3.62.24 \(\int \frac {-7 x^2+e^9 x^2-3 x^4+(14 x-2 e^9 x+6 x^3) \log (2)+(-8+e^9-3 x^2) \log ^2(2)}{x^2-2 x \log (2)+\log ^2(2)} \, dx\)

Optimal. Leaf size=23 \[ x \left (-8+e^9-x^2-\frac {x}{-x+\log (2)}\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {6, 27, 1850} \begin {gather*} -x^3-\left (7-e^9\right ) x+\frac {\log ^2(2)}{x-\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-7*x^2 + E^9*x^2 - 3*x^4 + (14*x - 2*E^9*x + 6*x^3)*Log[2] + (-8 + E^9 - 3*x^2)*Log[2]^2)/(x^2 - 2*x*Log[
2] + Log[2]^2),x]

[Out]

-((7 - E^9)*x) - x^3 + Log[2]^2/(x - Log[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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-7+e^9\right ) x^2-3 x^4+\left (14 x-2 e^9 x+6 x^3\right ) \log (2)+\left (-8+e^9-3 x^2\right ) \log ^2(2)}{x^2-2 x \log (2)+\log ^2(2)} \, dx\\ &=\int \frac {\left (-7+e^9\right ) x^2-3 x^4+\left (14 x-2 e^9 x+6 x^3\right ) \log (2)+\left (-8+e^9-3 x^2\right ) \log ^2(2)}{(x-\log (2))^2} \, dx\\ &=\int \left (-7 \left (1-\frac {e^9}{7}\right )-3 x^2-\frac {\log ^2(2)}{(x-\log (2))^2}\right ) \, dx\\ &=-\left (\left (7-e^9\right ) x\right )-x^3+\frac {\log ^2(2)}{x-\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.10, size = 75, normalized size = 3.26 \begin {gather*} -(x-\log (2))^3-\frac {1}{2} (x-\log (2))^2 \log (64)+\frac {\log ^2(2) \left (1+6 \log ^2(2)-\log (2) \log (64)\right )}{x-\log (2)}+(x-\log (2)) \left (-7+e^9-21 \log ^2(2)+\log (8) \log (64)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-7*x^2 + E^9*x^2 - 3*x^4 + (14*x - 2*E^9*x + 6*x^3)*Log[2] + (-8 + E^9 - 3*x^2)*Log[2]^2)/(x^2 - 2*
x*Log[2] + Log[2]^2),x]

[Out]

-(x - Log[2])^3 - ((x - Log[2])^2*Log[64])/2 + (Log[2]^2*(1 + 6*Log[2]^2 - Log[2]*Log[64]))/(x - Log[2]) + (x
- Log[2])*(-7 + E^9 - 21*Log[2]^2 + Log[8]*Log[64])

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fricas [B]  time = 0.54, size = 48, normalized size = 2.09 \begin {gather*} -\frac {x^{4} - x^{2} e^{9} + 7 \, x^{2} - {\left (x^{3} - x e^{9} + 7 \, x\right )} \log \relax (2) - \log \relax (2)^{2}}{x - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(9)-3*x^2-8)*log(2)^2+(-2*x*exp(9)+6*x^3+14*x)*log(2)+x^2*exp(9)-3*x^4-7*x^2)/(log(2)^2-2*x*log
(2)+x^2),x, algorithm="fricas")

[Out]

-(x^4 - x^2*e^9 + 7*x^2 - (x^3 - x*e^9 + 7*x)*log(2) - log(2)^2)/(x - log(2))

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giac [A]  time = 0.12, size = 26, normalized size = 1.13 \begin {gather*} -x^{3} + x e^{9} - 7 \, x + \frac {\log \relax (2)^{2}}{x - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(9)-3*x^2-8)*log(2)^2+(-2*x*exp(9)+6*x^3+14*x)*log(2)+x^2*exp(9)-3*x^4-7*x^2)/(log(2)^2-2*x*log
(2)+x^2),x, algorithm="giac")

[Out]

-x^3 + x*e^9 - 7*x + log(2)^2/(x - log(2))

