\(\int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} (-9+18 e^{25+x}+27 x)}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx\) [7491]

Optimal result

Integrand size = 67, antiderivative size = 21 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=-3+x+\frac {9 e^{3 x}}{5 \left (e^{25+x}+x\right )} \]

[Out]

x-3+9*exp(3*x)/(5*exp(x+25)+5*x)

Rubi [F]

\[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=\int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx \]

[In]

Int[(5*E^(50 + 2*x) + 10*E^(25 + x)*x + 5*x^2 + E^(3*x)*(-9 + 18*E^(25 + x) + 27*x))/(5*E^(50 + 2*x) + 10*E^(2
5 + x)*x + 5*x^2),x]

[Out]

(9*E^(-50 + x))/5 + (9*E^(-25 + 2*x))/5 - (9*E^(-50 + x)*(1 + x))/5 + (5*E^75 + 18*x)^2/(180*E^75) + (9*Defer[
Int][x^3/(E^(25 + x) + x)^2, x])/(5*E^75) - (9*Defer[Int][x^4/(E^(25 + x) + x)^2, x])/(5*E^75) - (27*Defer[Int
][x^2/(E^(25 + x) + x), x])/(5*E^75) + (9*Defer[Int][x^3/(E^(25 + x) + x), x])/(5*E^75)

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 \left (e^{25+x}+x\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{\left (e^{25+x}+x\right )^2} \, dx \\ & = \frac {1}{5} \int \left (18 e^{-25+2 x}-9 e^{-50+x} (1+x)-\frac {9 (-1+x) x^3}{e^{75} \left (e^{25+x}+x\right )^2}+\frac {9 (-3+x) x^2}{e^{75} \left (e^{25+x}+x\right )}+\frac {5 e^{75}+18 x}{e^{75}}\right ) \, dx \\ & = \frac {\left (5 e^{75}+18 x\right )^2}{180 e^{75}}-\frac {9}{5} \int e^{-50+x} (1+x) \, dx+\frac {18}{5} \int e^{-25+2 x} \, dx-\frac {9 \int \frac {(-1+x) x^3}{\left (e^{25+x}+x\right )^2} \, dx}{5 e^{75}}+\frac {9 \int \frac {(-3+x) x^2}{e^{25+x}+x} \, dx}{5 e^{75}} \\ & = \frac {9}{5} e^{-25+2 x}-\frac {9}{5} e^{-50+x} (1+x)+\frac {\left (5 e^{75}+18 x\right )^2}{180 e^{75}}+\frac {9}{5} \int e^{-50+x} \, dx-\frac {9 \int \left (-\frac {x^3}{\left (e^{25+x}+x\right )^2}+\frac {x^4}{\left (e^{25+x}+x\right )^2}\right ) \, dx}{5 e^{75}}+\frac {9 \int \left (-\frac {3 x^2}{e^{25+x}+x}+\frac {x^3}{e^{25+x}+x}\right ) \, dx}{5 e^{75}} \\ & = \frac {9 e^{-50+x}}{5}+\frac {9}{5} e^{-25+2 x}-\frac {9}{5} e^{-50+x} (1+x)+\frac {\left (5 e^{75}+18 x\right )^2}{180 e^{75}}+\frac {9 \int \frac {x^3}{\left (e^{25+x}+x\right )^2} \, dx}{5 e^{75}}-\frac {9 \int \frac {x^4}{\left (e^{25+x}+x\right )^2} \, dx}{5 e^{75}}+\frac {9 \int \frac {x^3}{e^{25+x}+x} \, dx}{5 e^{75}}-\frac {27 \int \frac {x^2}{e^{25+x}+x} \, dx}{5 e^{75}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=\frac {9 e^{3 x}+5 e^{25+x} x+5 x^2}{5 \left (e^{25+x}+x\right )} \]

[In]

Integrate[(5*E^(50 + 2*x) + 10*E^(25 + x)*x + 5*x^2 + E^(3*x)*(-9 + 18*E^(25 + x) + 27*x))/(5*E^(50 + 2*x) + 1
0*E^(25 + x)*x + 5*x^2),x]

[Out]

(9*E^(3*x) + 5*E^(25 + x)*x + 5*x^2)/(5*(E^(25 + x) + x))

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
parts	\(x +\frac {9 \,{\mathrm e}^{3 x}}{5 \left ({\mathrm e}^{x} {\mathrm e}^{25}+x \right )}\)	\(18\)
norman	\(\frac {x^{2}+x \,{\mathrm e}^{x} {\mathrm e}^{25}+\frac {9 \,{\mathrm e}^{3 x}}{5}}{{\mathrm e}^{x} {\mathrm e}^{25}+x}\)	\(27\)
risch	\(\frac {5 x \,{\mathrm e}^{x +25}+5 x^{2}+9 \,{\mathrm e}^{3 x}}{5 \,{\mathrm e}^{x +25}+5 x}\)	\(30\)
parallelrisch	\(\frac {5 x \,{\mathrm e}^{x +25}+5 x^{2}+9 \,{\mathrm e}^{3 x}}{5 \,{\mathrm e}^{x +25}+5 x}\)	\(30\)

