3.77.38 \(\int \frac {8+(-3+2 x+x^2) \log (4)+e^{2 x} (4+8 x+4 x^2+(1+4 x+5 x^2+2 x^3) \log (4))}{(1+2 x+x^2) \log (4)} \, dx\)

Optimal. Leaf size=25 \[ 9+x+\left (e^{2 x}-\frac {4}{1+x}\right ) \left (x+\frac {2}{\log (4)}\right ) \]

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Rubi [B]  time = 0.26, antiderivative size = 55, normalized size of antiderivative = 2.20, number of steps used = 8, number of rules used = 6, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 27, 6742, 2176, 2194, 683} \begin {gather*} x+\frac {e^{2 x} (x \log (16)+4+\log (4))}{2 \log (4)}-\frac {e^{2 x} \log (16)}{4 \log (4)}-\frac {2 (4-\log (16))}{(x+1) \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + (-3 + 2*x + x^2)*Log[4] + E^(2*x)*(4 + 8*x + 4*x^2 + (1 + 4*x + 5*x^2 + 2*x^3)*Log[4]))/((1 + 2*x + x
^2)*Log[4]),x]

[Out]

x - (2*(4 - Log[16]))/((1 + x)*Log[4]) - (E^(2*x)*Log[16])/(4*Log[4]) + (E^(2*x)*(4 + Log[4] + x*Log[16]))/(2*
Log[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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !$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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{1+2 x+x^2} \, dx}{\log (4)}\\ &=\frac {\int \frac {8+\left (-3+2 x+x^2\right ) \log (4)+e^{2 x} \left (4+8 x+4 x^2+\left (1+4 x+5 x^2+2 x^3\right ) \log (4)\right )}{(1+x)^2} \, dx}{\log (4)}\\ &=\frac {\int \left (e^{2 x} (4+\log (4)+x \log (16))+\frac {8-3 \log (4)+x^2 \log (4)+x \log (16)}{(1+x)^2}\right ) \, dx}{\log (4)}\\ &=\frac {\int e^{2 x} (4+\log (4)+x \log (16)) \, dx}{\log (4)}+\frac {\int \frac {8-3 \log (4)+x^2 \log (4)+x \log (16)}{(1+x)^2} \, dx}{\log (4)}\\ &=\frac {e^{2 x} (4+\log (4)+x \log (16))}{2 \log (4)}+\frac {\int \left (\log (4)-\frac {2 (-4+\log (16))}{(1+x)^2}\right ) \, dx}{\log (4)}-\frac {\log (16) \int e^{2 x} \, dx}{2 \log (4)}\\ &=x-\frac {2 (4-\log (16))}{(1+x) \log (4)}-\frac {e^{2 x} \log (16)}{4 \log (4)}+\frac {e^{2 x} (4+\log (4)+x \log (16))}{2 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 35, normalized size = 1.40 \begin {gather*} \frac {x \log (4)+\frac {1}{2} e^{2 x} (4+x \log (16))+\frac {-8+\log (256)}{1+x}}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + (-3 + 2*x + x^2)*Log[4] + E^(2*x)*(4 + 8*x + 4*x^2 + (1 + 4*x + 5*x^2 + 2*x^3)*Log[4]))/((1 + 2
*x + x^2)*Log[4]),x]

[Out]

(x*Log[4] + (E^(2*x)*(4 + x*Log[16]))/2 + (-8 + Log[256])/(1 + x))/Log[4]

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fricas [A]  time = 0.77, size = 37, normalized size = 1.48 \begin {gather*} \frac {{\left ({\left (x^{2} + x\right )} \log \relax (2) + x + 1\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + x + 4\right )} \log \relax (2) - 4}{{\left (x + 1\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),
x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 54, normalized size = 2.16 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} \log \relax (2) + x^{2} \log \relax (2) + x e^{\left (2 \, x\right )} \log \relax (2) + x e^{\left (2 \, x\right )} + x \log \relax (2) + e^{\left (2 \, x\right )} + 4 \, \log \relax (2) - 4}{{\left (x + 1\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),
x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 38, normalized size = 1.52

