3.7.3 \(\int \frac {1+e^x (3-x)+e^{288-432 x+162 x^2} (6485+e^x (15-5 x)-7020 x+1620 x^2)}{-3+x+e^{288-432 x+162 x^2} (-15+5 x)} \, dx\)

Optimal. Leaf size=30 \[ -e^x+\log \left (3+5 e^{18 (-4+3 x)^2} (3-x)-x\right ) \]

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Rubi [F]  time = 0.69, 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 {1+e^x (3-x)+e^{288-432 x+162 x^2} \left (6485+e^x (15-5 x)-7020 x+1620 x^2\right )}{-3+x+e^{288-432 x+162 x^2} (-15+5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + E^x*(3 - x) + E^(288 - 432*x + 162*x^2)*(6485 + E^x*(15 - 5*x) - 7020*x + 1620*x^2))/(-3 + x + E^(288
- 432*x + 162*x^2)*(-15 + 5*x)),x]

[Out]

-E^x - 432*x + 162*x^2 + Log[3 - x] + 432*Defer[Int][E^(432*x)/(E^(432*x) + 5*E^(288 + 162*x^2)), x] - 324*Def
er[Int][(E^(432*x)*x)/(E^(432*x) + 5*E^(288 + 162*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {108 e^{432 x} (-4+3 x)}{e^{432 x}+5 e^{288+162 x^2}}+\frac {1297+3 e^x-1404 x-e^x x+324 x^2}{-3+x}\right ) \, dx\\ &=-\left (108 \int \frac {e^{432 x} (-4+3 x)}{e^{432 x}+5 e^{288+162 x^2}} \, dx\right )+\int \frac {1297+3 e^x-1404 x-e^x x+324 x^2}{-3+x} \, dx\\ &=-\left (108 \int \left (-\frac {4 e^{432 x}}{e^{432 x}+5 e^{288+162 x^2}}+\frac {3 e^{432 x} x}{e^{432 x}+5 e^{288+162 x^2}}\right ) \, dx\right )+\int \left (-e^x+\frac {1297-1404 x+324 x^2}{-3+x}\right ) \, dx\\ &=-\left (324 \int \frac {e^{432 x} x}{e^{432 x}+5 e^{288+162 x^2}} \, dx\right )+432 \int \frac {e^{432 x}}{e^{432 x}+5 e^{288+162 x^2}} \, dx-\int e^x \, dx+\int \frac {1297-1404 x+324 x^2}{-3+x} \, dx\\ &=-e^x-324 \int \frac {e^{432 x} x}{e^{432 x}+5 e^{288+162 x^2}} \, dx+432 \int \frac {e^{432 x}}{e^{432 x}+5 e^{288+162 x^2}} \, dx+\int \left (-432+\frac {1}{-3+x}+324 x\right ) \, dx\\ &=-e^x-432 x+162 x^2+\log (3-x)-324 \int \frac {e^{432 x} x}{e^{432 x}+5 e^{288+162 x^2}} \, dx+432 \int \frac {e^{432 x}}{e^{432 x}+5 e^{288+162 x^2}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 29, normalized size = 0.97 \begin {gather*} -e^x+\log \left (1+5 e^{288-432 x+162 x^2}\right )+\log (3-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^x*(3 - x) + E^(288 - 432*x + 162*x^2)*(6485 + E^x*(15 - 5*x) - 7020*x + 1620*x^2))/(-3 + x +
E^(288 - 432*x + 162*x^2)*(-15 + 5*x)),x]

[Out]

-E^x + Log[1 + 5*E^(288 - 432*x + 162*x^2)] + Log[3 - x]

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fricas [A]  time = 0.73, size = 25, normalized size = 0.83 \begin {gather*} -e^{x} + \log \left (x - 3\right ) + \log \left (5 \, e^{\left (162 \, x^{2} - 432 \, x + 288\right )} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15-5*x)*exp(x)+1620*x^2-7020*x+6485)*exp(81*x^2-216*x+144)^2+(3-x)*exp(x)+1)/((5*x-15)*exp(81*x^2
-216*x+144)^2+x-3),x, algorithm="fricas")

