3.26.53 \(\int \frac {-560+280 x-35 x^2+e^{\frac {-35-4 x+x^2+(-28+7 x) \log (x)}{-140+35 x}} (112-5 x-x^2+x^3)}{560 x-280 x^2+35 x^3} \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {1}{4-x}+\frac {1}{5} \left (\frac {x}{7}+\log (x)\right )}-\log (x) \]

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Rubi [F]  time = 11.09, 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 {-560+280 x-35 x^2+\exp \left (\frac {-35-4 x+x^2+(-28+7 x) \log (x)}{-140+35 x}\right ) \left (112-5 x-x^2+x^3\right )}{560 x-280 x^2+35 x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-560 + 280*x - 35*x^2 + E^((-35 - 4*x + x^2 + (-28 + 7*x)*Log[x])/(-140 + 35*x))*(112 - 5*x - x^2 + x^3))
/(560*x - 280*x^2 + 35*x^3),x]

[Out]

-Log[x] + Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5))), x], x, x^(1/5)] + Defer[Subst][Def
er[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) - x)^2, x], x, x^(1/5)]/(2*2^(1/5)*(4 + (-1)^(1/5) -
(-1)^(2/5) + (-1)^(3/5) - (-1)^(4/5))) - Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^
(2/5) - x), x], x, x^(1/5)]/(2*2^(3/5)) + (2^(2/5)*(4 - 3*(-1)^(1/5) + 2*(-1)^(2/5) - (-1)^(3/5))*Defer[Subst]
[Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) - x), x], x, x^(1/5)])/(5*(2 - (-1)^(1/5) + (-1)
^(3/5) - 2*(-1)^(4/5))) + Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))*x^5, x], x, x^(1/5)
]/7 - ((-1)^(2/5)*Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) + (-1)^(1/5)*x)^2,
x], x, x^(1/5)])/(2*2^(1/5)*(1 + (-1)^(1/5))^2*(1 - 3*(-1)^(1/5) + (-1)^(2/5))) - Defer[Subst][Defer[Int][E^(
(-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) + (-1)^(1/5)*x), x], x, x^(1/5)]/(2*2^(3/5)) + ((-1)^(1/5)*2^(2
/5)*(1 - 3*(-1)^(1/5) + (-1)^(2/5))*Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5)
+ (-1)^(1/5)*x), x], x, x^(1/5)])/(5*(1 - (-1)^(2/5))^2) - Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35
*(-4 + x^5)))/(2^(2/5) - (-1)^(2/5)*x), x], x, x^(1/5)]/(2*2^(3/5)) - (5*(-1/2)^(1/5)*Defer[Subst][Defer[Int][
E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(-2^(2/5) + (-1)^(2/5)*x)^2, x], x, x^(1/5)])/(2*(1 - (-1)^(1/5))^4*(
1 + (-1)^(1/5))^8) - (5*(-1)^(1/5)*2^(2/5)*(1 - 2*(-1)^(1/5) + 3*(-1)^(2/5) + (-1)^(3/5))*Defer[Subst][Defer[I
nt][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(-2^(2/5) + (-1)^(2/5)*x), x], x, x^(1/5)])/((1 - (-1)^(1/5))^6*(
1 + (-1)^(1/5))^11) - ((-1)^(2/5)*Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) +
(-1)^(3/5)*x)^2, x], x, x^(1/5)])/(2*2^(1/5)*(1 + (-1)^(1/5))^2*(1 - 3*(-1)^(1/5) + (-1)^(2/5))) - Defer[Subst
][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) + (-1)^(3/5)*x), x], x, x^(1/5)]/(2*2^(3/5)) -
Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2^(2/5) - (-1)^(4/5)*x), x], x, x^(1/5)]/(2*
2^(3/5)) + Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(-2^(2/5) + (-1)^(4/5)*x)^2, x], x
, x^(1/5)]/(2*2^(1/5)*(4 + (-1)^(1/5) - (-1)^(2/5) + (-1)^(3/5) - (-1)^(4/5))) - (2^(2/5)*(3 - 2*(-1)^(1/5) +
(-1)^(2/5) - (-1)^(4/5))*Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(-2^(2/5) + (-1)^(4/
