Integrand size = 33, antiderivative size = 110 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\sqrt {c} \sqrt {2 \pi } \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {c} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )-\frac {\sin \left (a+b x+c x^2\right )}{x} \] Output:
c^(1/2)*2^(1/2)*Pi^(1/2)*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)-sin(c*x^2+b*x+a)/x
Time = 0.89 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\sqrt {c} \sqrt {2 \pi } \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\frac {\sqrt {c} \sqrt {2 \pi } x \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a-\frac {b^2}{4 c}\right )+\sin (a+x (b+c x))}{x} \] Input:
Integrate[-((b*Cos[a + b*x + c*x^2])/x) + Sin[a + b*x + c*x^2]/x^2,x]
Output:
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*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]* Sin[a - b^2/(4*c)] + Sin[a + x*(b + c*x)])/x
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (\frac {\sin \left (a+b x+c x^2\right )}{x^2}-\frac {b \cos \left (a+b x+c x^2\right )}{x}\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \sqrt {2 \pi } \sqrt {c} \cos \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {2 \pi } \sqrt {c} \sin \left (a-\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )-\frac {\sin \left (a+b x+c x^2\right )}{x}\) |
Input:
Int[-((b*Cos[a + b*x + c*x^2])/x) + Sin[a + b*x + c*x^2]/x^2,x]
Output:
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
\[\int \left (-\frac {b \cos \left (c \,x^{2}+b x +a \right )}{x}+\frac {\sin \left (c \,x^{2}+b x +a \right )}{x^{2}}\right )d x\]
Input:
int(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x)
Output:
int(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x)
Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.06 \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\frac {\sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi x \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - \sin \left (c x^{2} + b x + a\right )}{x} \] Input:
integrate(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x, algorithm="fricas" )
Output:
(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
\[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=- \int \left (- \frac {\sin {\left (a + b x + c x^{2} \right )}}{x^{2}}\right )\, dx - \int \frac {b \cos {\left (a + b x + c x^{2} \right )}}{x}\, dx \] Input:
integrate(-b*cos(c*x**2+b*x+a)/x+sin(c*x**2+b*x+a)/x**2,x)
Output:
-Integral(-sin(a + b*x + c*x**2)/x**2, x) - Integral(b*cos(a + b*x + c*x** 2)/x, x)
\[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\int { -\frac {b \cos \left (c x^{2} + b x + a\right )}{x} + \frac {\sin \left (c x^{2} + b x + a\right )}{x^{2}} \,d x } \] Input:
integrate(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x, algorithm="maxima" )
Output:
integrate(-b*cos(c*x^2 + b*x + a)/x + sin(c*x^2 + b*x + a)/x^2, x)
\[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\int { -\frac {b \cos \left (c x^{2} + b x + a\right )}{x} + \frac {\sin \left (c x^{2} + b x + a\right )}{x^{2}} \,d x } \] Input:
integrate(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x, algorithm="giac")
Output:
integrate(-b*cos(c*x^2 + b*x + a)/x + sin(c*x^2 + b*x + a)/x^2, x)
Timed out. \[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\int \frac {\sin \left (c\,x^2+b\,x+a\right )}{x^2}-\frac {b\,\cos \left (c\,x^2+b\,x+a\right )}{x} \,d x \] Input:
int(sin(a + b*x + c*x^2)/x^2 - (b*cos(a + b*x + c*x^2))/x,x)
Output:
int(sin(a + b*x + c*x^2)/x^2 - (b*cos(a + b*x + c*x^2))/x, x)
\[ \int \left (-\frac {b \cos \left (a+b x+c x^2\right )}{x}+\frac {\sin \left (a+b x+c x^2\right )}{x^2}\right ) \, dx=\frac {2 \left (\int \cos \left (c \,x^{2}+b x +a \right )d x \right ) b c x -\sin \left (c \,x^{2}+b x +a \right ) b -2 a c x}{b x} \] Input:
int(-b*cos(c*x^2+b*x+a)/x+sin(c*x^2+b*x+a)/x^2,x)
Output:
(2*int(cos(a + b*x + c*x**2),x)*b*c*x - sin(a + b*x + c*x**2)*b - 2*a*c*x) /(b*x)