3.5.21 \(\int \frac {24+e^2 (24-x)-4 e^2 \log (3)}{x+e^2 x+(4+4 e^2) \log (3)} \, dx\)

Optimal. Leaf size=24 \[ -x+\frac {x}{1+e^2}+24 \log \left (\frac {x}{4}+\log (3)\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 186, 43} \begin {gather*} 24 \log (x+\log (81))-\frac {e^2 x}{1+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24 + E^2*(24 - x) - 4*E^2*Log[3])/(x + E^2*x + (4 + 4*E^2)*Log[3]),x]

[Out]

-((E^2*x)/(1 + E^2)) + 24*Log[x + Log[81]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {24+e^2 (24-x)-4 e^2 \log (3)}{\left (1+e^2\right ) x+\left (4+4 e^2\right ) \log (3)} \, dx\\ &=\int \frac {-e^2 x+4 \left (6+e^2 (6-\log (3))\right )}{\left (1+e^2\right ) x+4 \left (1+e^2\right ) \log (3)} \, dx\\ &=\int \left (-\frac {e^2}{1+e^2}+\frac {24}{x+\log (81)}\right ) \, dx\\ &=-\frac {e^2 x}{1+e^2}+24 \log (x+\log (81))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.25 \begin {gather*} \frac {-e^2 (x+\log (81))+24 \left (1+e^2\right ) \log (x+\log (81))}{1+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 + E^2*(24 - x) - 4*E^2*Log[3])/(x + E^2*x + (4 + 4*E^2)*Log[3]),x]

[Out]

(-(E^2*(x + Log[81])) + 24*(1 + E^2)*Log[x + Log[81]])/(1 + E^2)

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fricas [A]  time = 0.64, size = 26, normalized size = 1.08 \begin {gather*} -\frac {x e^{2} - 24 \, {\left (e^{2} + 1\right )} \log \left (x + 4 \, \log \relax (3)\right )}{e^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)*log(3)+(-x+24)*exp(2)+24)/((4*exp(2)+4)*log(3)+x+exp(2)*x),x, algorithm="fricas")

[Out]

-(x*e^2 - 24*(e^2 + 1)*log(x + 4*log(3)))/(e^2 + 1)

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giac [A]  time = 0.35, size = 22, normalized size = 0.92 \begin {gather*} -\frac {x e^{2}}{e^{2} + 1} + 24 \, \log \left ({\left | x + 4 \, \log \relax (3) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)*log(3)+(-x+24)*exp(2)+24)/((4*exp(2)+4)*log(3)+x+exp(2)*x),x, algorithm="giac")

[Out]

-x*e^2/(e^2 + 1) + 24*log(abs(x + 4*log(3)))

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maple [A]  time = 0.27, size = 22, normalized size = 0.92

method	result	size
norman	\(-\frac {{\mathrm e}^{2} x}{{\mathrm e}^{2}+1}+24 \ln \left (4 \ln \relax (3)+x \right )\)	\(22\)
risch	\(-\frac {{\mathrm e}^{2} x}{{\mathrm e}^{2}+1}+24 \ln \left (4 \ln \relax (3)+x \right )\)	\(22\)
default	\(\frac {-{\mathrm e}^{2} x +\left (24 \,{\mathrm e}^{2}+24\right ) \ln \left (4 \ln \relax (3)+x \right )}{{\mathrm e}^{2}+1}\)	\(28\)
meijerg	\(\frac {24 \ln \left (1+\frac {x}{4 \ln \relax (3)}\right )}{{\mathrm e}^{2}+1}-\frac {4 \,{\mathrm e}^{2} \ln \relax (3) \ln \left (1+\frac {x}{4 \ln \relax (3)}\right )}{{\mathrm e}^{2}+1}-\frac {4 \,{\mathrm e}^{2} \ln \relax (3) \left (\frac {x}{4 \ln \relax (3)}-\ln \left (1+\frac {x}{4 \ln \relax (3)}\right )\right )}{{\mathrm e}^{2}+1}+\frac {24 \,{\mathrm e}^{2} \ln \left (1+\frac {x}{4 \ln \relax (3)}\right )}{{\mathrm e}^{2}+1}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(2)*ln(3)+(-x+24)*exp(2)+24)/((4*exp(2)+4)*ln(3)+x+exp(2)*x),x,method=_RETURNVERBOSE)

[Out]

-1/(exp(2)+1)*exp(2)*x+24*ln(4*ln(3)+x)

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maxima [A]  time = 0.46, size = 21, normalized size = 0.88 \begin {gather*} -\frac {x e^{2}}{e^{2} + 1} + 24 \, \log \left (x + 4 \, \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)*log(3)+(-x+24)*exp(2)+24)/((4*exp(2)+4)*log(3)+x+exp(2)*x),x, algorithm="maxima")

[Out]

-x*e^2/(e^2 + 1) + 24*log(x + 4*log(3))

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mupad [B]  time = 0.15, size = 19, normalized size = 0.79 \begin {gather*} 24\,\ln \left (x+\ln \left (81\right )\right )-\frac {x\,{\mathrm {e}}^2}{{\mathrm {e}}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(2)*log(3) + exp(2)*(x - 24) - 24)/(x + x*exp(2) + log(3)*(4*exp(2) + 4)),x)

[Out]

24*log(x + log(81)) - (x*exp(2))/(exp(2) + 1)

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sympy [A]  time = 0.17, size = 19, normalized size = 0.79 \begin {gather*} - \frac {x e^{2}}{1 + e^{2}} + 24 \log {\left (x + 4 \log {\relax (3 )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(2)*ln(3)+(-x+24)*exp(2)+24)/((4*exp(2)+4)*ln(3)+x+exp(2)*x),x)

[Out]

-x*exp(2)/(1 + exp(2)) + 24*log(x + 4*log(3))

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