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

Optimal. Leaf size=24 \[ -5+\frac {1}{2} \left (-2+\left (e^{-1+2 x}+\log \left (\frac {3 x}{2}\right )\right )^2\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {14, 2194, 2301, 2288} \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[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 2194

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 2288

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 2301

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.96 \begin {gather*} \frac {\left (e^{2 x}+e \log \left (\frac {3 x}{2}\right )\right )^2}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.57, size = 28, normalized size = 1.17 \begin {gather*} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.13, size = 64, normalized size = 2.67 \begin {gather*} \frac {1}{2} \, {\left (2 \, e^{3} \log \relax (3) \log \relax (x) - 2 \, e^{3} \log \relax (2) \log \relax (x) + e^{3} \log \relax (x)^{2} + 2 \, e^{\left (2 \, x + 2\right )} \log \relax (3) - 2 \, e^{\left (2 \, x + 2\right )} \log \relax (2) + 2 \, e^{\left (2 \, x + 2\right )} \log \relax (x) + e^{\left (4 \, x + 1\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 29, normalized size = 1.21

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 28, normalized size = 1.17 \begin {gather*} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 4.29, size = 15, normalized size = 0.62 \begin {gather*} \frac {{\left (\ln \left (\frac {3\,x}{2}\right )+{\mathrm {e}}^{2\,x-1}\right )}^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.32, size = 31, normalized size = 1.29 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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