3.162 \(\int \sin (\frac{b}{(c+d x)^2}) \, dx\)

Optimal. Leaf size=60 \[ \frac{(c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d}-\frac{\sqrt{2 \pi } \sqrt{b} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d} \]

[Out]

-((Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d) + ((c + d*x)*Sin[b/(c + d*x)^2])/d

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Rubi [A]  time = 0.0335677, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3359, 3387, 3352} \[ \frac{(c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d}-\frac{\sqrt{2 \pi } \sqrt{b} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[b/(c + d*x)^2],x]

[Out]

-((Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d) + ((c + d*x)*Sin[b/(c + d*x)^2])/d

Rule 3359

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(
a + b*Sin[c + d/x^n])^p/x^2, x], x, 1/(e + f*x)], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && ILtQ[n,
0] && EqQ[n, -2]

Rule 3387

Int[((e_.)*(x_))^(m_)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[((e*x)^(m + 1)*Sin[c + d*x^n])/(e*(m + 1
)), x] - Dist[(d*n)/(e^n*(m + 1)), Int[(e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n,
0] && LtQ[m, -1]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[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 (\frac{b}{(c+d x)^2}\right ) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sin \left (b x^2\right )}{x^2} \, dx,x,\frac{1}{c+d x}\right )}{d}\\ &=\frac{(c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d}-\frac{(2 b) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac{1}{c+d x}\right )}{d}\\ &=-\frac{\sqrt{b} \sqrt{2 \pi } C\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{c+d x}\right )}{d}+\frac{(c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.0328287, size = 60, normalized size = 1. \[ \frac{(c+d x) \sin \left (\frac{b}{(c+d x)^2}\right )}{d}-\frac{\sqrt{2 \pi } \sqrt{b} \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[b/(c + d*x)^2],x]

[Out]

-((Sqrt[b]*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/d) + ((c + d*x)*Sin[b/(c + d*x)^2])/d

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Maple [A]  time = 0.007, size = 52, normalized size = 0.9 \begin{align*} -{\frac{1}{d} \left ( - \left ( dx+c \right ) \sin \left ({\frac{b}{ \left ( dx+c \right ) ^{2}}} \right ) +\sqrt{b}\sqrt{2}\sqrt{\pi }{\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi } \left ( dx+c \right ) }\sqrt{b}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(-(d*x+c)*sin(b/(d*x+c)^2)+b^(1/2)*2^(1/2)*Pi^(1/2)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b d \int \frac{x \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\,{d x} + b d \int \frac{x \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \cos \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} +{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}}\,{d x} + x \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\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]

b*d*integrate(x*cos(b/(d^2*x^2 + 2*c*d*x + c^2))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x) + b*d*integrate
(x*cos(b/(d^2*x^2 + 2*c*d*x + c^2))/((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*cos(b/(d^2*x^2 + 2*c*d*x + c^2)
)^2 + (d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*sin(b/(d^2*x^2 + 2*c*d*x + c^2))^2), x) + x*sin(b/(d^2*x^2 + 2
*c*d*x + c^2))

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Fricas [A]  time = 1.69066, size = 177, normalized size = 2.95 \begin{align*} -\frac{\sqrt{2} \pi d \sqrt{\frac{b}{\pi d^{2}}} \operatorname{C}\left (\frac{\sqrt{2} d \sqrt{\frac{b}{\pi d^{2}}}}{d x + c}\right ) -{\left (d x + c\right )} \sin \left (\frac{b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*pi*d*sqrt(b/(pi*d^2))*fresnel_cos(sqrt(2)*d*sqrt(b/(pi*d^2))/(d*x + c)) - (d*x + c)*sin(b/(d^2*x^2 +
2*c*d*x + c^2)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b/(d*x + c)^2), x)