\(\int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} (-32 x+8 x^2+14 x^3+2 x^4))}{16+8 x+x^2} \, dx\) [9055]

Optimal result

Integrand size = 92, antiderivative size = 34 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=x^2 \left (7-x-e^x x \left (\frac {e^{x \left (-3+\frac {2}{4+x}\right )}}{x}+x\right )\right ) \]

[Out]

(7-exp(x)*(x+exp((2/(4+x)-3)*x)/x)*x-x)*x^2

Rubi [F]

\[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=\int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx \]

[In]

Int[(224*x + 64*x^2 - 10*x^3 - 3*x^4 + E^x*(-64*x^3 - 48*x^4 - 12*x^5 - x^6 + E^((-10*x - 3*x^2)/(4 + x))*(-32
*x + 8*x^2 + 14*x^3 + 2*x^4)))/(16 + 8*x + x^2),x]

[Out]

7*x^2 - x^3 - E^x*x^4 - 8*Defer[Int][E^((-2*x*(3 + x))/(4 + x)), x] - 2*Defer[Int][x/E^((2*x*(3 + x))/(4 + x))
, x] + 2*Defer[Int][x^2/E^((2*x*(3 + x))/(4 + x)), x] - 128*Defer[Int][1/(E^((2*x*(3 + x))/(4 + x))*(4 + x)^2)
, x] + 64*Defer[Int][1/(E^((2*x*(3 + x))/(4 + x))*(4 + x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{(4+x)^2} \, dx \\ & = \int \left (\frac {224 x}{(4+x)^2}+\frac {64 x^2}{(4+x)^2}-\frac {10 x^3}{(4+x)^2}-\frac {64 e^x x^3}{(4+x)^2}-\frac {3 x^4}{(4+x)^2}-\frac {48 e^x x^4}{(4+x)^2}-\frac {12 e^x x^5}{(4+x)^2}-\frac {e^x x^6}{(4+x)^2}+\frac {2 e^{-\frac {2 x (3+x)}{4+x}} x \left (-16+4 x+7 x^2+x^3\right )}{(4+x)^2}\right ) \, dx \\ & = 2 \int \frac {e^{-\frac {2 x (3+x)}{4+x}} x \left (-16+4 x+7 x^2+x^3\right )}{(4+x)^2} \, dx-3 \int \frac {x^4}{(4+x)^2} \, dx-10 \int \frac {x^3}{(4+x)^2} \, dx-12 \int \frac {e^x x^5}{(4+x)^2} \, dx-48 \int \frac {e^x x^4}{(4+x)^2} \, dx+64 \int \frac {x^2}{(4+x)^2} \, dx-64 \int \frac {e^x x^3}{(4+x)^2} \, dx+224 \int \frac {x}{(4+x)^2} \, dx-\int \frac {e^x x^6}{(4+x)^2} \, dx \\ & = 2 \int \left (-4 e^{-\frac {2 x (3+x)}{4+x}}-e^{-\frac {2 x (3+x)}{4+x}} x+e^{-\frac {2 x (3+x)}{4+x}} x^2-\frac {64 e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2}+\frac {32 e^{-\frac {2 x (3+x)}{4+x}}}{4+x}\right ) \, dx-3 \int \left (48-8 x+x^2+\frac {256}{(4+x)^2}-\frac {256}{4+x}\right ) \, dx-10 \int \left (-8+x-\frac {64}{(4+x)^2}+\frac {48}{4+x}\right ) \, dx-12 \int \left (-256 e^x+48 e^x x-8 e^x x^2+e^x x^3-\frac {1024 e^x}{(4+x)^2}+\frac {1280 e^x}{4+x}\right ) \, dx-48 \int \left (48 e^x-8 e^x x+e^x x^2+\frac {256 e^x}{(4+x)^2}-\frac {256 e^x}{4+x}\right ) \, dx+64 \int \left (1+\frac {16}{(4+x)^2}-\frac {8}{4+x}\right ) \, dx-64 \int \left (-8 e^x+e^x x-\frac {64 e^x}{(4+x)^2}+\frac {48 e^x}{4+x}\right ) \, dx+224 \int \left (-\frac {4}{(4+x)^2}+\frac {1}{4+x}\right ) \, dx-\int \left (1280 e^x-256 e^x x+48 e^x x^2-8 e^x x^3+e^x x^4+\frac {4096 e^x}{(4+x)^2}-\frac {6144 e^x}{4+x}\right ) \, dx \\ & = 7 x^2-x^3-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx+8 \int e^x x^3 \, dx-12 \int e^x x^3 \, dx-2 \left (48 \int e^x x^2 \, dx\right )-64 \int e^x x \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx+96 \int e^x x^2 \, dx-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx+256 \int e^x x \, dx+384 \int e^x x \, dx+512 \int e^x \, dx-576 \int e^x x \, dx-1280 \int e^x \, dx-2304 \int e^x \, dx+3072 \int e^x \, dx-3072 \int \frac {e^x}{4+x} \, dx+6144 \int \frac {e^x}{4+x} \, dx+12288 \int \frac {e^x}{4+x} \, dx-15360 \int \frac {e^x}{4+x} \, dx-\int e^x x^4 \, dx \\ & = 7 x^2+96 e^x x^2-x^3-4 e^x x^3-e^x x^4-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx+4 \int e^x x^3 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx-24 \int e^x x^2 \, dx+36 \int e^x x^2 \, dx+64 \int e^x \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx-2 \left (48 e^x x^2-96 \int e^x x \, dx\right )-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx-192 \int e^x x \, dx-256 \int e^x \, dx-384 \int e^x \, dx+576 \int e^x \, dx \\ & = -192 e^x x+7 x^2+108 e^x x^2-x^3-e^x x^4-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx-12 \int e^x x^2 \, dx+48 \int e^x x \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx-72 \int e^x x \, dx-2 \left (-96 e^x x+48 e^x x^2+96 \int e^x \, dx\right )-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx+192 \int e^x \, dx \\ & = 192 e^x-216 e^x x+7 x^2+96 e^x x^2-x^3-e^x x^4-2 \left (96 e^x-96 e^x x+48 e^x x^2\right )-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx+24 \int e^x x \, dx-48 \int e^x \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx+72 \int e^x \, dx-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx \\ & = 216 e^x-192 e^x x+7 x^2+96 e^x x^2-x^3-e^x x^4-2 \left (96 e^x-96 e^x x+48 e^x x^2\right )-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx-24 \int e^x \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx \\ & = 192 e^x-192 e^x x+7 x^2+96 e^x x^2-x^3-e^x x^4-2 \left (96 e^x-96 e^x x+48 e^x x^2\right )-2 \int e^{-\frac {2 x (3+x)}{4+x}} x \, dx+2 \int e^{-\frac {2 x (3+x)}{4+x}} x^2 \, dx-8 \int e^{-\frac {2 x (3+x)}{4+x}} \, dx+64 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{4+x} \, dx-128 \int \frac {e^{-\frac {2 x (3+x)}{4+x}}}{(4+x)^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=x^2 \left (7-e^{-\frac {2 x (3+x)}{4+x}}-x-e^x x^2\right ) \]