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maple [A]  time = 0.50, size = 27, normalized size = 1.17

method	result	size
default	\(-x^{3}+x \,{\mathrm e}^{9}-7 x +\frac {\ln \relax (2)^{2}}{x -\ln \relax (2)}\)	\(27\)
risch	\(-x^{3}+x \,{\mathrm e}^{9}-7 x -\frac {\ln \relax (2)^{2}}{\ln \relax (2)-x}\)	\(28\)
norman	\(\frac {x^{4}-x^{3} \ln \relax (2)+\left (-{\mathrm e}^{9}+7\right ) x^{2}+{\mathrm e}^{9} \ln \relax (2)^{2}-8 \ln \relax (2)^{2}}{\ln \relax (2)-x}\)	\(44\)
gosper	\(\frac {-x^{3} \ln \relax (2)+x^{4}+{\mathrm e}^{9} \ln \relax (2)^{2}-x^{2} {\mathrm e}^{9}-8 \ln \relax (2)^{2}+7 x^{2}}{\ln \relax (2)-x}\)	\(46\)
meijerg	\(-\frac {8 x}{1-\frac {x}{\ln \relax (2)}}+\frac {{\mathrm e}^{9} x}{1-\frac {x}{\ln \relax (2)}}-\left (2 \,{\mathrm e}^{9}-14\right ) \ln \relax (2) \left (\frac {x}{\ln \relax (2) \left (1-\frac {x}{\ln \relax (2)}\right )}+\ln \left (1-\frac {x}{\ln \relax (2)}\right )\right )+6 \ln \relax (2)^{3} \left (\frac {x \left (-\frac {2 x^{2}}{\ln \relax (2)^{2}}-\frac {6 x}{\ln \relax (2)}+12\right )}{4 \ln \relax (2) \left (1-\frac {x}{\ln \relax (2)}\right )}+3 \ln \left (1-\frac {x}{\ln \relax (2)}\right )\right )-\left (-3 \ln \relax (2)^{2}+{\mathrm e}^{9}-7\right ) \ln \relax (2) \left (-\frac {x \left (-\frac {3 x}{\ln \relax (2)}+6\right )}{3 \ln \relax (2) \left (1-\frac {x}{\ln \relax (2)}\right )}-2 \ln \left (1-\frac {x}{\ln \relax (2)}\right )\right )+3 \ln \relax (2)^{3} \left (-\frac {x \left (-\frac {5 x^{3}}{\ln \relax (2)^{3}}-\frac {10 x^{2}}{\ln \relax (2)^{2}}-\frac {30 x}{\ln \relax (2)}+60\right )}{15 \ln \relax (2) \left (1-\frac {x}{\ln \relax (2)}\right )}-4 \ln \left (1-\frac {x}{\ln \relax (2)}\right )\right )\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(9)-3*x^2-8)*ln(2)^2+(-2*x*exp(9)+6*x^3+14*x)*ln(2)+x^2*exp(9)-3*x^4-7*x^2)/(ln(2)^2-2*x*ln(2)+x^2),x
,method=_RETURNVERBOSE)

[Out]

-x^3+x*exp(9)-7*x+ln(2)^2/(x-ln(2))

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maxima [A]  time = 0.35, size = 25, normalized size = 1.09 \begin {gather*} -x^{3} + x {\left (e^{9} - 7\right )} + \frac {\log \relax (2)^{2}}{x - \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(9)-3*x^2-8)*log(2)^2+(-2*x*exp(9)+6*x^3+14*x)*log(2)+x^2*exp(9)-3*x^4-7*x^2)/(log(2)^2-2*x*log
(2)+x^2),x, algorithm="maxima")

[Out]

-x^3 + x*(e^9 - 7) + log(2)^2/(x - log(2))

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mupad [B]  time = 0.19, size = 25, normalized size = 1.09 \begin {gather*} \frac {{\ln \relax (2)}^2}{x-\ln \relax (2)}+x\,\left ({\mathrm {e}}^9-7\right )-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2)^2*(3*x^2 - exp(9) + 8) - log(2)*(14*x - 2*x*exp(9) + 6*x^3) - x^2*exp(9) + 7*x^2 + 3*x^4)/(log(2)
^2 - 2*x*log(2) + x^2),x)

[Out]

log(2)^2/(x - log(2)) + x*(exp(9) - 7) - x^3

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sympy [A]  time = 0.15, size = 19, normalized size = 0.83 \begin {gather*} - x^{3} - x \left (7 - e^{9}\right ) + \frac {\log {\relax (2 )}^{2}}{x - \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(9)-3*x**2-8)*ln(2)**2+(-2*x*exp(9)+6*x**3+14*x)*ln(2)+x**2*exp(9)-3*x**4-7*x**2)/(ln(2)**2-2*x
*ln(2)+x**2),x)

[Out]

-x**3 - x*(7 - exp(9)) + log(2)**2/(x - log(2))

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