3.156 \(\int \sin (b (c+d x)^2) \, dx\)

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]

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

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Rubi [A]  time = 0.0080351, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3351} \[ \frac{\sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d} \]

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 3351

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

Mathematica [A]  time = 0.0151665, size = 39, normalized size = 1. \[ \frac{\sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} (c+d x)\right )}{\sqrt{b} d} \]

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 = 0.007, size = 42, normalized size = 1.1 \begin{align*}{\frac{\sqrt{2}\sqrt{\pi }}{2}{\it FresnelS} \left ({\frac{\sqrt{2} \left ( b{d}^{2}x+bcd \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{d}^{2}b}}}} \right ){\frac{1}{\sqrt{{d}^{2}b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.83639, size = 193, normalized size = 4.95 \begin{align*} \frac{\sqrt{\pi }{\left ({\left (i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \operatorname{erf}\left (\frac{i \, b d x + i \, b c}{\sqrt{i \, b}}\right ) +{\left (i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \operatorname{erf}\left (\frac{i \, b d x + i \, b c}{\sqrt{-i \, b}}\right )\right )}}{8 \, d \sqrt{{\left | b \right |}}} \end{align*}

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(pi)*((I*cos(1/4*pi + 1/2*arctan2(0, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) + sin(1/4*pi + 1/2*arcta
n2(0, b)) - sin(-1/4*pi + 1/2*arctan2(0, b)))*erf((I*b*d*x + I*b*c)/sqrt(I*b)) + (I*cos(1/4*pi + 1/2*arctan2(0
, b)) + I*cos(-1/4*pi + 1/2*arctan2(0, b)) - sin(1/4*pi + 1/2*arctan2(0, b)) + sin(-1/4*pi + 1/2*arctan2(0, b)
))*erf((I*b*d*x + I*b*c)/sqrt(-I*b)))/(d*sqrt(abs(b)))

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Fricas [A]  time = 1.5917, size = 117, normalized size = 3. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (b \left (c + d x\right )^{2} \right )}\, dx \end{align*}

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]  time = 1.10767, size = 193, normalized size = 4.95 \begin{align*} -\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{align*}

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))