3.67.12 \(\int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx\)

Optimal. Leaf size=18 \[ \frac {3 e^{11/4} (x+\log (2 x))}{2 x^2} \]

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Rubi [B]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 2.67, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 14, 37, 2304} \begin {gather*} -\frac {3 e^{11/4} (1-x)^2}{4 x^2}+\frac {3 e^{11/4}}{4 x^2}+\frac {3 e^{11/4} \log (2 x)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(11/4)*(3 - 3*x) - 6*E^(11/4)*Log[2*x])/(2*x^3),x]

[Out]

(3*E^(11/4))/(4*x^2) - (3*E^(11/4)*(1 - x)^2)/(4*x^2) + (3*E^(11/4)*Log[2*x])/(2*x^2)

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]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{x^3} \, dx\\ &=\frac {1}{2} \int \left (-\frac {3 e^{11/4} (-1+x)}{x^3}-\frac {6 e^{11/4} \log (2 x)}{x^3}\right ) \, dx\\ &=-\left (\frac {1}{2} \left (3 e^{11/4}\right ) \int \frac {-1+x}{x^3} \, dx\right )-\left (3 e^{11/4}\right ) \int \frac {\log (2 x)}{x^3} \, dx\\ &=\frac {3 e^{11/4}}{4 x^2}-\frac {3 e^{11/4} (1-x)^2}{4 x^2}+\frac {3 e^{11/4} \log (2 x)}{2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.33 \begin {gather*} -\frac {3}{2} e^{11/4} \left (-\frac {1}{x}-\frac {\log (2 x)}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(11/4)*(3 - 3*x) - 6*E^(11/4)*Log[2*x])/(2*x^3),x]

[Out]

(-3*E^(11/4)*(-x^(-1) - Log[2*x]/x^2))/2

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fricas [A]  time = 0.55, size = 17, normalized size = 0.94 \begin {gather*} \frac {3 \, {\left (x e^{\frac {11}{4}} + e^{\frac {11}{4}} \log \left (2 \, x\right )\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-6*exp(11/4)*log(2*x)+(-3*x+3)*exp(11/4))/x^3,x, algorithm="fricas")

[Out]

3/2*(x*e^(11/4) + e^(11/4)*log(2*x))/x^2

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giac [A]  time = 0.13, size = 17, normalized size = 0.94 \begin {gather*} \frac {3 \, {\left (x e^{\frac {11}{4}} + e^{\frac {11}{4}} \log \left (2 \, x\right )\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-6*exp(11/4)*log(2*x)+(-3*x+3)*exp(11/4))/x^3,x, algorithm="giac")

[Out]

3/2*(x*e^(11/4) + e^(11/4)*log(2*x))/x^2

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maple [A]  time = 0.06, size = 19, normalized size = 1.06

method	result	size
norman	\(\frac {\frac {3 \,{\mathrm e}^{\frac {11}{4}} x}{2}+\frac {3 \,{\mathrm e}^{\frac {11}{4}} \ln \left (2 x \right )}{2}}{x^{2}}\)	\(19\)
risch	\(\frac {3 \,{\mathrm e}^{\frac {11}{4}} \ln \left (2 x \right )}{2 x^{2}}+\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{2 x}\)	\(20\)
derivativedivides	\(-12 \,{\mathrm e}^{\frac {11}{4}} \left (-\frac {\ln \left (2 x \right )}{8 x^{2}}-\frac {1}{16 x^{2}}\right )+\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{2 x}-\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{4 x^{2}}\)	\(35\)
default	\(-12 \,{\mathrm e}^{\frac {11}{4}} \left (-\frac {\ln \left (2 x \right )}{8 x^{2}}-\frac {1}{16 x^{2}}\right )+\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{2 x}-\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{4 x^{2}}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-6*exp(11/4)*ln(2*x)+(-3*x+3)*exp(11/4))/x^3,x,method=_RETURNVERBOSE)

[Out]

(3/2*exp(11/4)*x+3/2*exp(11/4)*ln(2*x))/x^2

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maxima [B]  time = 0.39, size = 32, normalized size = 1.78 \begin {gather*} \frac {3}{4} \, {\left (\frac {2 \, \log \left (2 \, x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} e^{\frac {11}{4}} + \frac {3 \, e^{\frac {11}{4}}}{2 \, x} - \frac {3 \, e^{\frac {11}{4}}}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-6*exp(11/4)*log(2*x)+(-3*x+3)*exp(11/4))/x^3,x, algorithm="maxima")

[Out]

3/4*(2*log(2*x)/x^2 + 1/x^2)*e^(11/4) + 3/2*e^(11/4)/x - 3/4*e^(11/4)/x^2

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mupad [B]  time = 4.22, size = 13, normalized size = 0.72 \begin {gather*} \frac {3\,{\mathrm {e}}^{11/4}\,\left (x+\ln \left (2\,x\right )\right )}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*log(2*x)*exp(11/4) + (exp(11/4)*(3*x - 3))/2)/x^3,x)

[Out]

(3*exp(11/4)*(x + log(2*x)))/(2*x^2)

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sympy [A]  time = 0.12, size = 26, normalized size = 1.44 \begin {gather*} \frac {3 e^{\frac {11}{4}}}{2 x} + \frac {3 e^{\frac {11}{4}} \log {\left (2 x \right )}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-6*exp(11/4)*ln(2*x)+(-3*x+3)*exp(11/4))/x**3,x)

[Out]

3*exp(11/4)/(2*x) + 3*exp(11/4)*log(2*x)/(2*x**2)

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