3.2.56 \(\int \sin (b (c+d x)^2) \, dx\) [156]

Optimal. Leaf size=39 \[ \frac {\sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d} \]

[Out]

1/2*FresnelS((d*x+c)*b^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/d/b^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3432} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[b*(c + d*x)^2],x]

[Out]

(Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(Sqrt[b]*d)

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin {align*} \int \sin \left (b (c+d x)^2\right ) \, dx &=\frac {\sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 39, normalized size = 1.00 \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} (c+d x)\right )}{\sqrt {b} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[b*(c + d*x)^2],x]

[Out]

(Sqrt[Pi/2]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*(c + d*x)])/(Sqrt[b]*d)

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Maple [A]
time = 0.03, size = 42, normalized size = 1.08

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b \,d^{2} x +b c d \right )}{\sqrt {\pi }\, \sqrt {b \,d^{2}}}\right )}{2 \sqrt {b \,d^{2}}}\)	\(42\)
risch	\(\frac {i \sqrt {\pi }\, \erf \left (d \sqrt {i b}\, x +\frac {i b c}{\sqrt {i b}}\right )}{4 d \sqrt {i b}}+\frac {i \sqrt {\pi }\, \erf \left (-d \sqrt {-i b}\, x +\frac {i b c}{\sqrt {-i b}}\right )}{4 d \sqrt {-i b}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)*Pi^(1/2)/(b*d^2)^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)/(b*d^2)^(1/2)*(b*d^2*x+b*c*d))

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Maxima [C] Result contains complex when optimal does not.
time = 0.30, size = 53, normalized size = 1.36 \begin {gather*} \frac {\sqrt {2} \sqrt {\pi } {\left (\left (i + 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {i \, b}}\right ) + \left (i - 1\right ) \, \operatorname {erf}\left (\frac {i \, b d x + i \, b c}{\sqrt {-i \, b}}\right )\right )}}{8 \, \sqrt {b} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*(d*x+c)^2),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*sqrt(pi)*((I + 1)*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I - 1)*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))/(s
qrt(b)*d)

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Fricas [A]
time = 0.39, size = 45, normalized size = 1.15 \begin {gather*} \frac {\sqrt {2} \pi \sqrt {\frac {b d^{2}}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b d^{2}}{\pi }} {\left (d x + c\right )}}{d}\right )}{2 \, b d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*(d*x+c)^2),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*pi*sqrt(b*d^2/pi)*fresnel_sin(sqrt(2)*sqrt(b*d^2/pi)*(d*x + c)/d)/(b*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (b \left (c + d x\right )^{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*(d*x+c)**2),x)

[Out]

Integral(sin(b*(c + d*x)**2), x)

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Giac [C] Result contains complex when optimal does not.
time = 3.28, size = 143, normalized size = 3.67 \begin {gather*} -\frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{4 \, \sqrt {b d^{2}} {\left (\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} + \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )} {\left (x + \frac {c}{d}\right )}\right )}{4 \, \sqrt {b d^{2}} {\left (-\frac {i \, b d^{2}}{\sqrt {b^{2} d^{4}}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*(d*x+c)^2),x, algorithm="giac")

[Out]

-1/4*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(I*b*d^2/sqrt(b^2*d^4) + 1)*(x + c/d))/(sqrt(b*d^2)*(I*b*
d^2/sqrt(b^2*d^4) + 1)) + 1/4*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1)*(x
+ c/d))/(sqrt(b*d^2)*(-I*b*d^2/sqrt(b^2*d^4) + 1))

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Mupad [B]
time = 0.08, size = 41, normalized size = 1.05 \begin {gather*} \frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,b\,d\,\sqrt {\frac {1}{b\,d^2}}\,\left (c+d\,x\right )}{\sqrt {\pi }}\right )\,\sqrt {\frac {1}{b\,d^2}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*(c + d*x)^2),x)

[Out]

(2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*b*d*(1/(b*d^2))^(1/2)*(c + d*x))/pi^(1/2))*(1/(b*d^2))^(1/2))/2

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