Optimal. Leaf size=110 \[ \sqrt{2 \pi } \sqrt{c} \cos \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )-\sqrt{2 \pi } \sqrt{c} \sin \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-\frac{\sin \left (a+b x+c x^2\right )}{x} \]
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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.
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Rule 3465
Rule 3448
Rule 3352
Rule 3351
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.
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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.
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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.
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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.
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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.
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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.
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