\(\int (c+d x)^3 \operatorname {FresnelS}(a+b x) \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 296 \[ \int (c+d x)^3 \operatorname {FresnelS}(a+b x) \, dx=\frac {(b c-a d)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {3 d (b c-a d)^2 (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{2 b^4 \pi }+\frac {d^2 (b c-a d) (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi }+\frac {d^3 (a+b x)^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi }-\frac {3 d (b c-a d)^2 \operatorname {FresnelC}(a+b x)}{2 b^4 \pi }-\frac {(b c-a d)^4 \operatorname {FresnelS}(a+b x)}{4 b^4 d}+\frac {3 d^3 \operatorname {FresnelS}(a+b x)}{4 b^4 \pi ^2}+\frac {(c+d x)^4 \operatorname {FresnelS}(a+b x)}{4 d}-\frac {2 d^2 (b c-a d) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^4 \pi ^2}-\frac {3 d^3 (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \] Output:

(-a*d+b*c)^3*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi+3/2*d*(-a*d+b*c)^2*(b*x+a)*cos(1 
/2*Pi*(b*x+a)^2)/b^4/Pi+d^2*(-a*d+b*c)*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)/b^4 
/Pi+1/4*d^3*(b*x+a)^3*cos(1/2*Pi*(b*x+a)^2)/b^4/Pi-3/2*d*(-a*d+b*c)^2*Fres 
nelC(b*x+a)/b^4/Pi-1/4*(-a*d+b*c)^4*FresnelS(b*x+a)/b^4/d+3/4*d^3*FresnelS 
(b*x+a)/b^4/Pi^2+1/4*(d*x+c)^4*FresnelS(b*x+a)/d-2*d^2*(-a*d+b*c)*sin(1/2* 
Pi*(b*x+a)^2)/b^4/Pi^2-3/4*d^3*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)/b^4/Pi^2
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.43 \[ \int (c+d x)^3 \operatorname {FresnelS}(a+b x) \, dx=\frac {4 b^3 c^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 a b^2 c^2 d \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+4 a^2 b c d^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a^3 d^3 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+6 b^3 c^2 d \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-4 a b^2 c d^2 \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+a^2 b d^3 \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+4 b^3 c d^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b^2 d^3 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^3 d^3 \pi x^3 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-6 d (b c-a d)^2 \pi \operatorname {FresnelC}(a+b x)+\left (4 b^3 c^3 \pi ^2 (a+b x)+6 b^2 c^2 d \pi ^2 \left (-a^2+b^2 x^2\right )+4 b c d^2 \pi ^2 \left (a^3+b^3 x^3\right )+d^3 \left (3-a^4 \pi ^2+b^4 \pi ^2 x^4\right )\right ) \operatorname {FresnelS}(a+b x)-8 b c d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+5 a d^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )-3 b d^3 x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{4 b^4 \pi ^2} \] Input:

Integrate[(c + d*x)^3*FresnelS[a + b*x],x]
 

Output:

(4*b^3*c^3*Pi*Cos[(Pi*(a + b*x)^2)/2] - 6*a*b^2*c^2*d*Pi*Cos[(Pi*(a + b*x) 
^2)/2] + 4*a^2*b*c*d^2*Pi*Cos[(Pi*(a + b*x)^2)/2] - a^3*d^3*Pi*Cos[(Pi*(a 
+ b*x)^2)/2] + 6*b^3*c^2*d*Pi*x*Cos[(Pi*(a + b*x)^2)/2] - 4*a*b^2*c*d^2*Pi 
*x*Cos[(Pi*(a + b*x)^2)/2] + a^2*b*d^3*Pi*x*Cos[(Pi*(a + b*x)^2)/2] + 4*b^ 
3*c*d^2*Pi*x^2*Cos[(Pi*(a + b*x)^2)/2] - a*b^2*d^3*Pi*x^2*Cos[(Pi*(a + b*x 
)^2)/2] + b^3*d^3*Pi*x^3*Cos[(Pi*(a + b*x)^2)/2] - 6*d*(b*c - a*d)^2*Pi*Fr 
esnelC[a + b*x] + (4*b^3*c^3*Pi^2*(a + b*x) + 6*b^2*c^2*d*Pi^2*(-a^2 + b^2 
*x^2) + 4*b*c*d^2*Pi^2*(a^3 + b^3*x^3) + d^3*(3 - a^4*Pi^2 + b^4*Pi^2*x^4) 
)*FresnelS[a + b*x] - 8*b*c*d^2*Sin[(Pi*(a + b*x)^2)/2] + 5*a*d^3*Sin[(Pi* 
(a + b*x)^2)/2] - 3*b*d^3*x*Sin[(Pi*(a + b*x)^2)/2])/(4*b^4*Pi^2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.92, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, 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.

