3.59.17 \(\int (1-40 x-15 x^2+60 x^2 \log ^2(3)) \, dx\) [5817]

Optimal. Leaf size=22 \[ -e^3+x-5 x^2 \left (4+x-4 x \log ^2(3)\right ) \]

[Out]

x-5*x^2*(4-4*x*ln(3)^2+x)-exp(3)

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Rubi [A]
time = 0.01, antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6} \begin {gather*} -5 x^3 \left (1-4 \log ^2(3)\right )-20 x^2+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 - 40*x - 15*x^2 + 60*x^2*Log[3]^2,x]

[Out]

x - 20*x^2 - 5*x^3*(1 - 4*Log[3]^2)

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-40 x+x^2 \left (-15+60 \log ^2(3)\right )\right ) \, dx\\ &=x-20 x^2-5 x^3 \left (1-4 \log ^2(3)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 21, normalized size = 0.95 \begin {gather*} x-20 x^2-5 x^3+20 x^3 \log ^2(3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 - 40*x - 15*x^2 + 60*x^2*Log[3]^2,x]

[Out]

x - 20*x^2 - 5*x^3 + 20*x^3*Log[3]^2

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Maple [A]
time = 0.28, size = 22, normalized size = 1.00

method	result	size
norman	\(\left (20 \ln \left (3\right )^{2}-5\right ) x^{3}-20 x^{2}+x\)	\(20\)
gosper	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)
default	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)
risch	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(60*x^2*ln(3)^2-15*x^2-40*x+1,x,method=_RETURNVERBOSE)

[Out]

20*x^3*ln(3)^2-5*x^3-20*x^2+x

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Maxima [A]
time = 0.37, size = 21, normalized size = 0.95 \begin {gather*} 20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(60*x^2*log(3)^2-15*x^2-40*x+1,x, algorithm="maxima")

[Out]

20*x^3*log(3)^2 - 5*x^3 - 20*x^2 + x

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Fricas [A]
time = 0.38, size = 21, normalized size = 0.95 \begin {gather*} 20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(60*x^2*log(3)^2-15*x^2-40*x+1,x, algorithm="fricas")

[Out]

20*x^3*log(3)^2 - 5*x^3 - 20*x^2 + x

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Sympy [A]
time = 0.01, size = 17, normalized size = 0.77 \begin {gather*} x^{3} \left (-5 + 20 \log {\left (3 \right )}^{2}\right ) - 20 x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(60*x**2*ln(3)**2-15*x**2-40*x+1,x)

[Out]

x**3*(-5 + 20*log(3)**2) - 20*x**2 + x

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Giac [A]
time = 0.39, size = 21, normalized size = 0.95 \begin {gather*} 20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(60*x^2*log(3)^2-15*x^2-40*x+1,x, algorithm="giac")

[Out]

20*x^3*log(3)^2 - 5*x^3 - 20*x^2 + x

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Mupad [B]
time = 3.99, size = 19, normalized size = 0.86 \begin {gather*} \left (20\,{\ln \left (3\right )}^2-5\right )\,x^3-20\,x^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(60*x^2*log(3)^2 - 40*x - 15*x^2 + 1,x)

[Out]

x + x^3*(20*log(3)^2 - 5) - 20*x^2

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