3.66.94 \(\int \frac {(75+3 e^3) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx\)

Optimal. Leaf size=13 \[ \frac {3 x \log (3)}{25+e^3+x} \]

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 1981, 27, 32} \begin {gather*} -\frac {3 \left (25+e^3\right ) \log (3)}{x+e^3+25} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((75 + 3*E^3)*Log[3])/(625 + E^6 + 50*x + x^2 + E^3*(50 + 2*x)),x]

[Out]

(-3*(25 + E^3)*Log[3])/(25 + E^3 + x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 1981

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (3 \left (25+e^3\right ) \log (3)\right ) \int \frac {1}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx\\ &=\left (3 \left (25+e^3\right ) \log (3)\right ) \int \frac {1}{\left (25+e^3\right )^2+2 \left (25+e^3\right ) x+x^2} \, dx\\ &=\left (3 \left (25+e^3\right ) \log (3)\right ) \int \frac {1}{\left (25+e^3+x\right )^2} \, dx\\ &=-\frac {3 \left (25+e^3\right ) \log (3)}{25+e^3+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.31 \begin {gather*} -\frac {3 \left (25+e^3\right ) \log (3)}{25+e^3+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((75 + 3*E^3)*Log[3])/(625 + E^6 + 50*x + x^2 + E^3*(50 + 2*x)),x]

[Out]

(-3*(25 + E^3)*Log[3])/(25 + E^3 + x)

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fricas [A]  time = 0.64, size = 15, normalized size = 1.15 \begin {gather*} -\frac {3 \, {\left (e^{3} + 25\right )} \log \relax (3)}{x + e^{3} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(3)+75)*log(3)/(exp(3)^2+(2*x+50)*exp(3)+x^2+50*x+625),x, algorithm="fricas")

[Out]

-3*(e^3 + 25)*log(3)/(x + e^3 + 25)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(3)+75)*log(3)/(exp(3)^2+(2*x+50)*exp(3)+x^2+50*x+625),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (exp(3)+25)*ln(3)*3*2*1/2/sqrt(-exp(3)^2
+exp(6))*atan((sageVARx+exp(3)+25)/sqrt(-exp(3)^2+exp(6)))

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maple [A]  time = 0.51, size = 16, normalized size = 1.23

method	result	size
gosper	\(-\frac {3 \ln \relax (3) \left ({\mathrm e}^{3}+25\right )}{x +{\mathrm e}^{3}+25}\)	\(16\)
norman	\(\frac {-3 \,{\mathrm e}^{3} \ln \relax (3)-75 \ln \relax (3)}{x +{\mathrm e}^{3}+25}\)	\(20\)
risch	\(-\frac {3 \ln \relax (3) {\mathrm e}^{3}}{x +{\mathrm e}^{3}+25}-\frac {75 \ln \relax (3)}{x +{\mathrm e}^{3}+25}\)	\(26\)
meijerg	\(\frac {3 \,{\mathrm e}^{3} \ln \relax (3) x}{\left ({\mathrm e}^{3}+25\right )^{2} \left (1+\frac {x}{{\mathrm e}^{3}+25}\right )}+\frac {75 \ln \relax (3) x}{\left ({\mathrm e}^{3}+25\right )^{2} \left (1+\frac {x}{{\mathrm e}^{3}+25}\right )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp(3)+75)*ln(3)/(exp(3)^2+(2*x+50)*exp(3)+x^2+50*x+625),x,method=_RETURNVERBOSE)

[Out]

-3*ln(3)*(exp(3)+25)/(x+exp(3)+25)

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maxima [A]  time = 0.44, size = 15, normalized size = 1.15 \begin {gather*} -\frac {3 \, {\left (e^{3} + 25\right )} \log \relax (3)}{x + e^{3} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(3)+75)*log(3)/(exp(3)^2+(2*x+50)*exp(3)+x^2+50*x+625),x, algorithm="maxima")

[Out]

-3*(e^3 + 25)*log(3)/(x + e^3 + 25)

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mupad [B]  time = 4.09, size = 15, normalized size = 1.15 \begin {gather*} -\frac {3\,\ln \relax (3)\,\left ({\mathrm {e}}^3+25\right )}{x+{\mathrm {e}}^3+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(3)*(3*exp(3) + 75))/(50*x + exp(6) + x^2 + exp(3)*(2*x + 50) + 625),x)

[Out]

-(3*log(3)*(exp(3) + 25))/(x + exp(3) + 25)

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sympy [A]  time = 0.14, size = 20, normalized size = 1.54 \begin {gather*} - \frac {3 e^{3} \log {\relax (3 )} + 75 \log {\relax (3 )}}{x + e^{3} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*exp(3)+75)*ln(3)/(exp(3)**2+(2*x+50)*exp(3)+x**2+50*x+625),x)

[Out]

-(3*exp(3)*log(3) + 75*log(3))/(x + exp(3) + 25)

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