\(\int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+(10 x+2 x^4) \log (5)+\log ^2(5)} \, dx\) [1356]

Optimal result

Integrand size = 40, antiderivative size = 19 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=1+\frac {1}{x+\frac {1}{3} \left (2 x+x^4+\log (5)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2099, 2124} \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^4+5 x+\log (5)} \]

[In]

Int[(-15 - 12*x^3)/(25*x^2 + 10*x^5 + x^8 + (10*x + 2*x^4)*Log[5] + Log[5]^2),x]

[Out]

3/(5*x + x^4 + Log[5])

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 2124

Int[(Pm_)*(Qn_)^(p_), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[Coeff[Pm, x, m]*(Qn^(p + 1)
/(n*(p + 1)*Coeff[Qn, x, n])), x] + Dist[Simplify[Pm - Coeff[Pm, x, m]*(D[Qn, x]/(n*Coeff[Qn, x, n]))], Int[Qn
^p, x], x] /; EqQ[m, n - 1] && EqQ[D[Simplify[Pm - (Coeff[Pm, x, m]/(n*Coeff[Qn, x, n]))*D[Qn, x]], x], 0]] /;
FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {15}{\left (5 x+x^4+\log (5)\right )^2}-\frac {12 x^3}{\left (5 x+x^4+\log (5)\right )^2}\right ) \, dx \\ & = -\left (12 \int \frac {x^3}{\left (5 x+x^4+\log (5)\right )^2} \, dx\right )-15 \int \frac {1}{\left (5 x+x^4+\log (5)\right )^2} \, dx \\ & = \frac {3}{5 x+x^4+\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{5 x+x^4+\log (5)} \]

[In]

Integrate[(-15 - 12*x^3)/(25*x^2 + 10*x^5 + x^8 + (10*x + 2*x^4)*Log[5] + Log[5]^2),x]

[Out]

3/(5*x + x^4 + Log[5])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
gosper	\(\frac {3}{x^{4}+\ln \left (5\right )+5 x}\)	\(14\)
default	\(\frac {3}{x^{4}+\ln \left (5\right )+5 x}\)	\(14\)
norman	\(\frac {3}{x^{4}+\ln \left (5\right )+5 x}\)	\(14\)
risch	\(\frac {3}{x^{4}+\ln \left (5\right )+5 x}\)	\(14\)
parallelrisch	\(\frac {3}{x^{4}+\ln \left (5\right )+5 x}\)	\(14\)

[In]

int((-12*x^3-15)/(ln(5)^2+(2*x^4+10*x)*ln(5)+x^8+10*x^5+25*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^{4} + 5 \, x + \log \left (5\right )} \]

[In]

integrate((-12*x^3-15)/(log(5)^2+(2*x^4+10*x)*log(5)+x^8+10*x^5+25*x^2),x, algorithm="fricas")

[Out]

3/(x^4 + 5*x + log(5))

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^{4} + 5 x + \log {\left (5 \right )}} \]

[In]

integrate((-12*x**3-15)/(ln(5)**2+(2*x**4+10*x)*ln(5)+x**8+10*x**5+25*x**2),x)

[Out]

3/(x**4 + 5*x + log(5))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^{4} + 5 \, x + \log \left (5\right )} \]

[In]

integrate((-12*x^3-15)/(log(5)^2+(2*x^4+10*x)*log(5)+x^8+10*x^5+25*x^2),x, algorithm="maxima")

[Out]

3/(x^4 + 5*x + log(5))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^{4} + 5 \, x + \log \left (5\right )} \]

[In]

integrate((-12*x^3-15)/(log(5)^2+(2*x^4+10*x)*log(5)+x^8+10*x^5+25*x^2),x, algorithm="giac")

[Out]

3/(x^4 + 5*x + log(5))

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {-15-12 x^3}{25 x^2+10 x^5+x^8+\left (10 x+2 x^4\right ) \log (5)+\log ^2(5)} \, dx=\frac {3}{x^4+5\,x+\ln \left (5\right )} \]

[In]

int(-(12*x^3 + 15)/(log(5)*(10*x + 2*x^4) + log(5)^2 + 25*x^2 + 10*x^5 + x^8),x)

[Out]

3/(5*x + log(5) + x^4)