3.82.9 \(\int \frac {-15 e^2+e^{\frac {2 (6 e^2 x+x^2-3 e^2 x \log (5))}{3 e^2}} (-60 e^2-20 x+30 e^2 \log (5))}{3 e^{2+\frac {4 (6 e^2 x+x^2-3 e^2 x \log (5))}{3 e^2}}+e^{2+\frac {2 (6 e^2 x+x^2-3 e^2 x \log (5))}{3 e^2}} (-6+6 x)+e^2 (3-6 x+3 x^2)} \, dx\)

Optimal. Leaf size=26 \[ \frac {5}{-1+e^{2 x \left (2+\frac {x}{3 e^2}-\log (5)\right )}+x} \]

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Rubi [F]  time = 4.63, 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 {-15 e^2+e^{\frac {2 \left (6 e^2 x+x^2-3 e^2 x \log (5)\right )}{3 e^2}} \left (-60 e^2-20 x+30 e^2 \log (5)\right )}{3 \exp \left (2+\frac {4 \left (6 e^2 x+x^2-3 e^2 x \log (5)\right )}{3 e^2}\right )+\exp \left (2+\frac {2 \left (6 e^2 x+x^2-3 e^2 x \log (5)\right )}{3 e^2}\right ) (-6+6 x)+e^2 \left (3-6 x+3 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-15*E^2 + E^((2*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2))*(-60*E^2 - 20*x + 30*E^2*Log[5]))/(3*E^(2 + (4
*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2)) + E^(2 + (2*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2))*(-6 + 6*x)
+ E^2*(3 - 6*x + 3*x^2)),x]

[Out]

