3.45.97 \(\int \frac {e^{2 x} (-5+x)^2 (30+42 x-12 x^2+e^{-4+x} (-15-6 x+3 x^2))}{4 (-20+e^{-4+x} (20-4 x)+e^{-8+2 x} (-5+x)+4 x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {3 e^{2 \left (x-\log \left (\frac {2}{-5+x}\right )\right )} x}{-2+e^{-4+x}} \]

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Rubi [F]  time = 3.03, 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^{2 x} (-5+x)^2 \left (30+42 x-12 x^2+e^{-4+x} \left (-15-6 x+3 x^2\right )\right )}{4 \left (-20+e^{-4+x} (20-4 x)+e^{-8+2 x} (-5+x)+4 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(-5 + x)^2*(30 + 42*x - 12*x^2 + E^(-4 + x)*(-15 - 6*x + 3*x^2)))/(4*(-20 + E^(-4 + x)*(20 - 4*x)
+ E^(-8 + 2*x)*(-5 + x) + 4*x)),x]

[Out]

(75*E^(4 + x))/4 + (75*E^8*Log[2*E^4 - E^x])/2 - (75*Defer[Int][(E^(8 + 2*x)*x)/(-2*E^4 + E^x)^2, x])/2 + (15*
Defer[Int][(E^(4 + 2*x)*x)/(-2*E^4 + E^x), x])/4 + 15*Defer[Int][(E^(8 + 2*x)*x^2)/(-2*E^4 + E^x)^2, x] - (21*
Defer[Int][(E^(4 + 2*x)*x^2)/(-2*E^4 + E^x), x])/4 - (3*Defer[Int][(E^(8 + 2*x)*x^3)/(-2*E^4 + E^x)^2, x])/2 +
(3*Defer[Int][(E^(4 + 2*x)*x^3)/(-2*E^4 + E^x), x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{2 x} (-5+x)^2 \left (30+42 x-12 x^2+e^{-4+x} \left (-15-6 x+3 x^2\right )\right )}{-20+e^{-4+x} (20-4 x)+e^{-8+2 x} (-5+x)+4 x} \, dx\\ &=\frac {1}{4} \int \frac {e^{8+2 x} (5-x) \left (-30-42 x+12 x^2-e^{-4+x} \left (-15-6 x+3 x^2\right )\right )}{\left (2 e^4-e^x\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {6 e^{8+2 x} (-5+x)^2 x}{\left (-2 e^4+e^x\right )^2}-\frac {3 e^{4+2 x} (-5+x) \left (-5-2 x+x^2\right )}{2 e^4-e^x}\right ) \, dx\\ &=-\left (\frac {3}{4} \int \frac {e^{4+2 x} (-5+x) \left (-5-2 x+x^2\right )}{2 e^4-e^x} \, dx\right )-\frac {3}{2} \int \frac {e^{8+2 x} (-5+x)^2 x}{\left (-2 e^4+e^x\right )^2} \, dx\\ &=-\left (\frac {3}{4} \int \left (-\frac {25 e^{4+2 x}}{-2 e^4+e^x}-\frac {5 e^{4+2 x} x}{-2 e^4+e^x}+\frac {7 e^{4+2 x} x^2}{-2 e^4+e^x}-\frac {e^{4+2 x} x^3}{-2 e^4+e^x}\right ) \, dx\right )-\frac {3}{2} \int \left (\frac {25 e^{8+2 x} x}{\left (-2 e^4+e^x\right )^2}-\frac {10 e^{8+2 x} x^2}{\left (-2 e^4+e^x\right )^2}+\frac {e^{8+2 x} x^3}{\left (-2 e^4+e^x\right )^2}\right ) \, dx\\ &=\frac {3}{4} \int \frac {e^{4+2 x} x^3}{-2 e^4+e^x} \, dx-\frac {3}{2} \int \frac {e^{8+2 x} x^3}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {15}{4} \int \frac {e^{4+2 x} x}{-2 e^4+e^x} \, dx-\frac {21}{4} \int \frac {e^{4+2 x} x^2}{-2 e^4+e^x} \, dx+15 \int \frac {e^{8+2 x} x^2}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {75}{4} \int \frac {e^{4+2 x}}{-2 e^4+e^x} \, dx-\frac {75}{2} \int \frac {e^{8+2 x} x}{\left (-2 e^4+e^x\right )^2} \, dx\\ &=\frac {3}{4} \int \frac {e^{4+2 x} x^3}{-2 e^4+e^x} \, dx-\frac {3}{2} \int \frac {e^{8+2 x} x^3}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {15}{4} \int \frac {e^{4+2 x} x}{-2 e^4+e^x} \, dx-\frac {21}{4} \int \frac {e^{4+2 x} x^2}{-2 e^4+e^x} \, dx+15 \int \frac {e^{8+2 x} x^2}{\left (-2 e^4+e^x\right )^2} \, dx-\frac {75}{2} \int \frac {e^{8+2 x} x}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {1}{4} \left (75 e^4\right ) \operatorname {Subst}\left (\int \frac {x}{-2 e^4+x} \, dx,x,e^x\right )\\ &=\frac {3}{4} \int \frac {e^{4+2 x} x^3}{-2 e^4+e^x} \, dx-\frac {3}{2} \int \frac {e^{8+2 x} x^3}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {15}{4} \int \frac {e^{4+2 x} x}{-2 e^4+e^x} \, dx-\frac {21}{4} \int \frac {e^{4+2 x} x^2}{-2 e^4+e^x} \, dx+15 \int \frac {e^{8+2 x} x^2}{\left (-2 e^4+e^x\right )^2} \, dx-\frac {75}{2} \int \frac {e^{8+2 x} x}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {1}{4} \left (75 e^4\right ) \operatorname {Subst}\left (\int \left (1-\frac {2 e^4}{2 e^4-x}\right ) \, dx,x,e^x\right )\\ &=\frac {75 e^{4+x}}{4}+\frac {75}{2} e^8 \log \left (2 e^4-e^x\right )+\frac {3}{4} \int \frac {e^{4+2 x} x^3}{-2 e^4+e^x} \, dx-\frac {3}{2} \int \frac {e^{8+2 x} x^3}{\left (-2 e^4+e^x\right )^2} \, dx+\frac {15}{4} \int \frac {e^{4+2 x} x}{-2 e^4+e^x} \, dx-\frac {21}{4} \int \frac {e^{4+2 x} x^2}{-2 e^4+e^x} \, dx+15 \int \frac {e^{8+2 x} x^2}{\left (-2 e^4+e^x\right )^2} \, dx-\frac {75}{2} \int \frac {e^{8+2 x} x}{\left (-2 e^4+e^x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.41, size = 28, normalized size = 1.00 \begin {gather*} \frac {3 e^{4+2 x} (-5+x)^2 x}{4 \left (-2 e^4+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-5 + x)^2*(30 + 42*x - 12*x^2 + E^(-4 + x)*(-15 - 6*x + 3*x^2)))/(4*(-20 + E^(-4 + x)*(20
- 4*x) + E^(-8 + 2*x)*(-5 + x) + 4*x)),x]

