\(\int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx\) [2505]

Optimal result

Integrand size = 15, antiderivative size = 21 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=3-\frac {1}{x}+x^2-2 \left (4+\frac {5 \log (3)}{x}\right ) \]

[Out]

x^2-5-1/x-10*ln(3)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14} \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^2-\frac {1+10 \log (3)}{x} \]

[In]

Int[(1 + 2*x^3 + 10*Log[3])/x^2,x]

[Out]

x^2 - (1 + 10*Log[3])/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (2 x+\frac {1+10 \log (3)}{x^2}\right ) \, dx \\ & = x^2-\frac {1+10 \log (3)}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^2+\frac {-1-10 \log (3)}{x} \]

[In]

Integrate[(1 + 2*x^3 + 10*Log[3])/x^2,x]

[Out]

x^2 + (-1 - 10*Log[3])/x

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
norman	\(\frac {x^{3}-1-10 \ln \left (3\right )}{x}\)	\(14\)
default	\(x^{2}-\frac {10 \ln \left (3\right )+1}{x}\)	\(16\)
gosper	\(-\frac {-x^{3}+10 \ln \left (3\right )+1}{x}\)	\(17\)
risch	\(x^{2}-\frac {10 \ln \left (3\right )}{x}-\frac {1}{x}\)	\(17\)
parallelrisch	\(-\frac {-x^{3}+10 \ln \left (3\right )+1}{x}\)	\(17\)

[In]

int((10*ln(3)+2*x^3+1)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/x*(x^3-1-10*ln(3))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=\frac {x^{3} - 10 \, \log \left (3\right ) - 1}{x} \]

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="fricas")

[Out]

(x^3 - 10*log(3) - 1)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^{2} + \frac {- 10 \log {\left (3 \right )} - 1}{x} \]

[In]

integrate((10*ln(3)+2*x**3+1)/x**2,x)

[Out]

x**2 + (-10*log(3) - 1)/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^{2} - \frac {10 \, \log \left (3\right ) + 1}{x} \]

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="maxima")

[Out]

x^2 - (10*log(3) + 1)/x

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^{2} - \frac {10 \, \log \left (3\right ) + 1}{x} \]

[In]

integrate((10*log(3)+2*x^3+1)/x^2,x, algorithm="giac")

[Out]

x^2 - (10*log(3) + 1)/x

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1+2 x^3+10 \log (3)}{x^2} \, dx=x^2-\frac {10\,\ln \left (3\right )+1}{x} \]

[In]

int((10*log(3) + 2*x^3 + 1)/x^2,x)

[Out]

x^2 - (10*log(3) + 1)/x