method	result	size
risch	\(x -\frac {4}{\ln \relax (2) \left (x +1\right )}+\frac {4}{x +1}+\frac {\left (2+2 x \ln \relax (2)\right ) {\mathrm e}^{2 x}}{2 \ln \relax (2)}\)	\(38\)
default	\(\frac {-\frac {8}{x +1}+\frac {8 \ln \relax (2)}{x +1}+2 x \ln \relax (2)+2 \,{\mathrm e}^{2 x}+2 x \ln \relax (2) {\mathrm e}^{2 x}}{2 \ln \relax (2)}\)	\(44\)
norman	\(\frac {x^{2}+{\mathrm e}^{2 x} x^{2}+\frac {{\mathrm e}^{2 x}}{\ln \relax (2)}+\frac {\left (1+\ln \relax (2)\right ) x \,{\mathrm e}^{2 x}}{\ln \relax (2)}+\frac {3 \ln \relax (2)-4}{\ln \relax (2)}}{x +1}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((2*(2*x^3+5*x^2+4*x+1)*ln(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3)*ln(2)+8)/(x^2+2*x+1)/ln(2),x,method=
_RETURNVERBOSE)

[Out]

x-4/ln(2)/(x+1)+4/(x+1)+1/2/ln(2)*(2+2*x*ln(2))*exp(2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left (x - \frac {1}{x + 1} - 2 \, \log \left (x + 1\right )\right )} \log \relax (2) + 2 \, {\left (\frac {1}{x + 1} + \log \left (x + 1\right )\right )} \log \relax (2) - \frac {e^{\left (-2\right )} E_{2}\left (-2 \, x - 2\right ) \log \relax (2)}{x + 1} + \frac {{\left (x^{3} \log \relax (2) + x^{2} {\left (2 \, \log \relax (2) + 1\right )} + x {\left (\log \relax (2) + 2\right )}\right )} e^{\left (2 \, x\right )}}{x^{2} + 2 \, x + 1} - \frac {2 \, e^{\left (-2\right )} E_{2}\left (-2 \, x - 2\right )}{x + 1} - \frac {e^{\left (2 \, x\right )} \log \relax (2)}{2 \, {\left (x^{2} + 2 \, x + 1\right )}} + \frac {e^{\left (-2\right )} E_{3}\left (-2 \, x - 2\right ) \log \relax (2)}{{\left (x + 1\right )}^{2}} + \frac {2 \, e^{\left (-2\right )} E_{3}\left (-2 \, x - 2\right )}{{\left (x + 1\right )}^{2}} + \frac {3 \, \log \relax (2)}{x + 1} - \frac {4}{x + 1}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(2*x^3+5*x^2+4*x+1)*log(2)+4*x^2+8*x+4)*exp(x)^2+2*(x^2+2*x-3)*log(2)+8)/(x^2+2*x+1)/log(2),
x, algorithm="maxima")

[Out]

((x - 1/(x + 1) - 2*log(x + 1))*log(2) + 2*(1/(x + 1) + log(x + 1))*log(2) - e^(-2)*exp_integral_e(2, -2*x - 2
)*log(2)/(x + 1) + (x^3*log(2) + x^2*(2*log(2) + 1) + x*(log(2) + 2))*e^(2*x)/(x^2 + 2*x + 1) - 2*e^(-2)*exp_i
ntegral_e(2, -2*x - 2)/(x + 1) + 3*log(2)/(x + 1) - 4/(x + 1) - integrate((x*log(2) + log(2) + 2)*e^(2*x)/(x^3
+ 3*x^2 + 3*x + 1), x))/log(2)

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mupad [B]  time = 0.21, size = 39, normalized size = 1.56 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}+x\,\ln \relax (2)+x\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)}{\ln \relax (2)}+\frac {\ln \relax (2)+\ln \relax (8)-4}{\ln \relax (2)\,\left (x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2*x)*(8*x + 2*log(2)*(4*x + 5*x^2 + 2*x^3 + 1) + 4*x^2 + 4))/2 + log(2)*(2*x + x^2 - 3) + 4)/(log(2)
*(2*x + x^2 + 1)),x)

[Out]

(exp(2*x) + x*log(2) + x*exp(2*x)*log(2))/log(2) + (log(2) + log(8) - 4)/(log(2)*(x + 1))

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sympy [A]  time = 0.27, size = 31, normalized size = 1.24 \begin {gather*} x + \frac {\left (x \log {\relax (2 )} + 1\right ) e^{2 x}}{\log {\relax (2 )}} + \frac {-4 + 4 \log {\relax (2 )}}{x \log {\relax (2 )} + \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((2*(2*x**3+5*x**2+4*x+1)*ln(2)+4*x**2+8*x+4)*exp(x)**2+2*(x**2+2*x-3)*ln(2)+8)/(x**2+2*x+1)/ln(
2),x)

[Out]

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

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