Optimal. Leaf size=69 \[ \frac {(a+b x) \text {FresnelC}(a+b x)^2}{b}+\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b \pi }-\frac {2 \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b \pi } \]
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Rubi [A]
time = 0.11, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6556, 6588,
3432} \begin {gather*} \frac {(a+b x) \text {FresnelC}(a+b x)^2}{b}-\frac {2 \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b}+\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b} \end {gather*}
Antiderivative was successfully verified.
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Rule 3432
Rule 6556
Rule 6588
Rubi steps
\begin {align*} \int C(a+b x)^2 \, dx &=\frac {(a+b x) C(a+b x)^2}{b}-2 \int (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) C(a+b x) \, dx\\ &=\frac {(a+b x) C(a+b x)^2}{b}-\frac {2 \text {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) C(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {(a+b x) C(a+b x)^2}{b}-\frac {2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b \pi }+\frac {\text {Subst}\left (\int \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b \pi }\\ &=\frac {(a+b x) C(a+b x)^2}{b}+\frac {S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b \pi }-\frac {2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b \pi }\\ \end {align*}
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Mathematica [A]
time = 0.01, size = 66, normalized size = 0.96 \begin {gather*} \frac {2 \pi (a+b x) \text {FresnelC}(a+b x)^2+\sqrt {2} S\left (\sqrt {2} (a+b x)\right )-4 \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b \pi } \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.48, size = 60, normalized size = 0.87
method | result | size |
derivativedivides | \(\frac {\FresnelC \left (b x +a \right )^{2} \left (b x +a \right )-\frac {2 \FresnelC \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \mathrm {S}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\) | \(60\) |
default | \(\frac {\FresnelC \left (b x +a \right )^{2} \left (b x +a \right )-\frac {2 \FresnelC \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\sqrt {2}\, \mathrm {S}\left (\left (b x +a \right ) \sqrt {2}\right )}{2 \pi }}{b}\) | \(60\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.35, size = 88, normalized size = 1.28 \begin {gather*} \frac {2 \, {\left (\pi b^{2} x + \pi a b\right )} \operatorname {C}\left (b x + a\right )^{2} - 4 \, b \operatorname {C}\left (b x + a\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) + \sqrt {2} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{2 \, \pi b^{2}} \end {gather*}
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 \begin {gather*} \int C^{2}\left (a + b x\right )\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {FresnelC}\left (a+b\,x\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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