^(x + 7)*log(3))*e^(-3)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx=2 e x^{3} + x^{2} \left (- 3 e + e^{4}\right ) + e^{4} e^{x} \log {\left (3 \right )} \]

[In]

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

[Out]

2*E*x**3 + x**2*(-3*E + exp(4)) + exp(4)*exp(x)*log(3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx={\left ({\left (2 \, x^{3} + x^{2} e^{3} - 3 \, x^{2}\right )} e^{4} + e^{\left (x + 7\right )} \log \left (3\right )\right )} e^{\left (-3\right )} \]

[In]

integrate((exp(3)*exp(4)*log(3)*exp(x)+(2*x*exp(3)+6*x^2-6*x)*exp(4))/exp(3),x, algorithm="maxima")

[Out]

((2*x^3 + x^2*e^3 - 3*x^2)*e^4 + e^(x + 7)*log(3))*e^(-3)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx={\left ({\left (2 \, x^{3} + x^{2} e^{3} - 3 \, x^{2}\right )} e^{4} + e^{\left (x + 7\right )} \log \left (3\right )\right )} e^{\left (-3\right )} \]

[In]

integrate((exp(3)*exp(4)*log(3)*exp(x)+(2*x*exp(3)+6*x^2-6*x)*exp(4))/exp(3),x, algorithm="giac")

[Out]

((2*x^3 + x^2*e^3 - 3*x^2)*e^4 + e^(x + 7)*log(3))*e^(-3)

Mupad [B] (verification not implemented)

Time = 12.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^4 \left (-6 x+2 e^3 x+6 x^2\right )+e^{7+x} \log (3)}{e^3} \, dx={\mathrm {e}}^{x+4}\,\ln \left (3\right )-x^2\,\left (3\,\mathrm {e}-{\mathrm {e}}^4\right )+2\,x^3\,\mathrm {e} \]

[In]

int(exp(-3)*(exp(4)*(2*x*exp(3) - 6*x + 6*x^2) + exp(7)*exp(x)*log(3)),x)

[Out]

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