\(\int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 (2 x+16 x^3)-e^2 \log (4)}{e^2} \, dx\) [5566]

Optimal result

Integrand size = 47, antiderivative size = 28 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx=x \left (-e+x+4 x \left (e^3+\left (x+\frac {2 x}{e}\right )^2\right )-\log (4)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 12} \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx=\frac {16 (1+e) x^4}{e^2}+4 x^4+4 e^3 x^2+x^2-x (e+\log (4)) \]

[In]

Int[(-E^3 + 8*E^5*x + 64*x^3 + 64*E*x^3 + E^2*(2*x + 16*x^3) - E^2*Log[4])/E^2,x]

[Out]

x^2 + 4*E^3*x^2 + 4*x^4 + (16*(1 + E)*x^4)/E^2 - x*(E + Log[4])

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]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^3+8 e^5 x+(64+64 e) x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx \\ & = \frac {\int \left (-e^3+8 e^5 x+(64+64 e) x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)\right ) \, dx}{e^2} \\ & = 4 e^3 x^2+\frac {16 (1+e) x^4}{e^2}-x (e+\log (4))+\int \left (2 x+16 x^3\right ) \, dx \\ & = x^2+4 e^3 x^2+4 x^4+\frac {16 (1+e) x^4}{e^2}-x (e+\log (4)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx=-e x+x^2+4 e^3 x^2+4 x^4+\frac {16 x^4}{e^2}+\frac {16 x^4}{e}-x \log (4) \]

[In]

Integrate[(-E^3 + 8*E^5*x + 64*x^3 + 64*E*x^3 + E^2*(2*x + 16*x^3) - E^2*Log[4])/E^2,x]

[Out]

-(E*x) + x^2 + 4*E^3*x^2 + 4*x^4 + (16*x^4)/E^2 + (16*x^4)/E - x*Log[4]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46

method	result	size
risch	\(4 x^{2} {\mathrm e}^{3}+4 x^{4}+16 x^{4} {\mathrm e}^{-1}-2 x \ln \left (2\right )-x \,{\mathrm e}+x^{2}+16 x^{4} {\mathrm e}^{-2}\)	\(41\)
norman	\(\left ({\mathrm e} \left (4 \,{\mathrm e}^{3}+1\right ) x^{2}+4 \left ({\mathrm e}^{2}+4 \,{\mathrm e}+4\right ) {\mathrm e}^{-1} x^{4}-{\mathrm e} \left (2 \ln \left (2\right )+{\mathrm e}\right ) x \right ) {\mathrm e}^{-1}\)	\(50\)
gosper	\(x \left (4 x^{3} {\mathrm e}^{2}+4 x \,{\mathrm e}^{2} {\mathrm e}^{3}+16 x^{3} {\mathrm e}-2 \,{\mathrm e}^{2} \ln \left (2\right )-{\mathrm e}^{3}+{\mathrm e}^{2} x +16 x^{3}\right ) {\mathrm e}^{-2}\)	\(58\)
default	\({\mathrm e}^{-2} \left (4 x^{4} {\mathrm e}^{2}+4 x^{2} {\mathrm e}^{2} {\mathrm e}^{3}+16 x^{4} {\mathrm e}-2 x \,{\mathrm e}^{2} \ln \left (2\right )-x \,{\mathrm e}^{3}+x^{2} {\mathrm e}^{2}+16 x^{4}\right )\)	\(63\)
parallelrisch	\({\mathrm e}^{-2} \left (4 x^{4} {\mathrm e}^{2}+4 x^{2} {\mathrm e}^{2} {\mathrm e}^{3}+16 x^{4} {\mathrm e}+x^{2} {\mathrm e}^{2}+16 x^{4}+\left (-2 \,{\mathrm e}^{2} \ln \left (2\right )-{\mathrm e}^{3}\right ) x \right )\)	\(64\)

[In]

int((-2*exp(1)^2*ln(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^2,x,met
hod=_RETURNVERBOSE)

[Out]

4*x^2*exp(3)+4*x^4+16*x^4*exp(-1)-2*x*ln(2)-x*exp(1)+x^2+16*x^4*exp(-2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx={\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \left (2\right ) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \]

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="fricas")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx=\frac {x^{4} \cdot \left (16 + 4 e^{2} + 16 e\right )}{e^{2}} + x^{2} \cdot \left (1 + 4 e^{3}\right ) + x \left (- e - 2 \log {\left (2 \right )}\right ) \]

[In]

integrate((-2*exp(1)**2*ln(2)+8*x*exp(1)**2*exp(3)-exp(1)**3+(16*x**3+2*x)*exp(1)**2+64*x**3*exp(1)+64*x**3)/e
xp(1)**2,x)

[Out]

x**4*(16 + 4*exp(2) + 16*E)*exp(-2) + x**2*(1 + 4*exp(3)) + x*(-E - 2*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx={\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \left (2\right ) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \]

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="maxima")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx={\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \left (2\right ) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \]

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="giac")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

Mupad [B] (verification not implemented)

Time = 11.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx=\left (16\,{\mathrm {e}}^{-1}+16\,{\mathrm {e}}^{-2}+4\right )\,x^4+\left (4\,{\mathrm {e}}^3+1\right )\,x^2+\left (-\mathrm {e}-\ln \left (4\right )\right )\,x \]

[In]

int(exp(-2)*(exp(2)*(2*x + 16*x^3) - 2*exp(2)*log(2) - exp(3) + 8*x*exp(5) + 64*x^3*exp(1) + 64*x^3),x)

[Out]

x^4*(16*exp(-1) + 16*exp(-2) + 4) - x*(exp(1) + log(4)) + x^2*(4*exp(3) + 1)