\(\int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+(2000+800 x^2+80 x^4) \log (2)}{25+10 x^2+x^4} \, dx\) [2766]

Optimal result

Integrand size = 51, antiderivative size = 23 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=x-\frac {x}{5+x^2}-20 x (-x-4 \log (2)) \]

[Out]

x-20*(-4*ln(2)-x)*x-x/(x^2+5)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {28, 1828, 1600} \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=20 x^2-\frac {x}{x^2+5}+x (1+80 \log (2)) \]

[In]

Int[(20 + 1000*x + 11*x^2 + 400*x^3 + x^4 + 40*x^5 + (2000 + 800*x^2 + 80*x^4)*Log[2])/(25 + 10*x^2 + x^4),x]

[Out]

20*x^2 - x/(5 + x^2) + x*(1 + 80*Log[2])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{\left (5+x^2\right )^2} \, dx \\ & = -\frac {x}{5+x^2}-\frac {1}{10} \int \frac {-2000 x-400 x^3-50 (1+80 \log (2))-10 x^2 (1+80 \log (2))}{5+x^2} \, dx \\ & = -\frac {x}{5+x^2}-\frac {1}{10} \int (-10-400 x-800 \log (2)) \, dx \\ & = 20 x^2-\frac {x}{5+x^2}+x (1+80 \log (2)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=x \left (1+20 x-\frac {1}{5+x^2}+80 \log (2)\right ) \]

[In]

Integrate[(20 + 1000*x + 11*x^2 + 400*x^3 + x^4 + 40*x^5 + (2000 + 800*x^2 + 80*x^4)*Log[2])/(25 + 10*x^2 + x^
4),x]

[Out]

x*(1 + 20*x - (5 + x^2)^(-1) + 80*Log[2])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
default	\(20 x^{2}+80 x \ln \left (2\right )+x -\frac {x}{x^{2}+5}\)	\(23\)
risch	\(20 x^{2}+80 x \ln \left (2\right )+x -\frac {x}{x^{2}+5}\)	\(23\)
gosper	\(\frac {80 x^{3} \ln \left (2\right )+20 x^{4}+x^{3}+400 x \ln \left (2\right )+4 x -500}{x^{2}+5}\)	\(34\)
norman	\(\frac {20 x^{4}+\left (80 \ln \left (2\right )+1\right ) x^{3}-500+\left (400 \ln \left (2\right )+4\right ) x}{x^{2}+5}\)	\(34\)
parallelrisch	\(\frac {80 x^{3} \ln \left (2\right )+20 x^{4}+x^{3}+400 x \ln \left (2\right )+4 x -500}{x^{2}+5}\)	\(34\)

[In]

int(((80*x^4+800*x^2+2000)*ln(2)+40*x^5+x^4+400*x^3+11*x^2+1000*x+20)/(x^4+10*x^2+25),x,method=_RETURNVERBOSE)

[Out]

20*x^2+80*x*ln(2)+x-x/(x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=\frac {20 \, x^{4} + x^{3} + 100 \, x^{2} + 80 \, {\left (x^{3} + 5 \, x\right )} \log \left (2\right ) + 4 \, x}{x^{2} + 5} \]

[In]

integrate(((80*x^4+800*x^2+2000)*log(2)+40*x^5+x^4+400*x^3+11*x^2+1000*x+20)/(x^4+10*x^2+25),x, algorithm="fri
cas")

[Out]

(20*x^4 + x^3 + 100*x^2 + 80*(x^3 + 5*x)*log(2) + 4*x)/(x^2 + 5)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=20 x^{2} + x \left (1 + 80 \log {\left (2 \right )}\right ) - \frac {x}{x^{2} + 5} \]

[In]

integrate(((80*x**4+800*x**2+2000)*ln(2)+40*x**5+x**4+400*x**3+11*x**2+1000*x+20)/(x**4+10*x**2+25),x)

[Out]

20*x**2 + x*(1 + 80*log(2)) - x/(x**2 + 5)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=20 \, x^{2} + x {\left (80 \, \log \left (2\right ) + 1\right )} - \frac {x}{x^{2} + 5} \]

[In]

integrate(((80*x^4+800*x^2+2000)*log(2)+40*x^5+x^4+400*x^3+11*x^2+1000*x+20)/(x^4+10*x^2+25),x, algorithm="max
ima")

[Out]

20*x^2 + x*(80*log(2) + 1) - x/(x^2 + 5)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=20 \, x^{2} + 80 \, x \log \left (2\right ) + x - \frac {x}{x^{2} + 5} \]

[In]

integrate(((80*x^4+800*x^2+2000)*log(2)+40*x^5+x^4+400*x^3+11*x^2+1000*x+20)/(x^4+10*x^2+25),x, algorithm="gia
c")

[Out]

20*x^2 + 80*x*log(2) + x - x/(x^2 + 5)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {20+1000 x+11 x^2+400 x^3+x^4+40 x^5+\left (2000+800 x^2+80 x^4\right ) \log (2)}{25+10 x^2+x^4} \, dx=x\,\left (80\,\ln \left (2\right )+1\right )-\frac {x}{x^2+5}+20\,x^2 \]

[In]

int((1000*x + log(2)*(800*x^2 + 80*x^4 + 2000) + 11*x^2 + 400*x^3 + x^4 + 40*x^5 + 20)/(10*x^2 + x^4 + 25),x)

[Out]

x*(80*log(2) + 1) - x/(x^2 + 5) + 20*x^2