3.922 \(\int \frac{\sin (2^x)}{1+2^x} \, dx\)

Optimal. Leaf size=37 \[ \frac{\sin (1) \text{CosIntegral}\left (2^x+1\right )}{\log (2)}+\frac{\text{Si}\left (2^x\right )}{\log (2)}-\frac{\cos (1) \text{Si}\left (1+2^x\right )}{\log (2)} \]

[Out]

(CosIntegral[1 + 2^x]*Sin[1])/Log[2] + SinIntegral[2^x]/Log[2] - (Cos[1]*SinIntegral[1 + 2^x])/Log[2]

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Rubi [A]  time = 0.173205, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2282, 6742, 3299, 3303, 3302} \[ \frac{\sin (1) \text{CosIntegral}\left (2^x+1\right )}{\log (2)}+\frac{\text{Si}\left (2^x\right )}{\log (2)}-\frac{\cos (1) \text{Si}\left (1+2^x\right )}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[2^x]/(1 + 2^x),x]

[Out]

(CosIntegral[1 + 2^x]*Sin[1])/Log[2] + SinIntegral[2^x]/Log[2] - (Cos[1]*SinIntegral[1 + 2^x])/Log[2]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6742

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\sin \left (2^x\right )}{1+2^x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{x (1+x)} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\sin (x)}{x}-\frac{\sin (x)}{1+x}\right ) \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{x} \, dx,x,2^x\right )}{\log (2)}-\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{1+x} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\text{Si}\left (2^x\right )}{\log (2)}-\frac{\cos (1) \operatorname{Subst}\left (\int \frac{\sin (1+x)}{1+x} \, dx,x,2^x\right )}{\log (2)}+\frac{\sin (1) \operatorname{Subst}\left (\int \frac{\cos (1+x)}{1+x} \, dx,x,2^x\right )}{\log (2)}\\ &=\frac{\text{Ci}\left (1+2^x\right ) \sin (1)}{\log (2)}+\frac{\text{Si}\left (2^x\right )}{\log (2)}-\frac{\cos (1) \text{Si}\left (1+2^x\right )}{\log (2)}\\ \end{align*}

Mathematica [A]  time = 0.0709211, size = 29, normalized size = 0.78 \[ \frac{\sin (1) \text{CosIntegral}\left (2^x+1\right )+\text{Si}\left (2^x\right )-\cos (1) \text{Si}\left (1+2^x\right )}{\log (2)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[2^x]/(1 + 2^x),x]

[Out]

(CosIntegral[1 + 2^x]*Sin[1] + SinIntegral[2^x] - Cos[1]*SinIntegral[1 + 2^x])/Log[2]

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Maple [A]  time = 0.01, size = 38, normalized size = 1. \begin{align*}{\frac{{\it Si} \left ({2}^{x} \right ) }{\ln \left ( 2 \right ) }}-{\frac{\cos \left ( 1 \right ){\it Si} \left ( 1+{2}^{x} \right ) }{\ln \left ( 2 \right ) }}+{\frac{{\it Ci} \left ( 1+{2}^{x} \right ) \sin \left ( 1 \right ) }{\ln \left ( 2 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2^x)/(1+2^x),x)

[Out]

Si(2^x)/ln(2)-cos(1)*Si(1+2^x)/ln(2)+Ci(1+2^x)*sin(1)/ln(2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (2^{x}\right )}{2^{x} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2^x)/(1+2^x),x, algorithm="maxima")

[Out]

integrate(sin(2^x)/(2^x + 1), x)

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Fricas [A]  time = 2.14152, size = 176, normalized size = 4.76 \begin{align*} \frac{\operatorname{Ci}\left (2^{x} + 1\right ) \sin \left (1\right ) + \operatorname{Ci}\left (-2^{x} - 1\right ) \sin \left (1\right ) - 2 \, \cos \left (1\right ) \operatorname{Si}\left (2^{x} + 1\right ) + 2 \, \operatorname{Si}\left (2^{x}\right )}{2 \, \log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2^x)/(1+2^x),x, algorithm="fricas")

[Out]

1/2*(cos_integral(2^x + 1)*sin(1) + cos_integral(-2^x - 1)*sin(1) - 2*cos(1)*sin_integral(2^x + 1) + 2*sin_int
egral(2^x))/log(2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (2^{x} \right )}}{2^{x} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2**x)/(1+2**x),x)

[Out]

Integral(sin(2**x)/(2**x + 1), x)

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Giac [A]  time = 1.09163, size = 39, normalized size = 1.05 \begin{align*} \frac{\operatorname{Ci}\left (2^{x} + 1\right ) \sin \left (1\right ) - \cos \left (1\right ) \operatorname{Si}\left (2^{x} + 1\right ) + \operatorname{Si}\left (2^{x}\right )}{\log \left (2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2^x)/(1+2^x),x, algorithm="giac")

[Out]

(cos_integral(2^x + 1)*sin(1) - cos(1)*sin_integral(2^x + 1) + sin_integral(2^x))/log(2)