\(\int (-\frac {b \cos (a+b x+c x^2)}{x}+\frac {\sin (a+b x+c x^2)}{x^2}) \, dx\) [5]

Optimal result

Integrand size = 33, antiderivative size = 110 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\sqrt {c} \sqrt {2 \pi } \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {c} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )-\frac {\sin \left (a+b x+c x^2\right )}{x} \]

[Out]

-sin(c*x^2+b*x+a)/x+cos(a-1/4*b^2/c)*FresnelC(1/2*(2*c*x+b)/c^(1/2)*2^(1/2)/Pi^(1/2))*c^(1/2)*2^(1/2)*Pi^(1/2)
-FresnelS(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 = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3546, 3529, 3433, 3432} \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\sqrt {2 \pi } \sqrt {c} \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {2 \pi } \sqrt {c} \sin \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\frac {\sin \left (a+b x+c x^2\right )}{x} \]

[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 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 3529

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 3546

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

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

Time = 0.53 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\sqrt {c} \sqrt {2 \pi } \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\frac {\sqrt {c} \sqrt {2 \pi } x \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )+\sin (a+x (b+c x))}{x} \]

[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]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\frac {\sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - \sin \left (c x^{2} + b x + a\right )}{x} \]

[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]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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