\(\int \frac {(75+3 e^3) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx\) [6594]

Optimal result

Integrand size = 32, antiderivative size = 13 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=\frac {3 x \log (3)}{25+e^3+x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2006, 27, 32} \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3 \left (25+e^3\right ) \log (3)}{x+e^3+25} \]

[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 2006

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3 \left (25+e^3\right ) \log (3)}{25+e^3+x} \]

[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)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23

method	result	size
gosper	\(-\frac {3 \ln \left (3\right ) \left ({\mathrm e}^{3}+25\right )}{x +{\mathrm e}^{3}+25}\)	\(16\)
parallelrisch	\(-\frac {\left (3 \,{\mathrm e}^{3}+75\right ) \ln \left (3\right )}{x +{\mathrm e}^{3}+25}\)	\(18\)
norman	\(\frac {-3 \,{\mathrm e}^{3} \ln \left (3\right )-75 \ln \left (3\right )}{x +{\mathrm e}^{3}+25}\)	\(20\)
risch	\(-\frac {3 \ln \left (3\right ) {\mathrm e}^{3}}{x +{\mathrm e}^{3}+25}-\frac {75 \ln \left (3\right )}{x +{\mathrm e}^{3}+25}\)	\(26\)
meijerg	\(\frac {3 \,{\mathrm e}^{3} \ln \left (3\right ) x}{\left ({\mathrm e}^{3}+25\right )^{2} \left (1+\frac {x}{{\mathrm e}^{3}+25}\right )}+\frac {75 \ln \left (3\right ) x}{\left ({\mathrm e}^{3}+25\right )^{2} \left (1+\frac {x}{{\mathrm e}^{3}+25}\right )}\)	\(50\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3 \, {\left (e^{3} + 25\right )} \log \left (3\right )}{x + e^{3} + 25} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.54 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=- \frac {3 e^{3} \log {\left (3 \right )} + 75 \log {\left (3 \right )}}{x + e^{3} + 25} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3 \, {\left (e^{3} + 25\right )} \log \left (3\right )}{x + e^{3} + 25} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3 \, {\left (e^{3} + 25\right )} \log \left (3\right )}{x + e^{3} + 25} \]

[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]

-3*(e^3 + 25)*log(3)/(x + e^3 + 25)

Mupad [B] (verification not implemented)

Time = 11.58 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\left (75+3 e^3\right ) \log (3)}{625+e^6+50 x+x^2+e^3 (50+2 x)} \, dx=-\frac {3\,\ln \left (3\right )\,\left ({\mathrm {e}}^3+25\right )}{x+{\mathrm {e}}^3+25} \]

[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)