\(\int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 (2 \sqrt {e} x^2+2 x^4+8 x^5)}{e^3 (\sqrt {e} x+x^3+4 x^4)} \, dx\) [7821]

Optimal result

Integrand size = 65, antiderivative size = 26 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx=x^2+\frac {5 \log \left (\frac {\sqrt {e}}{x}+x+4 x^2\right )}{e^3} \]

[Out]

x^2+5/exp(3)*ln(exp(1/2)/x+4*x^2+x)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 1608, 6874, 1601} \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx=x^2+\frac {5 \log \left (4 x^3+x^2+\sqrt {e}\right )}{e^3}-\frac {5 \log (x)}{e^3} \]

[In]

Int[(-5*Sqrt[E] + 5*x^2 + 40*x^3 + E^3*(2*Sqrt[E]*x^2 + 2*x^4 + 8*x^5))/(E^3*(Sqrt[E]*x + x^3 + 4*x^4)),x]

[Out]

x^2 - (5*Log[x])/E^3 + (5*Log[Sqrt[E] + x^2 + 4*x^3])/E^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{\sqrt {e} x+x^3+4 x^4} \, dx}{e^3} \\ & = \frac {\int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{x \left (\sqrt {e}+x^2+4 x^3\right )} \, dx}{e^3} \\ & = \frac {\int \left (-\frac {5}{x}+2 e^3 x+\frac {10 x (1+6 x)}{\sqrt {e}+x^2+4 x^3}\right ) \, dx}{e^3} \\ & = x^2-\frac {5 \log (x)}{e^3}+\frac {10 \int \frac {x (1+6 x)}{\sqrt {e}+x^2+4 x^3} \, dx}{e^3} \\ & = x^2-\frac {5 \log (x)}{e^3}+\frac {5 \log \left (\sqrt {e}+x^2+4 x^3\right )}{e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx=\frac {e^3 x^2-5 \log (x)+5 \log \left (\sqrt {e}+x^2+4 x^3\right )}{e^3} \]

[In]

Integrate[(-5*Sqrt[E] + 5*x^2 + 40*x^3 + E^3*(2*Sqrt[E]*x^2 + 2*x^4 + 8*x^5))/(E^3*(Sqrt[E]*x + x^3 + 4*x^4)),
x]

[Out]

(E^3*x^2 - 5*Log[x] + 5*Log[Sqrt[E] + x^2 + 4*x^3])/E^3

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04

method	result	size
risch	\(x^{2}-5 \,{\mathrm e}^{-3} \ln \left (x \right )+5 \,{\mathrm e}^{-3} \ln \left (4 x^{3}+x^{2}+{\mathrm e}^{\frac {1}{2}}\right )\)	\(27\)
default	\({\mathrm e}^{-3} \left (x^{2} {\mathrm e}^{3}-5 \ln \left (x \right )+5 \ln \left (4 x^{3}+x^{2}+{\mathrm e}^{\frac {1}{2}}\right )\right )\)	\(31\)
norman	\(x^{2}-5 \,{\mathrm e}^{-3} \ln \left (x \right )+5 \,{\mathrm e}^{-3} \ln \left (4 x^{3}+x^{2}+{\mathrm e}^{\frac {1}{2}}\right )\)	\(31\)
parallelrisch	\({\mathrm e}^{-3} \left (x^{2} {\mathrm e}^{3}-5 \ln \left (x \right )+5 \ln \left (x^{3}+\frac {x^{2}}{4}+\frac {{\mathrm e}^{\frac {1}{2}}}{4}\right )\right )\)	\(33\)

[In]

int(((2*x^2*exp(1/2)+8*x^5+2*x^4)*exp(3)-5*exp(1/2)+40*x^3+5*x^2)/(x*exp(1/2)+4*x^4+x^3)/exp(3),x,method=_RETU
RNVERBOSE)

[Out]

x^2-5*exp(-3)*ln(x)+5*exp(-3)*ln(4*x^3+x^2+exp(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx={\left (x^{2} e^{3} + 5 \, \log \left (4 \, x^{3} + x^{2} + e^{\frac {1}{2}}\right ) - 5 \, \log \left (x\right )\right )} e^{\left (-3\right )} \]

[In]

integrate(((2*x^2*exp(1/2)+8*x^5+2*x^4)*exp(3)-5*exp(1/2)+40*x^3+5*x^2)/(x*exp(1/2)+4*x^4+x^3)/exp(3),x, algor
ithm="fricas")

[Out]

(x^2*e^3 + 5*log(4*x^3 + x^2 + e^(1/2)) - 5*log(x))*e^(-3)

Sympy [F(-1)]

Timed out. \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx=\text {Timed out} \]

[In]

integrate(((2*x**2*exp(1/2)+8*x**5+2*x**4)*exp(3)-5*exp(1/2)+40*x**3+5*x**2)/(x*exp(1/2)+4*x**4+x**3)/exp(3),x
)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx={\left (x^{2} e^{3} + 5 \, \log \left (4 \, x^{3} + x^{2} + e^{\frac {1}{2}}\right ) - 5 \, \log \left (x\right )\right )} e^{\left (-3\right )} \]

[In]

integrate(((2*x^2*exp(1/2)+8*x^5+2*x^4)*exp(3)-5*exp(1/2)+40*x^3+5*x^2)/(x*exp(1/2)+4*x^4+x^3)/exp(3),x, algor
ithm="maxima")

[Out]

(x^2*e^3 + 5*log(4*x^3 + x^2 + e^(1/2)) - 5*log(x))*e^(-3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx={\left (x^{2} e^{3} + 5 \, \log \left ({\left | 4 \, x^{3} + x^{2} + e^{\frac {1}{2}} \right |}\right ) - 5 \, \log \left ({\left | x \right |}\right )\right )} e^{\left (-3\right )} \]

[In]

integrate(((2*x^2*exp(1/2)+8*x^5+2*x^4)*exp(3)-5*exp(1/2)+40*x^3+5*x^2)/(x*exp(1/2)+4*x^4+x^3)/exp(3),x, algor
ithm="giac")

[Out]

(x^2*e^3 + 5*log(abs(4*x^3 + x^2 + e^(1/2))) - 5*log(abs(x)))*e^(-3)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {-5 \sqrt {e}+5 x^2+40 x^3+e^3 \left (2 \sqrt {e} x^2+2 x^4+8 x^5\right )}{e^3 \left (\sqrt {e} x+x^3+4 x^4\right )} \, dx=5\,{\mathrm {e}}^{-3}\,\ln \left (x^3+\frac {x^2}{4}+\frac {\sqrt {\mathrm {e}}}{4}\right )-5\,{\mathrm {e}}^{-3}\,\ln \left (x\right )+x^2 \]

[In]

int((exp(-3)*(exp(3)*(2*x^2*exp(1/2) + 2*x^4 + 8*x^5) - 5*exp(1/2) + 5*x^2 + 40*x^3))/(x*exp(1/2) + x^3 + 4*x^
4),x)

[Out]

5*exp(-3)*log(exp(1/2)/4 + x^2/4 + x^3) - 5*exp(-3)*log(x) + x^2