3.34.64 \(\int \frac {e^{\frac {2 (-10+(13 x+4 x^2) \log (4))}{x \log (4)}} (20+8 x^2 \log (4))+e^{\frac {-10+(13 x+4 x^2) \log (4)}{x \log (4)}} (100+40 x^2 \log (4))}{x^2 \log (4)} \, dx\)

Optimal. Leaf size=22 \[ \left (5+e^{9+4 (1+x)-\frac {10}{x \log (4)}}\right )^2 \]

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Rubi [F]  time = 0.74, 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 (-10+\left (13 x+4 x^2\right ) \log (4)\right )}{x \log (4)}\right ) \left (20+8 x^2 \log (4)\right )+e^{\frac {-10+\left (13 x+4 x^2\right ) \log (4)}{x \log (4)}} \left (100+40 x^2 \log (4)\right )}{x^2 \log (4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(-10 + (13*x + 4*x^2)*Log[4]))/(x*Log[4]))*(20 + 8*x^2*Log[4]) + E^((-10 + (13*x + 4*x^2)*Log[4])/(
x*Log[4]))*(100 + 40*x^2*Log[4]))/(x^2*Log[4]),x]

[Out]

(4*Log[16]*Defer[Int][E^(26 + 8*x - 20/(x*Log[4])), x])/Log[4] + (20*Log[16]*Defer[Int][E^(13 + 4*x - 10/(x*Lo
g[4])), x])/Log[4] + (20*Defer[Int][E^(26 + 8*x - 20/(x*Log[4]))/x^2, x])/Log[4] + (100*Defer[Int][E^(13 + 4*x
- 10/(x*Log[4]))/x^2, x])/Log[4]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (\frac {2 \left (-10+\left (13 x+4 x^2\right ) \log (4)\right )}{x \log (4)}\right ) \left (20+8 x^2 \log (4)\right )+e^{\frac {-10+\left (13 x+4 x^2\right ) \log (4)}{x \log (4)}} \left (100+40 x^2 \log (4)\right )}{x^2} \, dx}{\log (4)}\\ &=\frac {\int \left (\frac {4 e^{26+8 x-\frac {20}{x \log (4)}} \left (5+x^2 \log (16)\right )}{x^2}+\frac {20 e^{13+4 x-\frac {10}{x \log (4)}} \left (5+x^2 \log (16)\right )}{x^2}\right ) \, dx}{\log (4)}\\ &=\frac {4 \int \frac {e^{26+8 x-\frac {20}{x \log (4)}} \left (5+x^2 \log (16)\right )}{x^2} \, dx}{\log (4)}+\frac {20 \int \frac {e^{13+4 x-\frac {10}{x \log (4)}} \left (5+x^2 \log (16)\right )}{x^2} \, dx}{\log (4)}\\ &=\frac {4 \int \left (\frac {5 e^{26+8 x-\frac {20}{x \log (4)}}}{x^2}+e^{26+8 x-\frac {20}{x \log (4)}} \log (16)\right ) \, dx}{\log (4)}+\frac {20 \int \left (\frac {5 e^{13+4 x-\frac {10}{x \log (4)}}}{x^2}+e^{13+4 x-\frac {10}{x \log (4)}} \log (16)\right ) \, dx}{\log (4)}\\ &=\frac {20 \int \frac {e^{26+8 x-\frac {20}{x \log (4)}}}{x^2} \, dx}{\log (4)}+\frac {100 \int \frac {e^{13+4 x-\frac {10}{x \log (4)}}}{x^2} \, dx}{\log (4)}+\frac {(4 \log (16)) \int e^{26+8 x-\frac {20}{x \log (4)}} \, dx}{\log (4)}+\frac {(20 \log (16)) \int e^{13+4 x-\frac {10}{x \log (4)}} \, dx}{\log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.62, size = 81, normalized size = 3.68 \begin {gather*} \frac {\int \frac {e^{\frac {2 \left (-10+\left (13 x+4 x^2\right ) \log (4)\right )}{x \log (4)}} \left (20+8 x^2 \log (4)\right )+e^{\frac {-10+\left (13 x+4 x^2\right ) \log (4)}{x \log (4)}} \left (100+40 x^2 \log (4)\right )}{x^2} \, dx}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(-10 + (13*x + 4*x^2)*Log[4]))/(x*Log[4]))*(20 + 8*x^2*Log[4]) + E^((-10 + (13*x + 4*x^2)*Log
[4])/(x*Log[4]))*(100 + 40*x^2*Log[4]))/(x^2*Log[4]),x]

[Out]

Integrate[(E^((2*(-10 + (13*x + 4*x^2)*Log[4]))/(x*Log[4]))*(20 + 8*x^2*Log[4]) + E^((-10 + (13*x + 4*x^2)*Log
[4])/(x*Log[4]))*(100 + 40*x^2*Log[4]))/x^2, x]/Log[4]

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fricas [B]  time = 0.85, size = 50, normalized size = 2.27 \begin {gather*} e^{\left (\frac {2 \, {\left ({\left (4 \, x^{2} + 13 \, x\right )} \log \relax (2) - 5\right )}}{x \log \relax (2)}\right )} + 10 \, e^{\left (\frac {{\left (4 \, x^{2} + 13 \, x\right )} \log \relax (2) - 5}{x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((16*x^2*log(2)+20)*exp(1/2*(2*(4*x^2+13*x)*log(2)-10)/x/log(2))^2+(80*x^2*log(2)+100)*exp(1/2*(
2*(4*x^2+13*x)*log(2)-10)/x/log(2)))/x^2/log(2),x, algorithm="fricas")

