3.103.7 \(\int \frac {e^x (5-5 x)+4 x^2+4 x^3-2 x^2 \log (\frac {\log ^2(2)}{x^2})}{4 x^2} \, dx\)

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

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Rubi [A]  time = 0.05, antiderivative size = 32, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {12, 14, 2197, 2295} \begin {gather*} \frac {x^2}{2}-\frac {1}{2} x \log \left (\frac {\log ^2(2)}{x^2}\right )-\frac {5 e^x}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(5 - 5*x) + 4*x^2 + 4*x^3 - 2*x^2*Log[Log[2]^2/x^2])/(4*x^2),x]

[Out]

(-5*E^x)/(4*x) + x^2/2 - (x*Log[Log[2]^2/x^2])/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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^x (5-5 x)+4 x^2+4 x^3-2 x^2 \log \left (\frac {\log ^2(2)}{x^2}\right )}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {5 e^x (-1+x)}{x^2}+2 \left (2+2 x-\log \left (\frac {\log ^2(2)}{x^2}\right )\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (2+2 x-\log \left (\frac {\log ^2(2)}{x^2}\right )\right ) \, dx-\frac {5}{4} \int \frac {e^x (-1+x)}{x^2} \, dx\\ &=-\frac {5 e^x}{4 x}+x+\frac {x^2}{2}-\frac {1}{2} \int \log \left (\frac {\log ^2(2)}{x^2}\right ) \, dx\\ &=-\frac {5 e^x}{4 x}+\frac {x^2}{2}-\frac {1}{2} x \log \left (\frac {\log ^2(2)}{x^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 1.10 \begin {gather*} -\frac {5 e^x}{4 x}+\frac {x^2}{2}-\frac {1}{2} x \log \left (\frac {\log ^2(2)}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(5 - 5*x) + 4*x^2 + 4*x^3 - 2*x^2*Log[Log[2]^2/x^2])/(4*x^2),x]

[Out]

(-5*E^x)/(4*x) + x^2/2 - (x*Log[Log[2]^2/x^2])/2

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fricas [A]  time = 0.73, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \, x^{3} - 2 \, x^{2} \log \left (\frac {\log \relax (2)^{2}}{x^{2}}\right ) - 5 \, e^{x}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2*log(log(2)^2/x^2)+(-5*x+5)*exp(x)+4*x^3+4*x^2)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 32, normalized size = 1.10 \begin {gather*} \frac {2 \, x^{3} + 2 \, x^{2} \log \left (x^{2}\right ) - 4 \, x^{2} \log \left (\log \relax (2)\right ) - 5 \, e^{x}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2*log(log(2)^2/x^2)+(-5*x+5)*exp(x)+4*x^3+4*x^2)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 27, normalized size = 0.93

method	result	size
default	\(\frac {x^{2}}{2}-\frac {5 \,{\mathrm e}^{x}}{4 x}-\frac {x \ln \left (\frac {1}{x^{2}}\right )}{2}-x \ln \left (\ln \relax (2)\right )\)	\(27\)
norman	\(\frac {\frac {x^{3}}{2}-\frac {x^{2} \ln \left (\frac {\ln \relax (2)^{2}}{x^{2}}\right )}{2}-\frac {5 \,{\mathrm e}^{x}}{4}}{x}\)	\(29\)
risch	\(x \ln \relax (x )+\frac {-i x^{2} \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i x^{2} \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i x^{2} \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-4 x^{2} \ln \left (\ln \relax (2)\right )+2 x^{3}-5 \,{\mathrm e}^{x}}{4 x}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-2*x^2*ln(ln(2)^2/x^2)+(-5*x+5)*exp(x)+4*x^3+4*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [C]  time = 0.38, size = 29, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, x^{2} - \frac {1}{2} \, x \log \left (\frac {\log \relax (2)^{2}}{x^{2}}\right ) - \frac {5}{4} \, {\rm Ei}\relax (x) + \frac {5}{4} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x^2*log(log(2)^2/x^2)+(-5*x+5)*exp(x)+4*x^3+4*x^2)/x^2,x, algorithm="maxima")

[Out]

1/2*x^2 - 1/2*x*log(log(2)^2/x^2) - 5/4*Ei(x) + 5/4*gamma(-1, -x)

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mupad [B]  time = 6.46, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^2}{2}-x\,\left (\frac {\ln \left (\frac {1}{x^2}\right )}{2}+\ln \left (\ln \relax (2)\right )\right )-\frac {5\,{\mathrm {e}}^x}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(x)*(5*x - 5))/4 + (x^2*log(log(2)^2/x^2))/2 - x^2 - x^3)/x^2,x)

[Out]

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

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sympy [A]  time = 0.30, size = 26, normalized size = 0.90 \begin {gather*} \frac {x^{2}}{2} - \frac {x \log {\left (\frac {\log {\relax (2 )}^{2}}{x^{2}} \right )}}{2} - \frac {5 e^{x}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-2*x**2*ln(ln(2)**2/x**2)+(-5*x+5)*exp(x)+4*x**3+4*x**2)/x**2,x)

[Out]

x**2/2 - x*log(log(2)**2/x**2)/2 - 5*exp(x)/(4*x)

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