Optimal. Leaf size=193 \[ -\frac{d (b c-a d) \text{FresnelC}(a+b x)}{\pi b^3}-\frac{(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac{(b c-a d)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{d (a+b x) (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{2 d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{d^2 (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac{(c+d x)^3 S(a+b x)}{3 d} \]
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Rubi [A] time = 0.225743, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6428, 3433, 3351, 3379, 2638, 3385, 3352, 3296, 2637} \[ -\frac{d (b c-a d) \text{FresnelC}(a+b x)}{\pi b^3}-\frac{(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac{(b c-a d)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{d (a+b x) (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{2 d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{d^2 (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac{(c+d x)^3 S(a+b x)}{3 d} \]
Antiderivative was successfully verified.
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Rule 6428
Rule 3433
Rule 3351
Rule 3379
Rule 2638
Rule 3385
Rule 3352
Rule 3296
Rule 2637
Rubi steps
\begin{align*} \int (c+d x)^2 S(a+b x) \, dx &=\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{b \int (c+d x)^3 \sin \left (\frac{1}{2} \pi (a+b x)^2\right ) \, dx}{3 d}\\ &=\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{\operatorname{Subst}\left (\int \left (b^3 c^3 \left (1-\frac{a d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )}{b^3 c^3}\right ) \sin \left (\frac{\pi x^2}{2}\right )+3 b^2 c^2 d \left (1+\frac{a d (-2 b c+a d)}{b^2 c^2}\right ) x \sin \left (\frac{\pi x^2}{2}\right )+3 b c d^2 \left (1-\frac{a d}{b c}\right ) x^2 \sin \left (\frac{\pi x^2}{2}\right )+d^3 x^3 \sin \left (\frac{\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{3 b^3 d}\\ &=\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{d^2 \operatorname{Subst}\left (\int x^3 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}-\frac{(d (b c-a d)) \operatorname{Subst}\left (\int x^2 \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac{(b c-a d)^2 \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac{(b c-a d)^3 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3 d}\\ &=\frac{d (b c-a d) (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac{(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{d^2 \operatorname{Subst}\left (\int x \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{6 b^3}-\frac{(b c-a d)^2 \operatorname{Subst}\left (\int \sin \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^3}-\frac{(d (b c-a d)) \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=\frac{(b c-a d)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{d (b c-a d) (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{d^2 (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }-\frac{d (b c-a d) C(a+b x)}{b^3 \pi }-\frac{(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{d^2 \operatorname{Subst}\left (\int \cos \left (\frac{\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{3 b^3 \pi }\\ &=\frac{(b c-a d)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{d (b c-a d) (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac{d^2 (a+b x)^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }-\frac{d (b c-a d) C(a+b x)}{b^3 \pi }-\frac{(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac{(c+d x)^3 S(a+b x)}{3 d}-\frac{2 d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}\\ \end{align*}
Mathematica [A] time = 0.457134, size = 236, normalized size = 1.22 \[ \frac{\pi ^2 S(a+b x) \left (-3 a^2 b c d+a^3 d^2+3 a b^2 c^2+b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )\right )+\pi a^2 d^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi b^2 c^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi b^2 c d x \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^2 d^2 x^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi d (a d-b c) \text{FresnelC}(a+b x)-3 \pi a b c d \cos \left (\frac{1}{2} \pi (a+b x)^2\right )-2 d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-\pi a b d^2 x \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.055, size = 251, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelS} \left ( bx+a \right ) \left ( d \left ( bx+a \right ) -ad+bc \right ) ^{3}}{3\,d{b}^{2}}}-{\frac{1}{3\,d{b}^{2}} \left ( -{\frac{{d}^{3} \left ( bx+a \right ) ^{2}}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+2\,{\frac{{d}^{3}\sin \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{{\pi }^{2}}}-{\frac{ \left ( -3\,a{d}^{3}+3\,bc{d}^{2} \right ) \left ( bx+a \right ) }{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+{\frac{ \left ( -3\,a{d}^{3}+3\,bc{d}^{2} \right ){\it FresnelC} \left ( bx+a \right ) }{\pi }}-{\frac{3\,{a}^{2}{d}^{3}-6\,abc{d}^{2}+3\,{b}^{2}{c}^{2}d}{\pi }\cos \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{a}^{3}{d}^{3}{\it FresnelS} \left ( bx+a \right ) +3\,{a}^{2}bc{d}^{2}{\it FresnelS} \left ( bx+a \right ) -3\,a{b}^{2}{c}^{2}d{\it FresnelS} \left ( bx+a \right ) +{b}^{3}{c}^{3}{\it FresnelS} \left ( bx+a \right ) \right ) } \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{\left (d x + c\right )}^{2}{\rm fresnels}\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 ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )}{\rm fresnels}\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 \left (c + d x\right )^{2} S\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{\left (d x + c\right )}^{2}{\rm fresnels}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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