[Out]

-e^x + log(x - 3) + log(5*e^(162*x^2 - 432*x + 288) + 1)

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giac [A]  time = 0.36, size = 29, normalized size = 0.97 \begin {gather*} -x - e^{x} + \log \left (x - 3\right ) + \log \left (5 \, e^{\left (162 \, x^{2} - 431 \, x + 288\right )} + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15-5*x)*exp(x)+1620*x^2-7020*x+6485)*exp(81*x^2-216*x+144)^2+(3-x)*exp(x)+1)/((5*x-15)*exp(81*x^2
-216*x+144)^2+x-3),x, algorithm="giac")

[Out]

-x - e^x + log(x - 3) + log(5*e^(162*x^2 - 431*x + 288) + e^x)

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maple [A]  time = 0.10, size = 24, normalized size = 0.80

method	result	size
risch	\(\ln \left (x -3\right )-{\mathrm e}^{x}-288+\ln \left ({\mathrm e}^{18 \left (3 x -4\right )^{2}}+\frac {1}{5}\right )\)	\(24\)
norman	\(-{\mathrm e}^{x}+\ln \left (x -3\right )+\ln \left (5 \,{\mathrm e}^{162 x^{2}-432 x +288}+1\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((15-5*x)*exp(x)+1620*x^2-7020*x+6485)*exp(81*x^2-216*x+144)^2+(3-x)*exp(x)+1)/((5*x-15)*exp(81*x^2-216*x
+144)^2+x-3),x,method=_RETURNVERBOSE)

[Out]

ln(x-3)-exp(x)-288+ln(exp(18*(3*x-4)^2)+1/5)

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maxima [A]  time = 1.87, size = 32, normalized size = 1.07 \begin {gather*} -432 \, x - e^{x} + \log \left (\frac {1}{5} \, {\left (5 \, e^{\left (162 \, x^{2} + 288\right )} + e^{\left (432 \, x\right )}\right )} e^{\left (-288\right )}\right ) + \log \left (x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15-5*x)*exp(x)+1620*x^2-7020*x+6485)*exp(81*x^2-216*x+144)^2+(3-x)*exp(x)+1)/((5*x-15)*exp(81*x^2
-216*x+144)^2+x-3),x, algorithm="maxima")

[Out]

-432*x - e^x + log(1/5*(5*e^(162*x^2 + 288) + e^(432*x))*e^(-288)) + log(x - 3)

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mupad [B]  time = 0.16, size = 26, normalized size = 0.87 \begin {gather*} \ln \left (\left (x-3\right )\,\left (5\,{\mathrm {e}}^{-432\,x}\,{\mathrm {e}}^{288}\,{\mathrm {e}}^{162\,x^2}+1\right )\right )-{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(162*x^2 - 432*x + 288)*(7020*x + exp(x)*(5*x - 15) - 1620*x^2 - 6485) + exp(x)*(x - 3) - 1)/(x + exp
(162*x^2 - 432*x + 288)*(5*x - 15) - 3),x)

[Out]

log((x - 3)*(5*exp(-432*x)*exp(288)*exp(162*x^2) + 1)) - exp(x)

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sympy [A]  time = 0.23, size = 24, normalized size = 0.80 \begin {gather*} - e^{x} + \log {\left (x - 3 \right )} + \log {\left (e^{162 x^{2} - 432 x + 288} + \frac {1}{5} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((15-5*x)*exp(x)+1620*x**2-7020*x+6485)*exp(81*x**2-216*x+144)**2+(3-x)*exp(x)+1)/((5*x-15)*exp(81*
x**2-216*x+144)**2+x-3),x)

[Out]

-exp(x) + log(x - 3) + log(exp(162*x**2 - 432*x + 288) + 1/5)

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