3.40.41 \(\int \frac {-105 x-x^2-2 x^3-2 x^5+(-105-210 x^3) \log (4)+(-105-210 x^3) \log (x)}{x^4+105 x^2 \log (4)+105 x^2 \log (x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{x}-x^2-\log \left (x^2+105 (\log (4)+\log (x))\right ) \]

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Rubi [A]  time = 0.24, antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 3, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6742, 14, 6684} \begin {gather*} -x^2-\log \left (x^2+105 \log (x)+105 \log (4)\right )+\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-105*x - x^2 - 2*x^3 - 2*x^5 + (-105 - 210*x^3)*Log[4] + (-105 - 210*x^3)*Log[x])/(x^4 + 105*x^2*Log[4] +
105*x^2*Log[x]),x]

[Out]

x^(-1) - x^2 - Log[x^2 + 105*Log[4] + 105*Log[x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1-2 x^3}{x^2}+\frac {-105-2 x^2}{x \left (x^2+105 \log (4)+105 \log (x)\right )}\right ) \, dx\\ &=\int \frac {-1-2 x^3}{x^2} \, dx+\int \frac {-105-2 x^2}{x \left (x^2+105 \log (4)+105 \log (x)\right )} \, dx\\ &=-\log \left (x^2+105 \log (4)+105 \log (x)\right )+\int \left (-\frac {1}{x^2}-2 x\right ) \, dx\\ &=\frac {1}{x}-x^2-\log \left (x^2+105 \log (4)+105 \log (x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 24, normalized size = 1.04 \begin {gather*} \frac {1}{x}-x^2-\log \left (x^2+105 \log (4)+105 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-105*x - x^2 - 2*x^3 - 2*x^5 + (-105 - 210*x^3)*Log[4] + (-105 - 210*x^3)*Log[x])/(x^4 + 105*x^2*Lo
g[4] + 105*x^2*Log[x]),x]

[Out]

x^(-1) - x^2 - Log[x^2 + 105*Log[4] + 105*Log[x]]

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fricas [A]  time = 1.02, size = 25, normalized size = 1.09 \begin {gather*} -\frac {x^{3} + x \log \left (x^{2} + 210 \, \log \relax (2) + 105 \, \log \relax (x)\right ) - 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-210*x^3-105)*log(x)+2*(-210*x^3-105)*log(2)-2*x^5-2*x^3-x^2-105*x)/(105*x^2*log(x)+210*x^2*log(2)
+x^4),x, algorithm="fricas")

[Out]

-(x^3 + x*log(x^2 + 210*log(2) + 105*log(x)) - 1)/x

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giac [A]  time = 0.13, size = 24, normalized size = 1.04 \begin {gather*} -x^{2} + \frac {1}{x} - \log \left (x^{2} + 210 \, \log \relax (2) + 105 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-210*x^3-105)*log(x)+2*(-210*x^3-105)*log(2)-2*x^5-2*x^3-x^2-105*x)/(105*x^2*log(x)+210*x^2*log(2)
+x^4),x, algorithm="giac")

[Out]

-x^2 + 1/x - log(x^2 + 210*log(2) + 105*log(x))

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maple [A]  time = 0.05, size = 27, normalized size = 1.17

method	result	size
risch	\(-\frac {x^{3}-1}{x}-\ln \left (\frac {x^{2}}{105}+2 \ln \relax (2)+\ln \relax (x )\right )\)	\(27\)
norman	\(\frac {-x^{3}+1}{x}-\ln \left (x^{2}+105 \ln \relax (x )+210 \ln \relax (2)\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-210*x^3-105)*ln(x)+2*(-210*x^3-105)*ln(2)-2*x^5-2*x^3-x^2-105*x)/(105*x^2*ln(x)+210*x^2*ln(2)+x^4),x,me
thod=_RETURNVERBOSE)

[Out]

-(x^3-1)/x-ln(1/105*x^2+2*ln(2)+ln(x))

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maxima [A]  time = 0.47, size = 26, normalized size = 1.13 \begin {gather*} -\frac {x^{3} - 1}{x} - \log \left (\frac {1}{105} \, x^{2} + 2 \, \log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-210*x^3-105)*log(x)+2*(-210*x^3-105)*log(2)-2*x^5-2*x^3-x^2-105*x)/(105*x^2*log(x)+210*x^2*log(2)
+x^4),x, algorithm="maxima")

[Out]

-(x^3 - 1)/x - log(1/105*x^2 + 2*log(2) + log(x))

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mupad [B]  time = 2.30, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{x}-\ln \left (\ln \left (4\,x\right )+\frac {x^2}{105}\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(105*x + 2*log(2)*(210*x^3 + 105) + x^2 + 2*x^3 + 2*x^5 + log(x)*(210*x^3 + 105))/(105*x^2*log(x) + 210*x
^2*log(2) + x^4),x)

[Out]

1/x - log(log(4*x) + x^2/105) - x^2

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sympy [A]  time = 0.16, size = 20, normalized size = 0.87 \begin {gather*} - x^{2} - \log {\left (\frac {x^{2}}{105} + \log {\relax (x )} + 2 \log {\relax (2 )} \right )} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-210*x**3-105)*ln(x)+2*(-210*x**3-105)*ln(2)-2*x**5-2*x**3-x**2-105*x)/(105*x**2*ln(x)+210*x**2*ln
(2)+x**4),x)

[Out]

-x**2 - log(x**2/105 + log(x) + 2*log(2)) + 1/x

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