[Out]

(3*E^(4 + 2*x)*(-5 + x)^2*x)/(4*(-2*E^4 + E^x))

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fricas [A]  time = 0.54, size = 26, normalized size = 0.93 \begin {gather*} \frac {3 \, {\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )}}{4 \, {\left (e^{\left (x - 4\right )} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x-15)*exp(x-4)-12*x^2+42*x+30)*exp(-2*log(2/(x-5))+2*x)/((x-5)*exp(x-4)^2+(-4*x+20)*exp(x-
4)+4*x-20),x, algorithm="fricas")

[Out]

3/4*(x^3 - 10*x^2 + 25*x)*e^(2*x)/(e^(x - 4) - 2)

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giac [A]  time = 0.15, size = 44, normalized size = 1.57 \begin {gather*} -\frac {3 \, {\left (x^{3} e^{\left (2 \, x + 4\right )} - 10 \, x^{2} e^{\left (2 \, x + 4\right )} + 25 \, x e^{\left (2 \, x + 4\right )}\right )}}{4 \, {\left (2 \, e^{4} - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x-15)*exp(x-4)-12*x^2+42*x+30)*exp(-2*log(2/(x-5))+2*x)/((x-5)*exp(x-4)^2+(-4*x+20)*exp(x-
4)+4*x-20),x, algorithm="giac")

