3.50.87 \(\int \frac {e^6-4 x^2-2 e^8 x^2-6 x^3+e^4 (6 x^2+3 x^3)}{e^6 x} \, dx\)

Optimal. Leaf size=24 \[ -\frac {\left (-2+e^4\right ) \left (-1+e^4-x\right ) x^2}{e^6}+\log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.50, number of steps used = 4, number of rules used = 3, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 12, 14} \begin {gather*} -\frac {\left (2-e^4\right ) x^3}{e^6}-\frac {\left (2-3 e^4+e^8\right ) x^2}{e^6}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^6 - 4*x^2 - 2*E^8*x^2 - 6*x^3 + E^4*(6*x^2 + 3*x^3))/(E^6*x),x]

[Out]

-(((2 - 3*E^4 + E^8)*x^2)/E^6) - ((2 - E^4)*x^3)/E^6 + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^6+\left (-4-2 e^8\right ) x^2-6 x^3+e^4 \left (6 x^2+3 x^3\right )}{e^6 x} \, dx\\ &=\frac {\int \frac {e^6+\left (-4-2 e^8\right ) x^2-6 x^3+e^4 \left (6 x^2+3 x^3\right )}{x} \, dx}{e^6}\\ &=\frac {\int \left (\frac {e^6}{x}-2 \left (2-3 e^4+e^8\right ) x-3 \left (2-e^4\right ) x^2\right ) \, dx}{e^6}\\ &=-\frac {\left (2-3 e^4+e^8\right ) x^2}{e^6}-\frac {\left (2-e^4\right ) x^3}{e^6}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.75 \begin {gather*} -\frac {2 x^2}{e^6}+\frac {3 x^2}{e^2}-e^2 x^2-\frac {2 x^3}{e^6}+\frac {x^3}{e^2}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^6 - 4*x^2 - 2*E^8*x^2 - 6*x^3 + E^4*(6*x^2 + 3*x^3))/(E^6*x),x]

[Out]

(-2*x^2)/E^6 + (3*x^2)/E^2 - E^2*x^2 - (2*x^3)/E^6 + x^3/E^2 + Log[x]

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fricas [A]  time = 1.01, size = 40, normalized size = 1.67 \begin {gather*} -{\left (2 \, x^{3} + x^{2} e^{8} + 2 \, x^{2} - {\left (x^{3} + 3 \, x^{2}\right )} e^{4} - e^{6} \log \relax (x)\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(4)^2+(3*x^3+6*x^2)*exp(4)+exp(3)^2-6*x^3-4*x^2)/x/exp(3)^2,x, algorithm="fricas")

[Out]

-(2*x^3 + x^2*e^8 + 2*x^2 - (x^3 + 3*x^2)*e^4 - e^6*log(x))*e^(-6)

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giac [A]  time = 0.15, size = 40, normalized size = 1.67 \begin {gather*} {\left (x^{3} e^{4} - 2 \, x^{3} - x^{2} e^{8} + 3 \, x^{2} e^{4} - 2 \, x^{2} + e^{6} \log \left ({\left | x \right |}\right )\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(4)^2+(3*x^3+6*x^2)*exp(4)+exp(3)^2-6*x^3-4*x^2)/x/exp(3)^2,x, algorithm="giac")

[Out]

(x^3*e^4 - 2*x^3 - x^2*e^8 + 3*x^2*e^4 - 2*x^2 + e^6*log(abs(x)))*e^(-6)

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maple [A]  time = 0.06, size = 38, normalized size = 1.58

method	result	size
risch	\(-x^{2} {\mathrm e}^{2}+x^{3} {\mathrm e}^{-2}+3 x^{2} {\mathrm e}^{-2}-2 \,{\mathrm e}^{-6} x^{3}-2 \,{\mathrm e}^{-6} x^{2}+\ln \relax (x )\)	\(38\)
norman	\(\left (\left ({\mathrm e}^{4}-2\right ) {\mathrm e}^{-3} x^{3}-\left ({\mathrm e}^{8}-3 \,{\mathrm e}^{4}+2\right ) {\mathrm e}^{-3} x^{2}\right ) {\mathrm e}^{-3}+\ln \relax (x )\)	\(41\)
default	\({\mathrm e}^{-6} \left (x^{3} {\mathrm e}^{4}+3 x^{2} {\mathrm e}^{4}-x^{2} {\mathrm e}^{8}-2 x^{3}-2 x^{2}+{\mathrm e}^{6} \ln \relax (x )\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x^2*exp(4)^2+(3*x^3+6*x^2)*exp(4)+exp(3)^2-6*x^3-4*x^2)/x/exp(3)^2,x,method=_RETURNVERBOSE)

[Out]

-x^2*exp(2)+x^3*exp(-2)+3*x^2*exp(-2)-2*exp(-6)*x^3-2*exp(-6)*x^2+ln(x)

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maxima [A]  time = 0.35, size = 30, normalized size = 1.25 \begin {gather*} {\left (x^{3} {\left (e^{4} - 2\right )} - x^{2} {\left (e^{8} - 3 \, e^{4} + 2\right )} + e^{6} \log \relax (x)\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x^2*exp(4)^2+(3*x^3+6*x^2)*exp(4)+exp(3)^2-6*x^3-4*x^2)/x/exp(3)^2,x, algorithm="maxima")

[Out]

(x^3*(e^4 - 2) - x^2*(e^8 - 3*e^4 + 2) + e^6*log(x))*e^(-6)

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mupad [B]  time = 0.09, size = 33, normalized size = 1.38 \begin {gather*} \ln \relax (x)-\frac {x^2\,{\mathrm {e}}^{-6}\,\left (2\,{\mathrm {e}}^8-6\,{\mathrm {e}}^4+4\right )}{2}+\frac {x^3\,{\mathrm {e}}^{-6}\,\left (3\,{\mathrm {e}}^4-6\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-6)*(2*x^2*exp(8) - exp(4)*(6*x^2 + 3*x^3) - exp(6) + 4*x^2 + 6*x^3))/x,x)

[Out]

log(x) - (x^2*exp(-6)*(2*exp(8) - 6*exp(4) + 4))/2 + (x^3*exp(-6)*(3*exp(4) - 6))/3

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sympy [A]  time = 0.12, size = 31, normalized size = 1.29 \begin {gather*} \frac {- x^{3} \left (2 - e^{4}\right ) - x^{2} \left (- 3 e^{4} + 2 + e^{8}\right ) + e^{6} \log {\relax (x )}}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x**2*exp(4)**2+(3*x**3+6*x**2)*exp(4)+exp(3)**2-6*x**3-4*x**2)/x/exp(3)**2,x)

[Out]

(-x**3*(2 - exp(4)) - x**2*(-3*exp(4) + 2 + exp(8)) + exp(6)*log(x))*exp(-6)

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