5)*x), x], x, x^(1/5)])/(5*(2 - (-1)^(1/5) + (-1)^(3/5) - 2*(-1)^(4/5))) + (2*(-1)^(1/5)*(1 - 3*(-1)^(1/5) + (
-1)^(2/5))*Defer[Subst][Defer[Int][E^((-35 - 4*x^5 + x^10)/(35*(-4 + x^5)))/(2*(1 - (-1)^(1/5))^2*(1 + (-1)^(1
/5)) + 2^(3/5)*(2 - 2*(-1)^(1/5) + (-1)^(2/5))*x), x], x, x^(1/5)])/(5*(1 + (-1)^(1/5)))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-560+280 x-35 x^2+\exp \left (\frac {-35-4 x+x^2+(-28+7 x) \log (x)}{-140+35 x}\right ) \left (112-5 x-x^2+x^3\right )}{x \left (560-280 x+35 x^2\right )} \, dx\\ &=\int \frac {-560+280 x-35 x^2+\exp \left (\frac {-35-4 x+x^2+(-28+7 x) \log (x)}{-140+35 x}\right ) \left (112-5 x-x^2+x^3\right )}{35 (-4+x)^2 x} \, dx\\ &=\frac {1}{35} \int \frac {-560+280 x-35 x^2+\exp \left (\frac {-35-4 x+x^2+(-28+7 x) \log (x)}{-140+35 x}\right ) \left (112-5 x-x^2+x^3\right )}{(-4+x)^2 x} \, dx\\ &=\frac {1}{35} \int \left (-\frac {35}{x}+\frac {e^{\frac {-35-4 x+x^2}{35 (-4+x)}} \left (112-5 x-x^2+x^3\right )}{(4-x)^2 x^{4/5}}\right ) \, dx\\ &=-\log (x)+\frac {1}{35} \int \frac {e^{\frac {-35-4 x+x^2}{35 (-4+x)}} \left (112-5 x-x^2+x^3\right )}{(4-x)^2 x^{4/5}} \, dx\\ &=-\log (x)+\frac {1}{7} \operatorname {Subst}\left (\int \frac {e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}} \left (112-5 x^5-x^{10}+x^{15}\right )}{\left (-4+x^5\right )^2} \, dx,x,\sqrt [5]{x}\right )\\ &=-\log (x)+\frac {1}{7} \operatorname {Subst}\left (\int \left (7 e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}}+e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}} x^5+\frac {140 e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}}}{\left (-4+x^5\right )^2}+\frac {35 e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}}}{-4+x^5}\right ) \, dx,x,\sqrt [5]{x}\right )\\ &=-\log (x)+\frac {1}{7} \operatorname {Subst}\left (\int e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}} x^5 \, dx,x,\sqrt [5]{x}\right )+5 \operatorname {Subst}\left (\int \frac {e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}}}{-4+x^5} \, dx,x,\sqrt [5]{x}\right )+20 \operatorname {Subst}\left (\int \frac {e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}}}{\left (-4+x^5\right )^2} \, dx,x,\sqrt [5]{x}\right )+\operatorname {Subst}\left (\int e^{\frac {-35-4 x^5+x^{10}}{35 \left (-4+x^5\right )}} \, dx,x,\sqrt [5]{x}\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.60, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {1}{4-x}+\frac {x}{35}} \sqrt [5]{x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-560 + 280*x - 35*x^2 + E^((-35 - 4*x + x^2 + (-28 + 7*x)*Log[x])/(-140 + 35*x))*(112 - 5*x - x^2 +
x^3))/(560*x - 280*x^2 + 35*x^3),x]

[Out]

E^((4 - x)^(-1) + x/35)*x^(1/5) - Log[x]

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fricas [A]  time = 0.71, size = 28, normalized size = 1.04 \begin {gather*} e^{\left (\frac {x^{2} + 7 \, {\left (x - 4\right )} \log \relax (x) - 4 \, x - 35}{35 \, {\left (x - 4\right )}}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2-5*x+112)*exp(((7*x-28)*log(x)+x^2-4*x-35)/(35*x-140))-35*x^2+280*x-560)/(35*x^3-280*x^2+56
0*x),x, algorithm="fricas")

[Out]

e^(1/35*(x^2 + 7*(x - 4)*log(x) - 4*x - 35)/(x - 4)) - log(x)

