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

Optimal. Leaf size=111 \[ -\sqrt{2 \pi } \sqrt{c} \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\frac{\cos \left (a+b x+c x^2\right )}{x} \]

[Out]

-(Cos[a + b*x + c*x^2]/x) - Sqrt[c]*Sqrt[2*Pi]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])] -
 Sqrt[c]*Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)]

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Rubi [A]  time = 0.0935354, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3466, 3447, 3351, 3352} \[ -\sqrt{2 \pi } \sqrt{c} \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\frac{\cos \left (a+b x+c x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x + c*x^2]/x^2 + (b*Sin[a + b*x + c*x^2])/x,x]

[Out]

-(Cos[a + b*x + c*x^2]/x) - Sqrt[c]*Sqrt[2*Pi]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])] -
 Sqrt[c]*Sqrt[2*Pi]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)]

Rule 3466

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*Cos
[a + b*x + c*x^2])/(e*(m + 1)), x] + (Dist[(2*c)/(e^2*(m + 1)), Int[(d + e*x)^(m + 2)*Sin[a + b*x + c*x^2], x]
, x] + Dist[(b*e - 2*c*d)/(e^2*(m + 1)), Int[(d + e*x)^(m + 1)*Sin[a + b*x + c*x^2], x], x]) /; FreeQ[{a, b, c
, d, e}, x] && NeQ[b*e - 2*c*d, 0] && LtQ[m, -1]

Rule 3447

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/
(4*c)], x], x] - Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
 NeQ[b^2 - 4*a*c, 0]

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]

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]

Rubi steps

\begin{align*} \int \left (\frac{\cos \left (a+b x+c x^2\right )}{x^2}+\frac{b \sin \left (a+b x+c x^2\right )}{x}\right ) \, dx &=b \int \frac{\sin \left (a+b x+c x^2\right )}{x} \, dx+\int \frac{\cos \left (a+b x+c x^2\right )}{x^2} \, dx\\ &=-\frac{\cos \left (a+b x+c x^2\right )}{x}-(2 c) \int \sin \left (a+b x+c x^2\right ) \, dx\\ &=-\frac{\cos \left (a+b x+c x^2\right )}{x}-\left (2 c \cos \left (a-\frac{b^2}{4 c}\right )\right ) \int \sin \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx-\left (2 c \sin \left (a-\frac{b^2}{4 c}\right )\right ) \int \cos \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx\\ &=-\frac{\cos \left (a+b x+c x^2\right )}{x}-\sqrt{c} \sqrt{2 \pi } \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\sqrt{c} \sqrt{2 \pi } C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \sin \left (a-\frac{b^2}{4 c}\right )\\ \end{align*}

Mathematica [A]  time = 3.74542, size = 110, normalized size = 0.99 \[ -\frac{\sqrt{2 \pi } \sqrt{c} x \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )+\sqrt{2 \pi } \sqrt{c} x \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )+\cos (a+x (b+c x))}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x + c*x^2]/x^2 + (b*Sin[a + b*x + c*x^2])/x,x]

[Out]

-((Cos[a + x*(b + c*x)] + Sqrt[c]*Sqrt[2*Pi]*x*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])] +
 Sqrt[c]*Sqrt[2*Pi]*x*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/x)

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Maple [F]  time = 0.224, size = 0, normalized size = 0. \begin{align*} \int{\frac{\cos \left ( c{x}^{2}+bx+a \right ) }{{x}^{2}}}+{\frac{b\sin \left ( c{x}^{2}+bx+a \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.40712, size = 312, normalized size = 2.81 \begin{align*} -\frac{\sqrt{2} \pi x \sqrt{\frac{c}{\pi }} \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2} \pi x \sqrt{\frac{c}{\pi }} \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) + \cos \left (c x^{2} + b x + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*pi*x*sqrt(c/pi)*cos(-1/4*(b^2 - 4*a*c)/c)*fresnel_sin(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c) + sqrt(2
)*pi*x*sqrt(c/pi)*fresnel_cos(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c)*sin(-1/4*(b^2 - 4*a*c)/c) + cos(c*x^2 + b*
x + a))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x*sin(a + b*x + c*x**2) + cos(a + b*x + c*x**2))/x**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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