\(\int \frac {2 e+32 x^3+(e+16 x^3) (i \pi +\log (-((-3+e) e^3)))}{e} \, dx\) [4354]

Optimal result

Integrand size = 36, antiderivative size = 27 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx=\left (x+\frac {4 x^4}{e}\right ) \left (2+i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {12} \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx=\frac {8 x^4}{e}+\frac {4 x^4 (3+i \pi +\log (3-e))}{e}+2 x+x (3+i \pi +\log (3-e)) \]

[In]

Int[(2*E + 32*x^3 + (E + 16*x^3)*(I*Pi + Log[-((-3 + E)*E^3)]))/E,x]

[Out]

2*x + (8*x^4)/E + x*(3 + I*Pi + Log[3 - E]) + (4*x^4*(3 + I*Pi + Log[3 - E]))/E

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}& = \frac {\int \left (2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )\right ) \, dx}{e} \\ & = 2 x+\frac {8 x^4}{e}+\frac {(3+i \pi +\log (3-e)) \int \left (e+16 x^3\right ) \, dx}{e} \\ & = 2 x+\frac {8 x^4}{e}+x (3+i \pi +\log (3-e))+\frac {4 x^4 (3+i \pi +\log (3-e))}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx=\frac {\left (e x+4 x^4\right ) (5+i \pi +\log (3-e))}{e} \]

[In]

Integrate[(2*E + 32*x^3 + (E + 16*x^3)*(I*Pi + Log[-((-3 + E)*E^3)]))/E,x]

[Out]

((E*x + 4*x^4)*(5 + I*Pi + Log[3 - E]))/E

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93

method	result	size
gosper	\(\left (2+\ln \left (\left ({\mathrm e}-3\right ) {\mathrm e}^{3}\right )\right ) x \left (4 x^{3}+{\mathrm e}\right ) {\mathrm e}^{-1}\)	\(25\)
default	\({\mathrm e}^{-1} \left (2+\ln \left (\left ({\mathrm e}-3\right ) {\mathrm e}^{3}\right )\right ) \left (4 x^{4}+x \,{\mathrm e}\right )\)	\(26\)
norman	\(\left (\ln \left ({\mathrm e}-3\right )+5\right ) x +4 \left (\ln \left ({\mathrm e}-3\right )+5\right ) {\mathrm e}^{-1} x^{4}\)	\(27\)
parallelrisch	\({\mathrm e}^{-1} \left (4 x^{4} \ln \left (\left ({\mathrm e}-3\right ) {\mathrm e}^{3}\right )+8 x^{4}+\ln \left (\left ({\mathrm e}-3\right ) {\mathrm e}^{3}\right ) {\mathrm e} x +2 x \,{\mathrm e}\right )\)	\(42\)
risch	\(4 i {\mathrm e}^{-1} \pi \,x^{4}+4 \ln \left (3-{\mathrm e}\right ) {\mathrm e}^{-1} x^{4}+i {\mathrm e}^{-1} \pi \,{\mathrm e} x +20 \,{\mathrm e}^{-1} x^{4}+\ln \left (3-{\mathrm e}\right ) {\mathrm e}^{-1} {\mathrm e} x +5 \,{\mathrm e}^{-1} {\mathrm e} x\)	\(61\)

[In]

int(((exp(1)+16*x^3)*ln((exp(1)-3)*exp(3))+2*exp(1)+32*x^3)/exp(1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx={\left (8 \, x^{4} + 2 \, x e + {\left (4 \, x^{4} + x e\right )} \log \left (e^{4} - 3 \, e^{3}\right )\right )} e^{\left (-1\right )} \]

[In]

integrate(((exp(1)+16*x^3)*log((exp(1)-3)*exp(3))+2*exp(1)+32*x^3)/exp(1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx=\frac {x^{4} \cdot \left (4 \log {\left (3 - e \right )} + 20 + 4 i \pi \right )}{e} + x \left (\log {\left (3 - e \right )} + 5 + i \pi \right ) \]

[In]

integrate(((exp(1)+16*x**3)*ln((exp(1)-3)*exp(3))+2*exp(1)+32*x**3)/exp(1),x)

[Out]

x**4*(4*log(3 - E) + 20 + 4*I*pi)*exp(-1) + x*(log(3 - E) + 5 + I*pi)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx={\left (8 \, x^{4} + 2 \, x e + {\left (4 \, x^{4} + x e\right )} \log \left ({\left (e - 3\right )} e^{3}\right )\right )} e^{\left (-1\right )} \]

[In]

integrate(((exp(1)+16*x^3)*log((exp(1)-3)*exp(3))+2*exp(1)+32*x^3)/exp(1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx={\left (8 \, x^{4} + 2 \, x e + {\left (4 \, x^{4} + x e\right )} \log \left ({\left (e - 3\right )} e^{3}\right )\right )} e^{\left (-1\right )} \]

[In]

integrate(((exp(1)+16*x^3)*log((exp(1)-3)*exp(3))+2*exp(1)+32*x^3)/exp(1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {2 e+32 x^3+\left (e+16 x^3\right ) \left (i \pi +\log \left (-\left ((-3+e) e^3\right )\right )\right )}{e} \, dx=x\,{\mathrm {e}}^{-1}\,\left (\ln \left ({\mathrm {e}}^3\,\left (\mathrm {e}-3\right )\right )+2\right )\,\left (4\,x^3+\mathrm {e}\right ) \]

[In]

int(exp(-1)*(2*exp(1) + log(exp(3)*(exp(1) - 3))*(exp(1) + 16*x^3) + 32*x^3),x)

[Out]

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