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

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=-e^3+x-5 x^2 \left (4+x-4 x \log ^2(3)\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {6} \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=-5 x^3 \left (1-4 \log ^2(3)\right )-20 x^2+x \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=x-20 x^2-5 x^3+20 x^3 \log ^2(3) \]

[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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

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	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)
parallelrisch	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)
parts	\(20 x^{3} \ln \left (3\right )^{2}-5 x^{3}-20 x^{2}+x\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \]

[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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=x^{3} \left (-5 + 20 \log {\left (3 \right )}^{2}\right ) - 20 x^{2} + x \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=20 \, x^{3} \log \left (3\right )^{2} - 5 \, x^{3} - 20 \, x^{2} + x \]

[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

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \left (1-40 x-15 x^2+60 x^2 \log ^2(3)\right ) \, dx=\left (20\,{\ln \left (3\right )}^2-5\right )\,x^3-20\,x^2+x \]

[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