3.38.19 \(\int \frac {e^{\frac {4 (15 x^2+15 x \log (5))}{8+12 x}} (-60 x-45 x^2-30 \log (5))}{4+12 x+9 x^2} \, dx\)

Optimal. Leaf size=26 \[ 3-e^{\frac {20 x (x+\log (5))}{\frac {2}{3}+2 (1+2 x)}} \]

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Rubi [A]  time = 0.44, antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {27, 6741, 6706} \begin {gather*} -5^{\frac {60 x}{12 x+8}} e^{\frac {15 x^2}{3 x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4*(15*x^2 + 15*x*Log[5]))/(8 + 12*x))*(-60*x - 45*x^2 - 30*Log[5]))/(4 + 12*x + 9*x^2),x]

[Out]

-(5^((60*x)/(8 + 12*x))*E^((15*x^2)/(2 + 3*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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {4 \left (15 x^2+15 x \log (5)\right )}{8+12 x}} \left (-60 x-45 x^2-30 \log (5)\right )}{(2+3 x)^2} \, dx\\ &=\int \frac {e^{\frac {4 x (15 x+15 \log (5))}{8+12 x}} \left (-60 x-45 x^2-30 \log (5)\right )}{(2+3 x)^2} \, dx\\ &=-5^{\frac {60 x}{8+12 x}} e^{\frac {15 x^2}{2+3 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 36, normalized size = 1.38 \begin {gather*} -\frac {5^{\frac {15 x}{2+3 x}} e^{\frac {15 x^2}{2+3 x}} \log (25)}{2 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4*(15*x^2 + 15*x*Log[5]))/(8 + 12*x))*(-60*x - 45*x^2 - 30*Log[5]))/(4 + 12*x + 9*x^2),x]

[Out]

-1/2*(5^((15*x)/(2 + 3*x))*E^((15*x^2)/(2 + 3*x))*Log[25])/Log[5]

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fricas [A]  time = 0.59, size = 20, normalized size = 0.77 \begin {gather*} -e^{\left (\frac {15 \, {\left (x^{2} + x \log \relax (5)\right )}}{3 \, x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*log(5)-45*x^2-60*x)*exp((15*x*log(5)+15*x^2)/(12*x+8))^4/(9*x^2+12*x+4),x, algorithm="fricas")

[Out]

-e^(15*(x^2 + x*log(5))/(3*x + 2))

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giac [A]  time = 0.15, size = 28, normalized size = 1.08 \begin {gather*} -e^{\left (\frac {15 \, x^{2}}{3 \, x + 2} + \frac {15 \, x \log \relax (5)}{3 \, x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*log(5)-45*x^2-60*x)*exp((15*x*log(5)+15*x^2)/(12*x+8))^4/(9*x^2+12*x+4),x, algorithm="giac")

[Out]

-e^(15*x^2/(3*x + 2) + 15*x*log(5)/(3*x + 2))

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maple [A]  time = 0.14, size = 18, normalized size = 0.69

method	result	size
risch	\(-{\mathrm e}^{\frac {15 x \left (\ln \relax (5)+x \right )}{3 x +2}}\)	\(18\)
gosper	\(-{\mathrm e}^{\frac {15 x \left (\ln \relax (5)+x \right )}{3 x +2}}\)	\(20\)
norman	\(\frac {-2 \,{\mathrm e}^{\frac {60 x \ln \relax (5)+60 x^{2}}{12 x +8}}-3 x \,{\mathrm e}^{\frac {60 x \ln \relax (5)+60 x^{2}}{12 x +8}}}{3 x +2}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-30*ln(5)-45*x^2-60*x)*exp((15*x*ln(5)+15*x^2)/(12*x+8))^4/(9*x^2+12*x+4),x,method=_RETURNVERBOSE)

[Out]

-exp(15*x*(ln(5)+x)/(3*x+2))

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maxima [A]  time = 0.58, size = 28, normalized size = 1.08 \begin {gather*} -3125 \, e^{\left (5 \, x - \frac {10 \, \log \relax (5)}{3 \, x + 2} + \frac {20}{3 \, {\left (3 \, x + 2\right )}} - \frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*log(5)-45*x^2-60*x)*exp((15*x*log(5)+15*x^2)/(12*x+8))^4/(9*x^2+12*x+4),x, algorithm="maxima")

[Out]

-3125*e^(5*x - 10*log(5)/(3*x + 2) + 20/3/(3*x + 2) - 10/3)

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mupad [B]  time = 2.29, size = 27, normalized size = 1.04 \begin {gather*} -5^{\frac {15\,x}{3\,x+2}}\,{\mathrm {e}}^{\frac {15\,x^2}{3\,x+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((4*(15*x*log(5) + 15*x^2))/(12*x + 8))*(60*x + 30*log(5) + 45*x^2))/(12*x + 9*x^2 + 4),x)

[Out]

-5^((15*x)/(3*x + 2))*exp((15*x^2)/(3*x + 2))

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sympy [A]  time = 0.25, size = 19, normalized size = 0.73 \begin {gather*} - e^{\frac {4 \left (15 x^{2} + 15 x \log {\relax (5 )}\right )}{12 x + 8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*ln(5)-45*x**2-60*x)*exp((15*x*ln(5)+15*x**2)/(12*x+8))**4/(9*x**2+12*x+4),x)

[Out]

-exp(4*(15*x**2 + 15*x*log(5))/(12*x + 8))

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