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giac [B]  time = 0.60, size = 51, normalized size = 1.89 \begin {gather*} e^{\left (\frac {x^{2}}{35 \, {\left (x - 4\right )}} + \frac {x \log \relax (x)}{5 \, {\left (x - 4\right )}} - \frac {4 \, x}{35 \, {\left (x - 4\right )}} - \frac {4 \, \log \relax (x)}{5 \, {\left (x - 4\right )}} - \frac {1}{x - 4}\right )} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2-5*x+112)*exp(((7*x-28)*log(x)+x^2-4*x-35)/(35*x-140))-35*x^2+280*x-560)/(35*x^3-280*x^2+56
0*x),x, algorithm="giac")

[Out]

e^(1/35*x^2/(x - 4) + 1/5*x*log(x)/(x - 4) - 4/35*x/(x - 4) - 4/5*log(x)/(x - 4) - 1/(x - 4)) - log(x)

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maple [A]  time = 0.07, size = 31, normalized size = 1.15

method	result	size
risch	\(-\ln \relax (x )+{\mathrm e}^{\frac {7 x \ln \relax (x )+x^{2}-28 \ln \relax (x )-4 x -35}{35 x -140}}\)	\(31\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (7 x -28\right ) \ln \relax (x )+x^{2}-4 x -35}{35 x -140}}-4 \,{\mathrm e}^{\frac {\left (7 x -28\right ) \ln \relax (x )+x^{2}-4 x -35}{35 x -140}}}{x -4}-\ln \relax (x )\)	\(67\)
default	\(\frac {35 x \,{\mathrm e}^{\frac {\left (7 x -28\right ) \ln \relax (x )+x^{2}-4 x -35}{35 x -140}}-140 \,{\mathrm e}^{\frac {\left (7 x -28\right ) \ln \relax (x )+x^{2}-4 x -35}{35 x -140}}}{35 x -140}-\ln \relax (x )\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-x^2-5*x+112)*exp(((7*x-28)*ln(x)+x^2-4*x-35)/(35*x-140))-35*x^2+280*x-560)/(35*x^3-280*x^2+560*x),x,
method=_RETURNVERBOSE)

[Out]

-ln(x)+exp(1/35*(7*x*ln(x)+x^2-28*ln(x)-4*x-35)/(x-4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{35} \, \int \frac {{\left (x^{3} e^{\left (\frac {1}{35} \, x\right )} - x^{2} e^{\left (\frac {1}{35} \, x\right )} - 5 \, x e^{\left (\frac {1}{35} \, x\right )} + 112 \, e^{\left (\frac {1}{35} \, x\right )}\right )} e^{\left (-\frac {1}{x - 4}\right )}}{x^{\frac {14}{5}} - 8 \, x^{\frac {9}{5}} + 16 \, x^{\frac {4}{5}}}\,{d x} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2-5*x+112)*exp(((7*x-28)*log(x)+x^2-4*x-35)/(35*x-140))-35*x^2+280*x-560)/(35*x^3-280*x^2+56
0*x),x, algorithm="maxima")

[Out]

1/35*integrate((x^3*e^(1/35*x) - x^2*e^(1/35*x) - 5*x*e^(1/35*x) + 112*e^(1/35*x))*e^(-1/(x - 4))/(x^(14/5) -
8*x^(9/5) + 16*x^(4/5)), x) - log(x)

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mupad [B]  time = 1.71, size = 64, normalized size = 2.37 \begin {gather*} \frac {x^{\frac {7\,x}{35\,x-140}}\,{\mathrm {e}}^{\frac {x^2}{35\,x-140}}\,{\mathrm {e}}^{-\frac {35}{35\,x-140}}\,{\mathrm {e}}^{-\frac {4\,x}{35\,x-140}}}{x^{\frac {28}{35\,x-140}}}-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(35*x^2 - 280*x + exp(-(4*x - log(x)*(7*x - 28) - x^2 + 35)/(35*x - 140))*(5*x + x^2 - x^3 - 112) + 560)/
(560*x - 280*x^2 + 35*x^3),x)

[Out]

(x^((7*x)/(35*x - 140))*exp(x^2/(35*x - 140))*exp(-35/(35*x - 140))*exp(-(4*x)/(35*x - 140)))/x^(28/(35*x - 14
0)) - log(x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-x**2-5*x+112)*exp(((7*x-28)*ln(x)+x**2-4*x-35)/(35*x-140))-35*x**2+280*x-560)/(35*x**3-280*x*
*2+560*x),x)

[Out]

Timed out

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