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

Optimal. Leaf size=20 \[ \frac {1}{9} 4^{2 x^3} 5^{2-2 x^3} \]

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Rubi [C]  time = 41.11, antiderivative size = 67, normalized size of antiderivative = 3.35, number of steps used = 986, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2287, 2209} \begin {gather*} \frac {25 \log (2) \gcd \left (1,\frac {\log (5)}{\log (2)}\right ) \exp \left (-\frac {2 x^3 \log \left (\frac {5}{4}\right ) \log \left (2^{\gcd \left (1,\frac {\log (5)}{\log (2)}\right )}\right )}{\log (2) \gcd \left (1,\frac {\log (5)}{\log (2)}\right )}\right )}{9 \log \left (2^{\gcd \left (1,\frac {\log (5)}{\log (2)}\right )}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(25*GCD[1, Log[5]/Log[2]]*Log[2])/(9*E^((2*x^3*Log[5/4]*Log[2^GCD[1, Log[5]/Log[2]]])/(GCD[1, Log[5]/Log[2]]*L
og[2]))*Log[2^GCD[1, Log[5]/Log[2]]])

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.40 \begin {gather*} \frac {16^{x^3} 25^{1-x^3} \log \left (\frac {5}{4}\right )}{3 \log \left (\frac {125}{64}\right )} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.71, size = 11, normalized size = 0.55 \begin {gather*} \frac {25}{9 \, \left (\frac {5}{4}\right )^{2 \, x^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.12, size = 22, normalized size = 1.10 \begin {gather*} \frac {25 \, \log \left (\frac {5}{4}\right )}{9 \, \left (\frac {5}{4}\right )^{2 \, x^{3}} {\left (\log \relax (5) - 2 \, \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.05, size = 12, normalized size = 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	\(\frac {25 \,{\mathrm e}^{-2 x^{3} \ln \left (\frac {5}{4}\right )}}{9}\)	\(12\)
risch	\(-\frac {\left (-\frac {50 \ln \relax (5)}{3}+\frac {100 \ln \relax (2)}{3}\right ) 5^{-2 x^{3}} 4^{2 x^{3}}}{6 \left (\ln \relax (5)-2 \ln \relax (2)\right )}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [B]  time = 0.37, size = 48, normalized size = 2.40 \begin {gather*} -\frac {5^{2 \, {\left (x^{3} - 1\right )} {\left (\frac {2 \, \log \relax (2)}{\log \relax (5)} - 1\right )} + \frac {4 \, \log \relax (2)}{\log \relax (5)}} \log \left (\frac {5}{4}\right )}{9 \, {\left (\frac {2 \, \log \relax (2)}{\log \relax (5)} - 1\right )} \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.08, size = 11, normalized size = 0.55 \begin {gather*} \frac {25}{9\,{\left (\frac {5}{4}\right )}^{2\,x^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.17, size = 14, normalized size = 0.70 \begin {gather*} \frac {25 e^{- 2 x^{3} \log {\left (\frac {5}{4} \right )}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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