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

Optimal. Leaf size=80 \[ -\sqrt{2 \pi } \sqrt{b} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\sqrt{2 \pi } \sqrt{b} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{\cos \left (a+b x^2\right )}{x} \]

[Out]

-(Cos[a + b*x^2]/x) - Sqrt[b]*Sqrt[2*Pi]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x] - Sqrt[b]*Sqrt[2*Pi]*FresnelC[S
qrt[b]*Sqrt[2/Pi]*x]*Sin[a]

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Rubi [A]  time = 0.0445045, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3388, 3353, 3352, 3351} \[ -\sqrt{2 \pi } \sqrt{b} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\sqrt{2 \pi } \sqrt{b} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{\cos \left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/x^2,x]

[Out]

-(Cos[a + b*x^2]/x) - Sqrt[b]*Sqrt[2*Pi]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x] - Sqrt[b]*Sqrt[2*Pi]*FresnelC[S
qrt[b]*Sqrt[2/Pi]*x]*Sin[a]

Rule 3388

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[((e*x)^(m + 1)*Cos[c + d*x^n])/(e*(m + 1
)), x] + Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{\cos \left (a+b x^2\right )}{x^2} \, dx &=-\frac{\cos \left (a+b x^2\right )}{x}-(2 b) \int \sin \left (a+b x^2\right ) \, dx\\ &=-\frac{\cos \left (a+b x^2\right )}{x}-(2 b \cos (a)) \int \sin \left (b x^2\right ) \, dx-(2 b \sin (a)) \int \cos \left (b x^2\right ) \, dx\\ &=-\frac{\cos \left (a+b x^2\right )}{x}-\sqrt{b} \sqrt{2 \pi } \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\sqrt{b} \sqrt{2 \pi } C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a)\\ \end{align*}

Mathematica [A]  time = 0.201662, size = 81, normalized size = 1.01 \[ -\sqrt{2 \pi } \sqrt{b} \left (\sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )+\cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )\right )+\frac{\sin (a) \sin \left (b x^2\right )}{x}-\frac{\cos (a) \cos \left (b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x^2,x]

[Out]

-((Cos[a]*Cos[b*x^2])/x) - Sqrt[b]*Sqrt[2*Pi]*(Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x] + FresnelC[Sqrt[b]*Sqrt[2
/Pi]*x]*Sin[a]) + (Sin[a]*Sin[b*x^2])/x

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Maple [A]  time = 0.024, size = 57, normalized size = 0.7 \begin{align*} -{\frac{\cos \left ( b{x}^{2}+a \right ) }{x}}-\sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(b*x^2+a)/x-b^(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelS(x*b^(1/2)*2^(1/2)/Pi^(1/2))+sin(a)*FresnelC(x*b^(1/2
)*2^(1/2)/Pi^(1/2)))

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Maxima [C]  time = 1.73907, size = 355, normalized size = 4.44 \begin{align*} -\frac{\sqrt{x^{2}{\left | b \right |}}{\left ({\left ({\left (\Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) - i \, \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) +{\left ({\left (-i \, \Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (-\frac{1}{2}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{2}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )}}{8 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(x^2*abs(b))*(((gamma(-1/2, I*b*x^2) + gamma(-1/2, -I*b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) + (gamm
a(-1/2, I*b*x^2) + gamma(-1/2, -I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (I*gamma(-1/2, I*b*x^2) - I*gamma
(-1/2, -I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) + (-I*gamma(-1/2, I*b*x^2) + I*gamma(-1/2, -I*b*x^2))*sin(-1
/4*pi + 1/2*arctan2(0, b)))*cos(a) + ((-I*gamma(-1/2, I*b*x^2) + I*gamma(-1/2, -I*b*x^2))*cos(1/4*pi + 1/2*arc
tan2(0, b)) + (-I*gamma(-1/2, I*b*x^2) + I*gamma(-1/2, -I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (gamma(-1
/2, I*b*x^2) + gamma(-1/2, -I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) - (gamma(-1/2, I*b*x^2) + gamma(-1/2, -I
*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))/x

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Fricas [A]  time = 1.68656, size = 209, normalized size = 2.61 \begin{align*} -\frac{\sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) + \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) + \cos \left (b x^{2} + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*x*sqrt(b/pi)) + sqrt(2)*pi*x*sqrt(b/pi)*fresnel_cos(sqrt(
2)*x*sqrt(b/pi))*sin(a) + cos(b*x^2 + a))/x

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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^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x**2)/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x^2 + a)/x^2, x)