[In]

Integrate[(224*x + 64*x^2 - 10*x^3 - 3*x^4 + E^x*(-64*x^3 - 48*x^4 - 12*x^5 - x^6 + E^((-10*x - 3*x^2)/(4 + x)
)*(-32*x + 8*x^2 + 14*x^3 + 2*x^4)))/(16 + 8*x + x^2),x]

[Out]

x^2*(7 - E^((-2*x*(3 + x))/(4 + x)) - x - E^x*x^2)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06

method	result	size
risch	\(-{\mathrm e}^{x} x^{4}-x^{2} {\mathrm e}^{-\frac {2 \left (3+x \right ) x}{4+x}}-x^{3}+7 x^{2}\)	\(36\)
parallelrisch	\(-{\mathrm e}^{x} x^{4}-{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-x^{3}+7 x^{2}-\frac {56}{3}\)	\(43\)
parts	\(\frac {-4 \,{\mathrm e}^{x} x^{4}-x^{5} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-{\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}}{4+x}-x^{3}+7 x^{2}\)	\(79\)
norman	\(\frac {28 x^{2}+3 x^{3}-x^{4}-x^{5} {\mathrm e}^{x}-4 \,{\mathrm e}^{x} x^{4}-4 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}-{\mathrm e}^{x} x^{3} {\mathrm e}^{\frac {-3 x^{2}-10 x}{4+x}}}{4+x}\)	\(83\)

