3.55.35 \(\int \frac {e^5 (-9-6 x)}{(15 x+5 x^2) \log ^2(\frac {1}{2} (3 x+x^2))} \, dx\) [5435]

Optimal. Leaf size=22 \[ \frac {1}{5} e^5 \left (-3+\frac {3}{\log \left (\frac {1}{2} x (3+x)\right )}\right ) \]

[Out]

1/5*(3/ln(1/2*(3+x)*x)-3)*exp(5)

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Rubi [A]
time = 0.12, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {12, 1607, 6818} \begin {gather*} \frac {3 e^5}{5 \log \left (\frac {1}{2} \left (x^2+3 x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(-9 - 6*x))/((15*x + 5*x^2)*Log[(3*x + x^2)/2]^2),x]

[Out]

(3*E^5)/(5*Log[(3*x + x^2)/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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^5 \int \frac {-9-6 x}{\left (15 x+5 x^2\right ) \log ^2\left (\frac {1}{2} \left (3 x+x^2\right )\right )} \, dx\\ &=e^5 \int \frac {-9-6 x}{x (15+5 x) \log ^2\left (\frac {1}{2} \left (3 x+x^2\right )\right )} \, dx\\ &=\frac {3 e^5}{5 \log \left (\frac {1}{2} \left (3 x+x^2\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 18, normalized size = 0.82 \begin {gather*} \frac {3 e^5}{5 \log \left (\frac {1}{2} x (3+x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-9 - 6*x))/((15*x + 5*x^2)*Log[(3*x + x^2)/2]^2),x]

[Out]

(3*E^5)/(5*Log[(x*(3 + x))/2])

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Maple [A]
time = 1.58, size = 20, normalized size = 0.91

method	result	size
norman	\(\frac {3 \,{\mathrm e}^{5}}{5 \ln \left (\frac {1}{2} x^{2}+\frac {3}{2} x \right )}\)	\(17\)
risch	\(\frac {3 \,{\mathrm e}^{5}}{5 \ln \left (\frac {1}{2} x^{2}+\frac {3}{2} x \right )}\)	\(17\)
default	\(-\frac {3 \,{\mathrm e}^{5}}{5 \left (\ln \left (2\right )-\ln \left (x^{2}+3 x \right )\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-6*x-9)*exp(5)/(5*x^2+15*x)/ln(1/2*x^2+3/2*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.54, size = 19, normalized size = 0.86 \begin {gather*} -\frac {3 \, e^{5}}{5 \, {\left (\log \left (2\right ) - \log \left (x + 3\right ) - \log \left (x\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x-9)*exp(5)/(5*x^2+15*x)/log(1/2*x^2+3/2*x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 16, normalized size = 0.73 \begin {gather*} \frac {3 \, e^{5}}{5 \, \log \left (\frac {1}{2} \, x^{2} + \frac {3}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x-9)*exp(5)/(5*x^2+15*x)/log(1/2*x^2+3/2*x)^2,x, algorithm="fricas")

[Out]

3/5*e^5/log(1/2*x^2 + 3/2*x)

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Sympy [A]
time = 0.05, size = 17, normalized size = 0.77 \begin {gather*} \frac {3 e^{5}}{5 \log {\left (\frac {x^{2}}{2} + \frac {3 x}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x-9)*exp(5)/(5*x**2+15*x)/ln(1/2*x**2+3/2*x)**2,x)

[Out]

3*exp(5)/(5*log(x**2/2 + 3*x/2))

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Giac [A]
time = 0.39, size = 16, normalized size = 0.73 \begin {gather*} \frac {3 \, e^{5}}{5 \, \log \left (\frac {1}{2} \, x^{2} + \frac {3}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-6*x-9)*exp(5)/(5*x^2+15*x)/log(1/2*x^2+3/2*x)^2,x, algorithm="giac")

[Out]

3/5*e^5/log(1/2*x^2 + 3/2*x)

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Mupad [B]
time = 3.70, size = 16, normalized size = 0.73 \begin {gather*} \frac {3\,{\mathrm {e}}^5}{5\,\ln \left (\frac {x^2}{2}+\frac {3\,x}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(5)*(6*x + 9))/(log((3*x)/2 + x^2/2)^2*(15*x + 5*x^2)),x)

[Out]

(3*exp(5))/(5*log((3*x)/2 + x^2/2))

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