\(\int \frac {\sin ^3(a+b x^2)}{x^2} \, dx\) [30]

Optimal result

Integrand size = 14, antiderivative size = 168 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\frac {3}{2} \sqrt {b} \sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\frac {1}{2} \sqrt {b} \sqrt {\frac {3 \pi }{2}} \cos (3 a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )-\frac {3}{2} \sqrt {b} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)+\frac {1}{2} \sqrt {b} \sqrt {\frac {3 \pi }{2}} \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right ) \sin (3 a)-\frac {\sin ^3\left (a+b x^2\right )}{x} \] Output:

3/4*b^(1/2)*2^(1/2)*Pi^(1/2)*cos(a)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*x)-1 
/4*b^(1/2)*6^(1/2)*Pi^(1/2)*cos(3*a)*FresnelC(b^(1/2)*6^(1/2)/Pi^(1/2)*x)- 
3/4*b^(1/2)*2^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*x)*sin(a)+1 
/4*b^(1/2)*6^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*6^(1/2)/Pi^(1/2)*x)*sin(3*a)- 
sin(b*x^2+a)^3/x
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\frac {3 \sqrt {b} \sqrt {2 \pi } x \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\sqrt {b} \sqrt {6 \pi } x \cos (3 a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )-3 \sqrt {b} \sqrt {2 \pi } x \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)+\sqrt {b} \sqrt {6 \pi } x \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right ) \sin (3 a)-3 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )}{4 x} \] Input:

Integrate[Sin[a + b*x^2]^3/x^2,x]
 

Output:

(3*Sqrt[b]*Sqrt[2*Pi]*x*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x] - Sqrt[b]*Sq 
rt[6*Pi]*x*Cos[3*a]*FresnelC[Sqrt[b]*Sqrt[6/Pi]*x] - 3*Sqrt[b]*Sqrt[2*Pi]* 
x*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a] + Sqrt[b]*Sqrt[6*Pi]*x*FresnelS[Sq 
rt[b]*Sqrt[6/Pi]*x]*Sin[3*a] - 3*Sin[a + b*x^2] + Sin[3*(a + b*x^2)])/(4*x 
)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3874, 5085, 2009}

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.

\(\displaystyle \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx\)

\(\Big \downarrow \) 3874

\(\displaystyle 6 b \int \cos \left (b x^2+a\right ) \sin ^2\left (b x^2+a\right )dx-\frac {\sin ^3\left (a+b x^2\right )}{x}\)

\(\Big \downarrow \) 5085

\(\displaystyle 6 b \int \left (\frac {1}{4} \cos \left (b x^2+a\right )-\frac {1}{4} \cos \left (3 b x^2+3 a\right )\right )dx-\frac {\sin ^3\left (a+b x^2\right )}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 6 b \left (\frac {\sqrt {\frac {\pi }{2}} \cos (a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{6}} \cos (3 a) \operatorname {FresnelC}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{2}} \sin (a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{4 \sqrt {b}}+\frac {\sqrt {\frac {\pi }{6}} \sin (3 a) \operatorname {FresnelS}\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )}{4 \sqrt {b}}\right )-\frac {\sin ^3\left (a+b x^2\right )}{x}\)

Input:

Int[Sin[a + b*x^2]^3/x^2,x]
 

Output:

6*b*((Sqrt[Pi/2]*Cos[a]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x])/(4*Sqrt[b]) - (Sqr 
t[Pi/6]*Cos[3*a]*FresnelC[Sqrt[b]*Sqrt[6/Pi]*x])/(4*Sqrt[b]) - (Sqrt[Pi/2] 
*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x]*Sin[a])/(4*Sqrt[b]) + (Sqrt[Pi/6]*FresnelS 
[Sqrt[b]*Sqrt[6/Pi]*x]*Sin[3*a])/(4*Sqrt[b])) - Sin[a + b*x^2]^3/x
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3874

