3.75.14 \(\int \frac {e^{\frac {2 (4 x^2+x^3+\log (\frac {15}{40+3 e^x}))}{4 x^2+x^3}} (e^x (-24 x-6 x^2)+(-640+e^x (-48-18 x)-240 x) \log (\frac {15}{40+3 e^x}))}{640 x^3+320 x^4+40 x^5+e^x (48 x^3+24 x^4+3 x^5)} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {2 \left (x+\frac {\log \left (\frac {5}{\frac {40}{3}+e^x}\right )}{x (4+x)}\right )}{x}} \]

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Rubi [F]  time = 27.47, 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 {2 \left (4 x^2+x^3+\log \left (\frac {15}{40+3 e^x}\right )\right )}{4 x^2+x^3}\right ) \left (e^x \left (-24 x-6 x^2\right )+\left (-640+e^x (-48-18 x)-240 x\right ) \log \left (\frac {15}{40+3 e^x}\right )\right )}{640 x^3+320 x^4+40 x^5+e^x \left (48 x^3+24 x^4+3 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(4*x^2 + x^3 + Log[15/(40 + 3*E^x)]))/(4*x^2 + x^3))*(E^x*(-24*x - 6*x^2) + (-640 + E^x*(-48 - 18*x
) - 240*x)*Log[15/(40 + 3*E^x)]))/(640*x^3 + 320*x^4 + 40*x^5 + E^x*(48*x^3 + 24*x^4 + 3*x^5)),x]

[Out]

