3.88.94 \(\int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} (6-3 x-2 x^2)+5 \log ^2(2)}{2 x^2 \log ^2(2)} \, dx\)

Optimal. Leaf size=32 \[ \frac {-5+\frac {\left (3-e^{\frac {1+x}{2}}+2 x\right )^2}{\log ^2(2)}}{2 x} \]

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Rubi [B]  time = 0.19, antiderivative size = 76, normalized size of antiderivative = 2.38, number of steps used = 12, number of rules used = 7, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {12, 14, 2197, 2199, 2194, 2177, 2178} \begin {gather*} \frac {2 x}{\log ^2(2)}-\frac {2 e^{\frac {x}{2}+\frac {1}{2}}}{\log ^2(2)}-\frac {3 e^{\frac {x}{2}+\frac {1}{2}}}{x \log ^2(2)}+\frac {e^{x+1}}{2 x \log ^2(2)}-\frac {5-\frac {9}{\log ^2(2)}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 + E^(1 + x)*(-1 + x) + 4*x^2 + E^((1 + x)/2)*(6 - 3*x - 2*x^2) + 5*Log[2]^2)/(2*x^2*Log[2]^2),x]

[Out]

-1/2*(5 - 9/Log[2]^2)/x - (2*E^(1/2 + x/2))/Log[2]^2 - (3*E^(1/2 + x/2))/(x*Log[2]^2) + E^(1 + x)/(2*x*Log[2]^
2) + (2*x)/Log[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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-9+e^{1+x} (-1+x)+4 x^2+e^{\frac {1+x}{2}} \left (6-3 x-2 x^2\right )+5 \log ^2(2)}{x^2} \, dx}{2 \log ^2(2)}\\ &=\frac {\int \left (\frac {e^{1+x} (-1+x)}{x^2}-\frac {e^{\frac {1}{2}+\frac {x}{2}} \left (-6+3 x+2 x^2\right )}{x^2}+\frac {-9+4 x^2+5 \log ^2(2)}{x^2}\right ) \, dx}{2 \log ^2(2)}\\ &=\frac {\int \frac {e^{1+x} (-1+x)}{x^2} \, dx}{2 \log ^2(2)}-\frac {\int \frac {e^{\frac {1}{2}+\frac {x}{2}} \left (-6+3 x+2 x^2\right )}{x^2} \, dx}{2 \log ^2(2)}+\frac {\int \frac {-9+4 x^2+5 \log ^2(2)}{x^2} \, dx}{2 \log ^2(2)}\\ &=\frac {e^{1+x}}{2 x \log ^2(2)}-\frac {\int \left (2 e^{\frac {1}{2}+\frac {x}{2}}-\frac {6 e^{\frac {1}{2}+\frac {x}{2}}}{x^2}+\frac {3 e^{\frac {1}{2}+\frac {x}{2}}}{x}\right ) \, dx}{2 \log ^2(2)}+\frac {\int \left (4+\frac {-9+5 \log ^2(2)}{x^2}\right ) \, dx}{2 \log ^2(2)}\\ &=-\frac {5-\frac {9}{\log ^2(2)}}{2 x}+\frac {e^{1+x}}{2 x \log ^2(2)}+\frac {2 x}{\log ^2(2)}-\frac {\int e^{\frac {1}{2}+\frac {x}{2}} \, dx}{\log ^2(2)}-\frac {3 \int \frac {e^{\frac {1}{2}+\frac {x}{2}}}{x} \, dx}{2 \log ^2(2)}+\frac {3 \int \frac {e^{\frac {1}{2}+\frac {x}{2}}}{x^2} \, dx}{\log ^2(2)}\\ &=-\frac {5-\frac {9}{\log ^2(2)}}{2 x}-\frac {2 e^{\frac {1}{2}+\frac {x}{2}}}{\log ^2(2)}-\frac {3 e^{\frac {1}{2}+\frac {x}{2}}}{x \log ^2(2)}+\frac {e^{1+x}}{2 x \log ^2(2)}+\frac {2 x}{\log ^2(2)}-\frac {3 \sqrt {e} \text {Ei}\left (\frac {x}{2}\right )}{2 \log ^2(2)}+\frac {3 \int \frac {e^{\frac {1}{2}+\frac {x}{2}}}{x} \, dx}{2 \log ^2(2)}\\ &=-\frac {5-\frac {9}{\log ^2(2)}}{2 x}-\frac {2 e^{\frac {1}{2}+\frac {x}{2}}}{\log ^2(2)}-\frac {3 e^{\frac {1}{2}+\frac {x}{2}}}{x \log ^2(2)}+\frac {e^{1+x}}{2 x \log ^2(2)}+\frac {2 x}{\log ^2(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 45, normalized size = 1.41 \begin {gather*} \frac {9+e^{1+x}+4 x^2-2 e^{\frac {1+x}{2}} (3+2 x)-5 \log ^2(2)}{2 x \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + E^(1 + x)*(-1 + x) + 4*x^2 + E^((1 + x)/2)*(6 - 3*x - 2*x^2) + 5*Log[2]^2)/(2*x^2*Log[2]^2),x]

