\(\int \frac {e^x (-150-265 x-136 x^2-16 x^3)+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} (-175+16 x^2+e^4 (200 x+120 x^2+16 x^3))}{25+40 x+16 x^2} \, dx\) [1032]

Optimal result

Integrand size = 81, antiderivative size = 28 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=\left (-e^x+e^{\frac {x \left (-2+e^4 x\right )}{\frac {5}{4}+x}}\right ) (5+x) \]

[Out]

(exp((x*exp(4)-2)*x/(x+5/4))-exp(x))*(5+x)

Rubi [F]

\[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=\int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx \]

[In]

Int[(E^x*(-150 - 265*x - 136*x^2 - 16*x^3) + E^((-8*x + 4*E^4*x^2)/(5 + 4*x))*(-175 + 16*x^2 + E^4*(200*x + 12
0*x^2 + 16*x^3)))/(25 + 40*x + 16*x^2),x]

[Out]

E^x - E^x*(6 + x) + (1 + 5*E^4)*Defer[Int][E^((4*x*(-2 + E^4*x))/(5 + 4*x)), x] + Defer[Int][E^((4*(5 + 2*x +
E^4*x^2))/(5 + 4*x))*x, x] - (75*(8 + 5*E^4)*Defer[Int][E^((4*x*(-2 + E^4*x))/(5 + 4*x))/(5 + 4*x)^2, x])/4 -
(5*(8 + 5*E^4)*Defer[Int][E^((4*x*(-2 + E^4*x))/(5 + 4*x))/(5 + 4*x), x])/4

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{(5+4 x)^2} \, dx \\ & = \int \left (-e^x (6+x)+\frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \left (-175+200 e^4 x+8 \left (2+15 e^4\right ) x^2+16 e^4 x^3\right )}{(5+4 x)^2}\right ) \, dx \\ & = -\int e^x (6+x) \, dx+\int \frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \left (-175+200 e^4 x+8 \left (2+15 e^4\right ) x^2+16 e^4 x^3\right )}{(5+4 x)^2} \, dx \\ & = -e^x (6+x)+\int e^x \, dx+\int \left (e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \left (1+5 e^4\right )+e^{4+\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} x-\frac {75 e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \left (8+5 e^4\right )}{4 (5+4 x)^2}-\frac {5 e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \left (8+5 e^4\right )}{4 (5+4 x)}\right ) \, dx \\ & = e^x-e^x (6+x)+\left (1+5 e^4\right ) \int e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \, dx-\frac {1}{4} \left (5 \left (8+5 e^4\right )\right ) \int \frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}}}{5+4 x} \, dx-\frac {1}{4} \left (75 \left (8+5 e^4\right )\right ) \int \frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}}}{(5+4 x)^2} \, dx+\int e^{4+\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} x \, dx \\ & = e^x-e^x (6+x)+\left (1+5 e^4\right ) \int e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}} \, dx-\frac {1}{4} \left (5 \left (8+5 e^4\right )\right ) \int \frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}}}{5+4 x} \, dx-\frac {1}{4} \left (75 \left (8+5 e^4\right )\right ) \int \frac {e^{\frac {4 x \left (-2+e^4 x\right )}{5+4 x}}}{(5+4 x)^2} \, dx+\int e^{\frac {4 \left (5+2 x+e^4 x^2\right )}{5+4 x}} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=e^{-2-\frac {5 e^4}{2}} \left (-e^{2+\frac {5 e^4}{2}+x}+e^{\frac {20+e^4 \left (25+20 x+8 x^2\right )}{10+8 x}}\right ) (5+x) \]

[In]

Integrate[(E^x*(-150 - 265*x - 136*x^2 - 16*x^3) + E^((-8*x + 4*E^4*x^2)/(5 + 4*x))*(-175 + 16*x^2 + E^4*(200*
x + 120*x^2 + 16*x^3)))/(25 + 40*x + 16*x^2),x]

[Out]

E^(-2 - (5*E^4)/2)*(-E^(2 + (5*E^4)/2 + x) + E^((20 + E^4*(25 + 20*x + 8*x^2))/(10 + 8*x)))*(5 + x)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11

method	result	size
risch	\(\left (-x -5\right ) {\mathrm e}^{x}+\left (5+x \right ) {\mathrm e}^{\frac {4 x \left (x \,{\mathrm e}^{4}-2\right )}{5+4 x}}\)	\(31\)
parallelrisch	\(-{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}-5 \,{\mathrm e}^{x}+5 \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}\)	\(55\)
parts	\(-5 \,{\mathrm e}^{x}-{\mathrm e}^{x} x +\frac {25 x \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}+4 x^{2} {\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}+25 \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}}{5+4 x}\)	\(90\)
norman	\(\frac {25 x \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}+4 x^{2} {\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}-25 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x} x^{2}-25 \,{\mathrm e}^{x}+25 \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{5+4 x}}}{5+4 x}\)	\(96\)

