\(\int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx\) [7850]

Optimal result

Integrand size = 26, antiderivative size = 19 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {1}{3} x \left (\frac {27}{5}+\frac {8}{x^2}+2 x+\log (4)\right ) \]

[Out]

1/3*x*(2*ln(2)+27/5+2*x+8/x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6, 12, 14} \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {2 x^2}{3}+\frac {8}{3 x}+\frac {1}{15} x (27+5 \log (4)) \]

[In]

Int[(-40 + 27*x^2 + 20*x^3 + 5*x^2*Log[4])/(15*x^2),x]

[Out]

8/(3*x) + (2*x^2)/3 + (x*(27 + 5*Log[4]))/15

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]]

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 \frac {-40+20 x^3+x^2 (27+5 \log (4))}{15 x^2} \, dx \\ & = \frac {1}{15} \int \frac {-40+20 x^3+x^2 (27+5 \log (4))}{x^2} \, dx \\ & = \frac {1}{15} \int \left (-\frac {40}{x^2}+20 x+27 \left (1+\frac {5 \log (4)}{27}\right )\right ) \, dx \\ & = \frac {8}{3 x}+\frac {2 x^2}{3}+\frac {1}{15} x (27+5 \log (4)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {8}{3 x}+\frac {9 x}{5}+\frac {2 x^2}{3}+\frac {2}{3} x \log (2) \]

[In]

Integrate[(-40 + 27*x^2 + 20*x^3 + 5*x^2*Log[4])/(15*x^2),x]

[Out]

8/(3*x) + (9*x)/5 + (2*x^2)/3 + (2*x*Log[2])/3

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {2 x^{2}}{3}+\frac {2 x \ln \left (2\right )}{3}+\frac {9 x}{5}+\frac {8}{3 x}\)	\(20\)
risch	\(\frac {2 x^{2}}{3}+\frac {2 x \ln \left (2\right )}{3}+\frac {9 x}{5}+\frac {8}{3 x}\)	\(20\)
norman	\(\frac {\frac {8}{3}+\left (\frac {2 \ln \left (2\right )}{3}+\frac {9}{5}\right ) x^{2}+\frac {2 x^{3}}{3}}{x}\)	\(22\)
gosper	\(\frac {10 x^{2} \ln \left (2\right )+10 x^{3}+27 x^{2}+40}{15 x}\)	\(25\)
parallelrisch	\(\frac {10 x^{2} \ln \left (2\right )+10 x^{3}+27 x^{2}+40}{15 x}\)	\(25\)

[In]

int(1/15*(10*x^2*ln(2)+20*x^3+27*x^2-40)/x^2,x,method=_RETURNVERBOSE)

[Out]

2/3*x^2+2/3*x*ln(2)+9/5*x+8/3/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {10 \, x^{3} + 10 \, x^{2} \log \left (2\right ) + 27 \, x^{2} + 40}{15 \, x} \]

[In]

integrate(1/15*(10*x^2*log(2)+20*x^3+27*x^2-40)/x^2,x, algorithm="fricas")

[Out]

1/15*(10*x^3 + 10*x^2*log(2) + 27*x^2 + 40)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {2 x^{2}}{3} + \frac {x \left (10 \log {\left (2 \right )} + 27\right )}{15} + \frac {8}{3 x} \]

[In]

integrate(1/15*(10*x**2*ln(2)+20*x**3+27*x**2-40)/x**2,x)

[Out]

2*x**2/3 + x*(10*log(2) + 27)/15 + 8/(3*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {2}{3} \, x^{2} + \frac {1}{15} \, x {\left (10 \, \log \left (2\right ) + 27\right )} + \frac {8}{3 \, x} \]

[In]

integrate(1/15*(10*x^2*log(2)+20*x^3+27*x^2-40)/x^2,x, algorithm="maxima")

[Out]

2/3*x^2 + 1/15*x*(10*log(2) + 27) + 8/3/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=\frac {2}{3} \, x^{2} + \frac {2}{3} \, x \log \left (2\right ) + \frac {9}{5} \, x + \frac {8}{3 \, x} \]

[In]

integrate(1/15*(10*x^2*log(2)+20*x^3+27*x^2-40)/x^2,x, algorithm="giac")

[Out]

2/3*x^2 + 2/3*x*log(2) + 9/5*x + 8/3/x

Mupad [B] (verification not implemented)

Time = 13.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-40+27 x^2+20 x^3+5 x^2 \log (4)}{15 x^2} \, dx=x\,\left (\frac {2\,\ln \left (2\right )}{3}+\frac {9}{5}\right )+\frac {8}{3\,x}+\frac {2\,x^2}{3} \]

[In]

int(((2*x^2*log(2))/3 + (9*x^2)/5 + (4*x^3)/3 - 8/3)/x^2,x)

[Out]

x*((2*log(2))/3 + 9/5) + 8/(3*x) + (2*x^2)/3