3.101.95 \(\int \frac {-5-6 x+48 e^2 x^2}{-1-e+\log (5)} \, dx\) [10095]

Optimal. Leaf size=28 \[ \frac {x \left (5-x+4 \left (x-4 e^2 x^2\right )\right )}{1+e-\log (5)} \]

[Out]

(3*x-16*x^2*exp(1)^2+5)/(-ln(5)+1+exp(1))*x

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.57, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12} \begin {gather*} -\frac {16 e^2 x^3}{1+e-\log (5)}+\frac {3 x^2}{1+e-\log (5)}+\frac {5 x}{1+e-\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 6*x + 48*E^2*x^2)/(-1 - E + Log[5]),x]

[Out]

(5*x)/(1 + E - Log[5]) + (3*x^2)/(1 + E - Log[5]) - (16*E^2*x^3)/(1 + E - Log[5])

Rule 12

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-5-6 x+48 e^2 x^2\right ) \, dx}{-1-e+\log (5)}\\ &=\frac {5 x}{1+e-\log (5)}+\frac {3 x^2}{1+e-\log (5)}-\frac {16 e^2 x^3}{1+e-\log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 27, normalized size = 0.96 \begin {gather*} \frac {5 x+3 x^2-16 e^2 x^3}{1+e-\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 6*x + 48*E^2*x^2)/(-1 - E + Log[5]),x]

[Out]

(5*x + 3*x^2 - 16*E^2*x^3)/(1 + E - Log[5])

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Maple [A]
time = 0.03, size = 30, normalized size = 1.07

method	result	size
gosper	\(\frac {x \left (16 x^{2} {\mathrm e}^{2}-3 x -5\right )}{\ln \left (5\right )-{\mathrm e}-1}\)	\(27\)
default	\(\frac {16 x^{3} {\mathrm e}^{2}-3 x^{2}-5 x}{\ln \left (5\right )-{\mathrm e}-1}\)	\(30\)
risch	\(-\frac {5 x}{\ln \left (5\right )-{\mathrm e}-1}-\frac {3 x^{2}}{\ln \left (5\right )-{\mathrm e}-1}+\frac {16 \,{\mathrm e}^{2} x^{3}}{\ln \left (5\right )-{\mathrm e}-1}\)	\(47\)
norman	\(-\frac {5 x}{\ln \left (5\right )-{\mathrm e}-1}-\frac {3 x^{2}}{\ln \left (5\right )-{\mathrm e}-1}+\frac {16 \,{\mathrm e}^{2} x^{3}}{\ln \left (5\right )-{\mathrm e}-1}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x^2*exp(1)^2-6*x-5)/(ln(5)-exp(1)-1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \left (5\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x^2*exp(1)^2-6*x-5)/(log(5)-exp(1)-1),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.54, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \left (5\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x^2*exp(1)^2-6*x-5)/(log(5)-exp(1)-1),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.01, size = 41, normalized size = 1.46 \begin {gather*} - \frac {16 x^{3} e^{2}}{- \log {\left (5 \right )} + 1 + e} + \frac {3 x^{2}}{- \log {\left (5 \right )} + 1 + e} + \frac {5 x}{- \log {\left (5 \right )} + 1 + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x**2*exp(1)**2-6*x-5)/(ln(5)-exp(1)-1),x)

[Out]

-16*x**3*exp(2)/(-log(5) + 1 + E) + 3*x**2/(-log(5) + 1 + E) + 5*x/(-log(5) + 1 + E)

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Giac [A]
time = 0.41, size = 28, normalized size = 1.00 \begin {gather*} -\frac {16 \, x^{3} e^{2} - 3 \, x^{2} - 5 \, x}{e - \log \left (5\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((48*x^2*exp(1)^2-6*x-5)/(log(5)-exp(1)-1),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.34, size = 24, normalized size = 0.86 \begin {gather*} \frac {x\,\left (-16\,{\mathrm {e}}^2\,x^2+3\,x+5\right )}{\mathrm {e}-\ln \left (5\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x - 48*x^2*exp(2) + 5)/(exp(1) - log(5) + 1),x)

[Out]

(x*(3*x - 16*x^2*exp(2) + 5))/(exp(1) - log(5) + 1)

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