\(\int \frac {5-10 x^2+(10 x^2+10 x^3) \log (3)}{x^2} \, dx\) [5571]

Optimal result

Integrand size = 25, antiderivative size = 20 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 \left (4-\frac {1}{x}-2 x+(1+x)^2 \log (3)\right ) \]

[Out]

5*(1+x)^2*ln(3)+20-10*x-5/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {14} \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 x^2 \log (3)-\frac {5}{x}-10 x (1-\log (3)) \]

[In]

Int[(5 - 10*x^2 + (10*x^2 + 10*x^3)*Log[3])/x^2,x]

[Out]

-5/x - 10*x*(1 - Log[3]) + 5*x^2*Log[3]

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 (\frac {5}{x^2}+10 (-1+\log (3))+10 x \log (3)\right ) \, dx \\ & = -\frac {5}{x}-10 x (1-\log (3))+5 x^2 \log (3) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 \left (-\frac {1}{x}-2 x+x \log (9)+\frac {1}{2} x^2 \log (9)\right ) \]

[In]

Integrate[(5 - 10*x^2 + (10*x^2 + 10*x^3)*Log[3])/x^2,x]

[Out]

5*(-x^(-1) - 2*x + x*Log[9] + (x^2*Log[9])/2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
default	\(5 x^{2} \ln \left (3\right )+10 x \ln \left (3\right )-10 x -\frac {5}{x}\)	\(22\)
risch	\(5 x^{2} \ln \left (3\right )+10 x \ln \left (3\right )-10 x -\frac {5}{x}\)	\(22\)
norman	\(\frac {-5+\left (10 \ln \left (3\right )-10\right ) x^{2}+5 x^{3} \ln \left (3\right )}{x}\)	\(24\)
gosper	\(\frac {5 x^{3} \ln \left (3\right )+10 x^{2} \ln \left (3\right )-10 x^{2}-5}{x}\)	\(26\)
parallelrisch	\(\frac {5 x^{3} \ln \left (3\right )+10 x^{2} \ln \left (3\right )-10 x^{2}-5}{x}\)	\(26\)

[In]

int(((10*x^3+10*x^2)*ln(3)-10*x^2+5)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=-\frac {5 \, {\left (2 \, x^{2} - {\left (x^{3} + 2 \, x^{2}\right )} \log \left (3\right ) + 1\right )}}{x} \]

[In]

integrate(((10*x^3+10*x^2)*log(3)-10*x^2+5)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 x^{2} \log {\left (3 \right )} + x \left (-10 + 10 \log {\left (3 \right )}\right ) - \frac {5}{x} \]

[In]

integrate(((10*x**3+10*x**2)*ln(3)-10*x**2+5)/x**2,x)

[Out]

5*x**2*log(3) + x*(-10 + 10*log(3)) - 5/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 \, x^{2} \log \left (3\right ) + 10 \, x {\left (\log \left (3\right ) - 1\right )} - \frac {5}{x} \]

[In]

integrate(((10*x^3+10*x^2)*log(3)-10*x^2+5)/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=5 \, x^{2} \log \left (3\right ) + 10 \, x \log \left (3\right ) - 10 \, x - \frac {5}{x} \]

[In]

integrate(((10*x^3+10*x^2)*log(3)-10*x^2+5)/x^2,x, algorithm="giac")

[Out]

5*x^2*log(3) + 10*x*log(3) - 10*x - 5/x

Mupad [B] (verification not implemented)

Time = 10.85 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {5-10 x^2+\left (10 x^2+10 x^3\right ) \log (3)}{x^2} \, dx=x\,\left (10\,\ln \left (3\right )-10\right )+5\,x^2\,\ln \left (3\right )-\frac {5}{x} \]

[In]

int((log(3)*(10*x^2 + 10*x^3) - 10*x^2 + 5)/x^2,x)

[Out]

x*(10*log(3) - 10) + 5*x^2*log(3) - 5/x