Optimal. Leaf size=279 \[ \frac{i d (a+b x)^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}-\frac{i d (a+b x)^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}+\frac{(a+b x) (b c-a d) S(a+b x)^2}{b^2}-\frac{(b c-a d) S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b^2}+\frac{2 (b c-a d) S(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac{d \text{FresnelC}(a+b x) S(a+b x)}{2 \pi b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}+\frac{d (a+b x) S(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac{d \cos \left (\pi (a+b x)^2\right )}{4 \pi ^2 b^2} \]
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Rubi [A] time = 0.196207, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6432, 6420, 6452, 3351, 6430, 6454, 6446, 3379, 2638} \[ \frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}-\frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}+\frac{(a+b x) (b c-a d) S(a+b x)^2}{b^2}-\frac{(b c-a d) S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b^2}+\frac{2 (b c-a d) S(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac{d \text{FresnelC}(a+b x) S(a+b x)}{2 \pi b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}+\frac{d (a+b x) S(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac{d \cos \left (\pi (a+b x)^2\right )}{4 \pi ^2 b^2} \]
Antiderivative was successfully verified.
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Rule 6432
Rule 6420
Rule 6452
Rule 3351
Rule 6430
Rule 6454
Rule 6446
Rule 3379
Rule 2638
Rubi steps
\begin{align*} \int (c+d x) S(a+b x)^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (b c \left (1-\frac{a d}{b c}\right ) S(x)^2+d x S(x)^2\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac{d \operatorname{Subst}\left (\int x S(x)^2 \, dx,x,a+b x\right )}{b^2}+\frac{(b c-a d) \operatorname{Subst}\left (\int S(x)^2 \, dx,x,a+b x\right )}{b^2}\\ &=\frac{(b c-a d) (a+b x) S(a+b x)^2}{b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}-\frac{d \operatorname{Subst}\left (\int x^2 S(x) \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^2}-\frac{(2 (b c-a d)) \operatorname{Subst}\left (\int x S(x) \sin \left (\frac{\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^2}\\ &=\frac{2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }+\frac{d (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }+\frac{(b c-a d) (a+b x) S(a+b x)^2}{b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}-\frac{d \operatorname{Subst}\left (\int x \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{2 b^2 \pi }-\frac{d \operatorname{Subst}\left (\int \cos \left (\frac{\pi x^2}{2}\right ) S(x) \, dx,x,a+b x\right )}{b^2 \pi }-\frac{(b c-a d) \operatorname{Subst}\left (\int \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b^2 \pi }\\ &=\frac{2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }+\frac{d (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }-\frac{d C(a+b x) S(a+b x)}{2 b^2 \pi }+\frac{(b c-a d) (a+b x) S(a+b x)^2}{b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}-\frac{(b c-a d) S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} b^2 \pi }+\frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi }-\frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi }-\frac{d \operatorname{Subst}\left (\int \sin (\pi x) \, dx,x,(a+b x)^2\right )}{4 b^2 \pi }\\ &=\frac{d \cos \left (\pi (a+b x)^2\right )}{4 b^2 \pi ^2}+\frac{2 (b c-a d) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }+\frac{d (a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^2 \pi }-\frac{d C(a+b x) S(a+b x)}{2 b^2 \pi }+\frac{(b c-a d) (a+b x) S(a+b x)^2}{b^2}+\frac{d (a+b x)^2 S(a+b x)^2}{2 b^2}-\frac{(b c-a d) S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} b^2 \pi }+\frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi }-\frac{i d (a+b x)^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i \pi (a+b x)^2\right )}{8 b^2 \pi }\\ \end{align*}
Mathematica [F] time = 0.585015, size = 0, normalized size = 0. \[ \int (c+d x) S(a+b x)^2 \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) \left ({\it FresnelS} \left ( bx+a \right ) \right ) ^{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{\left (d x + c\right )}{\rm fresnels}\left (b x + a\right )^{2}\,{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 x + c\right )}{\rm fresnels}\left (b x + a\right )^{2}, 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 ) S^{2}\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 )}{\rm fresnels}\left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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