[Out]

-3/4*(x^3*e^(2*x + 4) - 10*x^2*e^(2*x + 4) + 25*x*e^(2*x + 4))/(2*e^4 - e^x)

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maple [A]  time = 0.32, size = 21, normalized size = 0.75

method	result	size
risch	\(\frac {3 x \left (x -5\right )^{2} {\mathrm e}^{2 x}}{4 \left ({\mathrm e}^{x -4}-2\right )}\)	\(21\)
norman	\(\frac {\frac {75 x \,{\mathrm e}^{8} {\mathrm e}^{2 x -8}}{4}-\frac {15 x^{2} {\mathrm e}^{8} {\mathrm e}^{2 x -8}}{2}+\frac {3 \,{\mathrm e}^{8} x^{3} {\mathrm e}^{2 x -8}}{4}}{{\mathrm e}^{x -4}-2}\)	\(48\)
default	\(\frac {\frac {75 x \,{\mathrm e}^{8} {\mathrm e}^{2 x -8}}{4}-\frac {15 x^{2} {\mathrm e}^{8} {\mathrm e}^{2 x -8}}{2}+\frac {3 \,{\mathrm e}^{8} x^{3} {\mathrm e}^{2 x -8}}{4}}{{\mathrm e}^{x -4}-2}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2-6*x-15)*exp(x-4)-12*x^2+42*x+30)*exp(-2*ln(2/(x-5))+2*x)/((x-5)*exp(x-4)^2+(-4*x+20)*exp(x-4)+4*x-
20),x,method=_RETURNVERBOSE)

[Out]

3/4*x/(exp(x-4)-2)*(x-5)^2*exp(2*x)

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maxima [A]  time = 0.43, size = 36, normalized size = 1.29 \begin {gather*} -\frac {3 \, {\left (x^{3} e^{4} - 10 \, x^{2} e^{4} + 25 \, x e^{4}\right )} e^{\left (2 \, x\right )}}{4 \, {\left (2 \, e^{4} - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-6*x-15)*exp(x-4)-12*x^2+42*x+30)*exp(-2*log(2/(x-5))+2*x)/((x-5)*exp(x-4)^2+(-4*x+20)*exp(x-
4)+4*x-20),x, algorithm="maxima")

[Out]

-3/4*(x^3*e^4 - 10*x^2*e^4 + 25*x*e^4)*e^(2*x)/(2*e^4 - e^x)

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mupad [B]  time = 3.37, size = 22, normalized size = 0.79 \begin {gather*} \frac {3\,x\,{\mathrm {e}}^{2\,x}\,{\left (x-5\right )}^2}{4\,\left ({\mathrm {e}}^{x-4}-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x - 2*log(2/(x - 5)))*(42*x - exp(x - 4)*(6*x - 3*x^2 + 15) - 12*x^2 + 30))/(4*x + exp(2*x - 8)*(x
- 5) - exp(x - 4)*(4*x - 20) - 20),x)

[Out]

(3*x*exp(2*x)*(x - 5)^2)/(4*(exp(x - 4) - 2))

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sympy [B]  time = 0.25, size = 97, normalized size = 3.46 \begin {gather*} \frac {3 x^{3} e^{8}}{2} - 15 x^{2} e^{8} + \frac {75 x e^{8}}{2} + \frac {\left (3 x^{3} e^{4} - 30 x^{2} e^{4} + 75 x e^{4}\right ) \sqrt {e^{2 x}}}{4} + \frac {3 x^{3} e^{12} - 30 x^{2} e^{12} + 75 x e^{12}}{\sqrt {e^{2 x}} - 2 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2-6*x-15)*exp(x-4)-12*x**2+42*x+30)*exp(-2*ln(2/(x-5))+2*x)/((x-5)*exp(x-4)**2+(-4*x+20)*exp(
x-4)+4*x-20),x)

[Out]

3*x**3*exp(8)/2 - 15*x**2*exp(8) + 75*x*exp(8)/2 + (3*x**3*exp(4) - 30*x**2*exp(4) + 75*x*exp(4))*sqrt(exp(2*x
))/4 + (3*x**3*exp(12) - 30*x**2*exp(12) + 75*x*exp(12))/(sqrt(exp(2*x)) - 2*exp(4))

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