[In]

int((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+64*x^2+
224*x)/(x^2+8*x+16),x,method=_RETURNVERBOSE)

[Out]

-exp(x)*x^4-x^2*exp(-2*(3+x)*x/(4+x))-x^3+7*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{3} + 7 \, x^{2} - {\left (x^{4} + x^{2} e^{\left (-\frac {3 \, x^{2} + 10 \, x}{x + 4}\right )}\right )} e^{x} \]

[In]

integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+6
4*x^2+224*x)/(x^2+8*x+16),x, algorithm="fricas")

[Out]

-x^3 + 7*x^2 - (x^4 + x^2*e^(-(3*x^2 + 10*x)/(x + 4)))*e^x

Sympy [A] (verification not implemented)

Time = 3.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=- x^{4} e^{x} - x^{3} - x^{2} e^{x} e^{\frac {- 3 x^{2} - 10 x}{x + 4}} + 7 x^{2} \]

[In]

integrate((((2*x**4+14*x**3+8*x**2-32*x)*exp((-3*x**2-10*x)/(4+x))-x**6-12*x**5-48*x**4-64*x**3)*exp(x)-3*x**4
-10*x**3+64*x**2+224*x)/(x**2+8*x+16),x)

[Out]

-x**4*exp(x) - x**3 - x**2*exp(x)*exp((-3*x**2 - 10*x)/(x + 4)) + 7*x**2

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{3} + 7 \, x^{2} - {\left (x^{4} e^{\left (3 \, x\right )} + x^{2} e^{\left (-\frac {8}{x + 4} + 2\right )}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+6
4*x^2+224*x)/(x^2+8*x+16),x, algorithm="maxima")

[Out]

-x^3 + 7*x^2 - (x^4*e^(3*x) + x^2*e^(-8/(x + 4) + 2))*e^(-2*x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=-x^{4} e^{x} - x^{3} - x^{2} e^{\left (-\frac {2 \, {\left (x^{2} + 3 \, x\right )}}{x + 4}\right )} + 7 \, x^{2} \]

[In]

integrate((((2*x^4+14*x^3+8*x^2-32*x)*exp((-3*x^2-10*x)/(4+x))-x^6-12*x^5-48*x^4-64*x^3)*exp(x)-3*x^4-10*x^3+6
4*x^2+224*x)/(x^2+8*x+16),x, algorithm="giac")

[Out]

-x^4*e^x - x^3 - x^2*e^(-2*(x^2 + 3*x)/(x + 4)) + 7*x^2

Mupad [B] (verification not implemented)

Time = 14.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {224 x+64 x^2-10 x^3-3 x^4+e^x \left (-64 x^3-48 x^4-12 x^5-x^6+e^{\frac {-10 x-3 x^2}{4+x}} \left (-32 x+8 x^2+14 x^3+2 x^4\right )\right )}{16+8 x+x^2} \, dx=7\,x^2-x^4\,{\mathrm {e}}^x-x^3-x^2\,{\mathrm {e}}^{x-\frac {10\,x}{x+4}-\frac {3\,x^2}{x+4}} \]

[In]

int(-(exp(x)*(64*x^3 - exp(-(10*x + 3*x^2)/(x + 4))*(8*x^2 - 32*x + 14*x^3 + 2*x^4) + 48*x^4 + 12*x^5 + x^6) -
224*x - 64*x^2 + 10*x^3 + 3*x^4)/(8*x + x^2 + 16),x)

[Out]

7*x^2 - x^4*exp(x) - x^3 - x^2*exp(x - (10*x)/(x + 4) - (3*x^2)/(x + 4))