-8*E^2*Log[15/(40 + 3*E^x)]*Defer[Int][(3^(2/(x^2*(4 + x)))*5^(1 + 2/(x^2*(4 + x)))*((40 + 3*E^x)^(-1))^(1 + 2
/(x^2*(4 + x))))/x^3, x] - E^2*Log[15/(40 + 3*E^x)]*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^
x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x^3, x] + E^2*Log[15/(40 + 3*E^x)]*Defer[Int][(3^(2/(x^2*(4 + x))
)*5^(1 + 2/(x^2*(4 + x)))*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x^2, x] - (E^2*Defer[Int][(3^(1 + 2/(x^2*
(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x^2, x])/2 + (E^2*Log[15/(40 + 3*
E^x)]*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x
^2, x])/8 + (E^2*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(
4 + x))))/x, x])/8 - E^2*Log[15/(40 + 3*E^x)]*Defer[Int][(3^(2/(x^2*(4 + x)))*5^(1 + 2/(x^2*(4 + x)))*((40 + 3
*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/(4 + x)^2, x] - (E^2*Log[15/(40 + 3*E^x)]*Defer[Int][(3^(1 + 2/(x^2*(4 + x)
))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/(4 + x)^2, x])/8 - (E^2*Defer[Int][(3^(1
+ 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/(4 + x), x])/8 - 24*E^2
*Defer[Int][(E^x*Defer[Int][(5^(1 + 2/(x^2*(4 + x)))*9^(1/(x^2*(4 + x)))*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 +
x))))/x^3, x])/(40 + 3*E^x), x] - 3*E^2*Defer[Int][(E^x*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x))
)*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x^3, x])/(40 + 3*E^x), x] + 3*E^2*Defer[Int][(E^x*Defer[Int][
(5^(1 + 2/(x^2*(4 + x)))*9^(1/(x^2*(4 + x)))*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/x^2, x])/(40 + 3*E^x),
x] + (3*E^2*Defer[Int][(E^x*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(
1 + 2/(x^2*(4 + x))))/x^2, x])/(40 + 3*E^x), x])/8 - 3*E^2*Defer[Int][(E^x*Defer[Int][(5^(1 + 2/(x^2*(4 + x)))
*9^(1/(x^2*(4 + x)))*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x))))/(4 + x)^2, x])/(40 + 3*E^x), x] - (3*E^2*Defer
[Int][(E^x*Defer[Int][(3^(1 + 2/(x^2*(4 + x)))*5^(2/(x^2*(4 + x)))*E^x*((40 + 3*E^x)^(-1))^(1 + 2/(x^2*(4 + x)
)))/(4 + x)^2, x])/(40 + 3*E^x), x])/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2\ 15^{\frac {2}{x^2 (4+x)}} e^2 \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} \left (-3 e^x x (4+x)-\left (40+3 e^x\right ) (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )\right )}{x^3 (4+x)^2} \, dx\\ &=\left (2 e^2\right ) \int \frac {15^{\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} \left (-3 e^x x (4+x)-\left (40+3 e^x\right ) (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )\right )}{x^3 (4+x)^2} \, dx\\ &=\left (2 e^2\right ) \int \left (-\frac {8\ 3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )}{x^3 (4+x)^2}-\frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} \left (4 x+x^2+8 \log \left (\frac {15}{40+3 e^x}\right )+3 x \log \left (\frac {15}{40+3 e^x}\right )\right )}{x^3 (4+x)^2}\right ) \, dx\\ &=-\left (\left (2 e^2\right ) \int \frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} \left (4 x+x^2+8 \log \left (\frac {15}{40+3 e^x}\right )+3 x \log \left (\frac {15}{40+3 e^x}\right )\right )}{x^3 (4+x)^2} \, dx\right )-\left (16 e^2\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )}{x^3 (4+x)^2} \, dx\\ &=-\left (\left (2 e^2\right ) \int \frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} \left (x (4+x)+(8+3 x) \log \left (\frac {15}{40+3 e^x}\right )\right )}{x^3 (4+x)^2} \, dx\right )+\left (16 e^2\right ) \int \frac {3 e^x \left (-8 \int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx+\int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx-\int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx\right )}{16 \left (40+3 e^x\right )} \, dx+\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx-\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx-\left (8 e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx\\ &=-\left (\left (2 e^2\right ) \int \left (\frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2 (4+x)}+\frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )}{x^3 (4+x)^2}\right ) \, dx\right )+\left (3 e^2\right ) \int \frac {e^x \left (-8 \int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx+\int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx-\int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx\right )}{40+3 e^x} \, dx+\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx-\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx-\left (8 e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx\\ &=-\left (\left (2 e^2\right ) \int \frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2 (4+x)} \, dx\right )-\left (2 e^2\right ) \int \frac {3^{1+\frac {2}{x^2 (4+x)}} 5^{\frac {2}{x^2 (4+x)}} e^x \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}} (8+3 x) \log \left (\frac {15}{40+3 e^x}\right )}{x^3 (4+x)^2} \, dx+\left (3 e^2\right ) \int \left (-\frac {8 e^x \int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx}{40+3 e^x}+\frac {e^x \int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx}{40+3 e^x}-\frac {e^x \int \frac {5^{1+\frac {2}{x^2 (4+x)}} 9^{\frac {1}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx}{40+3 e^x}\right ) \, dx+\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^2} \, dx-\left (e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{(4+x)^2} \, dx-\left (8 e^2 \log \left (\frac {15}{40+3 e^x}\right )\right ) \int \frac {3^{\frac {2}{x^2 (4+x)}} 5^{1+\frac {2}{x^2 (4+x)}} \left (\frac {1}{40+3 e^x}\right )^{1+\frac {2}{x^2 (4+x)}}}{x^3} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 6.25, size = 36, normalized size = 1.20 \begin {gather*} 15^{\frac {2}{x^2 (4+x)}} e^2 \left (\frac {1}{40+3 e^x}\right )^{\frac {2}{x^2 (4+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(4*x^2 + x^3 + Log[15/(40 + 3*E^x)]))/(4*x^2 + x^3))*(E^x*(-24*x - 6*x^2) + (-640 + E^x*(-48
- 18*x) - 240*x)*Log[15/(40 + 3*E^x)]))/(640*x^3 + 320*x^4 + 40*x^5 + E^x*(48*x^3 + 24*x^4 + 3*x^5)),x]

[Out]

15^(2/(x^2*(4 + x)))*E^2*((40 + 3*E^x)^(-1))^(2/(x^2*(4 + x)))

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fricas [A]  time = 0.70, size = 34, normalized size = 1.13 \begin {gather*} e^{\left (\frac {2 \, {\left (x^{3} + 4 \, x^{2} + \log \left (\frac {15}{3 \, e^{x} + 40}\right )\right )}}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp(x))*exp((log(15/(3*exp(x)+40)
)+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="fricas")

[Out]

e^(2*(x^3 + 4*x^2 + log(15/(3*e^x + 40)))/(x^3 + 4*x^2))

