\(\int \frac {-7+(-2 x-x^2) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx\) [1404]

Optimal result

Integrand size = 38, antiderivative size = 17 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x+\log (7+x (x \log (2)-\log (\log (375)))) \]

[Out]

x+ln(7+(x*ln(2)-ln(ln(375)))*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1671, 642} \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=\log \left (x^2 \log (2)-x \log (\log (375))+7\right )+x \]

[In]

Int[(-7 + (-2*x - x^2)*Log[2] + (1 + x)*Log[Log[375]])/(-7 - x^2*Log[2] + x*Log[Log[375]]),x]

[Out]

x + Log[7 + x^2*Log[2] - x*Log[Log[375]]]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1671

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {-2 x \log (2)+\log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))}\right ) \, dx \\ & = x+\int \frac {-2 x \log (2)+\log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx \\ & = x+\log \left (7+x^2 \log (2)-x \log (\log (375))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x+\log \left (7+x^2 \log (2)-x \log (\log (375))\right ) \]

[In]

Integrate[(-7 + (-2*x - x^2)*Log[2] + (1 + x)*Log[Log[375]])/(-7 - x^2*Log[2] + x*Log[Log[375]]),x]

[Out]

x + Log[7 + x^2*Log[2] - x*Log[Log[375]]]

Maple [A] (verified)

Time = 1.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
default	\(x +\ln \left (x^{2} \ln \left (2\right )-x \ln \left (\ln \left (375\right )\right )+7\right )\)	\(18\)
norman	\(x +\ln \left (x^{2} \ln \left (2\right )-x \ln \left (\ln \left (375\right )\right )+7\right )\)	\(18\)
risch	\(x +\ln \left (x^{2} \ln \left (2\right )-x \ln \left (\ln \left (3\right )+3 \ln \left (5\right )\right )+7\right )\)	\(23\)
parallelrisch	\(x +\ln \left (\frac {x^{2} \ln \left (2\right )-x \ln \left (\ln \left (375\right )\right )+7}{\ln \left (2\right )}\right )\)	\(23\)

[In]

int(((1+x)*ln(ln(375))+(-x^2-2*x)*ln(2)-7)/(x*ln(ln(375))-x^2*ln(2)-7),x,method=_RETURNVERBOSE)

[Out]

x+ln(x^2*ln(2)-x*ln(ln(375))+7)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x + \log \left (x^{2} \log \left (2\right ) - x \log \left (\log \left (375\right )\right ) + 7\right ) \]

[In]

integrate(((1+x)*log(log(375))+(-x^2-2*x)*log(2)-7)/(x*log(log(375))-x^2*log(2)-7),x, algorithm="fricas")

[Out]

x + log(x^2*log(2) - x*log(log(375)) + 7)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x + \log {\left (x^{2} \log {\left (2 \right )} - x \log {\left (\log {\left (375 \right )} \right )} + 7 \right )} \]

[In]

integrate(((1+x)*ln(ln(375))+(-x**2-2*x)*ln(2)-7)/(x*ln(ln(375))-x**2*ln(2)-7),x)

[Out]

x + log(x**2*log(2) - x*log(log(375)) + 7)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x + \log \left (x^{2} \log \left (2\right ) - x \log \left (\log \left (375\right )\right ) + 7\right ) \]

[In]

integrate(((1+x)*log(log(375))+(-x^2-2*x)*log(2)-7)/(x*log(log(375))-x^2*log(2)-7),x, algorithm="maxima")

[Out]

x + log(x^2*log(2) - x*log(log(375)) + 7)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x + \log \left (x^{2} \log \left (2\right ) - x \log \left (\log \left (375\right )\right ) + 7\right ) \]

[In]

integrate(((1+x)*log(log(375))+(-x^2-2*x)*log(2)-7)/(x*log(log(375))-x^2*log(2)-7),x, algorithm="giac")

[Out]

x + log(x^2*log(2) - x*log(log(375)) + 7)

Mupad [B] (verification not implemented)

Time = 8.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-7+\left (-2 x-x^2\right ) \log (2)+(1+x) \log (\log (375))}{-7-x^2 \log (2)+x \log (\log (375))} \, dx=x+\ln \left (\ln \left (2\right )\,x^2-\ln \left (\ln \left (375\right )\right )\,x+7\right ) \]

[In]

int((log(2)*(2*x + x^2) - log(log(375))*(x + 1) + 7)/(x^2*log(2) - x*log(log(375)) + 7),x)

[Out]

x + log(x^2*log(2) - x*log(log(375)) + 7)