[Out]

e^(2*((4*x^2 + 13*x)*log(2) - 5)/(x*log(2))) + 10*e^(((4*x^2 + 13*x)*log(2) - 5)/(x*log(2)))

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giac [B]  time = 0.31, size = 60, normalized size = 2.73 \begin {gather*} \frac {e^{\left (\frac {2 \, {\left (4 \, x^{2} \log \relax (2) + 13 \, x \log \relax (2) - 5\right )}}{x \log \relax (2)}\right )} \log \relax (2) + 10 \, e^{\left (\frac {4 \, x^{2} \log \relax (2) + 13 \, x \log \relax (2) - 5}{x \log \relax (2)}\right )} \log \relax (2)}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((16*x^2*log(2)+20)*exp(1/2*(2*(4*x^2+13*x)*log(2)-10)/x/log(2))^2+(80*x^2*log(2)+100)*exp(1/2*(
2*(4*x^2+13*x)*log(2)-10)/x/log(2)))/x^2/log(2),x, algorithm="giac")

[Out]

(e^(2*(4*x^2*log(2) + 13*x*log(2) - 5)/(x*log(2)))*log(2) + 10*e^((4*x^2*log(2) + 13*x*log(2) - 5)/(x*log(2)))
*log(2))/log(2)

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maple [B]  time = 0.08, size = 51, normalized size = 2.32

method	result	size
risch	\({\mathrm e}^{\frac {8 x^{2} \ln \relax (2)+26 x \ln \relax (2)-10}{x \ln \relax (2)}}+10 \,{\mathrm e}^{\frac {4 x^{2} \ln \relax (2)+13 x \ln \relax (2)-5}{x \ln \relax (2)}}\)	\(51\)
norman	\(\frac {x \,{\mathrm e}^{\frac {2 \left (4 x^{2}+13 x \right ) \ln \relax (2)-10}{x \ln \relax (2)}}+10 x \,{\mathrm e}^{\frac {2 \left (4 x^{2}+13 x \right ) \ln \relax (2)-10}{2 x \ln \relax (2)}}}{x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((16*x^2*ln(2)+20)*exp(1/2*(2*(4*x^2+13*x)*ln(2)-10)/x/ln(2))^2+(80*x^2*ln(2)+100)*exp(1/2*(2*(4*x^2+1
3*x)*ln(2)-10)/x/ln(2)))/x^2/ln(2),x,method=_RETURNVERBOSE)

[Out]

exp(2*(4*x^2*ln(2)+13*x*ln(2)-5)/x/ln(2))+10*exp((4*x^2*ln(2)+13*x*ln(2)-5)/x/ln(2))

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maxima [B]  time = 0.55, size = 43, normalized size = 1.95 \begin {gather*} \frac {e^{\left (8 \, x - \frac {10}{x \log \relax (2)} + 26\right )} \log \relax (2) + 10 \, e^{\left (4 \, x - \frac {5}{x \log \relax (2)} + 13\right )} \log \relax (2)}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((16*x^2*log(2)+20)*exp(1/2*(2*(4*x^2+13*x)*log(2)-10)/x/log(2))^2+(80*x^2*log(2)+100)*exp(1/2*(
2*(4*x^2+13*x)*log(2)-10)/x/log(2)))/x^2/log(2),x, algorithm="maxima")

[Out]

(e^(8*x - 10/(x*log(2)) + 26)*log(2) + 10*e^(4*x - 5/(x*log(2)) + 13)*log(2))/log(2)

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mupad [B]  time = 2.35, size = 37, normalized size = 1.68 \begin {gather*} {\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{13}\,{\mathrm {e}}^{-\frac {10}{x\,\ln \relax (2)}}\,\left (10\,{\mathrm {e}}^{\frac {5}{x\,\ln \relax (2)}}+{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{13}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp((2*(log(2)*(13*x + 4*x^2) - 5))/(x*log(2)))*(16*x^2*log(2) + 20))/2 + (exp((log(2)*(13*x + 4*x^2) -
5)/(x*log(2)))*(80*x^2*log(2) + 100))/2)/(x^2*log(2)),x)

[Out]

exp(4*x)*exp(13)*exp(-10/(x*log(2)))*(10*exp(5/(x*log(2))) + exp(4*x)*exp(13))

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sympy [B]  time = 0.22, size = 42, normalized size = 1.91 \begin {gather*} e^{\frac {2 \left (\frac {\left (8 x^{2} + 26 x\right ) \log {\relax (2 )}}{2} - 5\right )}{x \log {\relax (2 )}}} + 10 e^{\frac {\frac {\left (8 x^{2} + 26 x\right ) \log {\relax (2 )}}{2} - 5}{x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((16*x**2*ln(2)+20)*exp(1/2*(2*(4*x**2+13*x)*ln(2)-10)/x/ln(2))**2+(80*x**2*ln(2)+100)*exp(1/2*(
2*(4*x**2+13*x)*ln(2)-10)/x/ln(2)))/x**2/ln(2),x)

[Out]

exp(2*((8*x**2 + 26*x)*log(2)/2 - 5)/(x*log(2))) + 10*exp(((8*x**2 + 26*x)*log(2)/2 - 5)/(x*log(2)))

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