\(\int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx\) [6612]

Optimal result

Integrand size = 30, antiderivative size = 18 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3 e^{11/4} (x+\log (2 x))}{2 x^2} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(18)=36\).

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.67, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 14, 37, 2341} \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=-\frac {3 e^{11/4} (1-x)^2}{4 x^2}+\frac {3 e^{11/4}}{4 x^2}+\frac {3 e^{11/4} \log (2 x)}{2 x^2} \]

[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 2341

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=-\frac {3}{2} e^{11/4} \left (-\frac {1}{x}-\frac {\log (2 x)}{x^2}\right ) \]

[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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 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	\(\frac {3 \,{\mathrm e}^{\frac {11}{4}} \ln \left (2 x \right )}{2 x^{2}}+\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{2 x}\)	\(20\)
parallelrisch	\(\frac {3 \,{\mathrm e}^{\frac {11}{4}} x +3 \,{\mathrm e}^{\frac {11}{4}} \ln \left (2 x \right )}{2 x^{2}}\)	\(20\)
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	\(-12 \,{\mathrm e}^{\frac {11}{4}} \left (-\frac {\ln \left (2 x \right )}{8 x^{2}}-\frac {1}{16 x^{2}}\right )+\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{2 x}-\frac {3 \,{\mathrm e}^{\frac {11}{4}}}{4 x^{2}}\)	\(35\)
parts	\(-3 \,{\mathrm e}^{\frac {11}{4}} \left (-\frac {\ln \left (2 x \right )}{2 x^{2}}-\frac {1}{4 x^{2}}\right )-\frac {3 \,{\mathrm e}^{\frac {11}{4}} \left (-\frac {1}{x}+\frac {1}{2 x^{2}}\right )}{2}\)	\(36\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3 \, {\left (x e^{\frac {11}{4}} + e^{\frac {11}{4}} \log \left (2 \, x\right )\right )}}{2 \, x^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3 e^{\frac {11}{4}}}{2 x} + \frac {3 e^{\frac {11}{4}} \log {\left (2 x \right )}}{2 x^{2}} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (13) = 26\).

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3}{4} \, {\left (\frac {2 \, \log \left (2 \, x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} e^{\frac {11}{4}} + \frac {3 \, e^{\frac {11}{4}}}{2 \, x} - \frac {3 \, e^{\frac {11}{4}}}{4 \, x^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3 \, {\left (x e^{\frac {11}{4}} + e^{\frac {11}{4}} \log \left (2 \, x\right )\right )}}{2 \, x^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {e^{11/4} (3-3 x)-6 e^{11/4} \log (2 x)}{2 x^3} \, dx=\frac {3\,{\mathrm {e}}^{11/4}\,\left (x+\ln \left (2\,x\right )\right )}{2\,x^2} \]

[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)