3.86.39 \(\int \frac {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} (324-180 x+25 x^2)+e^x (465+303 x-285 x^2+50 x^3)}{e^{5+2 x} (324-180 x+25 x^2)+e^{5+x} (648 x-360 x^2+50 x^3)+e^5 (324 x^2-180 x^3+25 x^4)} \, dx\)

Optimal. Leaf size=34 \[ 10-\frac {-x+\frac {3 (1-x)}{\left (\frac {18}{5}-x\right ) \left (e^x+x\right )}}{e^5} \]

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Rubi [F]  time = 1.13, 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 {270-150 x+399 x^2-180 x^3+25 x^4+e^{2 x} \left (324-180 x+25 x^2\right )+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^{5+2 x} \left (324-180 x+25 x^2\right )+e^{5+x} \left (648 x-360 x^2+50 x^3\right )+e^5 \left (324 x^2-180 x^3+25 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(270 - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + E^(2*x)*(324 - 180*x + 25*x^2) + E^x*(465 + 303*x - 285*x^2 +
50*x^3))/(E^(5 + 2*x)*(324 - 180*x + 25*x^2) + E^(5 + x)*(648*x - 360*x^2 + 50*x^3) + E^5*(324*x^2 - 180*x^3 +
25*x^4)),x]

[Out]

x/E^5 - 3/(E^5*(E^x + x)) - (39*Defer[Int][(E^x + x)^(-2), x])/(5*E^5) + (195*Defer[Int][1/((E^x + x)*(-18 + 5
*x)^2), x])/E^5 - (507*Defer[Int][1/((E^x + x)^2*(-18 + 5*x)), x])/(5*E^5) + (39*Defer[Int][1/((E^x + x)*(-18
+ 5*x)), x])/E^5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {270+e^{2 x} (18-5 x)^2-150 x+399 x^2-180 x^3+25 x^4+e^x \left (465+303 x-285 x^2+50 x^3\right )}{e^5 (18-5 x)^2 \left (e^x+x\right )^2} \, dx\\ &=\frac {\int \frac {270+e^{2 x} (18-5 x)^2-150 x+399 x^2-180 x^3+25 x^4+e^x \left (465+303 x-285 x^2+50 x^3\right )}{(18-5 x)^2 \left (e^x+x\right )^2} \, dx}{e^5}\\ &=\frac {\int \left (1-\frac {15 (-1+x)^2}{\left (e^x+x\right )^2 (-18+5 x)}+\frac {15 \left (31-23 x+5 x^2\right )}{\left (e^x+x\right ) (-18+5 x)^2}\right ) \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {15 \int \frac {(-1+x)^2}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{e^5}+\frac {15 \int \frac {31-23 x+5 x^2}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {15 \int \left (\frac {8}{25 \left (e^x+x\right )^2}+\frac {x}{5 \left (e^x+x\right )^2}+\frac {169}{25 \left (e^x+x\right )^2 (-18+5 x)}\right ) \, dx}{e^5}+\frac {15 \int \left (\frac {1}{5 \left (e^x+x\right )}+\frac {13}{\left (e^x+x\right ) (-18+5 x)^2}+\frac {13}{5 \left (e^x+x\right ) (-18+5 x)}\right ) \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {3 \int \frac {x}{\left (e^x+x\right )^2} \, dx}{e^5}+\frac {3 \int \frac {1}{e^x+x} \, dx}{e^5}-\frac {24 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{5 e^5}+\frac {39 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)} \, dx}{e^5}-\frac {507 \int \frac {1}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{5 e^5}+\frac {195 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ &=\frac {x}{e^5}-\frac {3}{e^5 \left (e^x+x\right )}-\frac {3 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{e^5}-\frac {24 \int \frac {1}{\left (e^x+x\right )^2} \, dx}{5 e^5}+\frac {39 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)} \, dx}{e^5}-\frac {507 \int \frac {1}{\left (e^x+x\right )^2 (-18+5 x)} \, dx}{5 e^5}+\frac {195 \int \frac {1}{\left (e^x+x\right ) (-18+5 x)^2} \, dx}{e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 25, normalized size = 0.74 \begin {gather*} \frac {x-\frac {15 (-1+x)}{\left (e^x+x\right ) (-18+5 x)}}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(270 - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + E^(2*x)*(324 - 180*x + 25*x^2) + E^x*(465 + 303*x - 285*
x^2 + 50*x^3))/(E^(5 + 2*x)*(324 - 180*x + 25*x^2) + E^(5 + x)*(648*x - 360*x^2 + 50*x^3) + E^5*(324*x^2 - 180
*x^3 + 25*x^4)),x]

[Out]

(x - (15*(-1 + x))/((E^x + x)*(-18 + 5*x)))/E^5

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fricas [B]  time = 0.73, size = 59, normalized size = 1.74 \begin {gather*} \frac {{\left (5 \, x^{3} - 18 \, x^{2} - 15 \, x + 15\right )} e^{5} + {\left (5 \, x^{2} - 18 \, x\right )} e^{\left (x + 5\right )}}{{\left (5 \, x^{2} - 18 \, x\right )} e^{10} + {\left (5 \, x - 18\right )} e^{\left (x + 10\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4-180*x^3+399*x^2-150*x+270)/((2
5*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algo
rithm="fricas")

