Optimal. Leaf size=279 \[ \frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}-\frac {i d (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{8 \pi b^2}+\frac {(a+b x) (b c-a d) S(a+b x)^2}{b^2}-\frac {(b c-a d) S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b^2}+\frac {2 (b c-a d) S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac {d C(a+b x) S(a+b x)}{2 \pi b^2}+\frac {d (a+b x)^2 S(a+b x)^2}{2 b^2}+\frac {d (a+b x) S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}+\frac {d \cos \left (\pi (a+b x)^2\right )}{4 \pi ^2 b^2} \]
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Rubi [A] time = 0.20, antiderivative size = 279, normalized size of antiderivative = 1.00, 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 2638
Rule 3351
Rule 3379
Rule 6420
Rule 6430
Rule 6432
Rule 6446
Rule 6452
Rule 6454
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*}
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Mathematica [F] time = 0.62, size = 0, normalized size = 0.00 \[ \int (c+d x) S(a+b x)^2 \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d x + c\right )} {\rm fresnels}\left (b x + a\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right ) \mathrm {S}\left (b x +a \right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {FresnelS}\left (a+b\,x\right )}^2\,\left (c+d\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) S^{2}\left (a + b x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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