3.7.83 \(\int \frac {-189-24 e^{-8+2 x}-114 x+e^{-4+x} (135+47 x+35 x^2-4 x^3)}{27+324 x+1296 x^2+1728 x^3+e^{-4+x} (-18-216 x-864 x^2-1152 x^3)+e^{-8+2 x} (3+36 x+144 x^2+192 x^3)} \, dx\)

Optimal. Leaf size=30 \[ \frac {3-\frac {(-9+x) x}{3-e^{-4+x}}}{3 (1+4 x)^2} \]

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Rubi [F]  time = 1.40, 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 {-189-24 e^{-8+2 x}-114 x+e^{-4+x} \left (135+47 x+35 x^2-4 x^3\right )}{27+324 x+1296 x^2+1728 x^3+e^{-4+x} \left (-18-216 x-864 x^2-1152 x^3\right )+e^{-8+2 x} \left (3+36 x+144 x^2+192 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-189 - 24*E^(-8 + 2*x) - 114*x + E^(-4 + x)*(135 + 47*x + 35*x^2 - 4*x^3))/(27 + 324*x + 1296*x^2 + 1728*
x^3 + E^(-4 + x)*(-18 - 216*x - 864*x^2 - 1152*x^3) + E^(-8 + 2*x)*(3 + 36*x + 144*x^2 + 192*x^3)),x]

[Out]

-1/48*E^4/(3*E^4 - E^x) + (1 + 4*x)^(-2) - (37*E^4*Defer[Int][1/((-3*E^4 + E^x)*(1 + 4*x)^3), x])/6 - (37*E^8*
Defer[Int][1/((-3*E^4 + E^x)^2*(1 + 4*x)^2), x])/16 + (115*E^4*Defer[Int][1/((-3*E^4 + E^x)*(1 + 4*x)^2), x])/
48 + (19*E^8*Defer[Int][1/((-3*E^4 + E^x)^2*(1 + 4*x)), x])/8 + (19*E^4*Defer[Int][1/((-3*E^4 + E^x)*(1 + 4*x)
), x])/24

