\(\int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx\) [1932]

Optimal result

Integrand size = 36, antiderivative size = 20 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=\frac {5 (3-x) x}{\frac {1}{15}-\frac {x}{e^2}} \]

[Out]

5*(-x+3)*x/(1/15-x/exp(2))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {27, 1864} \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=5 e^2 x+\frac {e^4 \left (45-e^2\right )}{3 \left (e^2-15 x\right )} \]

[In]

Int[(E^4*(225 - 150*x) + 1125*E^2*x^2)/(E^4 - 30*E^2*x + 225*x^2),x]

[Out]

(E^4*(45 - E^2))/(3*(E^2 - 15*x)) + 5*E^2*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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{\left (e^2-15 x\right )^2} \, dx \\ & = \int \left (5 e^2-\frac {5 e^4 \left (-45+e^2\right )}{\left (e^2-15 x\right )^2}\right ) \, dx \\ & = \frac {e^4 \left (45-e^2\right )}{3 \left (e^2-15 x\right )}+5 e^2 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=-\frac {2 e^6+225 e^2 x^2-15 e^4 (3+2 x)}{3 \left (e^2-15 x\right )} \]

[In]

Integrate[(E^4*(225 - 150*x) + 1125*E^2*x^2)/(E^4 - 30*E^2*x + 225*x^2),x]

[Out]

-1/3*(2*E^6 + 225*E^2*x^2 - 15*E^4*(3 + 2*x))/(E^2 - 15*x)

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
gosper	\(\frac {15 \left (-5 x^{2}+{\mathrm e}^{2}\right ) {\mathrm e}^{2}}{{\mathrm e}^{2}-15 x}\)	\(21\)
norman	\(\frac {-75 x^{2} {\mathrm e}^{2}+15 \,{\mathrm e}^{4}}{{\mathrm e}^{2}-15 x}\)	\(24\)
parallelrisch	\(\frac {-1125 x^{2} {\mathrm e}^{2}+225 \,{\mathrm e}^{4}}{15 \,{\mathrm e}^{2}-225 x}\)	\(25\)
risch	\(5 \,{\mathrm e}^{2} x -\frac {{\mathrm e}^{6}}{3 \left ({\mathrm e}^{2}-15 x \right )}+\frac {15 \,{\mathrm e}^{4}}{{\mathrm e}^{2}-15 x}\)	\(31\)
meijerg	\(\frac {225 x}{1-15 x \,{\mathrm e}^{-2}}-\frac {2 \,{\mathrm e}^{4} \left (\frac {15 x \,{\mathrm e}^{-2}}{1-15 x \,{\mathrm e}^{-2}}+\ln \left (1-15 x \,{\mathrm e}^{-2}\right )\right )}{3}-\frac {{\mathrm e}^{4} \left (-\frac {5 x \,{\mathrm e}^{-2} \left (-45 x \,{\mathrm e}^{-2}+6\right )}{1-15 x \,{\mathrm e}^{-2}}-2 \ln \left (1-15 x \,{\mathrm e}^{-2}\right )\right )}{3}\)	\(77\)

[In]

int(((-150*x+225)*exp(2)^2+1125*x^2*exp(2))/(exp(2)^2-30*exp(2)*x+225*x^2),x,method=_RETURNVERBOSE)

[Out]

15*(-5*x^2+exp(2))*exp(2)/(exp(2)-15*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=\frac {225 \, x^{2} e^{2} - 15 \, {\left (x + 3\right )} e^{4} + e^{6}}{3 \, {\left (15 \, x - e^{2}\right )}} \]

[In]

integrate(((-150*x+225)*exp(2)^2+1125*x^2*exp(2))/(exp(2)^2-30*exp(2)*x+225*x^2),x, algorithm="fricas")

[Out]

1/3*(225*x^2*e^2 - 15*(x + 3)*e^4 + e^6)/(15*x - e^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=5 x e^{2} + \frac {- 45 e^{4} + e^{6}}{45 x - 3 e^{2}} \]

[In]

integrate(((-150*x+225)*exp(2)**2+1125*x**2*exp(2))/(exp(2)**2-30*exp(2)*x+225*x**2),x)

[Out]

5*x*exp(2) + (-45*exp(4) + exp(6))/(45*x - 3*exp(2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=5 \, x e^{2} + \frac {e^{6} - 45 \, e^{4}}{3 \, {\left (15 \, x - e^{2}\right )}} \]

[In]

integrate(((-150*x+225)*exp(2)^2+1125*x^2*exp(2))/(exp(2)^2-30*exp(2)*x+225*x^2),x, algorithm="maxima")

[Out]

5*x*e^2 + 1/3*(e^6 - 45*e^4)/(15*x - e^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=5 \, x e^{2} + \frac {e^{6} - 45 \, e^{4}}{3 \, {\left (15 \, x - e^{2}\right )}} \]

[In]

integrate(((-150*x+225)*exp(2)^2+1125*x^2*exp(2))/(exp(2)^2-30*exp(2)*x+225*x^2),x, algorithm="giac")

[Out]

5*x*e^2 + 1/3*(e^6 - 45*e^4)/(15*x - e^2)

Mupad [B] (verification not implemented)

Time = 9.96 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^4 (225-150 x)+1125 e^2 x^2}{e^4-30 e^2 x+225 x^2} \, dx=5\,x\,{\mathrm {e}}^2-\frac {15\,{\mathrm {e}}^4-\frac {{\mathrm {e}}^6}{3}}{15\,x-{\mathrm {e}}^2} \]

[In]

int((1125*x^2*exp(2) - exp(4)*(150*x - 225))/(exp(4) - 30*x*exp(2) + 225*x^2),x)

[Out]

5*x*exp(2) - (15*exp(4) - exp(6)/3)/(15*x - exp(2))