Integrand size = 12, antiderivative size = 105 \[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=-\frac {\sqrt {b} \sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{d}+\frac {\sqrt {b} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right ) \sin (a)}{d}+\frac {(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d} \]
(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+FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)) *sin(a)*b^(1/2)*2^(1/2)*Pi^(1/2)/d
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.95 \[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=\frac {-\sqrt {b} \sqrt {2 \pi } \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )+\sqrt {b} \sqrt {2 \pi } \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right ) \sin (a)+(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d} \]
(-(Sqrt[b]*Sqrt[2*Pi]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]) + S qrt[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
Time = 0.34 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3840, 3868, 3835, 3832, 3833}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx\) |
\(\Big \downarrow \) 3840 |
\(\displaystyle -\frac {\int (c+d x)^2 \sin \left (a+\frac {b}{(c+d x)^2}\right )d\frac {1}{c+d x}}{d}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle -\frac {2 b \int \cos \left (a+\frac {b}{(c+d x)^2}\right )d\frac {1}{c+d x}-(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}\) |
\(\Big \downarrow \) 3835 |
\(\displaystyle -\frac {2 b \left (\cos (a) \int \cos \left (\frac {b}{(c+d x)^2}\right )d\frac {1}{c+d x}-\sin (a) \int \sin \left (\frac {b}{(c+d x)^2}\right )d\frac {1}{c+d x}\right )-(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle -\frac {2 b \left (\cos (a) \int \cos \left (\frac {b}{(c+d x)^2}\right )d\frac {1}{c+d x}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{\sqrt {b}}\right )-(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle -\frac {2 b \left (\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{c+d x}\right )}{\sqrt {b}}\right )-(c+d x) \sin \left (a+\frac {b}{(c+d x)^2}\right )}{d}\) |
-((2*b*((Sqrt[Pi/2]*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)])/Sqrt[ b] - (Sqrt[Pi/2]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)]*Sin[a])/Sqrt[b]) - (c + d*x)*Sin[a + b/(c + d*x)^2])/d)
3.2.79.3.1 Defintions of rubi rules used
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]
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]
Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[Cos[c] In t[Cos[d*(e + f*x)^2], x], x] - Simp[Sin[c] Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]
Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_S ymbol] :> Simp[-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]
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] - Simp[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]
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(-\frac {-\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{2}}\right )+\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {C}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )-\sin \left (a \right ) \operatorname {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )}{d}\) | \(80\) |
default | \(-\frac {-\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{2}}\right )+\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {C}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )-\sin \left (a \right ) \operatorname {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )}\right )\right )}{d}\) | \(80\) |
risch | \(-\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{d x +c}\right ) {\mathrm e}^{-i a}}{2 d \sqrt {i b}}-\frac {b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{d x +c}\right ) {\mathrm e}^{i a}}{2 d \sqrt {-i b}}-\frac {\left (-d x -c \right ) \sin \left (\frac {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}{\left (d x +c \right )^{2}}\right )}{d}\) | \(115\) |
-1/d*(-(d*x+c)*sin(a+b/(d*x+c)^2)+b^(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*Fresnel C(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))))
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.30 \[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=-\frac {\sqrt {2} \pi d \sqrt {\frac {b}{\pi d^{2}}} \cos \left (a\right ) \operatorname {C}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) - \sqrt {2} \pi d \sqrt {\frac {b}{\pi d^{2}}} \operatorname {S}\left (\frac {\sqrt {2} d \sqrt {\frac {b}{\pi d^{2}}}}{d x + c}\right ) \sin \left (a\right ) - {\left (d x + c\right )} \sin \left (\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{d} \]
-(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
\[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=\int \sin {\left (a + \frac {b}{\left (c + d x\right )^{2}} \right )}\, dx \]
\[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=\int { \sin \left (a + \frac {b}{{\left (d x + c\right )}^{2}}\right ) \,d x } \]
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), x) + b*d*integrate(x*c os((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))
\[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=\int { \sin \left (a + \frac {b}{{\left (d x + c\right )}^{2}}\right ) \,d x } \]
Timed out. \[ \int \sin \left (a+\frac {b}{(c+d x)^2}\right ) \, dx=\int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^2}\right ) \,d x \]