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[x^(m + 
1)*(Sin[a + b*x^n]^p/(m + 1)), x] - Simp[b*n*(p/(m + 1))   Int[Sin[a + b*x^ 
n]^(p - 1)*Cos[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] && EqQ[ 
m + n, 0] && NeQ[n, 1] && IntegerQ[n]
 

rule 5085

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p 
*Cos[w]^q, x], x] /; IGtQ[p, 0] && IGtQ[q, 0] && ((PolynomialQ[v, x] && Pol 
ynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x]))
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {3 \sin \left (b \,x^{2}+a \right )}{4 x}+\frac {3 \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {b}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )-\sin \left (a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {b}\, \sqrt {2}\, x}{\sqrt {\pi }}\right )\right )}{4}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{4 x}-\frac {\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (3 a \right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )-\sin \left (3 a \right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )\right )}{4}\)	\(130\)
risch	\(-\frac {{\mathrm e}^{-3 i a} b \sqrt {\pi }\, \sqrt {3}\, \operatorname {erf}\left (\sqrt {3}\, \sqrt {i b}\, x \right )}{8 \sqrt {i b}}-\frac {3 \,{\mathrm e}^{3 i a} b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-3 i b}\, x \right )}{8 \sqrt {-3 i b}}+\frac {3 \,{\mathrm e}^{i a} b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i b}\, x \right )}{8 \sqrt {-i b}}+\frac {3 \,{\mathrm e}^{-i a} b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {i b}\, x \right )}{8 \sqrt {i b}}-\frac {3 \sin \left (b \,x^{2}+a \right )}{4 x}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{4 x}\)	\(141\)

Input:

int(sin(b*x^2+a)^3/x^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=-\frac {\sqrt {6} \pi x \sqrt {\frac {b}{\pi }} \cos \left (3 \, a\right ) \operatorname {C}\left (\sqrt {6} x \sqrt {\frac {b}{\pi }}\right ) - 3 \, \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) - \sqrt {6} \pi x \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {6} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (3 \, a\right ) + 3 \, \sqrt {2} \pi x \sqrt {\frac {b}{\pi }} \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (a\right ) - 4 \, {\left (\cos \left (b x^{2} + a\right )^{2} - 1\right )} \sin \left (b x^{2} + a\right )}{4 \, x} \] Input:

integrate(sin(b*x^2+a)^3/x^2,x, algorithm="fricas")
 

Output:

-1/4*(sqrt(6)*pi*x*sqrt(b/pi)*cos(3*a)*fresnel_cos(sqrt(6)*x*sqrt(b/pi)) - 
 3*sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fresnel_cos(sqrt(2)*x*sqrt(b/pi)) - sqrt 
(6)*pi*x*sqrt(b/pi)*fresnel_sin(sqrt(6)*x*sqrt(b/pi))*sin(3*a) + 3*sqrt(2) 
*pi*x*sqrt(b/pi)*fresnel_sin(sqrt(2)*x*sqrt(b/pi))*sin(a) - 4*(cos(b*x^2 + 
 a)^2 - 1)*sin(b*x^2 + a))/x
 

Sympy [F]

\[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {\sin ^{3}{\left (a + b x^{2} \right )}}{x^{2}}\, dx \] Input:

integrate(sin(b*x**2+a)**3/x**2,x)
 

Output:

Integral(sin(a + b*x**2)**3/x**2, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\frac {\sqrt {3} \sqrt {b x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 3 i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 3 i \, b x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right )\right )} - 3 \, \sqrt {b x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )}}{32 \, x} \] Input:

integrate(sin(b*x^2+a)^3/x^2,x, algorithm="maxima")
 

Output:

1/32*(sqrt(3)*sqrt(b*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, 3*I*b*x^2) - (I + 
1)*sqrt(2)*gamma(-1/2, -3*I*b*x^2))*cos(3*a) + ((I + 1)*sqrt(2)*gamma(-1/2 
, 3*I*b*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -3*I*b*x^2))*sin(3*a)) - 3*sqrt 
(b*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, I*b*x^2) - (I + 1)*sqrt(2)*gamma(-1/ 
2, -I*b*x^2))*cos(a) + ((I + 1)*sqrt(2)*gamma(-1/2, I*b*x^2) - (I - 1)*sqr 
t(2)*gamma(-1/2, -I*b*x^2))*sin(a)))/x
 

Giac [F]

\[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\int { \frac {\sin \left (b x^{2} + a\right )^{3}}{x^{2}} \,d x } \] Input:

integrate(sin(b*x^2+a)^3/x^2,x, algorithm="giac")
 

Output:

integrate(sin(b*x^2 + a)^3/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {{\sin \left (b\,x^2+a\right )}^3}{x^2} \,d x \] Input:

int(sin(a + b*x^2)^3/x^2,x)
 

Output:

int(sin(a + b*x^2)^3/x^2, x)
 

Reduce [F]

\[ \int \frac {\sin ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {\sin \left (b \,x^{2}+a \right )^{3}}{x^{2}}d x \] Input:

int(sin(b*x^2+a)^3/x^2,x)
 

Output:

int(sin(a + b*x**2)**3/x**2,x)