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24 e^{2 x}-3 e^8 (63+38 x)+e^{4+x} \left (135+47 x+35 x^2-4 x^3\right )}{3 \left (3 e^4-e^x\right )^2 (1+4 x)^3} \, dx\\ &=\frac {1}{3} \int \frac {-24 e^{2 x}-3 e^8 (63+38 x)+e^{4+x} \left (135+47 x+35 x^2-4 x^3\right )}{\left (3 e^4-e^x\right )^2 (1+4 x)^3} \, dx\\ &=\frac {1}{3} \int \left (-\frac {24}{(1+4 x)^3}-\frac {3 e^8 (-9+x) x}{\left (3 e^4-e^x\right )^2 (1+4 x)^2}+\frac {e^4 \left (9-47 x-35 x^2+4 x^3\right )}{\left (3 e^4-e^x\right ) (1+4 x)^3}\right ) \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{3} e^4 \int \frac {9-47 x-35 x^2+4 x^3}{\left (3 e^4-e^x\right ) (1+4 x)^3} \, dx-e^8 \int \frac {(-9+x) x}{\left (3 e^4-e^x\right )^2 (1+4 x)^2} \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{3} e^4 \int \left (-\frac {1}{16 \left (-3 e^4+e^x\right )}-\frac {37}{2 \left (-3 e^4+e^x\right ) (1+4 x)^3}+\frac {115}{16 \left (-3 e^4+e^x\right ) (1+4 x)^2}+\frac {19}{8 \left (-3 e^4+e^x\right ) (1+4 x)}\right ) \, dx-e^8 \int \left (\frac {1}{16 \left (-3 e^4+e^x\right )^2}+\frac {37}{16 \left (-3 e^4+e^x\right )^2 (1+4 x)^2}-\frac {19}{8 \left (-3 e^4+e^x\right )^2 (1+4 x)}\right ) \, dx\\ &=\frac {1}{(1+4 x)^2}-\frac {1}{48} e^4 \int \frac {1}{-3 e^4+e^x} \, dx+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \int \frac {1}{\left (-3 e^4+e^x\right )^2} \, dx-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=\frac {1}{(1+4 x)^2}-\frac {1}{48} e^4 \operatorname {Subst}\left (\int \frac {1}{x \left (-3 e^4+x\right )} \, dx,x,e^x\right )+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \operatorname {Subst}\left (\int \frac {1}{x \left (-3 e^4+x\right )^2} \, dx,x,e^x\right )-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=\frac {1}{(1+4 x)^2}+\frac {1}{144} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {1}{144} \operatorname {Subst}\left (\int \frac {1}{-3 e^4+x} \, dx,x,e^x\right )+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} e^8 \operatorname {Subst}\left (\int \left (\frac {1}{3 e^4 \left (3 e^4-x\right )^2}+\frac {1}{9 e^8 \left (3 e^4-x\right )}+\frac {1}{9 e^8 x}\right ) \, dx,x,e^x\right )-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ &=-\frac {e^4}{48 \left (3 e^4-e^x\right )}+\frac {1}{(1+4 x)^2}+\frac {1}{24} \left (19 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)} \, dx+\frac {1}{48} \left (115 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^2} \, dx-\frac {1}{6} \left (37 e^4\right ) \int \frac {1}{\left (-3 e^4+e^x\right ) (1+4 x)^3} \, dx-\frac {1}{16} \left (37 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)^2} \, dx+\frac {1}{8} \left (19 e^8\right ) \int \frac {1}{\left (-3 e^4+e^x\right )^2 (1+4 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 43, normalized size = 1.43 \begin {gather*} \frac {1}{3} \left (\frac {3}{(1+4 x)^2}+\frac {e^4 \left (-9 x+x^2\right )}{\left (-3 e^4+e^x\right ) (1+4 x)^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-189 - 24*E^(-8 + 2*x) - 114*x + E^(-4 + x)*(135 + 47*x + 35*x^2 - 4*x^3))/(27 + 324*x + 1296*x^2 +
 1728*x^3 + E^(-4 + x)*(-18 - 216*x - 864*x^2 - 1152*x^3) + E^(-8 + 2*x)*(3 + 36*x + 144*x^2 + 192*x^3)),x]

[Out]

(3/(1 + 4*x)^2 + (E^4*(-9*x + x^2))/((-3*E^4 + E^x)*(1 + 4*x)^2))/3

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fricas [A]  time = 0.98, size = 44, normalized size = 1.47 \begin {gather*} -\frac {x^{2} - 9 \, x + 3 \, e^{\left (x - 4\right )} - 9}{3 \, {\left (48 \, x^{2} - {\left (16 \, x^{2} + 8 \, x + 1\right )} e^{\left (x - 4\right )} + 24 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*exp(x-4)^2+(-4*x^3+35*x^2+47*x+135)*exp(x-4)-114*x-189)/((192*x^3+144*x^2+36*x+3)*exp(x-4)^2+(-
1152*x^3-864*x^2-216*x-18)*exp(x-4)+1728*x^3+1296*x^2+324*x+27),x, algorithm="fricas")

[Out]

-1/3*(x^2 - 9*x + 3*e^(x - 4) - 9)/(48*x^2 - (16*x^2 + 8*x + 1)*e^(x - 4) + 24*x + 3)

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giac [B]  time = 0.41, size = 57, normalized size = 1.90 \begin {gather*} -\frac {x^{2} e^{4} - 9 \, x e^{4} - 9 \, e^{4} + 3 \, e^{x}}{3 \, {\left (48 \, x^{2} e^{4} - 16 \, x^{2} e^{x} + 24 \, x e^{4} - 8 \, x e^{x} + 3 \, e^{4} - e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*exp(x-4)^2+(-4*x^3+35*x^2+47*x+135)*exp(x-4)-114*x-189)/((192*x^3+144*x^2+36*x+3)*exp(x-4)^2+(-
1152*x^3-864*x^2-216*x-18)*exp(x-4)+1728*x^3+1296*x^2+324*x+27),x, algorithm="giac")