[Out]

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

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fricas [A]  time = 0.45, size = 39, normalized size = 1.22 \begin {gather*} \frac {4 \, x^{2} - 2 \, {\left (2 \, x + 3\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} - 5 \, \log \relax (2)^{2} + e^{\left (x + 1\right )} + 9}{2 \, x \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-1)*exp(1/2*x+1/2)^2+(-2*x^2-3*x+6)*exp(1/2*x+1/2)+5*log(2)^2+4*x^2-9)/x^2/log(2)^2,x, algori
thm="fricas")

[Out]

1/2*(4*x^2 - 2*(2*x + 3)*e^(1/2*x + 1/2) - 5*log(2)^2 + e^(x + 1) + 9)/(x*log(2)^2)

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giac [A]  time = 0.12, size = 43, normalized size = 1.34 \begin {gather*} \frac {4 \, x^{2} - 4 \, x e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} - 5 \, \log \relax (2)^{2} + e^{\left (x + 1\right )} - 6 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )} + 9}{2 \, x \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-1)*exp(1/2*x+1/2)^2+(-2*x^2-3*x+6)*exp(1/2*x+1/2)+5*log(2)^2+4*x^2-9)/x^2/log(2)^2,x, algori
thm="giac")

[Out]

1/2*(4*x^2 - 4*x*e^(1/2*x + 1/2) - 5*log(2)^2 + e^(x + 1) - 6*e^(1/2*x + 1/2) + 9)/(x*log(2)^2)

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maple [B]  time = 0.11, size = 56, normalized size = 1.75

method	result	size
risch	\(\frac {2 x}{\ln \relax (2)^{2}}-\frac {5}{2 x}+\frac {9}{2 \ln \relax (2)^{2} x}+\frac {{\mathrm e}^{x +1}}{2 \ln \relax (2)^{2} x}-\frac {\left (2 x +3\right ) {\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \relax (2)^{2} x}\)	\(56\)
derivativedivides	\(\frac {\frac {9}{2 x}+2 x +2-\frac {3 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{x}+\frac {{\mathrm e}^{x +1}}{2 x}-\frac {5 \ln \relax (2)^{2}}{2 x}-2 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \relax (2)^{2}}\)	\(57\)
default	\(\frac {\frac {9}{x}+4 x +4-\frac {6 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{x}+\frac {{\mathrm e}^{x +1}}{x}-\frac {5 \ln \relax (2)^{2}}{x}-4 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{2 \ln \relax (2)^{2}}\)	\(57\)
norman	\(\frac {\frac {2 x^{2}}{\ln \relax (2)}-\frac {5 \ln \relax (2)^{2}-9}{2 \ln \relax (2)}-\frac {3 \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \relax (2)}+\frac {{\mathrm e}^{x +1}}{2 \ln \relax (2)}-\frac {2 x \,{\mathrm e}^{\frac {x}{2}+\frac {1}{2}}}{\ln \relax (2)}}{x \ln \relax (2)}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((x-1)*exp(1/2*x+1/2)^2+(-2*x^2-3*x+6)*exp(1/2*x+1/2)+5*ln(2)^2+4*x^2-9)/x^2/ln(2)^2,x,method=_RETURNV
ERBOSE)

