\(\int x^3 \operatorname {FresnelS}(a+b x) \, dx\) [25]

Optimal result

Integrand size = 10, antiderivative size = 229 \[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=-\frac {a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }-\frac {a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 a^2 \operatorname {FresnelC}(a+b x)}{2 b^4 \pi }-\frac {a^4 \operatorname {FresnelS}(a+b x)}{4 b^4}+\frac {3 \operatorname {FresnelS}(a+b x)}{4 b^4 \pi ^2}+\frac {1}{4} x^4 \operatorname {FresnelS}(a+b x)+\frac {2 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.72 \[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=\frac {-a^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+a^2 b \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^3 \pi x^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 a^2 \pi \operatorname {FresnelC}(a+b x)+\left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right ) \operatorname {FresnelS}(a+b x)+5 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \] Input:

Integrate[x^3*FresnelS[a + b*x],x]
 

Output:

(-(a^3*Pi*Cos[(Pi*(a + b*x)^2)/2]) + a^2*b*Pi*x*Cos[(Pi*(a + b*x)^2)/2] - 
a*b^2*Pi*x^2*Cos[(Pi*(a + b*x)^2)/2] + b^3*Pi*x^3*Cos[(Pi*(a + b*x)^2)/2] 
- 6*a^2*Pi*FresnelC[a + b*x] + (3 - a^4*Pi^2 + b^4*Pi^2*x^4)*FresnelS[a + 
b*x] + 5*a*Sin[(Pi*(a + b*x)^2)/2] - 3*b*x*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4 
*Pi^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6982, 3914, 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.

\(\displaystyle \int x^3 \operatorname {FresnelS}(a+b x) \, dx\)

\(\Big \downarrow \) 6982

\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelS}(a+b x)-\frac {1}{4} b \int x^4 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )dx\)

\(\Big \downarrow \) 3914

\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelS}(a+b x)-\frac {\int \left (\sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^4-4 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^3+6 (a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a^2-4 (a+b x)^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) a+(a+b x)^4 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )\right )d(a+b x)}{4 b^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} x^4 \operatorname {FresnelS}(a+b x)-\frac {a^4 \operatorname {FresnelS}(a+b x)+\frac {4 a^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }+\frac {6 a^2 \operatorname {FresnelC}(a+b x)}{\pi }-\frac {6 a^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {3 \operatorname {FresnelS}(a+b x)}{\pi ^2}-\frac {8 a \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2}+\frac {3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi ^2}+\frac {4 a (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }-\frac {(a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi }}{4 b^4}\)

Input:

Int[x^3*FresnelS[a + b*x],x]
 

Output:

(x^4*FresnelS[a + b*x])/4 - ((4*a^3*Cos[(Pi*(a + b*x)^2)/2])/Pi - (6*a^2*( 
a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/Pi + (4*a*(a + b*x)^2*Cos[(Pi*(a + b*x)^ 
2)/2])/Pi - ((a + b*x)^3*Cos[(Pi*(a + b*x)^2)/2])/Pi + (6*a^2*FresnelC[a + 
 b*x])/Pi + a^4*FresnelS[a + b*x] - (3*FresnelS[a + b*x])/Pi^2 - (8*a*Sin[ 
(Pi*(a + b*x)^2)/2])/Pi^2 + (3*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/Pi^2)/(4 
*b^4)
 

Defintions of rubi rules used

rule 2009

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

rule 3914

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Module[{k = If[FractionQ[n], Denominat 
or[n], 1]}, Simp[k/f^(m + 1)   Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x 
^(k*n)])^p, x^(k - 1)*(f*g - e*h + h*x^k)^m, x], x], x, (e + f*x)^(1/k)], x 
]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && IGtQ[p, 0] && IGtQ[m, 0]
 

rule 6982

Int[FresnelS[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> S 
imp[(c + d*x)^(m + 1)*(FresnelS[a + b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 
1))   Int[(c + d*x)^(m + 1)*Sin[(Pi/2)*(a + b*x)^2], x], x] /; FreeQ[{a, b, 
 c, d}, x] && IGtQ[m, 0]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {FresnelS}\left (b x +a \right )}{4}-\frac {a^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 a^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {3 a^{2} \operatorname {FresnelC}\left (b x +a \right )}{2 \pi }-\frac {a \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x +a \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(189\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) b^{4} x^{4}}{4}-\frac {a^{4} \operatorname {FresnelS}\left (b x +a \right )}{4}-\frac {a^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 a^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{2 \pi }-\frac {3 a^{2} \operatorname {FresnelC}\left (b x +a \right )}{2 \pi }-\frac {a \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {2 a \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}+\frac {\left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{4 \pi }-\frac {3 \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x +a \right )}{\pi }\right )}{4 \pi }}{b^{4}}\)	\(189\)
parts	\(\frac {x^{4} \operatorname {FresnelS}\left (b x +a \right )}{4}-\frac {b \left (-\frac {x^{3} \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {x^{2} \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {x \cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {a \left (-\frac {\cos \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}+\frac {\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b}+\frac {\frac {2 \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {2 \sqrt {\pi }\, a \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}}{b^{2} \pi }\right )}{b}+\frac {\frac {3 x \sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {3 a \left (\frac {\sin \left (\frac {1}{2} b^{2} \pi \,x^{2}+\pi a b x +\frac {1}{2} \pi \,a^{2}\right )}{b^{2} \pi }-\frac {\sqrt {\pi }\, a \,\operatorname {FresnelC}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b \sqrt {b^{2} \pi }}\right )}{b}-\frac {3 \,\operatorname {FresnelS}\left (\frac {b^{2} \pi x +\pi b a}{\sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}}{b^{2} \pi }\right )}{4}\)	\(485\)

