3.3.52 \(\int \frac {20+4 x}{6 x^2+e^x x^2+x^3+(-12 x-2 e^x x-2 x^2) \log (30+5 e^x+5 x)+(6+e^x+x) \log ^2(30+5 e^x+5 x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {4}{-x+\log \left (5 \left (6+e^x+x\right )\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {4}{x-\log \left (5 \left (x+e^x+6\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(20 + 4*x)/(6*x^2 + E^x*x^2 + x^3 + (-12*x - 2*E^x*x - 2*x^2)*Log[30 + 5*E^x + 5*x] + (6 + E^x + x)*Log[30
+ 5*E^x + 5*x]^2),x]

[Out]

-4/(x - Log[5*(6 + E^x + 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 6686

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]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 (5+x)}{\left (6+e^x+x\right ) \left (x-\log \left (5 \left (6+e^x+x\right )\right )\right )^2} \, dx\\ &=4 \int \frac {5+x}{\left (6+e^x+x\right ) \left (x-\log \left (5 \left (6+e^x+x\right )\right )\right )^2} \, dx\\ &=-\frac {4}{x-\log \left (5 \left (6+e^x+x\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} -\frac {4}{x-\log \left (5 \left (6+e^x+x\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 4*x)/(6*x^2 + E^x*x^2 + x^3 + (-12*x - 2*E^x*x - 2*x^2)*Log[30 + 5*E^x + 5*x] + (6 + E^x + x)*
Log[30 + 5*E^x + 5*x]^2),x]

[Out]

-4/(x - Log[5*(6 + E^x + x)])

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fricas [A]  time = 0.74, size = 18, normalized size = 1.06 \begin {gather*} -\frac {4}{x - \log \left (5 \, x + 5 \, e^{x} + 30\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="fricas")

[Out]

-4/(x - log(5*x + 5*e^x + 30))

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giac [A]  time = 0.51, size = 18, normalized size = 1.06 \begin {gather*} -\frac {4}{x - \log \left (5 \, x + 5 \, e^{x} + 30\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="giac")

[Out]

-4/(x - log(5*x + 5*e^x + 30))

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maple [A]  time = 0.04, size = 19, normalized size = 1.12

method	result	size
risch	\(-\frac {4}{x -\ln \left (5 \,{\mathrm e}^{x}+5 x +30\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20+4*x)/((exp(x)+x+6)*ln(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*ln(5*exp(x)+5*x+30)+exp(x)*x^2+x^3+6
*x^2),x,method=_RETURNVERBOSE)

[Out]

-4/(x-ln(5*exp(x)+5*x+30))

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maxima [A]  time = 0.62, size = 18, normalized size = 1.06 \begin {gather*} -\frac {4}{x - \log \relax (5) - \log \left (x + e^{x} + 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)/((exp(x)+x+6)*log(5*exp(x)+5*x+30)^2+(-2*exp(x)*x-2*x^2-12*x)*log(5*exp(x)+5*x+30)+exp(x)*x
^2+x^3+6*x^2),x, algorithm="maxima")

[Out]

-4/(x - log(5) - log(x + e^x + 6))

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mupad [B]  time = 0.39, size = 18, normalized size = 1.06 \begin {gather*} -\frac {4}{x-\ln \left (5\,x+5\,{\mathrm {e}}^x+30\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x + 20)/(x^2*exp(x) + log(5*x + 5*exp(x) + 30)^2*(x + exp(x) + 6) - log(5*x + 5*exp(x) + 30)*(12*x + 2*
x*exp(x) + 2*x^2) + 6*x^2 + x^3),x)

[Out]

-4/(x - log(5*x + 5*exp(x) + 30))

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sympy [A]  time = 0.17, size = 14, normalized size = 0.82 \begin {gather*} \frac {4}{- x + \log {\left (5 x + 5 e^{x} + 30 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)/((exp(x)+x+6)*ln(5*exp(x)+5*x+30)**2+(-2*exp(x)*x-2*x**2-12*x)*ln(5*exp(x)+5*x+30)+exp(x)*x
**2+x**3+6*x**2),x)

[Out]

4/(-x + log(5*x + 5*exp(x) + 30))

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