\(\int \frac {1}{2} (-1-30 x-60 x^2-75 x^2 \log (2)) \, dx\) [993]

Optimal result

Integrand size = 21, antiderivative size = 26 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=\frac {1}{2} \left (3-x-5 x^2 (3-x+5 x (1+\log (2)))\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6, 12} \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=-\frac {5}{2} x^3 (4+\log (32))-\frac {15 x^2}{2}-\frac {x}{2} \]

[In]

Int[(-1 - 30*x - 60*x^2 - 75*x^2*Log[2])/2,x]

[Out]

-1/2*x - (15*x^2)/2 - (5*x^3*(4 + Log[32]))/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]

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{align*} \text {integral}& = \int \frac {1}{2} \left (-1-30 x+x^2 (-60-75 \log (2))\right ) \, dx \\ & = \frac {1}{2} \int \left (-1-30 x+x^2 (-60-75 \log (2))\right ) \, dx \\ & = -\frac {x}{2}-\frac {15 x^2}{2}-\frac {5}{2} x^3 (4+\log (32)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=\frac {1}{2} \left (-x-15 x^2-5 x^3 (4+\log (32))\right ) \]

[In]

Integrate[(-1 - 30*x - 60*x^2 - 75*x^2*Log[2])/2,x]

[Out]

(-x - 15*x^2 - 5*x^3*(4 + Log[32]))/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77

method	result	size
norman	\(\left (-\frac {25 \ln \left (2\right )}{2}-10\right ) x^{3}-\frac {x}{2}-\frac {15 x^{2}}{2}\)	\(20\)
gosper	\(-\frac {x \left (25 x^{2} \ln \left (2\right )+20 x^{2}+15 x +1\right )}{2}\)	\(21\)
default	\(-\frac {25 x^{3} \ln \left (2\right )}{2}-10 x^{3}-\frac {15 x^{2}}{2}-\frac {x}{2}\)	\(22\)
risch	\(-\frac {25 x^{3} \ln \left (2\right )}{2}-10 x^{3}-\frac {15 x^{2}}{2}-\frac {x}{2}\)	\(22\)
parallelrisch	\(-\frac {25 x^{3} \ln \left (2\right )}{2}-10 x^{3}-\frac {15 x^{2}}{2}-\frac {x}{2}\)	\(22\)
parts	\(-\frac {25 x^{3} \ln \left (2\right )}{2}-10 x^{3}-\frac {15 x^{2}}{2}-\frac {x}{2}\)	\(22\)

[In]

int(-75/2*x^2*ln(2)-30*x^2-15*x-1/2,x,method=_RETURNVERBOSE)

[Out]

(-25/2*ln(2)-10)*x^3-1/2*x-15/2*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=-\frac {25}{2} \, x^{3} \log \left (2\right ) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \]

[In]

integrate(-75/2*x^2*log(2)-30*x^2-15*x-1/2,x, algorithm="fricas")

[Out]

-25/2*x^3*log(2) - 10*x^3 - 15/2*x^2 - 1/2*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=x^{3} \left (-10 - \frac {25 \log {\left (2 \right )}}{2}\right ) - \frac {15 x^{2}}{2} - \frac {x}{2} \]

[In]

integrate(-75/2*x**2*ln(2)-30*x**2-15*x-1/2,x)

[Out]

x**3*(-10 - 25*log(2)/2) - 15*x**2/2 - x/2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=-\frac {25}{2} \, x^{3} \log \left (2\right ) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \]

[In]

integrate(-75/2*x^2*log(2)-30*x^2-15*x-1/2,x, algorithm="maxima")

[Out]

-25/2*x^3*log(2) - 10*x^3 - 15/2*x^2 - 1/2*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=-\frac {25}{2} \, x^{3} \log \left (2\right ) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \]

[In]

integrate(-75/2*x^2*log(2)-30*x^2-15*x-1/2,x, algorithm="giac")

[Out]

-25/2*x^3*log(2) - 10*x^3 - 15/2*x^2 - 1/2*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1}{2} \left (-1-30 x-60 x^2-75 x^2 \log (2)\right ) \, dx=\left (-\frac {25\,\ln \left (2\right )}{2}-10\right )\,x^3-\frac {15\,x^2}{2}-\frac {x}{2} \]

[In]

int(- 15*x - (75*x^2*log(2))/2 - 30*x^2 - 1/2,x)

[Out]

- x/2 - x^3*((25*log(2))/2 + 10) - (15*x^2)/2