3.64.8 \(\int \frac {e^8 (-96-160 x-64 x^2)}{9 x^7} \, dx\)

Optimal. Leaf size=23 \[ \frac {16 e^8 \left (-3+\frac {-1+2 x}{x}\right )^2}{9 x^4} \]

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.35, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 14} \begin {gather*} \frac {16 e^8}{9 x^6}+\frac {32 e^8}{9 x^5}+\frac {16 e^8}{9 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^8*(-96 - 160*x - 64*x^2))/(9*x^7),x]

[Out]

(16*E^8)/(9*x^6) + (32*E^8)/(9*x^5) + (16*E^8)/(9*x^4)

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} &=\frac {1}{9} e^8 \int \frac {-96-160 x-64 x^2}{x^7} \, dx\\ &=\frac {1}{9} e^8 \int \left (-\frac {96}{x^7}-\frac {160}{x^6}-\frac {64}{x^5}\right ) \, dx\\ &=\frac {16 e^8}{9 x^6}+\frac {32 e^8}{9 x^5}+\frac {16 e^8}{9 x^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.17 \begin {gather*} -\frac {32}{9} e^8 \left (-\frac {1}{2 x^6}-\frac {1}{x^5}-\frac {1}{2 x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^8*(-96 - 160*x - 64*x^2))/(9*x^7),x]

[Out]

(-32*E^8*(-1/2*1/x^6 - x^(-5) - 1/(2*x^4)))/9

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fricas [A]  time = 0.69, size = 15, normalized size = 0.65 \begin {gather*} \frac {16 \, {\left (x^{2} + 2 \, x + 1\right )} e^{8}}{9 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x^2-160*x-96)*exp(4)^2/x^7,x, algorithm="fricas")

[Out]

16/9*(x^2 + 2*x + 1)*e^8/x^6

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giac [A]  time = 0.15, size = 15, normalized size = 0.65 \begin {gather*} \frac {16 \, {\left (x^{2} + 2 \, x + 1\right )} e^{8}}{9 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x^2-160*x-96)*exp(4)^2/x^7,x, algorithm="giac")

[Out]

16/9*(x^2 + 2*x + 1)*e^8/x^6

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maple [A]  time = 0.05, size = 18, normalized size = 0.78

method	result	size
gosper	\(\frac {16 \left (x^{2}+2 x +1\right ) {\mathrm e}^{8}}{9 x^{6}}\)	\(18\)
risch	\(\frac {{\mathrm e}^{8} \left (16 x^{2}+32 x +16\right )}{9 x^{6}}\)	\(18\)
default	\(\frac {32 \,{\mathrm e}^{8} \left (\frac {1}{x^{5}}+\frac {1}{2 x^{6}}+\frac {1}{2 x^{4}}\right )}{9}\)	\(21\)
norman	\(\frac {\frac {16 \,{\mathrm e}^{8}}{9}+\frac {32 x \,{\mathrm e}^{8}}{9}+\frac {16 x^{2} {\mathrm e}^{8}}{9}}{x^{6}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-64*x^2-160*x-96)*exp(4)^2/x^7,x,method=_RETURNVERBOSE)

[Out]

16/9*(x^2+2*x+1)*exp(4)^2/x^6

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maxima [A]  time = 0.37, size = 15, normalized size = 0.65 \begin {gather*} \frac {16 \, {\left (x^{2} + 2 \, x + 1\right )} e^{8}}{9 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x^2-160*x-96)*exp(4)^2/x^7,x, algorithm="maxima")

[Out]

16/9*(x^2 + 2*x + 1)*e^8/x^6

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mupad [B]  time = 0.05, size = 22, normalized size = 0.96 \begin {gather*} \frac {16\,{\mathrm {e}}^8\,x^2+32\,{\mathrm {e}}^8\,x+16\,{\mathrm {e}}^8}{9\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(8)*(160*x + 64*x^2 + 96))/(9*x^7),x)

[Out]

(16*exp(8) + 32*x*exp(8) + 16*x^2*exp(8))/(9*x^6)

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sympy [A]  time = 0.17, size = 27, normalized size = 1.17 \begin {gather*} - \frac {- 16 x^{2} e^{8} - 32 x e^{8} - 16 e^{8}}{9 x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-64*x**2-160*x-96)*exp(4)**2/x**7,x)

[Out]

-(-16*x**2*exp(8) - 32*x*exp(8) - 16*exp(8))/(9*x**6)

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