3.73.95 \(\int -\frac {3 e^{\frac {6+8 x-x \log (\frac {9}{4})}{3+4 x}} \log (\frac {9}{4})}{9+24 x+16 x^2} \, dx\)

Optimal. Leaf size=18 \[ e^{2-\frac {x \log \left (\frac {9}{4}\right )}{3+4 x}} \]

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Rubi [A]  time = 0.11, antiderivative size = 27, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 27, 6706} \begin {gather*} e^2 3^{-\frac {2 x}{4 x+3}} 4^{\frac {x}{4 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*E^((6 + 8*x - x*Log[9/4])/(3 + 4*x))*Log[9/4])/(9 + 24*x + 16*x^2),x]

[Out]

(4^(x/(3 + 4*x))*E^2)/3^((2*x)/(3 + 4*x))

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (\left (3 \log \left (\frac {9}{4}\right )\right ) \int \frac {e^{\frac {6+8 x-x \log \left (\frac {9}{4}\right )}{3+4 x}}}{9+24 x+16 x^2} \, dx\right )\\ &=-\left (\left (3 \log \left (\frac {9}{4}\right )\right ) \int \frac {e^{\frac {6+8 x-x \log \left (\frac {9}{4}\right )}{3+4 x}}}{(3+4 x)^2} \, dx\right )\\ &=3^{-\frac {2 x}{3+4 x}} 4^{\frac {x}{3+4 x}} e^2\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.10, size = 42, normalized size = 2.33 \begin {gather*} \frac {2^{-\frac {3+2 x}{3+4 x}} 9^{-\frac {x}{3+4 x}} e^2 \log \left (\frac {9}{4}\right )}{\log \left (\frac {3}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*E^((6 + 8*x - x*Log[9/4])/(3 + 4*x))*Log[9/4])/(9 + 24*x + 16*x^2),x]

[Out]

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

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fricas [A]  time = 0.42, size = 18, normalized size = 1.00 \begin {gather*} e^{\left (\frac {x \log \left (\frac {4}{9}\right ) + 8 \, x + 6}{4 \, x + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*log(4/9)*exp((x*log(4/9)+8*x+6)/(3+4*x))/(16*x^2+24*x+9),x, algorithm="fricas")

[Out]

e^((x*log(4/9) + 8*x + 6)/(4*x + 3))

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giac [B]  time = 0.14, size = 32, normalized size = 1.78 \begin {gather*} e^{\left (\frac {x \log \left (\frac {4}{9}\right )}{4 \, x + 3} + \frac {8 \, x}{4 \, x + 3} + \frac {6}{4 \, x + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*log(4/9)*exp((x*log(4/9)+8*x+6)/(3+4*x))/(16*x^2+24*x+9),x, algorithm="giac")

[Out]

e^(x*log(4/9)/(4*x + 3) + 8*x/(4*x + 3) + 6/(4*x + 3))

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maple [A]  time = 0.18, size = 19, normalized size = 1.06

method	result	size
gosper	\({\mathrm e}^{\frac {x \ln \left (\frac {4}{9}\right )+8 x +6}{3+4 x}}\)	\(19\)
derivativedivides	\(\frac {\ln \left (\frac {4}{9}\right )^{2} {\mathrm e}^{\frac {\ln \left (\frac {4}{9}\right )}{4}+2-\frac {3 \ln \left (\frac {4}{9}\right )}{4 \left (3+4 x \right )}}}{4 \ln \relax (2)^{2}-8 \ln \relax (2) \ln \relax (3)+4 \ln \relax (3)^{2}}\)	\(42\)
default	\(\frac {\ln \left (\frac {4}{9}\right )^{2} {\mathrm e}^{\frac {\ln \left (\frac {4}{9}\right )}{4}+2-\frac {3 \ln \left (\frac {4}{9}\right )}{4 \left (3+4 x \right )}}}{4 \ln \relax (2)^{2}-8 \ln \relax (2) \ln \relax (3)+4 \ln \relax (3)^{2}}\)	\(42\)
risch	\(4^{\frac {x}{3+4 x}} \left (\frac {1}{9}\right )^{\frac {x}{3+4 x}} {\mathrm e}^{2}\)	\(45\)
norman	\(\frac {4 x \,{\mathrm e}^{\frac {x \ln \left (\frac {4}{9}\right )+8 x +6}{3+4 x}}+3 \,{\mathrm e}^{\frac {x \ln \left (\frac {4}{9}\right )+8 x +6}{3+4 x}}}{3+4 x}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*ln(4/9)*exp((x*ln(4/9)+8*x+6)/(3+4*x))/(16*x^2+24*x+9),x,method=_RETURNVERBOSE)

[Out]

exp((x*ln(4/9)+8*x+6)/(3+4*x))

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maxima [B]  time = 0.46, size = 44, normalized size = 2.44 \begin {gather*} -\frac {\sqrt {3} \sqrt {2} e^{\left (\frac {3 \, \log \relax (3)}{2 \, {\left (4 \, x + 3\right )}} - \frac {3 \, \log \relax (2)}{2 \, {\left (4 \, x + 3\right )}} + 2\right )} \log \left (\frac {4}{9}\right )}{6 \, {\left (\log \relax (3) - \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*log(4/9)*exp((x*log(4/9)+8*x+6)/(3+4*x))/(16*x^2+24*x+9),x, algorithm="maxima")

[Out]

-1/6*sqrt(3)*sqrt(2)*e^(3/2*log(3)/(4*x + 3) - 3/2*log(2)/(4*x + 3) + 2)*log(4/9)/(log(3) - log(2))

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mupad [B]  time = 0.69, size = 33, normalized size = 1.83 \begin {gather*} {\left (\frac {4}{9}\right )}^{\frac {x}{4\,x+3}}\,{\mathrm {e}}^{\frac {6}{4\,x+3}}\,{\mathrm {e}}^{\frac {8\,x}{4\,x+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*exp((8*x + x*log(4/9) + 6)/(4*x + 3))*log(4/9))/(24*x + 16*x^2 + 9),x)

[Out]

(4/9)^(x/(4*x + 3))*exp(6/(4*x + 3))*exp((8*x)/(4*x + 3))

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sympy [A]  time = 0.22, size = 17, normalized size = 0.94 \begin {gather*} e^{\frac {x \log {\left (\frac {4}{9} \right )} + 8 x + 6}{4 x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(3*ln(4/9)*exp((x*ln(4/9)+8*x+6)/(3+4*x))/(16*x**2+24*x+9),x)

[Out]

exp((x*log(4/9) + 8*x + 6)/(4*x + 3))

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