3.1.26 \(\int -\frac {50 x^4}{-10-e+e^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 x^5}{2+\frac {1}{5} \left (e-e^2\right )} \]

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Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 30} \begin {gather*} \frac {10 x^5}{10+e-e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*x^4)/(-10 - E + E^2),x]

[Out]

(10*x^5)/(10 + E - E^2)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {50 \int x^4 \, dx}{10+e-e^2}\\ &=\frac {10 x^5}{10+e-e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.75 \begin {gather*} -\frac {10 x^5}{-10-e+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x^4)/(-10 - E + E^2),x]

[Out]

(-10*x^5)/(-10 - E + E^2)

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fricas [A]  time = 0.78, size = 15, normalized size = 0.75 \begin {gather*} -\frac {10 \, x^{5}}{e^{2} - e - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x^4/(exp(2)-exp(1)-10),x, algorithm="fricas")

[Out]

-10*x^5/(e^2 - e - 10)

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giac [A]  time = 0.35, size = 15, normalized size = 0.75 \begin {gather*} -\frac {10 \, x^{5}}{e^{2} - e - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x^4/(exp(2)-exp(1)-10),x, algorithm="giac")

[Out]

-10*x^5/(e^2 - e - 10)

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maple [A]  time = 0.04, size = 16, normalized size = 0.80

method	result	size
gosper	\(-\frac {10 x^{5}}{{\mathrm e}^{2}-{\mathrm e}-10}\)	\(16\)
default	\(-\frac {10 x^{5}}{{\mathrm e}^{2}-{\mathrm e}-10}\)	\(16\)
norman	\(-\frac {10 x^{5}}{{\mathrm e}^{2}-{\mathrm e}-10}\)	\(16\)
risch	\(-\frac {10 x^{5}}{{\mathrm e}^{2}-{\mathrm e}-10}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-50*x^4/(exp(2)-exp(1)-10),x,method=_RETURNVERBOSE)

[Out]

-10*x^5/(exp(2)-exp(1)-10)

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maxima [A]  time = 0.54, size = 15, normalized size = 0.75 \begin {gather*} -\frac {10 \, x^{5}}{e^{2} - e - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x^4/(exp(2)-exp(1)-10),x, algorithm="maxima")

[Out]

-10*x^5/(e^2 - e - 10)

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mupad [B]  time = 0.19, size = 15, normalized size = 0.75 \begin {gather*} \frac {10\,x^5}{\mathrm {e}-{\mathrm {e}}^2+10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x^4)/(exp(1) - exp(2) + 10),x)

[Out]

(10*x^5)/(exp(1) - exp(2) + 10)

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sympy [A]  time = 0.05, size = 14, normalized size = 0.70 \begin {gather*} - \frac {10 x^{5}}{-10 - e + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-50*x**4/(exp(2)-exp(1)-10),x)

[Out]

-10*x**5/(-10 - E + exp(2))

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