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

Optimal. Leaf size=110 \[ \sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\sqrt{2 \pi } \sqrt{c} \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\frac{\sin \left (a+b x+c x^2\right )}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0831945, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {3465, 3448, 3352, 3351} \[ \sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\sqrt{2 \pi } \sqrt{c} \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\frac{\sin \left (a+b x+c x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3465

Int[((d_.) + (e_.)*(x_))^(m_)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[((d + e*x)^(m + 1)*Sin
[a + b*x + c*x^2])/(e*(m + 1)), x] + (-Dist[(2*c)/(e^2*(m + 1)), Int[(d + e*x)^(m + 2)*Cos[a + b*x + c*x^2], x
], x] - Dist[(b*e - 2*c*d)/(e^2*(m + 1)), Int[(d + e*x)^(m + 1)*Cos[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 3448

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

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

Mathematica [A]  time = 0.792563, size = 110, normalized size = 1. \[ \sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\frac{\sqrt{2 \pi } \sqrt{c} x \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )+\sin (a+x (b+c x))}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.198, size = 0, normalized size = 0. \begin{align*} \int -{\frac{b\cos \left ( c{x}^{2}+bx+a \right ) }{x}}+{\frac{\sin \left ( c{x}^{2}+bx+a \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.41277, size = 311, normalized size = 2.83 \begin{align*} \frac{\sqrt{2} \pi x \sqrt{\frac{c}{\pi }} \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname{C}\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{S}\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 ) - \sin \left (c x^{2} + b x + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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