3.48.16 \(\int \frac {25+16 x^2+8 x^3}{16 x^2} \, dx\)

Optimal. Leaf size=28 \[ \log \left (\frac {3}{8} e^{-\frac {(5+2 x)^2 \left (x-x^2\right )}{16 x^2}}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.57, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 14} \begin {gather*} \frac {x^2}{4}+x-\frac {25}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25 + 16*x^2 + 8*x^3)/(16*x^2),x]

[Out]

-25/(16*x) + x + x^2/4

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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {25+16 x^2+8 x^3}{x^2} \, dx\\ &=\frac {1}{16} \int \left (16+\frac {25}{x^2}+8 x\right ) \, dx\\ &=-\frac {25}{16 x}+x+\frac {x^2}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.57 \begin {gather*} -\frac {25}{16 x}+x+\frac {x^2}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 16*x^2 + 8*x^3)/(16*x^2),x]

[Out]

-25/(16*x) + x + x^2/4

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fricas [A]  time = 0.56, size = 17, normalized size = 0.61 \begin {gather*} \frac {4 \, x^{3} + 16 \, x^{2} - 25}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(8*x^3+16*x^2+25)/x^2,x, algorithm="fricas")

[Out]

1/16*(4*x^3 + 16*x^2 - 25)/x

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giac [A]  time = 0.14, size = 12, normalized size = 0.43 \begin {gather*} \frac {1}{4} \, x^{2} + x - \frac {25}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(8*x^3+16*x^2+25)/x^2,x, algorithm="giac")

[Out]

1/4*x^2 + x - 25/16/x

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maple [A]  time = 0.03, size = 13, normalized size = 0.46

method	result	size
default	\(\frac {x^{2}}{4}+x -\frac {25}{16 x}\)	\(13\)
risch	\(\frac {x^{2}}{4}+x -\frac {25}{16 x}\)	\(13\)
norman	\(\frac {-\frac {25}{16}+x^{2}+\frac {1}{4} x^{3}}{x}\)	\(15\)
gosper	\(\frac {4 x^{3}+16 x^{2}-25}{16 x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(8*x^3+16*x^2+25)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x^2+x-25/16/x

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maxima [A]  time = 0.36, size = 12, normalized size = 0.43 \begin {gather*} \frac {1}{4} \, x^{2} + x - \frac {25}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(8*x^3+16*x^2+25)/x^2,x, algorithm="maxima")

[Out]

1/4*x^2 + x - 25/16/x

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mupad [B]  time = 0.03, size = 14, normalized size = 0.50 \begin {gather*} \frac {\frac {x^3}{4}+x^2-\frac {25}{16}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^3/2 + 25/16)/x^2,x)

[Out]

(x^2 + x^3/4 - 25/16)/x

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sympy [A]  time = 0.07, size = 10, normalized size = 0.36 \begin {gather*} \frac {x^{2}}{4} + x - \frac {25}{16 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(8*x**3+16*x**2+25)/x**2,x)

[Out]

x**2/4 + x - 25/(16*x)

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