3.53.37 \(\int \frac {e^2 (48+32 x+96 x^3+80 x^4+32 x^7)}{\log (\frac {e^{10}}{9})} \, dx\)

Optimal. Leaf size=27 \[ \frac {4 e^2 \left (1+\left (3+2 x+x^4\right )^2\right )}{\log \left (\frac {e^{10}}{9}\right )} \]

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Rubi [B]  time = 0.02, antiderivative size = 79, normalized size of antiderivative = 2.93, number of steps used = 2, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12} \begin {gather*} \frac {4 e^2 x^8}{10-\log (9)}+\frac {16 e^2 x^5}{10-\log (9)}+\frac {24 e^2 x^4}{10-\log (9)}+\frac {16 e^2 x^2}{10-\log (9)}+\frac {48 e^2 x}{10-\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(48 + 32*x + 96*x^3 + 80*x^4 + 32*x^7))/Log[E^10/9],x]

[Out]

(48*E^2*x)/(10 - Log[9]) + (16*E^2*x^2)/(10 - Log[9]) + (24*E^2*x^4)/(10 - Log[9]) + (16*E^2*x^5)/(10 - Log[9]
) + (4*E^2*x^8)/(10 - Log[9])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^2 \int \left (48+32 x+96 x^3+80 x^4+32 x^7\right ) \, dx}{10-\log (9)}\\ &=\frac {48 e^2 x}{10-\log (9)}+\frac {16 e^2 x^2}{10-\log (9)}+\frac {24 e^2 x^4}{10-\log (9)}+\frac {16 e^2 x^5}{10-\log (9)}+\frac {4 e^2 x^8}{10-\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 35, normalized size = 1.30 \begin {gather*} -\frac {16 e^2 \left (3 x+x^2+\frac {3 x^4}{2}+x^5+\frac {x^8}{4}\right )}{-10+\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(48 + 32*x + 96*x^3 + 80*x^4 + 32*x^7))/Log[E^10/9],x]

[Out]

(-16*E^2*(3*x + x^2 + (3*x^4)/2 + x^5 + x^8/4))/(-10 + Log[9])

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fricas [A]  time = 0.52, size = 32, normalized size = 1.19 \begin {gather*} -\frac {2 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \relax (3) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^7+80*x^4+96*x^3+32*x+48)*exp(2)/log(1/9*exp(5)^2),x, algorithm="fricas")

[Out]

-2*(x^8 + 4*x^5 + 6*x^4 + 4*x^2 + 12*x)*e^2/(log(3) - 5)

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giac [A]  time = 1.98, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \left (\frac {1}{9} \, e^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^7+80*x^4+96*x^3+32*x+48)*exp(2)/log(1/9*exp(5)^2),x, algorithm="giac")

[Out]

4*(x^8 + 4*x^5 + 6*x^4 + 4*x^2 + 12*x)*e^2/log(1/9*e^10)

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maple [A]  time = 0.06, size = 33, normalized size = 1.22

method	result	size
gosper	\(\frac {4 \left (x^{7}+4 x^{4}+6 x^{3}+4 x +12\right ) {\mathrm e}^{2} x}{\ln \left (\frac {{\mathrm e}^{10}}{9}\right )}\)	\(33\)
default	\(\frac {{\mathrm e}^{2} \left (4 x^{8}+16 x^{5}+24 x^{4}+16 x^{2}+48 x \right )}{\ln \left (\frac {{\mathrm e}^{10}}{9}\right )}\)	\(37\)
norman	\(-\frac {24 \,{\mathrm e}^{2} x}{\ln \relax (3)-5}-\frac {8 \,{\mathrm e}^{2} x^{2}}{\ln \relax (3)-5}-\frac {12 \,{\mathrm e}^{2} x^{4}}{\ln \relax (3)-5}-\frac {8 \,{\mathrm e}^{2} x^{5}}{\ln \relax (3)-5}-\frac {2 \,{\mathrm e}^{2} x^{8}}{\ln \relax (3)-5}\)	\(65\)
risch	\(\frac {4 \,{\mathrm e}^{2} x^{8}}{-2 \ln \relax (3)+10}+\frac {16 \,{\mathrm e}^{2} x^{5}}{-2 \ln \relax (3)+10}+\frac {24 \,{\mathrm e}^{2} x^{4}}{-2 \ln \relax (3)+10}+\frac {16 \,{\mathrm e}^{2} x^{2}}{-2 \ln \relax (3)+10}+\frac {48 \,{\mathrm e}^{2} x}{-2 \ln \relax (3)+10}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((32*x^7+80*x^4+96*x^3+32*x+48)*exp(2)/ln(1/9*exp(5)^2),x,method=_RETURNVERBOSE)

[Out]

4*(x^7+4*x^4+6*x^3+4*x+12)*exp(2)*x/ln(1/9*exp(5)^2)

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maxima [A]  time = 0.35, size = 33, normalized size = 1.22 \begin {gather*} \frac {4 \, {\left (x^{8} + 4 \, x^{5} + 6 \, x^{4} + 4 \, x^{2} + 12 \, x\right )} e^{2}}{\log \left (\frac {1}{9} \, e^{10}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x^7+80*x^4+96*x^3+32*x+48)*exp(2)/log(1/9*exp(5)^2),x, algorithm="maxima")

[Out]

4*(x^8 + 4*x^5 + 6*x^4 + 4*x^2 + 12*x)*e^2/log(1/9*e^10)

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mupad [B]  time = 3.32, size = 64, normalized size = 2.37 \begin {gather*} -\frac {4\,{\mathrm {e}}^2\,x^8}{\ln \relax (9)-10}-\frac {16\,{\mathrm {e}}^2\,x^5}{\ln \relax (9)-10}-\frac {24\,{\mathrm {e}}^2\,x^4}{\ln \relax (9)-10}-\frac {16\,{\mathrm {e}}^2\,x^2}{\ln \relax (9)-10}-\frac {48\,{\mathrm {e}}^2\,x}{\ln \relax (9)-10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)*(32*x + 96*x^3 + 80*x^4 + 32*x^7 + 48))/log(exp(10)/9),x)

[Out]

- (48*x*exp(2))/(log(9) - 10) - (16*x^2*exp(2))/(log(9) - 10) - (24*x^4*exp(2))/(log(9) - 10) - (16*x^5*exp(2)
)/(log(9) - 10) - (4*x^8*exp(2))/(log(9) - 10)

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sympy [B]  time = 0.07, size = 66, normalized size = 2.44 \begin {gather*} - \frac {2 x^{8} e^{2}}{-5 + \log {\relax (3 )}} - \frac {8 x^{5} e^{2}}{-5 + \log {\relax (3 )}} - \frac {12 x^{4} e^{2}}{-5 + \log {\relax (3 )}} - \frac {8 x^{2} e^{2}}{-5 + \log {\relax (3 )}} - \frac {24 x e^{2}}{-5 + \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((32*x**7+80*x**4+96*x**3+32*x+48)*exp(2)/ln(1/9*exp(5)**2),x)

[Out]

-2*x**8*exp(2)/(-5 + log(3)) - 8*x**5*exp(2)/(-5 + log(3)) - 12*x**4*exp(2)/(-5 + log(3)) - 8*x**2*exp(2)/(-5
+ log(3)) - 24*x*exp(2)/(-5 + log(3))

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