3.51.20 \(\int \frac {216+72 x+e^x (-54-45 x-9 x^2)}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x (150-60 x+6 x^2) \log ^4(5)+(250-150 x+30 x^2-2 x^3) \log ^4(5)} \, dx\)

Optimal. Leaf size=23 \[ \frac {9 (3+x)^2}{4 \left (5+e^x-x\right )^2 \log ^4(5)} \]

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Rubi [F]  time = 0.73, 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 {216+72 x+e^x \left (-54-45 x-9 x^2\right )}{2 e^{3 x} \log ^4(5)+e^{2 x} (30-6 x) \log ^4(5)+e^x \left (150-60 x+6 x^2\right ) \log ^4(5)+\left (250-150 x+30 x^2-2 x^3\right ) \log ^4(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(216 + 72*x + E^x*(-54 - 45*x - 9*x^2))/(2*E^(3*x)*Log[5]^4 + E^(2*x)*(30 - 6*x)*Log[5]^4 + E^x*(150 - 60*
x + 6*x^2)*Log[5]^4 + (250 - 150*x + 30*x^2 - 2*x^3)*Log[5]^4),x]

[Out]

(243*Defer[Int][(5 + E^x - x)^(-3), x])/Log[5]^4 - (27*Defer[Int][(5 + E^x - x)^(-2), x])/Log[5]^4 + (243*Defe
r[Int][x/(5 + E^x - x)^3, x])/(2*Log[5]^4) - (45*Defer[Int][x/(5 + E^x - x)^2, x])/(2*Log[5]^4) - (9*Defer[Int
][x^2/(5 + E^x - x)^2, x])/(2*Log[5]^4) - (9*Defer[Int][x^3/(5 + E^x - x)^3, x])/(2*Log[5]^4)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 (3+x) \left (8-e^x (2+x)\right )}{2 \left (5+e^x-x\right )^3 \log ^4(5)} \, dx\\ &=\frac {9 \int \frac {(3+x) \left (8-e^x (2+x)\right )}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}\\ &=\frac {9 \int \left (-\frac {(-6+x) (3+x)^2}{\left (5+e^x-x\right )^3}-\frac {6+5 x+x^2}{\left (5+e^x-x\right )^2}\right ) \, dx}{2 \log ^4(5)}\\ &=-\frac {9 \int \frac {(-6+x) (3+x)^2}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}-\frac {9 \int \frac {6+5 x+x^2}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)}\\ &=-\frac {9 \int \left (\frac {6}{\left (5+e^x-x\right )^2}+\frac {5 x}{\left (5+e^x-x\right )^2}+\frac {x^2}{\left (5+e^x-x\right )^2}\right ) \, dx}{2 \log ^4(5)}-\frac {9 \int \left (-\frac {54}{\left (5+e^x-x\right )^3}-\frac {27 x}{\left (5+e^x-x\right )^3}+\frac {x^3}{\left (5+e^x-x\right )^3}\right ) \, dx}{2 \log ^4(5)}\\ &=-\frac {9 \int \frac {x^2}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)}-\frac {9 \int \frac {x^3}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}-\frac {45 \int \frac {x}{\left (5+e^x-x\right )^2} \, dx}{2 \log ^4(5)}-\frac {27 \int \frac {1}{\left (5+e^x-x\right )^2} \, dx}{\log ^4(5)}+\frac {243 \int \frac {x}{\left (5+e^x-x\right )^3} \, dx}{2 \log ^4(5)}+\frac {243 \int \frac {1}{\left (5+e^x-x\right )^3} \, dx}{\log ^4(5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 1.00 \begin {gather*} \frac {9 (3+x)^2}{4 \left (5+e^x-x\right )^2 \log ^4(5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(216 + 72*x + E^x*(-54 - 45*x - 9*x^2))/(2*E^(3*x)*Log[5]^4 + E^(2*x)*(30 - 6*x)*Log[5]^4 + E^x*(150
- 60*x + 6*x^2)*Log[5]^4 + (250 - 150*x + 30*x^2 - 2*x^3)*Log[5]^4),x]

[Out]

(9*(3 + x)^2)/(4*(5 + E^x - x)^2*Log[5]^4)

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fricas [B]  time = 0.73, size = 48, normalized size = 2.09 \begin {gather*} -\frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (2 \, {\left (x - 5\right )} e^{x} \log \relax (5)^{4} - {\left (x^{2} - 10 \, x + 25\right )} \log \relax (5)^{4} - e^{\left (2 \, x\right )} \log \relax (5)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-45*x-54)*exp(x)+72*x+216)/(2*log(5)^4*exp(x)^3+(-6*x+30)*log(5)^4*exp(x)^2+(6*x^2-60*x+150)
*log(5)^4*exp(x)+(-2*x^3+30*x^2-150*x+250)*log(5)^4),x, algorithm="fricas")

[Out]

-9/4*(x^2 + 6*x + 9)/(2*(x - 5)*e^x*log(5)^4 - (x^2 - 10*x + 25)*log(5)^4 - e^(2*x)*log(5)^4)

