Optimal. Leaf size=95 \[ -\frac {a^2 C(a+b x)}{2 b^2}+\frac {S(a+b x)}{2 \pi b^2}+\frac {a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac {(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac {1}{2} x^2 C(a+b x) \]
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Rubi [A] time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351} \[ -\frac {a^2 \text {FresnelC}(a+b x)}{2 b^2}+\frac {S(a+b x)}{2 \pi b^2}+\frac {a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^2}-\frac {(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2}+\frac {1}{2} x^2 \text {FresnelC}(a+b x) \]
Antiderivative was successfully verified.
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Rule 2637
Rule 3351
Rule 3352
Rule 3380
Rule 3386
Rule 3434
Rule 6429
Rubi steps
\begin {align*} \int x C(a+b x) \, dx &=\frac {1}{2} x^2 C(a+b x)-\frac {1}{2} b \int x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac {1}{2} x^2 C(a+b x)-\frac {\operatorname {Subst}\left (\int \left (a^2 \cos \left (\frac {\pi x^2}{2}\right )-2 a x \cos \left (\frac {\pi x^2}{2}\right )+x^2 \cos \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{2 b^2}\\ &=\frac {1}{2} x^2 C(a+b x)-\frac {\operatorname {Subst}\left (\int x^2 \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2}+\frac {a \operatorname {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^2}-\frac {a^2 \operatorname {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2}\\ &=-\frac {a^2 C(a+b x)}{2 b^2}+\frac {1}{2} x^2 C(a+b x)-\frac {(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }+\frac {a \operatorname {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^2}+\frac {\operatorname {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{2 b^2 \pi }\\ &=-\frac {a^2 C(a+b x)}{2 b^2}+\frac {1}{2} x^2 C(a+b x)+\frac {S(a+b x)}{2 b^2 \pi }+\frac {a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^2 \pi }-\frac {(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^2 \pi }\\ \end {align*}
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Mathematica [A] time = 0.18, size = 59, normalized size = 0.62 \[ \frac {\left (\pi b^2 x^2-\pi a^2\right ) C(a+b x)+S(a+b x)+(a-b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 \pi b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x {\rm fresnelc}\left (b x + a\right ), x\right ) \]
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 \[ \int x {\rm fresnelc}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.02, size = 79, normalized size = 0.83 \[ \frac {\FresnelC \left (b x +a \right ) \left (\frac {\left (b x +a \right )^{2}}{2}-a \left (b x +a \right )\right )-\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }+\frac {\mathrm {S}\left (b x +a \right )}{2 \pi }+\frac {a \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }}{b^{2}} \]
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 \[ \int x {\rm fresnelc}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {FresnelC}\left (a+b\,x\right ) \,d x \]
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 \[ \int x C\left (a + b x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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