3.2.62 \(\int \sin (\frac {b}{(c+d x)^2}) \, dx\) [162]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3440, 3468, 3433} \begin {gather*} \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} \end {gather*}

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 3433

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

Rule 3440

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 3468

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]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 60, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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 = 0.04, size = 52, normalized size = 0.87

method	result	size
derivativedivides	\(-\frac {-\left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )}{d}\)	\(52\)
default	\(-\frac {-\left (d x +c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )+\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )}{d}\)	\(52\)
risch	\(-\frac {b \sqrt {\pi }\, \erf \left (\frac {\sqrt {i b}}{d x +c}\right )}{2 d \sqrt {i b}}-\frac {b \sqrt {\pi }\, \erf \left (\frac {\sqrt {-i b}}{d x +c}\right )}{2 d \sqrt {-i b}}-\frac {\left (-d x -c \right ) \sin \left (\frac {b}{\left (d x +c \right )^{2}}\right )}{d}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 = 0.37, size = 73, normalized size = 1.22 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (\frac {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 [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.

[In]

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

[Out]

integrate(sin(b/(d*x + c)^2), x)

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Mupad [B]
time = 5.20, size = 52, normalized size = 0.87 \begin {gather*} \frac {\sin \left (\frac {b}{{\left (c+d\,x\right )}^2}\right )\,\left (c+d\,x\right )}{d}-\frac {\sqrt {2}\,\sqrt {b}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\sqrt {b}}{\sqrt {\pi }\,\left (c+d\,x\right )}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b/(c + d*x)^2),x)

[Out]

(sin(b/(c + d*x)^2)*(c + d*x))/d - (2^(1/2)*b^(1/2)*pi^(1/2)*fresnelc((2^(1/2)*b^(1/2))/(pi^(1/2)*(c + d*x))))
/d

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