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

\(\Big \downarrow \) 6982

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

\(\Big \downarrow \) 3914

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

\(\Big \downarrow \) 2009

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

Input:

Int[(c + d*x)^3*FresnelS[a + b*x],x]
 

Output:

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

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 = 0.66 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {-\frac {d^{4} \left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 d^{4} \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x +a \right )}{\pi }\right )}{\pi }+\frac {4 \left (a d -c b \right ) d^{3} \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {8 \left (a d -c b \right ) d^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {6 \left (a d -c b \right )^{2} d^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {6 \left (a d -c b \right )^{2} d^{2} \operatorname {FresnelC}\left (b x +a \right )}{\pi }+\frac {4 \left (a d -c b \right )^{3} d \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\left (a d -c b \right )^{4} \operatorname {FresnelS}\left (b x +a \right )}{4 b^{3} d}}{b}\)	\(276\)
default	\(\frac {\frac {\operatorname {FresnelS}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{4}}{4 b^{3} d}-\frac {-\frac {d^{4} \left (b x +a \right )^{3} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {3 d^{4} \left (\frac {\left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {\operatorname {FresnelS}\left (b x +a \right )}{\pi }\right )}{\pi }+\frac {4 \left (a d -c b \right ) d^{3} \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {8 \left (a d -c b \right ) d^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {6 \left (a d -c b \right )^{2} d^{2} \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {6 \left (a d -c b \right )^{2} d^{2} \operatorname {FresnelC}\left (b x +a \right )}{\pi }+\frac {4 \left (a d -c b \right )^{3} d \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\left (a d -c b \right )^{4} \operatorname {FresnelS}\left (b x +a \right )}{4 b^{3} d}}{b}\)	\(276\)
parts	\(\text {Expression too large to display}\)	\(1125\)

Input:

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

Output:

1/b*(1/4*FresnelS(b*x+a)*(a*d-c*b-d*(b*x+a))^4/b^3/d-1/4/b^3/d*(-d^4/Pi*(b 
*x+a)^3*cos(1/2*Pi*(b*x+a)^2)+3*d^4/Pi*(1/Pi*(b*x+a)*sin(1/2*Pi*(b*x+a)^2) 
-1/Pi*FresnelS(b*x+a))+4*(a*d-b*c)*d^3/Pi*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)- 
8*(a*d-b*c)*d^3/Pi^2*sin(1/2*Pi*(b*x+a)^2)-6*(a*d-b*c)^2*d^2/Pi*(b*x+a)*co 
s(1/2*Pi*(b*x+a)^2)+6*(a*d-b*c)^2*d^2/Pi*FresnelC(b*x+a)+4*(a*d-b*c)^3*d/P 
i*cos(1/2*Pi*(b*x+a)^2)+(a*d-b*c)^4*FresnelS(b*x+a)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.27 \[ \int (c+d x)^3 \operatorname {FresnelS}(a+b x) \, dx=-\frac {6 \, \pi {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi ^{2} {\left (4 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 4 \, a^{3} b c d^{2} - a^{4} d^{3}\right )} + 3 \, d^{3}\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - {\left (\pi b^{4} d^{3} x^{3} + \pi {\left (4 \, b^{4} c d^{2} - a b^{3} d^{3}\right )} x^{2} + \pi {\left (6 \, b^{4} c^{2} d - 4 \, a b^{3} c d^{2} + a^{2} b^{2} d^{3}\right )} x + \pi {\left (4 \, b^{4} c^{3} - 6 \, a b^{3} c^{2} d + 4 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - {\left (\pi ^{2} b^{5} d^{3} x^{4} + 4 \, \pi ^{2} b^{5} c d^{2} x^{3} + 6 \, \pi ^{2} b^{5} c^{2} d x^{2} + 4 \, \pi ^{2} b^{5} c^{3} x\right )} \operatorname {S}\left (b x + a\right ) + {\left (3 \, b^{2} d^{3} x + 8 \, b^{2} c d^{2} - 5 \, a b d^{3}\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{4 \, \pi ^{2} b^{5}} \] Input:

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

Output:

-1/4*(6*pi*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sqrt(b^2)*fresnel_cos(sqrt( 
b^2)*(b*x + a)/b) - (pi^2*(4*a*b^3*c^3 - 6*a^2*b^2*c^2*d + 4*a^3*b*c*d^2 - 
 a^4*d^3) + 3*d^3)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) - (pi*b^4* 
d^3*x^3 + pi*(4*b^4*c*d^2 - a*b^3*d^3)*x^2 + pi*(6*b^4*c^2*d - 4*a*b^3*c*d 
^2 + a^2*b^2*d^3)*x + pi*(4*b^4*c^3 - 6*a*b^3*c^2*d + 4*a^2*b^2*c*d^2 - a^ 
3*b*d^3))*cos(1/2*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - (pi^2*b^5*d^3*x^4 
+ 4*pi^2*b^5*c*d^2*x^3 + 6*pi^2*b^5*c^2*d*x^2 + 4*pi^2*b^5*c^3*x)*fresnel_ 
sin(b*x + a) + (3*b^2*d^3*x + 8*b^2*c*d^2 - 5*a*b*d^3)*sin(1/2*pi*b^2*x^2 
+ pi*a*b*x + 1/2*pi*a^2))/(pi^2*b^5)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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