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

Optimal. Leaf size=105 \[ -\frac {\sqrt {2 \pi } \sqrt {b} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d}+\frac {\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d}+\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3359, 3387, 3354, 3352, 3351} \[ -\frac {\sqrt {2 \pi } \sqrt {b} \cos (a) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{c+d x}\right )}{d}+\frac {\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d}+\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^2],x]

[Out]

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

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]

Rule 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

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]

Rubi steps

\begin {align*} \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\sin \left (a+b x^2\right )}{x^2} \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}-\frac {(2 b) \operatorname {Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}-\frac {(2 b \cos (a)) \operatorname {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac {1}{c+d x}\right )}{d}+\frac {(2 b \sin (a)) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac {1}{c+d x}\right )}{d}\\ &=-\frac {\sqrt {b} \sqrt {2 \pi } \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d}+\frac {\sqrt {b} \sqrt {2 \pi } S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right ) \sin (a)}{d}+\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 100, normalized size = 0.95 \[ \frac {\sqrt {2 \pi } \left (-\sqrt {b}\right ) \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+\sqrt {2 \pi } \sqrt {b} \sin (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^2],x]

[Out]

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

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fricas [A]  time = 0.77, size = 137, normalized size = 1.30 \[ -\frac {\sqrt {2} \pi d \sqrt {\frac {b}{\pi d^{2}}} \cos \relax (a) \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) - \sqrt {2} \pi d \sqrt {\frac {b}{\pi d^{2}}} \operatorname {S}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) \sin \relax (a) - {\left (d x + c\right )} \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (a + \frac {b}{{\left (d x + c\right )}^{2}}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(a + b/(d*x + c)^2), x)

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maple [A]  time = 0.04, size = 80, normalized size = 0.76 \[ -\frac {-\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{2}}\right )+\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )-\sin \relax (a ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(-(d*x+c)*sin(a+b/(d*x+c)^2)+b^(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c))-
sin(a)*FresnelS(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 \[ b d \int \frac {x \cos \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\,{d x} + b d \int \frac {x \cos \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \cos \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2} + {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )^{2}}\,{d x} + x \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/(c + d*x)^2),x)

[Out]

int(sin(a + b/(c + d*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + \frac {b}{\left (c + d x\right )^{2}} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b/(c + d*x)**2), x)

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