3.100.91 \(\int \frac {e^{\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}} (6-100 x+200 x^4+e^x (3-49 x-25 x^2+104 x^4-100 x^5))}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx\)

Optimal. Leaf size=27 \[ e^{2+\frac {\left (4+\frac {1}{x^3}\right ) \left (5-\frac {1}{5 x}\right ) x}{2+e^x}} \]

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Rubi [F]  time = 14.59, 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 {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{10 x^3+5 e^x x^3}\right ) \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{20 x^4+20 e^x x^4+5 e^{2 x} x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(10*x^3 + 5*E^x*x^3))*(6 - 100*x + 200*x^4 + E^x*(3 - 49*x
- 25*x^2 + 104*x^4 - 100*x^5)))/(20*x^4 + 20*E^x*x^4 + 5*E^(2*x)*x^4),x]

[Out]

(-8*Defer[Int][E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/(2 + E^x)^2, x])/5 + (104*Def
er[Int][E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/(2 + E^x), x])/5 + (3*Defer[Int][E^(
(-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/((2 + E^x)*x^4), x])/5 - (2*Defer[Int][E^((-1 +
25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/((2 + E^x)^2*x^3), x])/5 - (49*Defer[Int][E^((-1 + 2
5*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/((2 + E^x)*x^3), x])/5 + 10*Defer[Int][E^((-1 + 25*x +
16*x^3 + 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/((2 + E^x)^2*x^2), x] - 5*Defer[Int][E^((-1 + 25*x + 16*x^3
+ 10*E^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))/((2 + E^x)*x^2), x] + 40*Defer[Int][(E^((-1 + 25*x + 16*x^3 + 10*E
^x*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))*x)/(2 + E^x)^2, x] - 20*Defer[Int][(E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 +
100*x^4)/(5*(2 + E^x)*x^3))*x)/(2 + E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{5 \left (2+e^x\right )^2 x^4} \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (6-100 x+200 x^4+e^x \left (3-49 x-25 x^2+104 x^4-100 x^5\right )\right )}{\left (2+e^x\right )^2 x^4} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (-1+25 x-4 x^3+100 x^4\right )}{\left (2+e^x\right )^2 x^3}-\frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (-3+49 x+25 x^2-104 x^4+100 x^5\right )}{\left (2+e^x\right ) x^4}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (-3+49 x+25 x^2-104 x^4+100 x^5\right )}{\left (2+e^x\right ) x^4} \, dx\right )+\frac {2}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) \left (-1+25 x-4 x^3+100 x^4\right )}{\left (2+e^x\right )^2 x^3} \, dx\\ &=-\left (\frac {1}{5} \int \left (-\frac {104 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{2+e^x}-\frac {3 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^4}+\frac {49 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^3}+\frac {25 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^2}+\frac {100 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) x}{2+e^x}\right ) \, dx\right )+\frac {2}{5} \int \left (-\frac {4 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2}-\frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^3}+\frac {25 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^2}+\frac {100 \exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) x}{\left (2+e^x\right )^2}\right ) \, dx\\ &=-\left (\frac {2}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^3} \, dx\right )+\frac {3}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^4} \, dx-\frac {8}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2} \, dx-5 \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^2} \, dx-\frac {49}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right ) x^3} \, dx+10 \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{\left (2+e^x\right )^2 x^2} \, dx-20 \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) x}{2+e^x} \, dx+\frac {104}{5} \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right )}{2+e^x} \, dx+40 \int \frac {\exp \left (\frac {-1+25 x+16 x^3+10 e^x x^3+100 x^4}{5 \left (2+e^x\right ) x^3}\right ) x}{\left (2+e^x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 33, normalized size = 1.22 \begin {gather*} e^{2+\frac {-1+25 x-4 x^3+100 x^4}{5 \left (2+e^x\right ) x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 + 25*x + 16*x^3 + 10*E^x*x^3 + 100*x^4)/(10*x^3 + 5*E^x*x^3))*(6 - 100*x + 200*x^4 + E^x*(3
- 49*x - 25*x^2 + 104*x^4 - 100*x^5)))/(20*x^4 + 20*E^x*x^4 + 5*E^(2*x)*x^4),x]

[Out]

E^(2 + (-1 + 25*x - 4*x^3 + 100*x^4)/(5*(2 + E^x)*x^3))

