\(\int \frac {\sin (a+\frac {b}{x^2})}{x^4} \, dx\) [123]

Optimal result

Integrand size = 12, antiderivative size = 97 \[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{2 b^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right ) \sin (a)}{2 b^{3/2}} \] Output:

1/2*cos(a+b/x^2)/b/x-1/4*2^(1/2)*Pi^(1/2)*cos(a)*FresnelC(b^(1/2)*2^(1/2)/ 
Pi^(1/2)/x)/b^(3/2)+1/4*2^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2) 
/x)*sin(a)/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\frac {2 \sqrt {b} \cos \left (a+\frac {b}{x^2}\right )-\sqrt {2 \pi } x \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )+\sqrt {2 \pi } x \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right ) \sin (a)}{4 b^{3/2} x} \] Input:

Integrate[Sin[a + b/x^2]/x^4,x]
 

Output:

(2*Sqrt[b]*Cos[a + b/x^2] - Sqrt[2*Pi]*x*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/P 
i])/x] + Sqrt[2*Pi]*x*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/x]*Sin[a])/(4*b^(3/2)* 
x)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3890, 3866, 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.

\(\displaystyle \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx\)

\(\Big \downarrow \) 3890

\(\displaystyle -\int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^2}d\frac {1}{x}\)

\(\Big \downarrow \) 3866

\(\displaystyle \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\int \cos \left (a+\frac {b}{x^2}\right )d\frac {1}{x}}{2 b}\)

\(\Big \downarrow \) 3835

\(\displaystyle \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\cos (a) \int \cos \left (\frac {b}{x^2}\right )d\frac {1}{x}-\sin (a) \int \sin \left (\frac {b}{x^2}\right )d\frac {1}{x}}{2 b}\)

\(\Big \downarrow \) 3832

\(\displaystyle \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\cos (a) \int \cos \left (\frac {b}{x^2}\right )d\frac {1}{x}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{\sqrt {b}}}{2 b}\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {\cos \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{\sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{x}\right )}{\sqrt {b}}}{2 b}\)

Input:

Int[Sin[a + b/x^2]/x^4,x]
 

Output:

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

Defintions of rubi rules used

rule 3832

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]
 

rule 3833

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 3835

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]
 

rule 3866

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

rule 3890

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

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67

method	result	size
derivativedivides	\(\frac {\cos \left (a +\frac {b}{x^{2}}\right )}{2 b x}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )-\sin \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right )}{4 b^{\frac {3}{2}}}\)	\(65\)
default	\(\frac {\cos \left (a +\frac {b}{x^{2}}\right )}{2 b x}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )-\sin \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )\right )}{4 b^{\frac {3}{2}}}\)	\(65\)
risch	\(-\frac {{\mathrm e}^{i a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-i b}}{x}\right )}{8 b \sqrt {-i b}}-\frac {{\mathrm e}^{-i a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {i b}}{x}\right )}{8 b \sqrt {i b}}+\frac {\cos \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{2 b x}\)	\(82\)
meijerg	\(-\frac {\sqrt {\pi }\, \cos \left (a \right ) \sqrt {2}\, \left (-\frac {\sqrt {2}\, \sqrt {b}\, \cos \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x}+\frac {\operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )}{2}\right )}{2 b^{\frac {3}{2}}}-\frac {\sqrt {\pi }\, \sin \left (a \right ) \sqrt {2}\, \left (b^{2}\right )^{\frac {1}{4}} \left (\frac {\sqrt {2}\, \left (b^{2}\right )^{\frac {3}{4}} \sin \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x b}-\frac {\left (b^{2}\right )^{\frac {3}{4}} \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, x}\right )}{2 b^{\frac {3}{2}}}\right )}{2 b^{2}}\)	\(120\)

Input:

int(sin(a+b/x^2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

1/2*cos(a+b/x^2)/b/x-1/4/b^(3/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelC(b^(1/2) 
*2^(1/2)/Pi^(1/2)/x)-sin(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {\sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) - \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{x}\right ) \sin \left (a\right ) - 2 \, b \cos \left (\frac {a x^{2} + b}{x^{2}}\right )}{4 \, b^{2} x} \] Input:

integrate(sin(a+b/x^2)/x^4,x, algorithm="fricas")
 

Output:

-1/4*(sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*sqrt(b/pi)/x) - s 
qrt(2)*pi*x*sqrt(b/pi)*fresnel_sin(sqrt(2)*sqrt(b/pi)/x)*sin(a) - 2*b*cos( 
(a*x^2 + b)/x^2))/(b^2*x)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\sin {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \] Input:

integrate(sin(a+b/x**2)/x**4,x)
 

Output:

Integral(sin(a + b/x**2)/x**4, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.76 \[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {\sqrt {2} {\left (x^{4}\right )}^{\frac {3}{2}} {\left ({\left (\left (i - 1\right ) \, \Gamma \left (\frac {3}{2}, \frac {i \, b}{x^{2}}\right ) - \left (i + 1\right ) \, \Gamma \left (\frac {3}{2}, -\frac {i \, b}{x^{2}}\right )\right )} \cos \left (a\right ) + {\left (\left (i + 1\right ) \, \Gamma \left (\frac {3}{2}, \frac {i \, b}{x^{2}}\right ) - \left (i - 1\right ) \, \Gamma \left (\frac {3}{2}, -\frac {i \, b}{x^{2}}\right )\right )} \sin \left (a\right )\right )} \left (\frac {b^{2}}{x^{4}}\right )^{\frac {3}{4}}}{8 \, b^{3} x^{3}} \] Input:

integrate(sin(a+b/x^2)/x^4,x, algorithm="maxima")
 

Output:

-1/8*sqrt(2)*(((I - 1)*gamma(3/2, I*b/x^2) - (I + 1)*gamma(3/2, -I*b/x^2)) 
*cos(a) + ((I + 1)*gamma(3/2, I*b/x^2) - (I - 1)*gamma(3/2, -I*b/x^2))*sin 
(a))*(x^4)^(3/2)*(b^2/x^4)^(3/4)/(b^3*x^3)
 

Giac [F]

\[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int { \frac {\sin \left (a + \frac {b}{x^{2}}\right )}{x^{4}} \,d x } \] Input:

integrate(sin(a+b/x^2)/x^4,x, algorithm="giac")
 

Output:

integrate(sin(a + b/x^2)/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \,d x \] Input:

int(sin(a + b/x^2)/x^4,x)
 

Output:

int(sin(a + b/x^2)/x^4, x)
 

Reduce [F]

\[ \int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\sin \left (\frac {a \,x^{2}+b}{x^{2}}\right )}{x^{4}}d x \] Input:

int(sin(a+b/x^2)/x^4,x)
 

Output:

int(sin((a*x**2 + b)/x**2)/x**4,x)