3.100.1 \(\int \frac {e^{\frac {1}{16} (33+16 x)} (1+2 x+x^2)-2^{1+\frac {4}{1+x}} \log (4)}{1+2 x+x^2} \, dx\)

Optimal. Leaf size=17 \[ 4^{\frac {2}{1+x}}+e^{\frac {33}{16}+x} \]

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Rubi [A]  time = 0.25, antiderivative size = 26, normalized size of antiderivative = 1.53, number of steps used = 6, number of rules used = 5, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {27, 6742, 2194, 2230, 2209} \begin {gather*} e^{x+\frac {33}{16}}+\frac {2^{\frac {4}{x+1}-1} \log (4)}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((33 + 16*x)/16)*(1 + 2*x + x^2) - 2^(1 + 4/(1 + x))*Log[4])/(1 + 2*x + x^2),x]

[Out]

E^(33/16 + x) + (2^(-1 + 4/(1 + x))*Log[4])/Log[2]

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 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 2209

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

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{16} (33+16 x)} \left (1+2 x+x^2\right )-2^{1+\frac {4}{1+x}} \log (4)}{(1+x)^2} \, dx\\ &=\int \left (e^{\frac {33}{16}+x}-\frac {2^{\frac {5+x}{1+x}} \log (4)}{(1+x)^2}\right ) \, dx\\ &=-\left (\log (4) \int \frac {2^{\frac {5+x}{1+x}}}{(1+x)^2} \, dx\right )+\int e^{\frac {33}{16}+x} \, dx\\ &=e^{\frac {33}{16}+x}-\log (4) \int \frac {2^{1+\frac {4}{1+x}}}{(1+x)^2} \, dx\\ &=e^{\frac {33}{16}+x}+\frac {2^{-1+\frac {4}{1+x}} \log (4)}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 15, normalized size = 0.88 \begin {gather*} 16^{\frac {1}{1+x}}+e^{\frac {33}{16}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((33 + 16*x)/16)*(1 + 2*x + x^2) - 2^(1 + 4/(1 + x))*Log[4])/(1 + 2*x + x^2),x]

[Out]

16^(1 + x)^(-1) + E^(33/16 + x)

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fricas [A]  time = 0.94, size = 14, normalized size = 0.82 \begin {gather*} 2^{\frac {4}{x + 1}} + e^{\left (x + \frac {33}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*exp(2*log(2)/(x+1))^2+(x^2+2*x+1)*exp(x+33/16))/(x^2+2*x+1),x, algorithm="fricas")

[Out]

2^(4/(x + 1)) + e^(x + 33/16)

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giac [A]  time = 0.22, size = 14, normalized size = 0.82 \begin {gather*} 2^{\frac {4}{x + 1}} + e^{\left (x + \frac {33}{16}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*exp(2*log(2)/(x+1))^2+(x^2+2*x+1)*exp(x+33/16))/(x^2+2*x+1),x, algorithm="giac")

[Out]

2^(4/(x + 1)) + e^(x + 33/16)

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maple [A]  time = 0.16, size = 15, normalized size = 0.88

method	result	size
risch	\(4^{\frac {2}{x +1}}+{\mathrm e}^{x +\frac {33}{16}}\)	\(15\)
norman	\(\frac {{\mathrm e}^{\frac {4 \ln \relax (2)}{x +1}}+x \,{\mathrm e}^{\frac {4 \ln \relax (2)}{x +1}}+{\mathrm e}^{x +\frac {33}{16}} x +{\mathrm e}^{x +\frac {33}{16}}}{x +1}\)	\(44\)
default	\({\mathrm e}^{\frac {33}{16}} \left (-\frac {{\mathrm e}^{x}}{x +1}-{\mathrm e}^{-1} \expIntegralEi \left (1, -x -1\right )\right )+{\mathrm e}^{\frac {33}{16}} \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{x +1}+{\mathrm e}^{-1} \expIntegralEi \left (1, -x -1\right )\right )+\frac {{\mathrm e}^{\frac {4 \ln \relax (2)}{x +1}}+x \,{\mathrm e}^{\frac {4 \ln \relax (2)}{x +1}}}{x +1}+\frac {2 \,{\mathrm e}^{\frac {33}{16}} {\mathrm e}^{x}}{x +1}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(2)*exp(2*ln(2)/(x+1))^2+(x^2+2*x+1)*exp(x+33/16))/(x^2+2*x+1),x,method=_RETURNVERBOSE)

[Out]

(4^(1/(x+1)))^2+exp(x+33/16)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {e^{\frac {17}{16}} E_{2}\left (-x - 1\right )}{x + 1} + \frac {x^{2} e^{\left (x + \frac {33}{16}\right )} + {\left (x^{2} + 2 \, x + 1\right )} 2^{\frac {4}{x + 1}}}{x^{2} + 2 \, x + 1} + \frac {2 \, e^{\left (x + \frac {33}{16}\right )}}{x + 1} - 2 \, \int \frac {x e^{\left (x + \frac {33}{16}\right )}}{x^{3} + 3 \, x^{2} + 3 \, x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*exp(2*log(2)/(x+1))^2+(x^2+2*x+1)*exp(x+33/16))/(x^2+2*x+1),x, algorithm="maxima")

[Out]

-e^(17/16)*exp_integral_e(2, -x - 1)/(x + 1) + (x^2*e^(x + 33/16) + (x^2 + 2*x + 1)*2^(4/(x + 1)))/(x^2 + 2*x
+ 1) + 2*e^(x + 33/16)/(x + 1) - 2*integrate(x*e^(x + 33/16)/(x^3 + 3*x^2 + 3*x + 1), x)

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mupad [B]  time = 5.62, size = 15, normalized size = 0.88 \begin {gather*} {\mathrm {e}}^{33/16}\,{\mathrm {e}}^x+2^{\frac {4}{x+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp((4*log(2))/(x + 1))*log(2) - exp(x + 33/16)*(2*x + x^2 + 1))/(2*x + x^2 + 1),x)

[Out]

exp(33/16)*exp(x) + 2^(4/(x + 1))

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sympy [A]  time = 0.43, size = 15, normalized size = 0.88 \begin {gather*} e^{\frac {4 \log {\relax (2 )}}{x + 1}} + e^{x + \frac {33}{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(2)*exp(2*ln(2)/(x+1))**2+(x**2+2*x+1)*exp(x+33/16))/(x**2+2*x+1),x)

[Out]

exp(4*log(2)/(x + 1)) + exp(x + 33/16)

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