[Out]

((5*x^3 - 18*x^2 - 15*x + 15)*e^5 + (5*x^2 - 18*x)*e^(x + 5))/((5*x^2 - 18*x)*e^10 + (5*x - 18)*e^(x + 10))

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giac [B]  time = 0.19, size = 56, normalized size = 1.65 \begin {gather*} \frac {5 \, x^{3} + 5 \, x^{2} e^{x} - 18 \, x^{2} - 18 \, x e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + 5 \, x e^{\left (x + 5\right )} - 18 \, e^{\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4-180*x^3+399*x^2-150*x+270)/((2
5*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algo
rithm="giac")

[Out]

(5*x^3 + 5*x^2*e^x - 18*x^2 - 18*x*e^x - 15*x + 15)/(5*x^2*e^5 - 18*x*e^5 + 5*x*e^(x + 5) - 18*e^(x + 5))

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maple [A]  time = 0.24, size = 26, normalized size = 0.76

method	result	size
risch	\(x \,{\mathrm e}^{-5}-\frac {15 \left (x -1\right ) {\mathrm e}^{-5}}{\left (5 x -18\right ) \left ({\mathrm e}^{x}+x \right )}\)	\(26\)
norman	\(\frac {-\frac {399 x \,{\mathrm e}^{-5}}{5}-\frac {324 \,{\mathrm e}^{-5} {\mathrm e}^{x}}{5}+15 \,{\mathrm e}^{-5}+5 x^{3} {\mathrm e}^{-5}+5 x^{2} {\mathrm e}^{-5} {\mathrm e}^{x}}{5 \,{\mathrm e}^{x} x +5 x^{2}-18 \,{\mathrm e}^{x}-18 x}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4-180*x^3+399*x^2-150*x+270)/((25*x^2-
180*x+324)*exp(5)*exp(x)^2+(50*x^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x,method=_RET
URNVERBOSE)

[Out]

x*exp(-5)-15*(x-1)*exp(-5)/(5*x-18)/(exp(x)+x)

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maxima [B]  time = 0.43, size = 56, normalized size = 1.65 \begin {gather*} \frac {5 \, x^{3} - 18 \, x^{2} + {\left (5 \, x^{2} - 18 \, x\right )} e^{x} - 15 \, x + 15}{5 \, x^{2} e^{5} - 18 \, x e^{5} + {\left (5 \, x e^{5} - 18 \, e^{5}\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x^2-180*x+324)*exp(x)^2+(50*x^3-285*x^2+303*x+465)*exp(x)+25*x^4-180*x^3+399*x^2-150*x+270)/((2
5*x^2-180*x+324)*exp(5)*exp(x)^2+(50*x^3-360*x^2+648*x)*exp(5)*exp(x)+(25*x^4-180*x^3+324*x^2)*exp(5)),x, algo
rithm="maxima")

[Out]

(5*x^3 - 18*x^2 + (5*x^2 - 18*x)*e^x - 15*x + 15)/(5*x^2*e^5 - 18*x*e^5 + (5*x*e^5 - 18*e^5)*e^x)

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mupad [B]  time = 0.27, size = 42, normalized size = 1.24 \begin {gather*} x\,{\mathrm {e}}^{-5}-\frac {15\,{\mathrm {e}}^{-5}\,\left (5\,x^3-28\,x^2+41\,x-18\right )}{{\left (5\,x-18\right )}^2\,\left (x+{\mathrm {e}}^x\right )\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(25*x^2 - 180*x + 324) - 150*x + 399*x^2 - 180*x^3 + 25*x^4 + exp(x)*(303*x - 285*x^2 + 50*x^3 +
465) + 270)/(exp(5)*(324*x^2 - 180*x^3 + 25*x^4) + exp(5)*exp(x)*(648*x - 360*x^2 + 50*x^3) + exp(2*x)*exp(5)
*(25*x^2 - 180*x + 324)),x)

[Out]

x*exp(-5) - (15*exp(-5)*(41*x - 28*x^2 + 5*x^3 - 18))/((5*x - 18)^2*(x + exp(x))*(x - 1))

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sympy [A]  time = 0.19, size = 39, normalized size = 1.15 \begin {gather*} \frac {x}{e^{5}} + \frac {15 - 15 x}{5 x^{2} e^{5} - 18 x e^{5} + \left (5 x e^{5} - 18 e^{5}\right ) e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((25*x**2-180*x+324)*exp(x)**2+(50*x**3-285*x**2+303*x+465)*exp(x)+25*x**4-180*x**3+399*x**2-150*x+2
70)/((25*x**2-180*x+324)*exp(5)*exp(x)**2+(50*x**3-360*x**2+648*x)*exp(5)*exp(x)+(25*x**4-180*x**3+324*x**2)*e
xp(5)),x)

[Out]

x*exp(-5) + (15 - 15*x)/(5*x**2*exp(5) - 18*x*exp(5) + (5*x*exp(5) - 18*exp(5))*exp(x))

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