3.35.70 \(\int \frac {-5+e^4 (-6-3 x) x}{15+5 x+e^4 x (3 x+x^2)} \, dx\)

Optimal. Leaf size=20 \[ \log \left (\frac {2}{(3+x) \left (-5-e^4 x^2\right )}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {1587} \begin {gather*} -\log \left (e^4 \left (x^2+3 x\right ) x+5 x+15\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + E^4*(-6 - 3*x)*x)/(15 + 5*x + E^4*x*(3*x + x^2)),x]

[Out]

-Log[15 + 5*x + E^4*x*(3*x + x^2)]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\log \left (15+5 x+e^4 x \left (3 x+x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.90 \begin {gather*} -\log \left (15+5 x+e^4 x^2 (3+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + E^4*(-6 - 3*x)*x)/(15 + 5*x + E^4*x*(3*x + x^2)),x]

[Out]

-Log[15 + 5*x + E^4*x^2*(3 + x)]

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fricas [A]  time = 1.23, size = 20, normalized size = 1.00 \begin {gather*} -\log \left ({\left (x^{3} + 3 \, x^{2}\right )} e^{4} + 5 \, x + 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-6)*exp(log(x)+4)-5)/((x^2+3*x)*exp(log(x)+4)+5*x+15),x, algorithm="fricas")

[Out]

-log((x^3 + 3*x^2)*e^4 + 5*x + 15)

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giac [A]  time = 0.22, size = 19, normalized size = 0.95 \begin {gather*} -\log \left (x^{2} e^{4} + 5\right ) - \log \left ({\left | x + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-6)*exp(log(x)+4)-5)/((x^2+3*x)*exp(log(x)+4)+5*x+15),x, algorithm="giac")

[Out]

-log(x^2*e^4 + 5) - log(abs(x + 3))

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maple [A]  time = 0.04, size = 19, normalized size = 0.95

method	result	size
norman	\(-\ln \left (3+x \right )-\ln \left (x^{2} {\mathrm e}^{4}+5\right )\)	\(19\)
default	\(-\ln \left (x^{3} {\mathrm e}^{4}+3 x^{2} {\mathrm e}^{4}+5 x +15\right )\)	\(22\)
risch	\(-\ln \left (x^{3} {\mathrm e}^{4}+3 x^{2} {\mathrm e}^{4}+5 x +15\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x-6)*exp(ln(x)+4)-5)/((x^2+3*x)*exp(ln(x)+4)+5*x+15),x,method=_RETURNVERBOSE)

[Out]

-ln(3+x)-ln(x^2*exp(4)+5)

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maxima [A]  time = 0.60, size = 18, normalized size = 0.90 \begin {gather*} -\log \left (x^{2} e^{4} + 5\right ) - \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-6)*exp(log(x)+4)-5)/((x^2+3*x)*exp(log(x)+4)+5*x+15),x, algorithm="maxima")

[Out]

-log(x^2*e^4 + 5) - log(x + 3)

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mupad [B]  time = 0.17, size = 15, normalized size = 0.75 \begin {gather*} -\ln \left (\left (x+3\right )\,\left ({\mathrm {e}}^4\,x^2+5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x) + 4)*(3*x + 6) + 5)/(5*x + exp(log(x) + 4)*(3*x + x^2) + 15),x)

[Out]

-log((x + 3)*(x^2*exp(4) + 5))

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sympy [A]  time = 0.36, size = 22, normalized size = 1.10 \begin {gather*} - \log {\left (x^{3} e^{4} + 3 x^{2} e^{4} + 5 x + 15 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x-6)*exp(ln(x)+4)-5)/((x**2+3*x)*exp(ln(x)+4)+5*x+15),x)

[Out]

-log(x**3*exp(4) + 3*x**2*exp(4) + 5*x + 15)

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