[Out]

2*x/ln(2)^2-5/2/x+9/2/ln(2)^2/x+1/2/ln(2)^2*exp(x+1)/x-1/ln(2)^2*(2*x+3)/x*exp(1/2*x+1/2)

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maxima [C]  time = 0.40, size = 63, normalized size = 1.97 \begin {gather*} \frac {{\rm Ei}\relax (x) e - 3 \, {\rm Ei}\left (\frac {1}{2} \, x\right ) e^{\frac {1}{2}} + 3 \, e^{\frac {1}{2}} \Gamma \left (-1, -\frac {1}{2} \, x\right ) - e \Gamma \left (-1, -x\right ) + 4 \, x - \frac {5 \, \log \relax (2)^{2}}{x} + \frac {9}{x} - 4 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2}\right )}}{2 \, \log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-1)*exp(1/2*x+1/2)^2+(-2*x^2-3*x+6)*exp(1/2*x+1/2)+5*log(2)^2+4*x^2-9)/x^2/log(2)^2,x, algori
thm="maxima")

[Out]

1/2*(Ei(x)*e - 3*Ei(1/2*x)*e^(1/2) + 3*e^(1/2)*gamma(-1, -1/2*x) - e*gamma(-1, -x) + 4*x - 5*log(2)^2/x + 9/x
- 4*e^(1/2*x + 1/2))/log(2)^2

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mupad [B]  time = 0.09, size = 43, normalized size = 1.34 \begin {gather*} \frac {{\mathrm {e}}^{x+1}-6\,{\mathrm {e}}^{\frac {x}{2}+\frac {1}{2}}-4\,x\,{\mathrm {e}}^{\frac {x}{2}+\frac {1}{2}}-5\,{\ln \relax (2)}^2+4\,x^2+9}{2\,x\,{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x + 1)*(x - 1))/2 - (exp(x/2 + 1/2)*(3*x + 2*x^2 - 6))/2 + (5*log(2)^2)/2 + 2*x^2 - 9/2)/(x^2*log(2)
^2),x)

[Out]

(exp(x + 1) - 6*exp(x/2 + 1/2) - 4*x*exp(x/2 + 1/2) - 5*log(2)^2 + 4*x^2 + 9)/(2*x*log(2)^2)

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sympy [B]  time = 0.19, size = 70, normalized size = 2.19 \begin {gather*} \frac {4 x + \frac {9 - 5 \log {\relax (2 )}^{2}}{x}}{2 \log {\relax (2 )}^{2}} + \frac {x e^{x + 1} \log {\relax (2 )}^{2} + \left (- 4 x^{2} \log {\relax (2 )}^{2} - 6 x \log {\relax (2 )}^{2}\right ) e^{\frac {x}{2} + \frac {1}{2}}}{2 x^{2} \log {\relax (2 )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((x-1)*exp(1/2*x+1/2)**2+(-2*x**2-3*x+6)*exp(1/2*x+1/2)+5*ln(2)**2+4*x**2-9)/x**2/ln(2)**2,x)

[Out]

(4*x + (9 - 5*log(2)**2)/x)/(2*log(2)**2) + (x*exp(x + 1)*log(2)**2 + (-4*x**2*log(2)**2 - 6*x*log(2)**2)*exp(
x/2 + 1/2))/(2*x**2*log(2)**4)

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