[Out]

-1/3*(x^2*e^4 - 9*x*e^4 - 9*e^4 + 3*e^x)/(48*x^2*e^4 - 16*x^2*e^x + 24*x*e^4 - 8*x*e^x + 3*e^4 - e^x)

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maple [A]  time = 0.23, size = 31, normalized size = 1.03




method result size



norman \(\frac {-3 x +{\mathrm e}^{x -4}+\frac {x^{2}}{3}-3}{\left ({\mathrm e}^{x -4}-3\right ) \left (4 x +1\right )^{2}}\) \(31\)
risch \(\frac {1}{16 x^{2}+8 x +1}+\frac {x \left (x -9\right )}{3 \left (16 x^{2}+8 x +1\right ) \left ({\mathrm e}^{x -4}-3\right )}\) \(40\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-24*exp(x-4)^2+(-4*x^3+35*x^2+47*x+135)*exp(x-4)-114*x-189)/((192*x^3+144*x^2+36*x+3)*exp(x-4)^2+(-1152*x
^3-864*x^2-216*x-18)*exp(x-4)+1728*x^3+1296*x^2+324*x+27),x,method=_RETURNVERBOSE)

[Out]

(-3*x+exp(x-4)+1/3*x^2-3)/(exp(x-4)-3)/(4*x+1)^2

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maxima [B]  time = 0.61, size = 55, normalized size = 1.83 \begin {gather*} -\frac {x^{2} e^{4} - 9 \, x e^{4} - 9 \, e^{4} + 3 \, e^{x}}{3 \, {\left (48 \, x^{2} e^{4} + 24 \, x e^{4} - {\left (16 \, x^{2} + 8 \, x + 1\right )} e^{x} + 3 \, e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*exp(x-4)^2+(-4*x^3+35*x^2+47*x+135)*exp(x-4)-114*x-189)/((192*x^3+144*x^2+36*x+3)*exp(x-4)^2+(-
1152*x^3-864*x^2-216*x-18)*exp(x-4)+1728*x^3+1296*x^2+324*x+27),x, algorithm="maxima")

[Out]

-1/3*(x^2*e^4 - 9*x*e^4 - 9*e^4 + 3*e^x)/(48*x^2*e^4 + 24*x*e^4 - (16*x^2 + 8*x + 1)*e^x + 3*e^4)

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mupad [B]  time = 0.76, size = 34, normalized size = 1.13 \begin {gather*} \frac {1}{{\left (4\,x+1\right )}^2}-\frac {3\,x-\frac {x^2}{3}}{{\left (4\,x+1\right )}^2\,\left ({\mathrm {e}}^{x-4}-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(114*x + 24*exp(2*x - 8) - exp(x - 4)*(47*x + 35*x^2 - 4*x^3 + 135) + 189)/(324*x - exp(x - 4)*(216*x + 8
64*x^2 + 1152*x^3 + 18) + exp(2*x - 8)*(36*x + 144*x^2 + 192*x^3 + 3) + 1296*x^2 + 1728*x^3 + 27),x)

[Out]

1/(4*x + 1)^2 - (3*x - x^2/3)/((4*x + 1)^2*(exp(x - 4) - 3))

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sympy [B]  time = 0.23, size = 42, normalized size = 1.40 \begin {gather*} \frac {x^{2} - 9 x}{- 144 x^{2} - 72 x + \left (48 x^{2} + 24 x + 3\right ) e^{x - 4} - 9} + \frac {8}{128 x^{2} + 64 x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*exp(x-4)**2+(-4*x**3+35*x**2+47*x+135)*exp(x-4)-114*x-189)/((192*x**3+144*x**2+36*x+3)*exp(x-4)
**2+(-1152*x**3-864*x**2-216*x-18)*exp(x-4)+1728*x**3+1296*x**2+324*x+27),x)

[Out]

(x**2 - 9*x)/(-144*x**2 - 72*x + (48*x**2 + 24*x + 3)*exp(x - 4) - 9) + 8/(128*x**2 + 64*x + 8)

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