[In]

int(((18*exp(x+25)+27*x-9)*exp(3*x)+5*exp(x+25)^2+10*x*exp(x+25)+5*x^2)/(5*exp(x+25)^2+10*x*exp(x+25)+5*x^2),x
,method=_RETURNVERBOSE)

[Out]

x+9/5*exp(x)^3/(exp(x)*exp(25)+x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=\frac {5 \, x^{2} e^{75} + 5 \, x e^{\left (x + 100\right )} + 9 \, e^{\left (3 \, x + 75\right )}}{5 \, {\left (x e^{75} + e^{\left (x + 100\right )}\right )}} \]

[In]

integrate(((18*exp(x+25)+27*x-9)*exp(3*x)+5*exp(x+25)^2+10*x*exp(x+25)+5*x^2)/(5*exp(x+25)^2+10*x*exp(x+25)+5*
x^2),x, algorithm="fricas")

[Out]

1/5*(5*x^2*e^75 + 5*x*e^(x + 100) + 9*e^(3*x + 75))/(x*e^75 + e^(x + 100))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (19) = 38\).

Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.33 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=- \frac {9 x^{3}}{5 x e^{75} + 5 e^{100} \sqrt [3]{e^{3 x}}} + \frac {9 x^{2}}{5 e^{75}} + x + \frac {- 45 x e^{25} \sqrt [3]{e^{3 x}} + 45 e^{50} \left (e^{3 x}\right )^{\frac {2}{3}}}{25 e^{75}} \]

[In]

integrate(((18*exp(x+25)+27*x-9)*exp(3*x)+5*exp(x+25)**2+10*x*exp(x+25)+5*x**2)/(5*exp(x+25)**2+10*x*exp(x+25)
+5*x**2),x)

[Out]

-9*x**3/(5*x*exp(75) + 5*exp(100)*exp(3*x)**(1/3)) + 9*x**2*exp(-75)/5 + x + (-45*x*exp(25)*exp(3*x)**(1/3) +
45*exp(50)*exp(3*x)**(2/3))*exp(-75)/25

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=\frac {5 \, x^{2} + 5 \, x e^{\left (x + 25\right )} + 9 \, e^{\left (3 \, x\right )}}{5 \, {\left (x + e^{\left (x + 25\right )}\right )}} \]

[In]

integrate(((18*exp(x+25)+27*x-9)*exp(3*x)+5*exp(x+25)^2+10*x*exp(x+25)+5*x^2)/(5*exp(x+25)^2+10*x*exp(x+25)+5*
x^2),x, algorithm="maxima")

[Out]

1/5*(5*x^2 + 5*x*e^(x + 25) + 9*e^(3*x))/(x + e^(x + 25))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.95 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=\frac {5 \, {\left (x + 25\right )}^{2} e^{75} - 125 \, {\left (x + 25\right )} e^{75} + 5 \, {\left (x + 25\right )} e^{\left (x + 100\right )} - 5625 \, x + 9 \, e^{\left (3 \, x + 75\right )} - 5625 \, e^{\left (x + 25\right )}}{5 \, {\left ({\left (x + 25\right )} e^{75} - 25 \, e^{75} + e^{\left (x + 100\right )}\right )}} \]

[In]

integrate(((18*exp(x+25)+27*x-9)*exp(3*x)+5*exp(x+25)^2+10*x*exp(x+25)+5*x^2)/(5*exp(x+25)^2+10*x*exp(x+25)+5*
x^2),x, algorithm="giac")

[Out]

1/5*(5*(x + 25)^2*e^75 - 125*(x + 25)*e^75 + 5*(x + 25)*e^(x + 100) - 5625*x + 9*e^(3*x + 75) - 5625*e^(x + 25
))/((x + 25)*e^75 - 25*e^75 + e^(x + 100))

Mupad [B] (verification not implemented)

Time = 13.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {5 e^{50+2 x}+10 e^{25+x} x+5 x^2+e^{3 x} \left (-9+18 e^{25+x}+27 x\right )}{5 e^{50+2 x}+10 e^{25+x} x+5 x^2} \, dx=x+\frac {9\,{\mathrm {e}}^{3\,x}}{5\,\left (x+{\mathrm {e}}^{x+25}\right )} \]

[In]

int((5*exp(2*x + 50) + 10*x*exp(x + 25) + exp(3*x)*(27*x + 18*exp(x + 25) - 9) + 5*x^2)/(5*exp(2*x + 50) + 10*
x*exp(x + 25) + 5*x^2),x)

[Out]

x + (9*exp(3*x))/(5*(x + exp(x + 25)))