[In]

int((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(5+4*x))+(-16*x^3-136*x^2-265*x-150)*ex
p(x))/(16*x^2+40*x+25),x,method=_RETURNVERBOSE)

[Out]

(-x-5)*exp(x)+(5+x)*exp(4*x*(x*exp(4)-2)/(5+4*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=-{\left (x + 5\right )} e^{x} + {\left (x + 5\right )} e^{\left (\frac {4 \, {\left (x^{2} e^{4} - 2 \, x\right )}}{4 \, x + 5}\right )} \]

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(5+4*x))+(-16*x^3-136*x^2-265*x-1
50)*exp(x))/(16*x^2+40*x+25),x, algorithm="fricas")

[Out]

-(x + 5)*e^x + (x + 5)*e^(4*(x^2*e^4 - 2*x)/(4*x + 5))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=\left (- x - 5\right ) e^{x} + \left (x + 5\right ) e^{\frac {4 x^{2} e^{4} - 8 x}{4 x + 5}} \]

[In]

integrate((((16*x**3+120*x**2+200*x)*exp(4)+16*x**2-175)*exp((4*x**2*exp(4)-8*x)/(5+4*x))+(-16*x**3-136*x**2-2
65*x-150)*exp(x))/(16*x**2+40*x+25),x)

[Out]

(-x - 5)*exp(x) + (x + 5)*exp((4*x**2*exp(4) - 8*x)/(4*x + 5))

Maxima [F]

\[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=\int { -\frac {{\left (16 \, x^{3} + 136 \, x^{2} + 265 \, x + 150\right )} e^{x} - {\left (16 \, x^{2} + 8 \, {\left (2 \, x^{3} + 15 \, x^{2} + 25 \, x\right )} e^{4} - 175\right )} e^{\left (\frac {4 \, {\left (x^{2} e^{4} - 2 \, x\right )}}{4 \, x + 5}\right )}}{16 \, x^{2} + 40 \, x + 25} \,d x } \]

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(5+4*x))+(-16*x^3-136*x^2-265*x-1
50)*exp(x))/(16*x^2+40*x+25),x, algorithm="maxima")

[Out]

75/2*e^(-5/4)*exp_integral_e(2, -x - 5/4)/(4*x + 5) + ((16*x^3 + 120*x^2 + 225*x + 125)*e^(x*e^4 + 25/4*e^4/(4
*x + 5) + 10/(4*x + 5)) - (16*x^3*e^(5/4*e^4 + 2) + 120*x^2*e^(5/4*e^4 + 2) + 225*x*e^(5/4*e^4 + 2))*e^x)/(16*
x^2*e^(5/4*e^4 + 2) + 40*x*e^(5/4*e^4 + 2) + 25*e^(5/4*e^4 + 2)) + integrate(25*(4*x + 45)*e^x/(64*x^3 + 240*x
^2 + 300*x + 125), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).

Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=-x e^{x} + x e^{\left (\frac {4 \, {\left (x^{2} e^{4} - 2 \, x\right )}}{4 \, x + 5}\right )} - 5 \, e^{x} + 5 \, e^{\left (\frac {4 \, {\left (x^{2} e^{4} - 2 \, x\right )}}{4 \, x + 5}\right )} \]

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(5+4*x))+(-16*x^3-136*x^2-265*x-1
50)*exp(x))/(16*x^2+40*x+25),x, algorithm="giac")

[Out]

-x*e^x + x*e^(4*(x^2*e^4 - 2*x)/(4*x + 5)) - 5*e^x + 5*e^(4*(x^2*e^4 - 2*x)/(4*x + 5))

Mupad [B] (verification not implemented)

Time = 9.65 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^x \left (-150-265 x-136 x^2-16 x^3\right )+e^{\frac {-8 x+4 e^4 x^2}{5+4 x}} \left (-175+16 x^2+e^4 \left (200 x+120 x^2+16 x^3\right )\right )}{25+40 x+16 x^2} \, dx=\left ({\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^4}{4\,x+5}-\frac {8\,x}{4\,x+5}}-{\mathrm {e}}^x\right )\,\left (x+5\right ) \]

[In]

int((exp(-(8*x - 4*x^2*exp(4))/(4*x + 5))*(exp(4)*(200*x + 120*x^2 + 16*x^3) + 16*x^2 - 175) - exp(x)*(265*x +
136*x^2 + 16*x^3 + 150))/(40*x + 16*x^2 + 25),x)

[Out]

(exp((4*x^2*exp(4))/(4*x + 5) - (8*x)/(4*x + 5)) - exp(x))*(x + 5)