3.58.2 \(\int \frac {4+e^{3 x} (12 x-60 \log (4))+e^{\frac {2+x}{4}} (-x+5 \log (4))}{-4 x+20 \log (4)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (\frac {e^{4-e^{3 x}+e^{\frac {2+x}{4}}}}{x-5 \log (4)}\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 28, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6688, 2194} \begin {gather*} e^{\frac {x}{4}+\frac {1}{2}}-e^{3 x}-\log (x-5 \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + E^(3*x)*(12*x - 60*Log[4]) + E^((2 + x)/4)*(-x + 5*Log[4]))/(-4*x + 20*Log[4]),x]

[Out]

E^(1/2 + x/4) - E^(3*x) - Log[x - 5*Log[4]]

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.00 \begin {gather*} e^{\frac {1}{2}+\frac {x}{4}}-e^{3 x}-\log (-x+5 \log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + E^(3*x)*(12*x - 60*Log[4]) + E^((2 + x)/4)*(-x + 5*Log[4]))/(-4*x + 20*Log[4]),x]

[Out]

E^(1/2 + x/4) - E^(3*x) - Log[-x + 5*Log[4]]

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fricas [A]  time = 1.63, size = 29, normalized size = 0.97 \begin {gather*} -{\left (e^{6} \log \left (x - 10 \, \log \relax (2)\right ) + e^{\left (3 \, x + 6\right )} - e^{\left (\frac {1}{4} \, x + \frac {13}{2}\right )}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-120*log(2)+12*x)*exp(3*x)+(10*log(2)-x)*exp(1/2+1/4*x)+4)/(40*log(2)-4*x),x, algorithm="fricas")

[Out]

-(e^6*log(x - 10*log(2)) + e^(3*x + 6) - e^(1/4*x + 13/2))*e^(-6)

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giac [A]  time = 0.15, size = 22, normalized size = 0.73 \begin {gather*} -e^{\left (3 \, x\right )} + e^{\left (\frac {1}{4} \, x + \frac {1}{2}\right )} - \log \left (x - 10 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-120*log(2)+12*x)*exp(3*x)+(10*log(2)-x)*exp(1/2+1/4*x)+4)/(40*log(2)-4*x),x, algorithm="giac")

[Out]

-e^(3*x) + e^(1/4*x + 1/2) - log(x - 10*log(2))

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maple [A]  time = 0.28, size = 23, normalized size = 0.77

method	result	size
risch	\(-\ln \left (x -10 \ln \relax (2)\right )-{\mathrm e}^{3 x}+{\mathrm e}^{\frac {1}{2}+\frac {x}{4}}\)	\(23\)
default	\(-\ln \left (10 \ln \relax (2)-x \right )-{\mathrm e}^{3 x}+{\mathrm e}^{\frac {1}{2}} {\mathrm e}^{\frac {x}{4}}\)	\(26\)
norman	\({\mathrm e}^{\frac {1}{2}} {\mathrm e}^{\frac {x}{4}}-{\mathrm e}^{3 x}-\ln \left (40 \ln \relax (2)-4 x \right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-120*ln(2)+12*x)*exp(3*x)+(10*ln(2)-x)*exp(1/2+1/4*x)+4)/(40*ln(2)-4*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x-10*ln(2))-exp(3*x)+exp(1/2+1/4*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{2} \, e^{\left (\frac {5}{2} \, \log \relax (2) + \frac {1}{2}\right )} E_{1}\left (-\frac {1}{4} \, x + \frac {5}{2} \, \log \relax (2)\right ) \log \relax (2) + 10 \, \int \frac {e^{\left (\frac {1}{4} \, x + \frac {1}{2}\right )}}{x^{2} - 20 \, x \log \relax (2) + 100 \, \log \relax (2)^{2}}\,{d x} \log \relax (2) - \frac {{\left (x - 10 \, \log \relax (2)\right )} e^{\left (3 \, x\right )} - x e^{\left (\frac {1}{4} \, x + \frac {1}{2}\right )}}{x - 10 \, \log \relax (2)} - \log \left (x - 10 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-120*log(2)+12*x)*exp(3*x)+(10*log(2)-x)*exp(1/2+1/4*x)+4)/(40*log(2)-4*x),x, algorithm="maxima")

[Out]

5/2*e^(5/2*log(2) + 1/2)*exp_integral_e(1, -1/4*x + 5/2*log(2))*log(2) + 10*integrate(e^(1/4*x + 1/2)/(x^2 - 2
0*x*log(2) + 100*log(2)^2), x)*log(2) - ((x - 10*log(2))*e^(3*x) - x*e^(1/4*x + 1/2))/(x - 10*log(2)) - log(x
- 10*log(2))

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mupad [B]  time = 3.85, size = 22, normalized size = 0.73 \begin {gather*} {\mathrm {e}}^{\frac {x}{4}+\frac {1}{2}}-{\mathrm {e}}^{3\,x}-\ln \left (x-10\,\ln \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3*x)*(12*x - 120*log(2)) - exp(x/4 + 1/2)*(x - 10*log(2)) + 4)/(4*x - 40*log(2)),x)

[Out]

exp(x/4 + 1/2) - exp(3*x) - log(x - 10*log(2))

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sympy [A]  time = 0.22, size = 26, normalized size = 0.87 \begin {gather*} - e^{3 x} + e^{\frac {1}{2}} \sqrt [12]{e^{3 x}} - \log {\left (x - 10 \log {\relax (2 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-120*ln(2)+12*x)*exp(3*x)+(10*ln(2)-x)*exp(1/2+1/4*x)+4)/(40*ln(2)-4*x),x)

[Out]

-exp(3*x) + exp(1/2)*exp(3*x)**(1/12) - log(x - 10*log(2))

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