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

Optimal. Leaf size=26 \[ \frac {1}{2} \left (3-x-5 x^2 (3-x+5 x (1+\log (2)))\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6, 12} \begin {gather*} -\frac {5}{2} x^3 (4+\log (32))-\frac {15 x^2}{2}-\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{2} \left (-x-15 x^2-5 x^3 (4+\log (32))\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.57, size = 21, normalized size = 0.81 \begin {gather*} -\frac {25}{2} \, x^{3} \log \relax (2) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.37, size = 21, normalized size = 0.81 \begin {gather*} -\frac {25}{2} \, x^{3} \log \relax (2) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 20, normalized size = 0.77

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.71, size = 21, normalized size = 0.81 \begin {gather*} -\frac {25}{2} \, x^{3} \log \relax (2) - 10 \, x^{3} - \frac {15}{2} \, x^{2} - \frac {1}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.03, size = 20, normalized size = 0.77 \begin {gather*} \left (-\frac {25\,\ln \relax (2)}{2}-10\right )\,x^3-\frac {15\,x^2}{2}-\frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.06, size = 22, normalized size = 0.85 \begin {gather*} x^{3} \left (-10 - \frac {25 \log {\relax (2 )}}{2}\right ) - \frac {15 x^{2}}{2} - \frac {x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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