\(\int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx\) [3708]

Optimal result

Integrand size = 19, antiderivative size = 19 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \log \left (\frac {5}{4} \left (-1-\frac {3 e^5}{2 x}\right )\right ) \]

[Out]

3*ln(-5/4-15/8*exp(5)/x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 629} \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \log \left (2 x+3 e^5\right )-3 \log (x) \]

[In]

Int[(-9*E^5)/(3*E^5*x + 2*x^2),x]

[Out]

-3*Log[x] + 3*Log[3*E^5 + 2*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 629

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[Log[x]/b, x] - Simp[Log[RemoveContent[b + c*x, x]]/b,
x] /; FreeQ[{b, c}, x]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (9 e^5\right ) \int \frac {1}{3 e^5 x+2 x^2} \, dx\right ) \\ & = -3 \log (x)+3 \log \left (3 e^5+2 x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=-9 e^5 \left (\frac {\log (x)}{3 e^5}-\frac {\log \left (3 e^5+2 x\right )}{3 e^5}\right ) \]

[In]

Integrate[(-9*E^5)/(3*E^5*x + 2*x^2),x]

[Out]

-9*E^5*(Log[x]/(3*E^5) - Log[3*E^5 + 2*x]/(3*E^5))

Maple [A] (verified)

Time = 1.70 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(-3 \ln \left (x \right )+3 \ln \left (\frac {3 \,{\mathrm e}^{5}}{2}+x \right )\)	\(15\)
norman	\(-3 \ln \left (x \right )+3 \ln \left (3 \,{\mathrm e}^{5}+2 x \right )\)	\(17\)
risch	\(-3 \ln \left (x \right )+3 \ln \left (3 \,{\mathrm e}^{5}+2 x \right )\)	\(17\)
meijerg	\(-3 \ln \left (x \right )+3 \ln \left (3\right )-3 \ln \left (2\right )+15+3 \ln \left (1+\frac {2 \,{\mathrm e}^{-5} x}{3}\right )\)	\(25\)
default	\(-9 \,{\mathrm e}^{5} \left (\frac {\ln \left (x \right ) {\mathrm e}^{-5}}{3}-\frac {{\mathrm e}^{-5} \ln \left (3 \,{\mathrm e}^{5}+2 x \right )}{3}\right )\)	\(29\)

[In]

int(-9*exp(5)/(3*x*exp(5)+2*x^2),x,method=_RETURNVERBOSE)

[Out]

-3*ln(x)+3*ln(3/2*exp(5)+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, \log \left (2 \, x + 3 \, e^{5}\right ) - 3 \, \log \left (x\right ) \]

[In]

integrate(-9*exp(5)/(3*x*exp(5)+2*x^2),x, algorithm="fricas")

[Out]

3*log(2*x + 3*e^5) - 3*log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=- 9 \left (\frac {\log {\left (x \right )}}{3 e^{5}} - \frac {\log {\left (x + \frac {3 e^{5}}{2} \right )}}{3 e^{5}}\right ) e^{5} \]

[In]

integrate(-9*exp(5)/(3*x*exp(5)+2*x**2),x)

[Out]

-9*(exp(-5)*log(x)/3 - exp(-5)*log(x + 3*exp(5)/2)/3)*exp(5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, {\left (e^{\left (-5\right )} \log \left (2 \, x + 3 \, e^{5}\right ) - e^{\left (-5\right )} \log \left (x\right )\right )} e^{5} \]

[In]

integrate(-9*exp(5)/(3*x*exp(5)+2*x^2),x, algorithm="maxima")

[Out]

3*(e^(-5)*log(2*x + 3*e^5) - e^(-5)*log(x))*e^5

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (12) = 24\).

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=3 \, {\left (e^{\left (-5\right )} \log \left ({\left | 2 \, x + 3 \, e^{5} \right |}\right ) - e^{\left (-5\right )} \log \left ({\left | x \right |}\right )\right )} e^{5} \]

[In]

integrate(-9*exp(5)/(3*x*exp(5)+2*x^2),x, algorithm="giac")

[Out]

3*(e^(-5)*log(abs(2*x + 3*e^5)) - e^(-5)*log(abs(x)))*e^5

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int -\frac {9 e^5}{3 e^5 x+2 x^2} \, dx=6\,\mathrm {atanh}\left (\frac {4\,x\,{\mathrm {e}}^{-5}}{3}+1\right ) \]

[In]

int(-(9*exp(5))/(3*x*exp(5) + 2*x^2),x)

[Out]

6*atanh((4*x*exp(-5))/3 + 1)