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giac [B]  time = 0.56, size = 58, normalized size = 1.93 \begin {gather*} e^{\left (\frac {2 \, x^{3}}{x^{3} + 4 \, x^{2}} + \frac {8 \, x^{2}}{x^{3} + 4 \, x^{2}} + \frac {2 \, \log \left (\frac {15}{3 \, e^{x} + 40}\right )}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp(x))*exp((log(15/(3*exp(x)+40)
)+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="giac")

[Out]

e^(2*x^3/(x^3 + 4*x^2) + 8*x^2/(x^3 + 4*x^2) + 2*log(15/(3*e^x + 40))/(x^3 + 4*x^2))

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maple [A]  time = 0.16, size = 34, normalized size = 1.13

method	result	size
risch	\(\left ({\mathrm e}^{x}+\frac {40}{3}\right )^{-\frac {2}{x^{2} \left (4+x \right )}} 5^{\frac {2}{x^{2} \left (4+x \right )}} {\mathrm e}^{2}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-18*x-48)*exp(x)-240*x-640)*ln(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp(x))*exp((ln(15/(3*exp(x)+40))+x^3+4*
x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x,method=_RETURNVERBOSE)

[Out]

((exp(x)+40/3)^(-1/x^2/(4+x)))^2*(5^(1/x^2/(4+x)))^2*exp(2)

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maxima [B]  time = 0.74, size = 87, normalized size = 2.90 \begin {gather*} e^{\left (\frac {\log \relax (5)}{8 \, {\left (x + 4\right )}} - \frac {\log \relax (5)}{8 \, x} + \frac {\log \relax (3)}{8 \, {\left (x + 4\right )}} - \frac {\log \relax (3)}{8 \, x} - \frac {\log \left (3 \, e^{x} + 40\right )}{8 \, {\left (x + 4\right )}} + \frac {\log \left (3 \, e^{x} + 40\right )}{8 \, x} + \frac {\log \relax (5)}{2 \, x^{2}} + \frac {\log \relax (3)}{2 \, x^{2}} - \frac {\log \left (3 \, e^{x} + 40\right )}{2 \, x^{2}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-18*x-48)*exp(x)-240*x-640)*log(15/(3*exp(x)+40))+(-6*x^2-24*x)*exp(x))*exp((log(15/(3*exp(x)+40)
)+x^3+4*x^2)/(x^3+4*x^2))^2/((3*x^5+24*x^4+48*x^3)*exp(x)+40*x^5+320*x^4+640*x^3),x, algorithm="maxima")

[Out]

e^(1/8*log(5)/(x + 4) - 1/8*log(5)/x + 1/8*log(3)/(x + 4) - 1/8*log(3)/x - 1/8*log(3*e^x + 40)/(x + 4) + 1/8*l
og(3*e^x + 40)/x + 1/2*log(5)/x^2 + 1/2*log(3)/x^2 - 1/2*log(3*e^x + 40)/x^2 + 2)

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mupad [B]  time = 5.18, size = 57, normalized size = 1.90 \begin {gather*} {\mathrm {e}}^{\frac {2\,x^3}{x^3+4\,x^2}}\,{\mathrm {e}}^{\frac {8\,x^2}{x^3+4\,x^2}}\,{\left (\frac {225}{{\left (3\,{\mathrm {e}}^x+40\right )}^2}\right )}^{\frac {1}{x^3+4\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*(log(15/(3*exp(x) + 40)) + 4*x^2 + x^3))/(4*x^2 + x^3))*(exp(x)*(24*x + 6*x^2) + log(15/(3*exp(x)
+ 40))*(240*x + exp(x)*(18*x + 48) + 640)))/(exp(x)*(48*x^3 + 24*x^4 + 3*x^5) + 640*x^3 + 320*x^4 + 40*x^5),x
)

[Out]

exp((2*x^3)/(4*x^2 + x^3))*exp((8*x^2)/(4*x^2 + x^3))*(225/(3*exp(x) + 40)^2)^(1/(4*x^2 + x^3))

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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((((-18*x-48)*exp(x)-240*x-640)*ln(15/(3*exp(x)+40))+(-6*x**2-24*x)*exp(x))*exp((ln(15/(3*exp(x)+40))
+x**3+4*x**2)/(x**3+4*x**2))**2/((3*x**5+24*x**4+48*x**3)*exp(x)+40*x**5+320*x**4+640*x**3),x)

[Out]

Timed out

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