3.30.20 \(\int \frac {-15-3 x-x^2-3 \log ^2(4)}{5 x^2+x^3+x^2 \log ^2(4)} \, dx\) [2920]

Optimal. Leaf size=28 \[ \frac {1875}{4}+\frac {3}{x}-\log (2)-\log \left (1+\frac {1}{5} \left (x+\log ^2(4)\right )\right ) \]

[Out]

1875/4-ln(2)-ln(1+1/5*x+4/5*ln(2)^2)+3/x

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Rubi [A]
time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6, 1607, 907} \begin {gather*} \frac {3}{x}-\log \left (x+5+\log ^2(4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3/x - Log[5 + x + Log[4]^2]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15-3 x-x^2-3 \log ^2(4)}{x^3+x^2 \left (5+\log ^2(4)\right )} \, dx\\ &=\int \frac {-15-3 x-x^2-3 \log ^2(4)}{x^2 \left (5+x+\log ^2(4)\right )} \, dx\\ &=\int \left (-\frac {3}{x^2}+\frac {1}{-5-x-\log ^2(4)}\right ) \, dx\\ &=\frac {3}{x}-\log \left (5+x+\log ^2(4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 16, normalized size = 0.57 \begin {gather*} \frac {3}{x}-\log \left (5+x+\log ^2(4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3/x - Log[5 + x + Log[4]^2]

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Maple [A]
time = 0.65, size = 19, normalized size = 0.68

method	result	size
default	\(-\ln \left (4 \ln \left (2\right )^{2}+x +5\right )+\frac {3}{x}\)	\(19\)
norman	\(-\ln \left (4 \ln \left (2\right )^{2}+x +5\right )+\frac {3}{x}\)	\(19\)
risch	\(-\ln \left (4 \ln \left (2\right )^{2}+x +5\right )+\frac {3}{x}\)	\(19\)
meijerg	\(-\frac {12 \ln \left (2\right )^{2} \left (\ln \left (1+\frac {x}{4 \ln \left (2\right )^{2}+5}\right )-\ln \left (x \right )+\ln \left (4 \ln \left (2\right )^{2}+5\right )-\frac {4 \ln \left (2\right )^{2}+5}{x}\right )}{\left (4 \ln \left (2\right )^{2}+5\right )^{2}}-\ln \left (1+\frac {x}{4 \ln \left (2\right )^{2}+5}\right )-\frac {3 \left (-\ln \left (1+\frac {x}{4 \ln \left (2\right )^{2}+5}\right )+\ln \left (x \right )-\ln \left (4 \ln \left (2\right )^{2}+5\right )\right )}{4 \ln \left (2\right )^{2}+5}-\frac {15 \left (\ln \left (1+\frac {x}{4 \ln \left (2\right )^{2}+5}\right )-\ln \left (x \right )+\ln \left (4 \ln \left (2\right )^{2}+5\right )-\frac {4 \ln \left (2\right )^{2}+5}{x}\right )}{\left (4 \ln \left (2\right )^{2}+5\right )^{2}}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*ln(2)^2-x^2-3*x-15)/(4*x^2*ln(2)^2+x^3+5*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 18, normalized size = 0.64 \begin {gather*} \frac {3}{x} - \log \left (4 \, \log \left (2\right )^{2} + x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(2)^2-x^2-3*x-15)/(4*x^2*log(2)^2+x^3+5*x^2),x, algorithm="maxima")

[Out]

3/x - log(4*log(2)^2 + x + 5)

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Fricas [A]
time = 0.40, size = 19, normalized size = 0.68 \begin {gather*} -\frac {x \log \left (4 \, \log \left (2\right )^{2} + x + 5\right ) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(2)^2-x^2-3*x-15)/(4*x^2*log(2)^2+x^3+5*x^2),x, algorithm="fricas")

[Out]

-(x*log(4*log(2)^2 + x + 5) - 3)/x

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Sympy [A]
time = 0.14, size = 14, normalized size = 0.50 \begin {gather*} - \log {\left (x + 4 \log {\left (2 \right )}^{2} + 5 \right )} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*ln(2)**2-x**2-3*x-15)/(4*x**2*ln(2)**2+x**3+5*x**2),x)

[Out]

-log(x + 4*log(2)**2 + 5) + 3/x

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Giac [A]
time = 0.39, size = 19, normalized size = 0.68 \begin {gather*} \frac {3}{x} - \log \left ({\left | 4 \, \log \left (2\right )^{2} + x + 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*log(2)^2-x^2-3*x-15)/(4*x^2*log(2)^2+x^3+5*x^2),x, algorithm="giac")

[Out]

3/x - log(abs(4*log(2)^2 + x + 5))

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Mupad [B]
time = 1.82, size = 18, normalized size = 0.64 \begin {gather*} \frac {3}{x}-\ln \left (x+4\,{\ln \left (2\right )}^2+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 12*log(2)^2 + x^2 + 15)/(4*x^2*log(2)^2 + 5*x^2 + x^3),x)

[Out]

3/x - log(x + 4*log(2)^2 + 5)

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