\(\int (-2+4 e^{1+4 x}+2 x+\log (x^3)) \, dx\) [1723]

Optimal result

Integrand size = 18, antiderivative size = 17 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=e^{1+4 x}+x \left (-5+x+\log \left (x^3\right )\right ) \]

[Out]

x*(x-5+ln(x^3))+exp(1+4*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2225, 2332} \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=x \log \left (x^3\right )+x^2-5 x+e^{4 x+1} \]

[In]

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

[Out]

E^(1 + 4*x) - 5*x + x^2 + x*Log[x^3]

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 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=e^{1+4 x}-5 x+x^2+x \log \left (x^3\right ) \]

[In]

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

[Out]

E^(1 + 4*x) - 5*x + x^2 + x*Log[x^3]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18

method	result	size
default	\(x^{2}-5 x +{\mathrm e}^{1+4 x}+x \ln \left (x^{3}\right )\)	\(20\)
norman	\(x^{2}-5 x +{\mathrm e}^{1+4 x}+x \ln \left (x^{3}\right )\)	\(20\)
risch	\(x^{2}-5 x +{\mathrm e}^{1+4 x}+x \ln \left (x^{3}\right )\)	\(20\)
parallelrisch	\(x^{2}-5 x +{\mathrm e}^{1+4 x}+x \ln \left (x^{3}\right )\)	\(20\)
parts	\(x^{2}-5 x +{\mathrm e}^{1+4 x}+x \ln \left (x^{3}\right )\)	\(20\)

[In]

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

[Out]

x^2-5*x+exp(1+4*x)+x*ln(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=x^{2} + x \log \left (x^{3}\right ) - 5 \, x + e^{\left (4 \, x + 1\right )} \]

[In]

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

[Out]

x^2 + x*log(x^3) - 5*x + e^(4*x + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=x^{2} + x \log {\left (x^{3} \right )} - 5 x + e^{4 x + 1} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=x^{2} + x \log \left (x^{3}\right ) - 5 \, x + e^{\left (4 \, x + 1\right )} \]

[In]

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

[Out]

x^2 + x*log(x^3) - 5*x + e^(4*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx=x^{2} + x \log \left (x^{3}\right ) - 5 \, x + e^{\left (4 \, x + 1\right )} \]

[In]

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

[Out]

x^2 + x*log(x^3) - 5*x + e^(4*x + 1)

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \left (-2+4 e^{1+4 x}+2 x+\log \left (x^3\right )\right ) \, dx={\mathrm {e}}^{4\,x+1}-5\,x+x\,\ln \left (x^3\right )+x^2 \]

[In]

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

[Out]

exp(4*x + 1) - 5*x + x*log(x^3) + x^2