Optimal. Leaf size=148 \[ \frac{a^3 \text{FresnelC}(a+b x)}{3 b^3}-\frac{a^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{a S(a+b x)}{\pi b^3}+\frac{a (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{(a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{1}{3} x^3 \text{FresnelC}(a+b x) \]
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Rubi [A] time = 0.122512, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638} \[ \frac{a^3 \text{FresnelC}(a+b x)}{3 b^3}-\frac{a^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{a S(a+b x)}{\pi b^3}+\frac{a (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{(a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{1}{3} x^3 \text{FresnelC}(a+b x) \]
Antiderivative was successfully verified.
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Rule 6429
Rule 3434
Rule 3352
Rule 3380
Rule 2637
Rule 3386
Rule 3351
Rule 3296
Rule 2638
Rubi steps
\begin{align*} \int x^2 C(a+b x) \, dx &=\frac{1}{3} x^3 C(a+b x)-\frac{1}{3} b \int x^3 \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac{1}{3} x^3 C(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-a^3 \cos \left (\frac{\pi x^2}{2}\right )+3 a^2 x \cos \left (\frac{\pi x^2}{2}\right )-3 a x^2 \cos \left (\frac{\pi x^2}{2}\right )+x^3 \cos \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac{1}{3} x^3 C(a+b x)-\frac{\operatorname{Subst}\left (\int x^3 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}+\frac{a \operatorname{Subst}\left (\int x^2 \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac{a^2 \operatorname{Subst}\left (\int x \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}+\frac{a^3 \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac{a^3 C(a+b x)}{3 b^3}+\frac{1}{3} x^3 C(a+b x)+\frac{a (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{\operatorname{Subst}\left (\int x \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{6 b^3}-\frac{a^2 \operatorname{Subst}\left (\int \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^3}-\frac{a \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=\frac{a^3 C(a+b x)}{3 b^3}+\frac{1}{3} x^3 C(a+b x)-\frac{a S(a+b x)}{b^3 \pi }-\frac{a^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{a (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{(a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac{\operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{3 b^3 \pi }\\ &=-\frac{2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}+\frac{a^3 C(a+b x)}{3 b^3}+\frac{1}{3} x^3 C(a+b x)-\frac{a S(a+b x)}{b^3 \pi }-\frac{a^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{a (a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{(a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }\\ \end{align*}
Mathematica [A] time = 0.29704, size = 116, normalized size = 0.78 \[ -\frac{-\pi ^2 \left (a^3+b^3 x^3\right ) \text{FresnelC}(a+b x)+\pi a^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^2 x^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi a S(a+b x)-\pi a b x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.052, size = 122, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3}{\it FresnelC} \left ( bx+a \right ) }{3}}-{\frac{ \left ( bx+a \right ) ^{2}}{3\,\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{2}{3\,{\pi }^{2}}\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{a \left ( bx+a \right ) }{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{a{\it FresnelS} \left ( bx+a \right ) }{\pi }}-{\frac{{a}^{2}}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{{a}^{3}{\it FresnelC} \left ( bx+a \right ) }{3}} \right ) } \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 x^{2}{\rm fresnelc}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2}{\rm fresnelc}\left (b x + a\right ), x\right ) \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 x^{2} C\left (a + b x\right )\, 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 x^{2}{\rm fresnelc}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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