Input:

int(x^3*FresnelS(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76 \[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=\frac {\pi ^{2} b^{5} x^{4} \operatorname {S}\left (b x + a\right ) - 6 \, \pi a^{2} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} a^{4} - 3\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{4} x^{3} - \pi a b^{3} x^{2} + \pi a^{2} b^{2} x - \pi a^{3} b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (3 \, b^{2} x - 5 \, a b\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \] Input:

integrate(x^3*fresnel_sin(b*x+a),x, algorithm="fricas")
 

Output:

1/4*(pi^2*b^5*x^4*fresnel_sin(b*x + a) - 6*pi*a^2*sqrt(b^2)*fresnel_cos(sq 
rt(b^2)*(b*x + a)/b) - (pi^2*a^4 - 3)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x 
 + a)/b) + (pi*b^4*x^3 - pi*a*b^3*x^2 + pi*a^2*b^2*x - pi*a^3*b)*cos(1/2*p 
i*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - (3*b^2*x - 5*a*b)*sin(1/2*pi*b^2*x^2 
+ pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^5)
 

Sympy [F]

\[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=\int x^{3} S\left (a + b x\right )\, dx \] Input:

integrate(x**3*fresnels(b*x+a),x)
 

Output:

Integral(x**3*fresnels(a + b*x), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.94 (sec) , antiderivative size = 503, normalized size of antiderivative = 2.20 \[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx =\text {Too large to display} \] Input:

integrate(x^3*fresnel_sin(b*x+a),x, algorithm="maxima")
 

Output:

1/4*x^4*fresnel_sin(b*x + a) - 1/32*(16*(pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi*a 
*b*x + 1/2*I*pi*a^2) + pi^2*e^(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a 
^2))*a^4 + 32*(-I*pi*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2 
) + I*pi*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 + 16 
*((pi^2*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi^2*e^(-1/2*I* 
pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^3 + 2*(-I*pi*gamma(2, 1/2*I*pi* 
b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + I*pi*gamma(2, -1/2*I*pi*b^2*x^2 - I 
*pi*a*b*x - 1/2*I*pi*a^2))*a)*b*x - ((-(I + 1)*sqrt(2)*pi^(5/2)*(erf(sqrt( 
1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1) + (I - 1)*sqrt(2)*pi^( 
5/2)*(erf(sqrt(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^4 - 
 12*((I - 1)*sqrt(2)*pi*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*p 
i*a^2) - (I + 1)*sqrt(2)*pi*gamma(3/2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/ 
2*I*pi*a^2))*a^2 - (4*I + 4)*sqrt(2)*gamma(5/2, 1/2*I*pi*b^2*x^2 + I*pi*a* 
b*x + 1/2*I*pi*a^2) + (4*I - 4)*sqrt(2)*gamma(5/2, -1/2*I*pi*b^2*x^2 - I*p 
i*a*b*x - 1/2*I*pi*a^2))*sqrt(2*pi*b^2*x^2 + 4*pi*a*b*x + 2*pi*a^2))*b/(pi 
^3*b^6*x + pi^3*a*b^5)
 

Giac [F]

\[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=\int { x^{3} \operatorname {S}\left (b x + a\right ) \,d x } \] Input:

integrate(x^3*fresnel_sin(b*x+a),x, algorithm="giac")
 

Output:

integrate(x^3*fresnel_sin(b*x + a), x)
 

Mupad [F(-1)]

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

int(x^3*FresnelS(a + b*x),x)
 

Output:

int(x^3*FresnelS(a + b*x), x)
 

Reduce [F]

\[ \int x^3 \operatorname {FresnelS}(a+b x) \, dx=\int x^{3} \mathrm {FresnelS}\left (b x +a \right )d x \] Input:

int(x^3*FresnelS(b*x+a),x)
 

Output:

int(x^3*FresnelS(b*x+a),x)