3.95.64 \(\int \frac {224 e^{10+2 x}+e^{5+x} (40-56 x+32 x^2-4 x^3)}{25 x^2-10 x^3+x^4+e^{10+2 x} (4096-512 x+16 x^2)+e^{5+x} (640 x-168 x^2+8 x^3)} \, dx\)

Optimal. Leaf size=28 \[ \frac {-2+x}{16-x+\frac {1}{4} e^{-5-x} (5-x) x} \]

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Rubi [F]  time = 3.20, 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 {224 e^{10+2 x}+e^{5+x} \left (40-56 x+32 x^2-4 x^3\right )}{25 x^2-10 x^3+x^4+e^{10+2 x} \left (4096-512 x+16 x^2\right )+e^{5+x} \left (640 x-168 x^2+8 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(224*E^(10 + 2*x) + E^(5 + x)*(40 - 56*x + 32*x^2 - 4*x^3))/(25*x^2 - 10*x^3 + x^4 + E^(10 + 2*x)*(4096 -
512*x + 16*x^2) + E^(5 + x)*(640*x - 168*x^2 + 8*x^3)),x]

[Out]

-576*Defer[Int][E^(5 + x)/(-64*E^(5 + x) - 5*x + 4*E^(5 + x)*x + x^2)^2, x] - 9856*Defer[Int][E^(5 + x)/((-16
+ x)*(-64*E^(5 + x) - 5*x + 4*E^(5 + x)*x + x^2)^2), x] - 112*Defer[Int][(E^(5 + x)*x)/(-64*E^(5 + x) - 5*x +
4*E^(5 + x)*x + x^2)^2, x] + 32*Defer[Int][(E^(5 + x)*x^2)/(-64*E^(5 + x) - 5*x + 4*E^(5 + x)*x + x^2)^2, x] -
4*Defer[Int][(E^(5 + x)*x^3)/(-64*E^(5 + x) - 5*x + 4*E^(5 + x)*x + x^2)^2, x] + 56*Defer[Int][E^(5 + x)/((-1
6 + x)*(-64*E^(5 + x) - 5*x + 4*E^(5 + x)*x + x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 e^{5+x} \left (10+56 e^{5+x}-14 x+8 x^2-x^3\right )}{\left (4 e^{5+x} (-16+x)+(-5+x) x\right )^2} \, dx\\ &=4 \int \frac {e^{5+x} \left (10+56 e^{5+x}-14 x+8 x^2-x^3\right )}{\left (4 e^{5+x} (-16+x)+(-5+x) x\right )^2} \, dx\\ &=4 \int \left (\frac {14 e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )}-\frac {e^{5+x} \left (160-304 x+156 x^2-24 x^3+x^4\right )}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {e^{5+x} \left (160-304 x+156 x^2-24 x^3+x^4\right )}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx\right )+56 \int \frac {e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )} \, dx\\ &=-\left (4 \int \left (\frac {144 e^{5+x}}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}+\frac {2464 e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}+\frac {28 e^{5+x} x}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}-\frac {8 e^{5+x} x^2}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}+\frac {e^{5+x} x^3}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2}\right ) \, dx\right )+56 \int \frac {e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )} \, dx\\ &=-\left (4 \int \frac {e^{5+x} x^3}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx\right )+32 \int \frac {e^{5+x} x^2}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx+56 \int \frac {e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )} \, dx-112 \int \frac {e^{5+x} x}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx-576 \int \frac {e^{5+x}}{\left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx-9856 \int \frac {e^{5+x}}{(-16+x) \left (-64 e^{5+x}-5 x+4 e^{5+x} x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.31, size = 32, normalized size = 1.14 \begin {gather*} \frac {-56 e^{5+x}+(-5+x) x}{4 e^{5+x} (-16+x)+(-5+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(224*E^(10 + 2*x) + E^(5 + x)*(40 - 56*x + 32*x^2 - 4*x^3))/(25*x^2 - 10*x^3 + x^4 + E^(10 + 2*x)*(4
096 - 512*x + 16*x^2) + E^(5 + x)*(640*x - 168*x^2 + 8*x^3)),x]

[Out]

(-56*E^(5 + x) + (-5 + x)*x)/(4*E^(5 + x)*(-16 + x) + (-5 + x)*x)

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fricas [A]  time = 1.00, size = 32, normalized size = 1.14 \begin {gather*} \frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, {\left (x - 16\right )} e^{\left (x + 5\right )} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2-512*x+4096)*exp(5)^2*exp(x)^2
+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10*x^3+25*x^2),x, algorithm="fricas")

