\(\int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log (\frac {5}{4}) \, dx\) [7608]

Optimal result

Integrand size = 29, antiderivative size = 20 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {1}{9} 4^{2 x^3} 5^{2-2 x^3} \]

[Out]

25/9/exp(x^3*ln(5/4))^2

Rubi [F(-1)]

Timed out. \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\text {\$Aborted} \]

[In]

Int[-1/3*(2^(1 + 4*x^3)*5^(2 - 2*x^3)*x^2*Log[5/4]),x]

[Out]

$Aborted

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \log \left (\frac {5}{4}\right ) \int 2^{1+4 x^3} 5^{2-2 x^3} x^2 \, dx\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {16^{x^3} 25^{1-x^3} \log \left (\frac {5}{4}\right )}{3 \log \left (\frac {125}{64}\right )} \]

[In]

Integrate[-1/3*(2^(1 + 4*x^3)*5^(2 - 2*x^3)*x^2*Log[5/4]),x]

[Out]

(16^x^3*25^(1 - x^3)*Log[5/4])/(3*Log[125/64])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60

method	result	size
gosper	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
derivativedivides	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
default	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
norman	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
parallelrisch	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
risch	\(-\frac {\left (-\frac {50 \ln \left (5\right )}{3}+\frac {100 \ln \left (2\right )}{3}\right ) 5^{-2 x^{3}} 4^{2 x^{3}}}{6 \left (\ln \left (5\right )-2 \ln \left (2\right )\right )}\)	\(35\)

[In]

int(-50/3*x^2*ln(5/4)/exp(x^3*ln(5/4))^2,x,method=_RETURNVERBOSE)

[Out]

25/9/exp(x^3*ln(5/4))^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {25}{9 \, \left (\frac {5}{4}\right )^{2 \, x^{3}}} \]

[In]

integrate(-50/3*x^2*log(5/4)/exp(x^3*log(5/4))^2,x, algorithm="fricas")

[Out]

25/9/(5/4)^(2*x^3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {25 e^{- 2 x^{3} \log {\left (\frac {5}{4} \right )}}}{9} \]

[In]

integrate(-50/3*x**2*ln(5/4)/exp(x**3*ln(5/4))**2,x)

[Out]

25*exp(-2*x**3*log(5/4))/9

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (11) = 22\).

Time = 0.18 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=-\frac {5^{2 \, {\left (x^{3} - 1\right )} {\left (\frac {2 \, \log \left (2\right )}{\log \left (5\right )} - 1\right )} + \frac {4 \, \log \left (2\right )}{\log \left (5\right )}} \log \left (\frac {5}{4}\right )}{9 \, {\left (\frac {2 \, \log \left (2\right )}{\log \left (5\right )} - 1\right )} \log \left (5\right )} \]

[In]

integrate(-50/3*x^2*log(5/4)/exp(x^3*log(5/4))^2,x, algorithm="maxima")

[Out]

-1/9*5^(2*(x^3 - 1)*(2*log(2)/log(5) - 1) + 4*log(2)/log(5))*log(5/4)/((2*log(2)/log(5) - 1)*log(5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {25 \, \log \left (\frac {5}{4}\right )}{9 \, \left (\frac {5}{4}\right )^{2 \, x^{3}} {\left (\log \left (5\right ) - 2 \, \log \left (2\right )\right )}} \]

[In]

integrate(-50/3*x^2*log(5/4)/exp(x^3*log(5/4))^2,x, algorithm="giac")

[Out]

25/9*log(5/4)/((5/4)^(2*x^3)*(log(5) - 2*log(2)))

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int -\frac {1}{3} 2^{1+4 x^3} 5^{2-2 x^3} x^2 \log \left (\frac {5}{4}\right ) \, dx=\frac {25}{9\,{\left (\frac {5}{4}\right )}^{2\,x^3}} \]

[In]

int(-(50*x^2*exp(-2*x^3*log(5/4))*log(5/4))/3,x)

[Out]

25/(9*(5/4)^(2*x^3))