-Defer[Int][5^(1 + 4*x)/(-25^x + E^((2*x*(6 + x/E^2))/3) + 25^x*x)^2, x] - (10*(2 - Log[5])*Defer[Int][E^(2 +
(2*x^2)/(3*E^2) + 2*x*(2 + Log[5]))/(-25^x + E^((2*x*(6 + x/E^2))/3) + 25^x*x)^2, x])/E^2 - (4*Defer[Int][(5^(
1 + 4*x)*x)/(-25^x + E^((2*x*(6 + x/E^2))/3) + 25^x*x)^2, x])/(3*E^2) + (4*Defer[Int][(5^(1 + 4*x)*x^2)/(-25^x
+ E^((2*x*(6 + x/E^2))/3) + 25^x*x)^2, x])/(3*E^2) - (4*Defer[Int][(5^(1 + 2*x)*x)/(-25^x + E^((2*x*(6 + x/E^
2))/3) + 25^x*x), x])/(3*E^2)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5^{1+2 x} \left (-3 25^x e^2-4 e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )} x+6 e^{2+4 x+\frac {2 x^2}{3 e^2}} (-2+\log (5))\right )}{3 e^2 \left (e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x (-1+x)\right )^2} \, dx\\ &=\frac {\int \frac {5^{1+2 x} \left (-3 25^x e^2-4 e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )} x+6 e^{2+4 x+\frac {2 x^2}{3 e^2}} (-2+\log (5))\right )}{\left (e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x (-1+x)\right )^2} \, dx}{3 e^2}\\ &=\frac {\int \left (-\frac {4\ 5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x}+\frac {5^{1+2 x} \left (-3 25^x e^2-4\ 25^x x+4\ 25^x x^2-12 e^{2+4 x+\frac {2 x^2}{3 e^2}} \left (1-\frac {\log (5)}{2}\right )\right )}{\left (25^x-e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}-25^x x\right )^2}\right ) \, dx}{3 e^2}\\ &=\frac {\int \frac {5^{1+2 x} \left (-3 25^x e^2-4\ 25^x x+4\ 25^x x^2-12 e^{2+4 x+\frac {2 x^2}{3 e^2}} \left (1-\frac {\log (5)}{2}\right )\right )}{\left (25^x-e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}-25^x x\right )^2} \, dx}{3 e^2}-\frac {4 \int \frac {5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x} \, dx}{3 e^2}\\ &=\frac {\int \left (-\frac {3\ 5^{1+4 x} e^2}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2}-\frac {4\ 5^{1+4 x} x}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2}+\frac {4\ 5^{1+4 x} x^2}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2}+\frac {6\ 5^{1+2 x} e^{2+4 x+\frac {2 x^2}{3 e^2}} (-2+\log (5))}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2}\right ) \, dx}{3 e^2}-\frac {4 \int \frac {5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x} \, dx}{3 e^2}\\ &=-\frac {4 \int \frac {5^{1+4 x} x}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}+\frac {4 \int \frac {5^{1+4 x} x^2}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}-\frac {4 \int \frac {5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x} \, dx}{3 e^2}-\frac {(2 (2-\log (5))) \int \frac {5^{1+2 x} e^{2+4 x+\frac {2 x^2}{3 e^2}}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{e^2}-\int \frac {5^{1+4 x}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx\\ &=-\frac {4 \int \frac {5^{1+4 x} x}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}+\frac {4 \int \frac {5^{1+4 x} x^2}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}-\frac {4 \int \frac {5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x} \, dx}{3 e^2}-\frac {(2 (2-\log (5))) \int \frac {5 e^{2+\frac {2 x^2}{3 e^2}+2 x (2+\log (5))}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{e^2}-\int \frac {5^{1+4 x}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx\\ &=-\frac {4 \int \frac {5^{1+4 x} x}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}+\frac {4 \int \frac {5^{1+4 x} x^2}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{3 e^2}-\frac {4 \int \frac {5^{1+2 x} x}{-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x} \, dx}{3 e^2}-\frac {(10 (2-\log (5))) \int \frac {e^{2+\frac {2 x^2}{3 e^2}+2 x (2+\log (5))}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx}{e^2}-\int \frac {5^{1+4 x}}{\left (-25^x+e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 32, normalized size = 1.23 \begin {gather*} \frac {5^{1+2 x}}{e^{\frac {2}{3} x \left (6+\frac {x}{e^2}\right )}+25^x (-1+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^2 + E^((2*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2))*(-60*E^2 - 20*x + 30*E^2*Log[5]))/(3*E^(
2 + (4*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2)) + E^(2 + (2*(6*E^2*x + x^2 - 3*E^2*x*Log[5]))/(3*E^2))*(-6 +
6*x) + E^2*(3 - 6*x + 3*x^2)),x]

[Out]

5^(1 + 2*x)/(E^((2*x*(6 + x/E^2))/3) + 25^x*(-1 + x))

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fricas [A]  time = 0.60, size = 40, normalized size = 1.54 \begin {gather*} \frac {5 \, e^{2}}{{\left (x - 1\right )} e^{2} + e^{\left (-\frac {2}{3} \, {\left (3 \, x e^{2} \log \relax (5) - x^{2} - 3 \, {\left (2 \, x + 1\right )} e^{2}\right )} e^{\left (-2\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*exp(2)*log(5)-60*exp(2)-20*x)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^2-15*exp(2))/
(3*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^4+(6*x-6)*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*e
xp(2)*x+x^2)/exp(2))^2+(3*x^2-6*x+3)*exp(2)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.46, size = 30, normalized size = 1.15 \begin {gather*} \frac {5}{x + e^{\left (-\frac {2}{3} \, {\left (3 \, x e^{2} \log \relax (5) - x^{2} - 6 \, x e^{2}\right )} e^{\left (-2\right )}\right )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*exp(2)*log(5)-60*exp(2)-20*x)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^2-15*exp(2))/
(3*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^4+(6*x-6)*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*e
xp(2)*x+x^2)/exp(2))^2+(3*x^2-6*x+3)*exp(2)),x, algorithm="giac")

[Out]

5/(x + e^(-2/3*(3*x*e^2*log(5) - x^2 - 6*x*e^2)*e^(-2)) - 1)

