\(\int \frac {25+x^3+2 x^4}{x^3} \, dx\) [3583]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=-\frac {25}{2 x^2}-x+(1+x)^2 \]

[Out]

(1+x)^2-x-25/2/x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14} \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=x^2-\frac {25}{2 x^2}+x \]

[In]

Int[(25 + x^3 + 2*x^4)/x^3,x]

[Out]

-25/(2*x^2) + x + x^2

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 (1+\frac {25}{x^3}+2 x\right ) \, dx \\ & = -\frac {25}{2 x^2}+x+x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=-\frac {25}{2 x^2}+x+x^2 \]

[In]

Integrate[(25 + x^3 + 2*x^4)/x^3,x]

[Out]

-25/(2*x^2) + x + x^2

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
default	\(x +x^{2}-\frac {25}{2 x^{2}}\)	\(11\)
risch	\(x +x^{2}-\frac {25}{2 x^{2}}\)	\(11\)
norman	\(\frac {x^{4}+x^{3}-\frac {25}{2}}{x^{2}}\)	\(13\)
gosper	\(\frac {2 x^{4}+2 x^{3}-25}{2 x^{2}}\)	\(18\)
parallelrisch	\(\frac {2 x^{4}+2 x^{3}-25}{2 x^{2}}\)	\(18\)

[In]

int((2*x^4+x^3+25)/x^3,x,method=_RETURNVERBOSE)

[Out]

x+x^2-25/2/x^2

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=\frac {2 \, x^{4} + 2 \, x^{3} - 25}{2 \, x^{2}} \]

[In]

integrate((2*x^4+x^3+25)/x^3,x, algorithm="fricas")

[Out]

1/2*(2*x^4 + 2*x^3 - 25)/x^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=x^{2} + x - \frac {25}{2 x^{2}} \]

[In]

integrate((2*x**4+x**3+25)/x**3,x)

[Out]

x**2 + x - 25/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=x^{2} + x - \frac {25}{2 \, x^{2}} \]

[In]

integrate((2*x^4+x^3+25)/x^3,x, algorithm="maxima")

[Out]

x^2 + x - 25/2/x^2

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=x^{2} + x - \frac {25}{2 \, x^{2}} \]

[In]

integrate((2*x^4+x^3+25)/x^3,x, algorithm="giac")

[Out]

x^2 + x - 25/2/x^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {25+x^3+2 x^4}{x^3} \, dx=\frac {x^4+x^3-\frac {25}{2}}{x^2} \]

[In]

int((x^3 + 2*x^4 + 25)/x^3,x)

[Out]

(x^3 + x^4 - 25/2)/x^2