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giac [B]  time = 0.17, size = 60, normalized size = 2.61 \begin {gather*} \frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (x^{2} \log \relax (5)^{4} - 2 \, x e^{x} \log \relax (5)^{4} - 10 \, x \log \relax (5)^{4} + e^{\left (2 \, x\right )} \log \relax (5)^{4} + 10 \, e^{x} \log \relax (5)^{4} + 25 \, \log \relax (5)^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-45*x-54)*exp(x)+72*x+216)/(2*log(5)^4*exp(x)^3+(-6*x+30)*log(5)^4*exp(x)^2+(6*x^2-60*x+150)
*log(5)^4*exp(x)+(-2*x^3+30*x^2-150*x+250)*log(5)^4),x, algorithm="giac")

[Out]

9/4*(x^2 + 6*x + 9)/(x^2*log(5)^4 - 2*x*e^x*log(5)^4 - 10*x*log(5)^4 + e^(2*x)*log(5)^4 + 10*e^x*log(5)^4 + 25
*log(5)^4)

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maple [A]  time = 0.32, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {\frac {9}{4} x^{2}+\frac {27}{2} x +\frac {81}{4}}{\ln \relax (5)^{4} \left (-{\mathrm e}^{x}+x -5\right )^{2}}\)	\(24\)
norman	\(\frac {\frac {36 x}{\ln \relax (5)}-\frac {9 \,{\mathrm e}^{2 x}}{4 \ln \relax (5)}-\frac {45 \,{\mathrm e}^{x}}{2 \ln \relax (5)}+\frac {9 \,{\mathrm e}^{x} x}{2 \ln \relax (5)}-\frac {36}{\ln \relax (5)}}{\ln \relax (5)^{3} \left (-{\mathrm e}^{x}+x -5\right )^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x^2-45*x-54)*exp(x)+72*x+216)/(2*ln(5)^4*exp(x)^3+(-6*x+30)*ln(5)^4*exp(x)^2+(6*x^2-60*x+150)*ln(5)^4
*exp(x)+(-2*x^3+30*x^2-150*x+250)*ln(5)^4),x,method=_RETURNVERBOSE)

[Out]

9/4*(x^2+6*x+9)/ln(5)^4/(-exp(x)+x-5)^2

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maxima [B]  time = 0.49, size = 60, normalized size = 2.61 \begin {gather*} \frac {9 \, {\left (x^{2} + 6 \, x + 9\right )}}{4 \, {\left (x^{2} \log \relax (5)^{4} - 10 \, x \log \relax (5)^{4} + e^{\left (2 \, x\right )} \log \relax (5)^{4} + 25 \, \log \relax (5)^{4} - 2 \, {\left (x \log \relax (5)^{4} - 5 \, \log \relax (5)^{4}\right )} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-45*x-54)*exp(x)+72*x+216)/(2*log(5)^4*exp(x)^3+(-6*x+30)*log(5)^4*exp(x)^2+(6*x^2-60*x+150)
*log(5)^4*exp(x)+(-2*x^3+30*x^2-150*x+250)*log(5)^4),x, algorithm="maxima")

[Out]

9/4*(x^2 + 6*x + 9)/(x^2*log(5)^4 - 10*x*log(5)^4 + e^(2*x)*log(5)^4 + 25*log(5)^4 - 2*(x*log(5)^4 - 5*log(5)^
4)*e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {72\,x-{\mathrm {e}}^x\,\left (9\,x^2+45\,x+54\right )+216}{2\,{\mathrm {e}}^{3\,x}\,{\ln \relax (5)}^4-{\ln \relax (5)}^4\,\left (2\,x^3-30\,x^2+150\,x-250\right )+{\mathrm {e}}^x\,{\ln \relax (5)}^4\,\left (6\,x^2-60\,x+150\right )-{\mathrm {e}}^{2\,x}\,{\ln \relax (5)}^4\,\left (6\,x-30\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((72*x - exp(x)*(45*x + 9*x^2 + 54) + 216)/(2*exp(3*x)*log(5)^4 - log(5)^4*(150*x - 30*x^2 + 2*x^3 - 250) +
exp(x)*log(5)^4*(6*x^2 - 60*x + 150) - exp(2*x)*log(5)^4*(6*x - 30)),x)

[Out]

int((72*x - exp(x)*(45*x + 9*x^2 + 54) + 216)/(2*exp(3*x)*log(5)^4 - log(5)^4*(150*x - 30*x^2 + 2*x^3 - 250) +
exp(x)*log(5)^4*(6*x^2 - 60*x + 150) - exp(2*x)*log(5)^4*(6*x - 30)), x)

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sympy [B]  time = 0.17, size = 65, normalized size = 2.83 \begin {gather*} \frac {9 x^{2} + 54 x + 81}{4 x^{2} \log {\relax (5 )}^{4} - 40 x \log {\relax (5 )}^{4} + \left (- 8 x \log {\relax (5 )}^{4} + 40 \log {\relax (5 )}^{4}\right ) e^{x} + 4 e^{2 x} \log {\relax (5 )}^{4} + 100 \log {\relax (5 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x**2-45*x-54)*exp(x)+72*x+216)/(2*ln(5)**4*exp(x)**3+(-6*x+30)*ln(5)**4*exp(x)**2+(6*x**2-60*x+
150)*ln(5)**4*exp(x)+(-2*x**3+30*x**2-150*x+250)*ln(5)**4),x)

[Out]

(9*x**2 + 54*x + 81)/(4*x**2*log(5)**4 - 40*x*log(5)**4 + (-8*x*log(5)**4 + 40*log(5)**4)*exp(x) + 4*exp(2*x)*
log(5)**4 + 100*log(5)**4)

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