\(\int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+(1+2 e^{-1+2 x} x) \log (\frac {3 x}{2})}{x} \, dx\) [6672]

Optimal result

Integrand size = 41, antiderivative size = 24 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=-5+\frac {1}{2} \left (-2+\left (e^{-1+2 x}+\log \left (\frac {3 x}{2}\right )\right )^2\right ) \]

[Out]

1/2*(exp(-1+2*x)+ln(3/2*x))^2-6

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {14, 2225, 2338, 2326} \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=\frac {1}{2} e^{4 x-2}+\frac {1}{2} \log ^2\left (\frac {3 x}{2}\right )+e^{2 x-1} \log \left (\frac {3 x}{2}\right ) \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=\frac {\left (e^{2 x}+e \log \left (\frac {3 x}{2}\right )\right )^2}{2 e^2} \]

[In]

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

[Out]

(E^(2*x) + E*Log[(3*x)/2])^2/(2*E^2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
risch	\(\ln \left (\frac {3 x}{2}\right ) {\mathrm e}^{-1+2 x}+\frac {\ln \left (\frac {3 x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{4 x -2}}{2}\)	\(29\)
default	\(\ln \left (\frac {3 x}{2}\right ) {\mathrm e}^{-1+2 x}+\frac {\ln \left (\frac {3 x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{4 x -2}}{2}\)	\(31\)
norman	\(\ln \left (\frac {3 x}{2}\right ) {\mathrm e}^{-1+2 x}+\frac {\ln \left (\frac {3 x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{4 x -2}}{2}\)	\(31\)
parallelrisch	\(\ln \left (\frac {3 x}{2}\right ) {\mathrm e}^{-1+2 x}+\frac {\ln \left (\frac {3 x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{4 x -2}}{2}\)	\(31\)
parts	\(\ln \left (\frac {3 x}{2}\right ) {\mathrm e}^{-1+2 x}+\frac {\ln \left (\frac {3 x}{2}\right )^{2}}{2}+\frac {{\mathrm e}^{4 x -2}}{2}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=e^{\left (2 \, x - 1\right )} \log \left (\frac {3}{2} \, x\right ) + \frac {1}{2} \, \log \left (\frac {3}{2} \, x\right )^{2} + \frac {1}{2} \, e^{\left (4 \, x - 2\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=e^{2 x - 1} \log {\left (\frac {3 x}{2} \right )} + \frac {e^{4 x - 2}}{2} + \frac {\log {\left (\frac {3 x}{2} \right )}^{2}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=e^{\left (2 \, x - 1\right )} \log \left (\frac {3}{2} \, x\right ) + \frac {1}{2} \, \log \left (\frac {3}{2} \, x\right )^{2} + \frac {1}{2} \, e^{\left (4 \, x - 2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=\frac {1}{2} \, {\left (e^{3} \log \left (\frac {3}{2} \, x\right )^{2} + 2 \, e^{\left (2 \, x + 2\right )} \log \left (\frac {3}{2} \, x\right ) + e^{\left (4 \, x + 1\right )}\right )} e^{\left (-3\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-1+2 x}+2 e^{-2+4 x} x+\left (1+2 e^{-1+2 x} x\right ) \log \left (\frac {3 x}{2}\right )}{x} \, dx=\frac {{\left (\ln \left (\frac {3\,x}{2}\right )+{\mathrm {e}}^{2\,x-1}\right )}^2}{2} \]

[In]

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

[Out]

(log((3*x)/2) + exp(2*x - 1))^2/2