3.84.95 \(\int \frac {12+19 x^2-8 e^{\frac {1}{4 \log (3)}} x^3-15 x^4}{4 x^2} \, dx\)

Optimal. Leaf size=29 \[ \left (e^{\frac {1}{4 \log (3)}}-\frac {1}{x}+\frac {5 x}{4}\right ) \left (3-x^2\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 14} \begin {gather*} -\frac {5 x^3}{4}-x^2 e^{\frac {1}{\log (81)}}+\frac {19 x}{4}-\frac {3}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 19*x^2 - 8*E^(1/(4*Log[3]))*x^3 - 15*x^4)/(4*x^2),x]

[Out]

-3/x + (19*x)/4 - E^Log[81]^(-1)*x^2 - (5*x^3)/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}{4} \int \frac {12+19 x^2-8 e^{\frac {1}{4 \log (3)}} x^3-15 x^4}{x^2} \, dx\\ &=\frac {1}{4} \int \left (19+\frac {12}{x^2}-8 e^{\frac {1}{\log (81)}} x-15 x^2\right ) \, dx\\ &=-\frac {3}{x}+\frac {19 x}{4}-e^{\frac {1}{\log (81)}} x^2-\frac {5 x^3}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.00 \begin {gather*} -\frac {3}{x}+\frac {19 x}{4}-e^{\frac {1}{\log (81)}} x^2-\frac {5 x^3}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 19*x^2 - 8*E^(1/(4*Log[3]))*x^3 - 15*x^4)/(4*x^2),x]

[Out]

-3/x + (19*x)/4 - E^Log[81]^(-1)*x^2 - (5*x^3)/4

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fricas [A]  time = 0.69, size = 29, normalized size = 1.00 \begin {gather*} -\frac {5 \, x^{4} + 4 \, x^{3} e^{\left (\frac {1}{4 \, \log \relax (3)}\right )} - 19 \, x^{2} + 12}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*x^3*exp(1/4/log(3))-15*x^4+19*x^2+12)/x^2,x, algorithm="fricas")

[Out]

-1/4*(5*x^4 + 4*x^3*e^(1/4/log(3)) - 19*x^2 + 12)/x

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giac [A]  time = 0.13, size = 26, normalized size = 0.90 \begin {gather*} -\frac {5}{4} \, x^{3} - x^{2} e^{\left (\frac {1}{4 \, \log \relax (3)}\right )} + \frac {19}{4} \, x - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*x^3*exp(1/4/log(3))-15*x^4+19*x^2+12)/x^2,x, algorithm="giac")

[Out]

-5/4*x^3 - x^2*e^(1/4/log(3)) + 19/4*x - 3/x

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maple [A]  time = 0.06, size = 27, normalized size = 0.93

method	result	size
default	\(\frac {19 x}{4}-\frac {5 x^{3}}{4}-x^{2} {\mathrm e}^{\frac {1}{4 \ln \relax (3)}}-\frac {3}{x}\)	\(27\)
risch	\(\frac {19 x}{4}-\frac {5 x^{3}}{4}-x^{2} {\mathrm e}^{\frac {1}{4 \ln \relax (3)}}-\frac {3}{x}\)	\(27\)
norman	\(\frac {-3+\frac {19 x^{2}}{4}-\frac {5 x^{4}}{4}-x^{3} {\mathrm e}^{\frac {1}{4 \ln \relax (3)}}}{x}\)	\(29\)
gosper	\(-\frac {5 x^{4}+4 x^{3} {\mathrm e}^{\frac {1}{4 \ln \relax (3)}}-19 x^{2}+12}{4 x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-8*x^3*exp(1/4/ln(3))-15*x^4+19*x^2+12)/x^2,x,method=_RETURNVERBOSE)

[Out]

19/4*x-5/4*x^3-x^2*exp(1/4/ln(3))-3/x

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maxima [A]  time = 0.36, size = 26, normalized size = 0.90 \begin {gather*} -\frac {5}{4} \, x^{3} - x^{2} e^{\left (\frac {1}{4 \, \log \relax (3)}\right )} + \frac {19}{4} \, x - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*x^3*exp(1/4/log(3))-15*x^4+19*x^2+12)/x^2,x, algorithm="maxima")

[Out]

-5/4*x^3 - x^2*e^(1/4/log(3)) + 19/4*x - 3/x

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mupad [B]  time = 5.25, size = 26, normalized size = 0.90 \begin {gather*} \frac {19\,x}{4}-x^2\,{\mathrm {e}}^{\frac {1}{4\,\ln \relax (3)}}-\frac {3}{x}-\frac {5\,x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3*exp(1/(4*log(3))) - (19*x^2)/4 + (15*x^4)/4 - 3)/x^2,x)

[Out]

(19*x)/4 - x^2*exp(1/(4*log(3))) - 3/x - (5*x^3)/4

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sympy [A]  time = 0.09, size = 26, normalized size = 0.90 \begin {gather*} - \frac {5 x^{3}}{4} - x^{2} e^{\frac {1}{4 \log {\relax (3 )}}} + \frac {19 x}{4} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-8*x**3*exp(1/4/ln(3))-15*x**4+19*x**2+12)/x**2,x)

[Out]

-5*x**3/4 - x**2*exp(1/(4*log(3))) + 19*x/4 - 3/x

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