[Out]

(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*(x - 16)*e^(x + 5) - 5*x)

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giac [A]  time = 0.20, size = 36, normalized size = 1.29 \begin {gather*} \frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, x e^{\left (x + 5\right )} - 5 \, x - 64 \, e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2-512*x+4096)*exp(5)^2*exp(x)^2
+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10*x^3+25*x^2),x, algorithm="giac")

[Out]

(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*x*e^(x + 5) - 5*x - 64*e^(x + 5))

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maple [A]  time = 0.24, size = 38, normalized size = 1.36

method	result	size
norman	\(\frac {8 \,{\mathrm e}^{5} {\mathrm e}^{x}-4 x \,{\mathrm e}^{5} {\mathrm e}^{x}}{4 x \,{\mathrm e}^{5} {\mathrm e}^{x}-64 \,{\mathrm e}^{5} {\mathrm e}^{x}+x^{2}-5 x}\)	\(38\)
risch	\(-\frac {14}{x -16}+\frac {\left (x^{2}-7 x +10\right ) x}{\left (x -16\right ) \left (4 x \,{\mathrm e}^{5+x}-64 \,{\mathrm e}^{5+x}+x^{2}-5 x \right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2-512*x+4096)*exp(5)^2*exp(x)^2+(8*x^
3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

(8*exp(5)*exp(x)-4*x*exp(5)*exp(x))/(4*x*exp(5)*exp(x)-64*exp(5)*exp(x)+x^2-5*x)

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maxima [A]  time = 0.41, size = 36, normalized size = 1.29 \begin {gather*} \frac {x^{2} - 5 \, x - 56 \, e^{\left (x + 5\right )}}{x^{2} + 4 \, {\left (x e^{5} - 16 \, e^{5}\right )} e^{x} - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((224*exp(5)^2*exp(x)^2+(-4*x^3+32*x^2-56*x+40)*exp(5)*exp(x))/((16*x^2-512*x+4096)*exp(5)^2*exp(x)^2
+(8*x^3-168*x^2+640*x)*exp(5)*exp(x)+x^4-10*x^3+25*x^2),x, algorithm="maxima")

[Out]

(x^2 - 5*x - 56*e^(x + 5))/(x^2 + 4*(x*e^5 - 16*e^5)*e^x - 5*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {224\,{\mathrm {e}}^{2\,x+10}-{\mathrm {e}}^{x+5}\,\left (4\,x^3-32\,x^2+56\,x-40\right )}{{\mathrm {e}}^{x+5}\,\left (8\,x^3-168\,x^2+640\,x\right )+{\mathrm {e}}^{2\,x+10}\,\left (16\,x^2-512\,x+4096\right )+25\,x^2-10\,x^3+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((224*exp(2*x)*exp(10) - exp(5)*exp(x)*(56*x - 32*x^2 + 4*x^3 - 40))/(25*x^2 - 10*x^3 + x^4 + exp(5)*exp(x)
*(640*x - 168*x^2 + 8*x^3) + exp(2*x)*exp(10)*(16*x^2 - 512*x + 4096)),x)

[Out]

int((224*exp(2*x + 10) - exp(x + 5)*(56*x - 32*x^2 + 4*x^3 - 40))/(exp(x + 5)*(640*x - 168*x^2 + 8*x^3) + exp(
2*x + 10)*(16*x^2 - 512*x + 4096) + 25*x^2 - 10*x^3 + x^4), x)

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sympy [B]  time = 0.33, size = 51, normalized size = 1.82 \begin {gather*} \frac {x^{3} - 7 x^{2} + 10 x}{x^{3} - 21 x^{2} + 80 x + \left (4 x^{2} e^{5} - 128 x e^{5} + 1024 e^{5}\right ) e^{x}} - \frac {14}{x - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((224*exp(5)**2*exp(x)**2+(-4*x**3+32*x**2-56*x+40)*exp(5)*exp(x))/((16*x**2-512*x+4096)*exp(5)**2*ex
p(x)**2+(8*x**3-168*x**2+640*x)*exp(5)*exp(x)+x**4-10*x**3+25*x**2),x)

[Out]

(x**3 - 7*x**2 + 10*x)/(x**3 - 21*x**2 + 80*x + (4*x**2*exp(5) - 128*x*exp(5) + 1024*exp(5))*exp(x)) - 14/(x -
16)

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