\(\int (\frac {\cos (a+b x-c x^2)}{x^2}+\frac {b \sin (a+b x-c x^2)}{x}) \, dx\) [10]

Optimal result

Integrand size = 34, antiderivative size = 111 \[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=-\frac {\cos \left (a+b x-c x^2\right )}{x}+\sqrt {c} \sqrt {2 \pi } \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {c} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right ) \]

[Out]

-cos(-c*x^2+b*x+a)/x+cos(a+1/4*b^2/c)*FresnelS(1/2*(-2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*c^(1/2)*2^(1/2)*Pi^(1/
2)-FresnelC(1/2*(-2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*sin(a+1/4*b^2/c)*c^(1/2)*2^(1/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3547, 3528, 3432, 3433} \[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=-\sqrt {2 \pi } \sqrt {c} \sin \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+\sqrt {2 \pi } \sqrt {c} \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\frac {\cos \left (a+b x-c x^2\right )}{x} \]

[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 3432

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

Rule 3433

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

Rule 3528

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 3547

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[(b*e - 2*c*d)/(e^2*(m + 1)), Int[(d + e*x)^(m + 1)*Sin[a + b*x + c*
x^2], x], x] + Dist[2*(c/(e^2*(m + 1))), Int[(d + e*x)^(m + 2)*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]

Rubi steps \begin{align*} \text {integral}& = 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 ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {c} \sqrt {2 \pi } \operatorname {FresnelC}\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] (verified)

Time = 2.42 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03 \[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=-\frac {\cos (a+x (b-c x))}{x}-\sqrt {c} \sqrt {2 \pi } \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+\sqrt {c} \sqrt {2 \pi } \operatorname {FresnelC}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right ) \]

[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)]/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)]

Maple [F]

\[\int \left (\frac {\cos \left (-c \,x^{2}+b x +a \right )}{x^{2}}+\frac {b \sin \left (-c \,x^{2}+b x +a \right )}{x}\right )d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.11 \[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=-\frac {\sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + \cos \left (c x^{2} - b x - a\right )}{x} \]

[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

Sympy [F]

\[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=\int \frac {b x \sin {\left (a + b x - c x^{2} \right )} + \cos {\left (a + b x - c x^{2} \right )}}{x^{2}}\, dx \]

[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)

Maxima [F]

\[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=\int { \frac {b \sin \left (-c x^{2} + b x + a\right )}{x} + \frac {\cos \left (-c x^{2} + b x + a\right )}{x^{2}} \,d x } \]

[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)

Giac [F]

\[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=\int { \frac {b \sin \left (-c x^{2} + b x + a\right )}{x} + \frac {\cos \left (-c x^{2} + b x + a\right )}{x^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (\frac {\cos \left (a+b x-c x^2\right )}{x^2}+\frac {b \sin \left (a+b x-c x^2\right )}{x}\right ) \, dx=\int \frac {\cos \left (-c\,x^2+b\,x+a\right )}{x^2}+\frac {b\,\sin \left (-c\,x^2+b\,x+a\right )}{x} \,d x \]

[In]

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

[Out]

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