3.5 \(\int \frac{\cos (a+b x^2)}{x} \, dx\)

Optimal. Leaf size=25 \[ \frac{1}{2} \cos (a) \text{CosIntegral}\left (b x^2\right )-\frac{1}{2} \sin (a) \text{Si}\left (b x^2\right ) \]

[Out]

(Cos[a]*CosIntegral[b*x^2])/2 - (Sin[a]*SinIntegral[b*x^2])/2

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Rubi [A]  time = 0.0289689, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3378, 3376, 3375} \[ \frac{1}{2} \cos (a) \text{CosIntegral}\left (b x^2\right )-\frac{1}{2} \sin (a) \text{Si}\left (b x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/x,x]

[Out]

(Cos[a]*CosIntegral[b*x^2])/2 - (Sin[a]*SinIntegral[b*x^2])/2

Rule 3378

Int[Cos[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Cos[c], Int[Cos[d*x^n]/x, x], x] - Dist[Sin[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 3376

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rubi steps

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

Mathematica [A]  time = 0.0544486, size = 24, normalized size = 0.96 \[ \frac{1}{2} \left (\cos (a) \text{CosIntegral}\left (b x^2\right )-\sin (a) \text{Si}\left (b x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x,x]

[Out]

(Cos[a]*CosIntegral[b*x^2] - Sin[a]*SinIntegral[b*x^2])/2

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Maple [A]  time = 0.025, size = 22, normalized size = 0.9 \begin{align*}{\frac{{\it Ci} \left ( b{x}^{2} \right ) \cos \left ( a \right ) }{2}}-{\frac{{\it Si} \left ( b{x}^{2} \right ) \sin \left ( a \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x^2+a)/x,x)

[Out]

1/2*Ci(b*x^2)*cos(a)-1/2*Si(b*x^2)*sin(a)

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Maxima [C]  time = 1.46817, size = 58, normalized size = 2.32 \begin{align*} \frac{1}{4} \,{\left ({\rm Ei}\left (i \, b x^{2}\right ) +{\rm Ei}\left (-i \, b x^{2}\right )\right )} \cos \left (a\right ) + \frac{1}{4} \,{\left (i \,{\rm Ei}\left (i \, b x^{2}\right ) - i \,{\rm Ei}\left (-i \, b x^{2}\right )\right )} \sin \left (a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x,x, algorithm="maxima")

[Out]

1/4*(Ei(I*b*x^2) + Ei(-I*b*x^2))*cos(a) + 1/4*(I*Ei(I*b*x^2) - I*Ei(-I*b*x^2))*sin(a)

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Fricas [A]  time = 1.53231, size = 122, normalized size = 4.88 \begin{align*} \frac{1}{4} \,{\left (\operatorname{Ci}\left (b x^{2}\right ) + \operatorname{Ci}\left (-b x^{2}\right )\right )} \cos \left (a\right ) - \frac{1}{2} \, \sin \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x,x, algorithm="fricas")

[Out]

1/4*(cos_integral(b*x^2) + cos_integral(-b*x^2))*cos(a) - 1/2*sin(a)*sin_integral(b*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)/x,x)

[Out]

Integral(cos(a + b*x**2)/x, x)

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Giac [A]  time = 1.1287, size = 28, normalized size = 1.12 \begin{align*} \frac{1}{2} \, \cos \left (a\right ) \operatorname{Ci}\left (b x^{2}\right ) - \frac{1}{2} \, \sin \left (a\right ) \operatorname{Si}\left (b x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)/x,x, algorithm="giac")

[Out]

1/2*cos(a)*cos_integral(b*x^2) - 1/2*sin(a)*sin_integral(b*x^2)