\(\int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 (36+6 x+45 x^2+9 x^3)}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx\) [866]

Optimal result

Integrand size = 71, antiderivative size = 25 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=-9+x^3+\log \left (3-e^{2 (x+\log (5+x))}-\frac {x}{3}\right ) \]

[Out]

x^3+ln(3-exp(2*ln(5+x)+2*x)-1/3*x)-9

Rubi [F]

\[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=\int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx \]

[In]

Int[(5 + x - 135*x^2 - 12*x^3 + 3*x^4 + E^(2*x)*(5 + x)^2*(36 + 6*x + 45*x^2 + 9*x^3))/(-45 - 4*x + x^2 + E^(2
*x)*(5 + x)^2*(15 + 3*x)),x]

[Out]

2*x + x^3 + 2*Log[5 + x] + 28*Defer[Int][1/((5 + x)*(-9 + 75*E^(2*x) + x + 30*E^(2*x)*x + 3*E^(2*x)*x^2)), x]
+ 17*Defer[Int][(-9 + x + 3*E^(2*x)*(5 + x)^2)^(-1), x] - 2*Defer[Int][x/(-9 + x + 3*E^(2*x)*(5 + x)^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {-113-7 x+2 x^2}{(5+x) \left (-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2\right )}+\frac {12+2 x+15 x^2+3 x^3}{5+x}\right ) \, dx \\ & = -\int \frac {-113-7 x+2 x^2}{(5+x) \left (-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2\right )} \, dx+\int \frac {12+2 x+15 x^2+3 x^3}{5+x} \, dx \\ & = \int \left (2+3 x^2+\frac {2}{5+x}\right ) \, dx-\int \frac {113+7 x-2 x^2}{(5+x) \left (9-x-3 e^{2 x} (5+x)^2\right )} \, dx \\ & = 2 x+x^3+2 \log (5+x)-\int \left (-\frac {17}{-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2}+\frac {2 x}{-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2}-\frac {28}{(5+x) \left (-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2\right )}\right ) \, dx \\ & = 2 x+x^3+2 \log (5+x)-2 \int \frac {x}{-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2} \, dx+17 \int \frac {1}{-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2} \, dx+28 \int \frac {1}{(5+x) \left (-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2\right )} \, dx \\ & = 2 x+x^3+2 \log (5+x)-2 \int \frac {x}{-9+x+3 e^{2 x} (5+x)^2} \, dx+17 \int \frac {1}{-9+x+3 e^{2 x} (5+x)^2} \, dx+28 \int \frac {1}{(5+x) \left (-9+75 e^{2 x}+x+30 e^{2 x} x+3 e^{2 x} x^2\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.53 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=x^3+\log \left (9-75 e^{2 x}-x-30 e^{2 x} x-3 e^{2 x} x^2\right ) \]

[In]

Integrate[(5 + x - 135*x^2 - 12*x^3 + 3*x^4 + E^(2*x)*(5 + x)^2*(36 + 6*x + 45*x^2 + 9*x^3))/(-45 - 4*x + x^2
+ E^(2*x)*(5 + x)^2*(15 + 3*x)),x]

[Out]

x^3 + Log[9 - 75*E^(2*x) - x - 30*E^(2*x)*x - 3*E^(2*x)*x^2]

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

method	result	size
risch	\(x^{3}+\ln \left (\left (5+x \right )^{2} {\mathrm e}^{2 x}+\frac {x}{3}-3\right )\)	\(21\)
norman	\(x^{3}+\ln \left (x +3 \,{\mathrm e}^{2 \ln \left (5+x \right )+2 x}-9\right )\)	\(22\)
parallelrisch	\(x^{3}+\ln \left (x +3 \,{\mathrm e}^{2 \ln \left (5+x \right )+2 x}-9\right )\)	\(22\)

[In]

int(((9*x^3+45*x^2+6*x+36)*exp(2*ln(5+x)+2*x)+3*x^4-12*x^3-135*x^2+x+5)/((15+3*x)*exp(2*ln(5+x)+2*x)+x^2-4*x-4
5),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=x^{3} + \log \left (x + 3 \, e^{\left (2 \, x + 2 \, \log \left (x + 5\right )\right )} - 9\right ) \]

[In]

integrate(((9*x^3+45*x^2+6*x+36)*exp(2*log(5+x)+2*x)+3*x^4-12*x^3-135*x^2+x+5)/((15+3*x)*exp(2*log(5+x)+2*x)+x
^2-4*x-45),x, algorithm="fricas")

[Out]

x^3 + log(x + 3*e^(2*x + 2*log(x + 5)) - 9)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=x^{3} + 2 \log {\left (x + 5 \right )} + \log {\left (\frac {x - 9}{3 x^{2} + 30 x + 75} + e^{2 x} \right )} \]

[In]

integrate(((9*x**3+45*x**2+6*x+36)*exp(2*ln(5+x)+2*x)+3*x**4-12*x**3-135*x**2+x+5)/((15+3*x)*exp(2*ln(5+x)+2*x
)+x**2-4*x-45),x)

[Out]

x**3 + 2*log(x + 5) + log((x - 9)/(3*x**2 + 30*x + 75) + exp(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=x^{3} + 2 \, \log \left (x + 5\right ) + \log \left (\frac {3 \, {\left (x^{2} + 10 \, x + 25\right )} e^{\left (2 \, x\right )} + x - 9}{3 \, {\left (x^{2} + 10 \, x + 25\right )}}\right ) \]

[In]

integrate(((9*x^3+45*x^2+6*x+36)*exp(2*log(5+x)+2*x)+3*x^4-12*x^3-135*x^2+x+5)/((15+3*x)*exp(2*log(5+x)+2*x)+x
^2-4*x-45),x, algorithm="maxima")

[Out]

x^3 + 2*log(x + 5) + log(1/3*(3*(x^2 + 10*x + 25)*e^(2*x) + x - 9)/(x^2 + 10*x + 25))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=x^{3} + \log \left (3 \, x^{2} e^{\left (2 \, x\right )} + 30 \, x e^{\left (2 \, x\right )} + x + 75 \, e^{\left (2 \, x\right )} - 9\right ) \]

[In]

integrate(((9*x^3+45*x^2+6*x+36)*exp(2*log(5+x)+2*x)+3*x^4-12*x^3-135*x^2+x+5)/((15+3*x)*exp(2*log(5+x)+2*x)+x
^2-4*x-45),x, algorithm="giac")

[Out]

x^3 + log(3*x^2*e^(2*x) + 30*x*e^(2*x) + x + 75*e^(2*x) - 9)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {5+x-135 x^2-12 x^3+3 x^4+e^{2 x} (5+x)^2 \left (36+6 x+45 x^2+9 x^3\right )}{-45-4 x+x^2+e^{2 x} (5+x)^2 (15+3 x)} \, dx=\ln \left (x+75\,{\mathrm {e}}^{2\,x}+30\,x\,{\mathrm {e}}^{2\,x}+3\,x^2\,{\mathrm {e}}^{2\,x}-9\right )+x^3 \]

[In]

int(-(x + exp(2*x + 2*log(x + 5))*(6*x + 45*x^2 + 9*x^3 + 36) - 135*x^2 - 12*x^3 + 3*x^4 + 5)/(4*x - exp(2*x +
2*log(x + 5))*(3*x + 15) - x^2 + 45),x)

[Out]

log(x + 75*exp(2*x) + 30*x*exp(2*x) + 3*x^2*exp(2*x) - 9) + x^3