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

Optimal. Leaf size=28 \[ \left (-e^x+e^{\frac {x \left (-2+e^4 x\right )}{\frac {5}{4}+x}}\right ) (5+x) \]

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Rubi [F]  time = 1.17, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 56, normalized size = 2.00 \begin {gather*} 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) \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.52, size = 32, normalized size = 1.14 \begin {gather*} -{\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(4*x+5))+(-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))

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giac [B]  time = 0.74, size = 54, normalized size = 1.93 \begin {gather*} -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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(4*x+5))+(-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))

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maple [A]  time = 0.19, size = 31, normalized size = 1.11




method result size



risch \(\left (-x -5\right ) {\mathrm e}^{x}+\left (5+x \right ) {\mathrm e}^{\frac {4 x \left (x \,{\mathrm e}^{4}-2\right )}{4 x +5}}\) \(31\)
norman \(\frac {25 x \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{4 x +5}}+4 x^{2} {\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{4 x +5}}-25 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{x} x^{2}-25 \,{\mathrm e}^{x}+25 \,{\mathrm e}^{\frac {4 x^{2} {\mathrm e}^{4}-8 x}{4 x +5}}}{4 x +5}\) \(96\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(4*x+5))+(-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)/(4*x+5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {75 \, e^{\left (-\frac {5}{4}\right )} E_{2}\left (-x - \frac {5}{4}\right )}{2 \, {\left (4 \, x + 5\right )}} + \frac {{\left (16 \, x^{3} + 120 \, x^{2} + 225 \, x + 125\right )} e^{\left (x e^{4} + \frac {25 \, e^{4}}{4 \, {\left (4 \, x + 5\right )}} + \frac {10}{4 \, x + 5}\right )} - {\left (16 \, x^{3} e^{\left (\frac {5}{4} \, e^{4} + 2\right )} + 120 \, x^{2} e^{\left (\frac {5}{4} \, e^{4} + 2\right )} + 225 \, x e^{\left (\frac {5}{4} \, e^{4} + 2\right )}\right )} e^{x}}{16 \, x^{2} e^{\left (\frac {5}{4} \, e^{4} + 2\right )} + 40 \, x e^{\left (\frac {5}{4} \, e^{4} + 2\right )} + 25 \, e^{\left (\frac {5}{4} \, e^{4} + 2\right )}} + \int \frac {25 \, {\left (4 \, x + 45\right )} e^{x}}{64 \, x^{3} + 240 \, x^{2} + 300 \, x + 125}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^3+120*x^2+200*x)*exp(4)+16*x^2-175)*exp((4*x^2*exp(4)-8*x)/(4*x+5))+(-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)

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mupad [B]  time = 1.11, size = 35, normalized size = 1.25 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.46, size = 29, normalized size = 1.04 \begin {gather*} \left (- x - 5\right ) e^{x} + \left (x + 5\right ) e^{\frac {4 x^{2} e^{4} - 8 x}{4 x + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**3+120*x**2+200*x)*exp(4)+16*x**2-175)*exp((4*x**2*exp(4)-8*x)/(4*x+5))+(-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))

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