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

Optimal. Leaf size=26 \[ x^2+\frac {5 \log \left (\frac {\sqrt {e}}{x}+x+4 x^2\right )}{e^3} \]

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Rubi [A]  time = 0.31, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 1594, 6742, 1587} \begin {gather*} x^2+\frac {5 \log \left (4 x^3+x^2+\sqrt {e}\right )}{e^3}-\frac {5 \log (x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[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 1587

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]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 1594

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.27 \begin {gather*} \frac {e^3 x^2-5 \log (x)+5 \log \left (\sqrt {e}+x^2+4 x^3\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.56, size = 28, normalized size = 1.08 \begin {gather*} {\left (x^{2} e^{3} + 5 \, \log \left (4 \, x^{3} + x^{2} + e^{\frac {1}{2}}\right ) - 5 \, \log \relax (x)\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.14, size = 30, normalized size = 1.15 \begin {gather*} {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 27, normalized size = 1.04

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.36, size = 28, normalized size = 1.08 \begin {gather*} {\left (x^{2} e^{3} + 5 \, \log \left (4 \, x^{3} + x^{2} + e^{\frac {1}{2}}\right ) - 5 \, \log \relax (x)\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.23, size = 28, normalized size = 1.08 \begin {gather*} 5\,{\mathrm {e}}^{-3}\,\ln \left (x^3+\frac {x^2}{4}+\frac {\sqrt {\mathrm {e}}}{4}\right )-5\,{\mathrm {e}}^{-3}\,\ln \relax (x)+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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