3.100.1 \(\int \frac {e^{\frac {1}{16} (33+16 x)} (1+2 x+x^2)-2^{1+\frac {4}{1+x}} \log (4)}{1+2 x+x^2} \, dx\) [9901]

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

[Out]

exp(2*ln(2)/(1+x))^2+exp(x+33/16)

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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, 6874, 2225, 2262, 2240} \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 2225

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 2240

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 2262

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 6874

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 = 0.06, 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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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.16, size = 93, normalized size = 5.47

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

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]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2)*exp(2*log(2)/(1+x))^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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Fricas [A]
time = 0.38, 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)/(1+x))^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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Sympy [A]
time = 0.12, size = 15, normalized size = 0.88 \begin {gather*} e^{\frac {4 \log {\left (2 \right )}}{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)/(1+x))**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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Giac [A]
time = 0.43, 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)/(1+x))^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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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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