3.17.78 \(\int (-4 e^x+2 e^{2 x}+\log (\frac {5 x}{2})) \, dx\)

Optimal. Leaf size=20 \[ 27+\left (-2+e^x\right )^2-x+x \log \left (\frac {5 x}{2}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2194, 2295} \begin {gather*} -x-4 e^x+e^{2 x}+x \log \left (\frac {5 x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*E^x + E^(2*x) - x + x*Log[(5*x)/2]

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 2295

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 {gather*} \begin {aligned} \text {integral} &=2 \int e^{2 x} \, dx-4 \int e^x \, dx+\int \log \left (\frac {5 x}{2}\right ) \, dx\\ &=-4 e^x+e^{2 x}-x+x \log \left (\frac {5 x}{2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.10 \begin {gather*} -4 e^x+e^{2 x}-x+x \log \left (\frac {5 x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*E^x + E^(2*x) - x + x*Log[(5*x)/2]

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fricas [A]  time = 0.81, size = 18, normalized size = 0.90 \begin {gather*} x \log \left (\frac {5}{2} \, x\right ) - x + e^{\left (2 \, x\right )} - 4 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/2*x)+2*exp(x)^2-4*exp(x),x, algorithm="fricas")

[Out]

x*log(5/2*x) - x + e^(2*x) - 4*e^x

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giac [A]  time = 0.48, size = 18, normalized size = 0.90 \begin {gather*} x \log \left (\frac {5}{2} \, x\right ) - x + e^{\left (2 \, x\right )} - 4 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/2*x)+2*exp(x)^2-4*exp(x),x, algorithm="giac")

[Out]

x*log(5/2*x) - x + e^(2*x) - 4*e^x

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maple [A]  time = 0.02, size = 19, normalized size = 0.95

method	result	size
default	\({\mathrm e}^{2 x}-4 \,{\mathrm e}^{x}+\ln \left (\frac {5 x}{2}\right ) x -x\)	\(19\)
norman	\({\mathrm e}^{2 x}-4 \,{\mathrm e}^{x}+\ln \left (\frac {5 x}{2}\right ) x -x\)	\(19\)
risch	\({\mathrm e}^{2 x}-4 \,{\mathrm e}^{x}+\ln \left (\frac {5 x}{2}\right ) x -x\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(5/2*x)+2*exp(x)^2-4*exp(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 18, normalized size = 0.90 \begin {gather*} x \log \left (\frac {5}{2} \, x\right ) - x + e^{\left (2 \, x\right )} - 4 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(5/2*x)+2*exp(x)^2-4*exp(x),x, algorithm="maxima")

[Out]

x*log(5/2*x) - x + e^(2*x) - 4*e^x

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mupad [B]  time = 1.12, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^x+x\,\left (\ln \left (\frac {5\,x}{2}\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(2*x) + log((5*x)/2) - 4*exp(x),x)

[Out]

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

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sympy [A]  time = 0.28, size = 19, normalized size = 0.95 \begin {gather*} x \log {\left (\frac {5 x}{2} \right )} - x + e^{2 x} - 4 e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(5/2*x)+2*exp(x)**2-4*exp(x),x)

[Out]

x*log(5*x/2) - x + exp(2*x) - 4*exp(x)

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