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maple [A]  time = 0.23, size = 28, normalized size = 1.08

method	result	size
risch	\(\frac {5}{{\mathrm e}^{-\frac {2 x \left (3 \,{\mathrm e}^{2} \ln \relax (5)-6 \,{\mathrm e}^{2}-x \right ) {\mathrm e}^{-2}}{3}}+x -1}\)	\(28\)
norman	\(\frac {5}{{\mathrm e}^{\frac {2 \left (-3 x \,{\mathrm e}^{2} \ln \relax (5)+6 \,{\mathrm e}^{2} x +x^{2}\right ) {\mathrm e}^{-2}}{3}}+x -1}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((30*exp(2)*ln(5)-60*exp(2)-20*x)*exp(1/3*(-3*x*exp(2)*ln(5)+6*exp(2)*x+x^2)/exp(2))^2-15*exp(2))/(3*exp(2
)*exp(1/3*(-3*x*exp(2)*ln(5)+6*exp(2)*x+x^2)/exp(2))^4+(6*x-6)*exp(2)*exp(1/3*(-3*x*exp(2)*ln(5)+6*exp(2)*x+x^
2)/exp(2))^2+(3*x^2-6*x+3)*exp(2)),x,method=_RETURNVERBOSE)

[Out]

5/(exp(-2/3*x*(3*exp(2)*ln(5)-6*exp(2)-x)*exp(-2))+x-1)

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maxima [A]  time = 0.54, size = 31, normalized size = 1.19 \begin {gather*} \frac {5 \cdot 5^{2 \, x}}{5^{2 \, x} {\left (x - 1\right )} + e^{\left (\frac {2}{3} \, x^{2} e^{\left (-2\right )} + 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*exp(2)*log(5)-60*exp(2)-20*x)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^2-15*exp(2))/
(3*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*exp(2)*x+x^2)/exp(2))^4+(6*x-6)*exp(2)*exp(1/3*(-3*x*exp(2)*log(5)+6*e
xp(2)*x+x^2)/exp(2))^2+(3*x^2-6*x+3)*exp(2)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.09, size = 38, normalized size = 1.46 \begin {gather*} \frac {15\,{\mathrm {e}}^2}{3\,x\,{\mathrm {e}}^2-3\,{\mathrm {e}}^2+\frac {3\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-2}\,x^2}{3}+4\,x+2}}{5^{2\,x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*exp(2) + exp(2*exp(-2)*(2*x*exp(2) + x^2/3 - x*exp(2)*log(5)))*(20*x + 60*exp(2) - 30*exp(2)*log(5)))
/(exp(2)*(3*x^2 - 6*x + 3) + 3*exp(2)*exp(4*exp(-2)*(2*x*exp(2) + x^2/3 - x*exp(2)*log(5))) + exp(2)*exp(2*exp
(-2)*(2*x*exp(2) + x^2/3 - x*exp(2)*log(5)))*(6*x - 6)),x)

[Out]

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

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sympy [A]  time = 0.22, size = 32, normalized size = 1.23 \begin {gather*} \frac {5}{x + e^{\frac {2 \left (\frac {x^{2}}{3} - x e^{2} \log {\relax (5 )} + 2 x e^{2}\right )}{e^{2}}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*exp(2)*ln(5)-60*exp(2)-20*x)*exp(1/3*(-3*x*exp(2)*ln(5)+6*exp(2)*x+x**2)/exp(2))**2-15*exp(2))/
(3*exp(2)*exp(1/3*(-3*x*exp(2)*ln(5)+6*exp(2)*x+x**2)/exp(2))**4+(6*x-6)*exp(2)*exp(1/3*(-3*x*exp(2)*ln(5)+6*e
xp(2)*x+x**2)/exp(2))**2+(3*x**2-6*x+3)*exp(2)),x)

[Out]

5/(x + exp(2*(x**2/3 - x*exp(2)*log(5) + 2*x*exp(2))*exp(-2)) - 1)

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