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fricas [A]  time = 0.84, size = 39, normalized size = 1.44 \begin {gather*} e^{\left (\frac {100 \, x^{4} + 10 \, x^{3} e^{x} + 16 \, x^{3} + 25 \, x - 1}{5 \, {\left (x^{3} e^{x} + 2 \, x^{3}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((10*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/
(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4+20*exp(x)*x^4+20*x^4),x, algorithm="fricas")

[Out]

e^(1/5*(100*x^4 + 10*x^3*e^x + 16*x^3 + 25*x - 1)/(x^3*e^x + 2*x^3))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((10*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/
(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4+20*exp(x)*x^4+20*x^4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1000000000000,[1,29]%%%}+%%%{-2200000000000,[1,28]%%%}+%
%%{13360000

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maple [A]  time = 0.15, size = 35, normalized size = 1.30

method	result	size
risch	\({\mathrm e}^{\frac {10 \,{\mathrm e}^{x} x^{3}+100 x^{4}+16 x^{3}+25 x -1}{5 x^{3} \left ({\mathrm e}^{x}+2\right )}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((10*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/(5*exp
(x)*x^3+10*x^3))/(5*exp(x)^2*x^4+20*exp(x)*x^4+20*x^4),x,method=_RETURNVERBOSE)

[Out]

exp(1/5*(10*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/x^3/(exp(x)+2))

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maxima [B]  time = 0.55, size = 61, normalized size = 2.26 \begin {gather*} e^{\left (\frac {20 \, x}{e^{x} + 2} + \frac {2 \, e^{x}}{e^{x} + 2} - \frac {1}{5 \, {\left (x^{3} e^{x} + 2 \, x^{3}\right )}} + \frac {5}{x^{2} e^{x} + 2 \, x^{2}} + \frac {16}{5 \, {\left (e^{x} + 2\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x^5+104*x^4-25*x^2-49*x+3)*exp(x)+200*x^4-100*x+6)*exp((10*exp(x)*x^3+100*x^4+16*x^3+25*x-1)/
(5*exp(x)*x^3+10*x^3))/(5*exp(x)^2*x^4+20*exp(x)*x^4+20*x^4),x, algorithm="maxima")

[Out]

e^(20*x/(e^x + 2) + 2*e^x/(e^x + 2) - 1/5/(x^3*e^x + 2*x^3) + 5/(x^2*e^x + 2*x^2) + 16/5/(e^x + 2))

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mupad [B]  time = 8.99, size = 68, normalized size = 2.52 \begin {gather*} {\mathrm {e}}^{\frac {2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+2}}\,{\mathrm {e}}^{\frac {16}{5\,{\mathrm {e}}^x+10}}\,{\mathrm {e}}^{\frac {5}{x^2\,{\mathrm {e}}^x+2\,x^2}}\,{\mathrm {e}}^{-\frac {1}{5\,x^3\,{\mathrm {e}}^x+10\,x^3}}\,{\mathrm {e}}^{\frac {20\,x}{{\mathrm {e}}^x+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((25*x + 10*x^3*exp(x) + 16*x^3 + 100*x^4 - 1)/(5*x^3*exp(x) + 10*x^3))*(100*x + exp(x)*(49*x + 25*x^
2 - 104*x^4 + 100*x^5 - 3) - 200*x^4 - 6))/(20*x^4*exp(x) + 5*x^4*exp(2*x) + 20*x^4),x)

[Out]

exp((2*exp(x))/(exp(x) + 2))*exp(16/(5*exp(x) + 10))*exp(5/(x^2*exp(x) + 2*x^2))*exp(-1/(5*x^3*exp(x) + 10*x^3
))*exp((20*x)/(exp(x) + 2))

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sympy [A]  time = 0.43, size = 37, normalized size = 1.37 \begin {gather*} e^{\frac {100 x^{4} + 10 x^{3} e^{x} + 16 x^{3} + 25 x - 1}{5 x^{3} e^{x} + 10 x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x**5+104*x**4-25*x**2-49*x+3)*exp(x)+200*x**4-100*x+6)*exp((10*exp(x)*x**3+100*x**4+16*x**3+2
5*x-1)/(5*exp(x)*x**3+10*x**3))/(5*exp(x)**2*x**4+20*exp(x)*x**4+20*x**4),x)

[Out]

exp((100*x**4 + 10*x**3*exp(x) + 16*x**3 + 25*x - 1)/(5*x**3*exp(x) + 10*x**3))

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