3.37.38 \(\int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x (10-8 x+9 x^2-3 x^3)+e^{2 x} (-10-2 e^x+8 x-9 x^2+6 x^3)}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} (-10-2 e^x+8 x)} \, dx\)

Optimal. Leaf size=29 \[ 3+x-\frac {x^3}{x+\frac {1}{3} \left (-5-e^x+e^{2 x}+x\right )} \]

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Rubi [F]  time = 1.52, 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 {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25 + E^(2*x) + E^(4*x) - 40*x + 61*x^2 - 24*x^3 + E^x*(10 - 8*x + 9*x^2 - 3*x^3) + E^(2*x)*(-10 - 2*E^x +
8*x - 9*x^2 + 6*x^3))/(25 + E^(2*x) + E^(4*x) + E^x*(10 - 8*x) - 40*x + 16*x^2 + E^(2*x)*(-10 - 2*E^x + 8*x))
,x]

[Out]

x + 42*Defer[Int][x^3/(-5 - E^x + E^(2*x) + 4*x)^2, x] + 3*Defer[Int][(E^x*x^3)/(-5 - E^x + E^(2*x) + 4*x)^2,
x] - 24*Defer[Int][x^4/(-5 - E^x + E^(2*x) + 4*x)^2, x] - 9*Defer[Int][x^2/(-5 - E^x + E^(2*x) + 4*x), x] + 6*
Defer[Int][x^3/(-5 - E^x + E^(2*x) + 4*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25-2 e^{3 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-9+8 x-9 x^2+6 x^3\right )}{\left (5+e^x-e^{2 x}-4 x\right )^2} \, dx\\ &=\int \left (1+\frac {3 \left (14+e^x-8 x\right ) x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2}+\frac {3 x^2 (-3+2 x)}{-5-e^x+e^{2 x}+4 x}\right ) \, dx\\ &=x+3 \int \frac {\left (14+e^x-8 x\right ) x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2} \, dx+3 \int \frac {x^2 (-3+2 x)}{-5-e^x+e^{2 x}+4 x} \, dx\\ &=x+3 \int \left (\frac {14 x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2}+\frac {e^x x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2}-\frac {8 x^4}{\left (-5-e^x+e^{2 x}+4 x\right )^2}\right ) \, dx+3 \int \left (-\frac {3 x^2}{-5-e^x+e^{2 x}+4 x}+\frac {2 x^3}{-5-e^x+e^{2 x}+4 x}\right ) \, dx\\ &=x+3 \int \frac {e^x x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2} \, dx+6 \int \frac {x^3}{-5-e^x+e^{2 x}+4 x} \, dx-9 \int \frac {x^2}{-5-e^x+e^{2 x}+4 x} \, dx-24 \int \frac {x^4}{\left (-5-e^x+e^{2 x}+4 x\right )^2} \, dx+42 \int \frac {x^3}{\left (-5-e^x+e^{2 x}+4 x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 24, normalized size = 0.83 \begin {gather*} x-\frac {3 x^3}{-5-e^x+e^{2 x}+4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + E^(2*x) + E^(4*x) - 40*x + 61*x^2 - 24*x^3 + E^x*(10 - 8*x + 9*x^2 - 3*x^3) + E^(2*x)*(-10 - 2
*E^x + 8*x - 9*x^2 + 6*x^3))/(25 + E^(2*x) + E^(4*x) + E^x*(10 - 8*x) - 40*x + 16*x^2 + E^(2*x)*(-10 - 2*E^x +
8*x)),x]

[Out]

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

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fricas [A]  time = 0.57, size = 42, normalized size = 1.45 \begin {gather*} -\frac {3 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*
x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="f
ricas")

[Out]

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

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giac [A]  time = 0.24, size = 42, normalized size = 1.45 \begin {gather*} -\frac {6 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*
x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="g
iac")

[Out]

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

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maple [A]  time = 0.08, size = 23, normalized size = 0.79

method	result	size
risch	\(x -\frac {3 x^{3}}{4 x -5+{\mathrm e}^{2 x}-{\mathrm e}^{x}}\)	\(23\)
norman	\(\frac {x \,{\mathrm e}^{2 x}-\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{2 x}}{4}+4 x^{2}-3 x^{3}-{\mathrm e}^{x} x -\frac {25}{4}}{4 x -5+{\mathrm e}^{2 x}-{\mathrm e}^{x}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40
*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x,method=_RETURNVERBO
SE)

[Out]

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

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maxima [A]  time = 0.70, size = 42, normalized size = 1.45 \begin {gather*} -\frac {3 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*
x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="m
axima")

[Out]

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

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mupad [B]  time = 2.38, size = 22, normalized size = 0.76 \begin {gather*} x-\frac {3\,x^3}{4\,x+{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*x - exp(2*x) - exp(4*x) + exp(2*x)*(2*exp(x) - 8*x + 9*x^2 - 6*x^3 + 10) - 61*x^2 + 24*x^3 + exp(x)*(
8*x - 9*x^2 + 3*x^3 - 10) - 25)/(exp(2*x) - 40*x + exp(4*x) - exp(2*x)*(2*exp(x) - 8*x + 10) - exp(x)*(8*x - 1
0) + 16*x^2 + 25),x)

[Out]

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

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sympy [A]  time = 0.16, size = 19, normalized size = 0.66 \begin {gather*} - \frac {3 x^{3}}{4 x + e^{2 x} - e^{x} - 5} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2*x)**2+(-2*exp(x)+6*x**3-9*x**2+8*x-10)*exp(2*x)+exp(x)**2+(-3*x**3+9*x**2-8*x+10)*exp(x)-24*x
**3+61*x**2-40*x+25)/(exp(2*x)**2+(-2*exp(x)+8*x-10)*exp(2*x)+exp(x)**2+(-8*x+10)*exp(x)+16*x**2-40*x+25),x)

[Out]

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

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