3.49 \(\int (c+d x)^2 S(a+b x)^2 \, dx\)

Optimal. Leaf size=497 \[ \frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {d (b c-a d) C(a+b x) S(a+b x)}{\pi b^3}+\frac {d (a+b x)^2 (b c-a d) S(a+b x)^2}{b^3}+\frac {(a+b x) (b c-a d)^2 S(a+b x)^2}{b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b^3}+\frac {2 d (a+b x) (b c-a d) S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {2 (b c-a d)^2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}-\frac {5 d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} \pi ^2 b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {2 d^2 (a+b x)^2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac {2 d^2 x}{3 \pi ^2 b^2} \]

[Out]

2/3*d^2*x/b^2/Pi^2+1/2*d*(-a*d+b*c)*cos(Pi*(b*x+a)^2)/b^3/Pi^2+1/6*d^2*(b*x+a)*cos(Pi*(b*x+a)^2)/b^3/Pi^2+2*(-
a*d+b*c)^2*cos(1/2*Pi*(b*x+a)^2)*FresnelS(b*x+a)/b^3/Pi+2*d*(-a*d+b*c)*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)*FresnelS(
b*x+a)/b^3/Pi+2/3*d^2*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)*FresnelS(b*x+a)/b^3/Pi-d*(-a*d+b*c)*FresnelC(b*x+a)*Fres
nelS(b*x+a)/b^3/Pi+(-a*d+b*c)^2*(b*x+a)*FresnelS(b*x+a)^2/b^3+d*(-a*d+b*c)*(b*x+a)^2*FresnelS(b*x+a)^2/b^3+1/3
*d^2*(b*x+a)^3*FresnelS(b*x+a)^2/b^3+1/4*I*d*(-a*d+b*c)*(b*x+a)^2*HypergeometricPFQ([1, 1],[3/2, 2],-1/2*I*Pi*
(b*x+a)^2)/b^3/Pi-1/4*I*d*(-a*d+b*c)*(b*x+a)^2*HypergeometricPFQ([1, 1],[3/2, 2],1/2*I*Pi*(b*x+a)^2)/b^3/Pi-4/
3*d^2*FresnelS(b*x+a)*sin(1/2*Pi*(b*x+a)^2)/b^3/Pi^2-5/12*d^2*FresnelC((b*x+a)*2^(1/2))/b^3/Pi^2*2^(1/2)-1/2*(
-a*d+b*c)^2*FresnelS((b*x+a)*2^(1/2))/b^3/Pi*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.41, antiderivative size = 497, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {6432, 6420, 6452, 3351, 6430, 6454, 6446, 3379, 2638, 6460, 3357, 3352, 3385} \[ \frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {d (b c-a d) \text {FresnelC}(a+b x) S(a+b x)}{\pi b^3}+\frac {d (a+b x)^2 (b c-a d) S(a+b x)^2}{b^3}+\frac {(a+b x) (b c-a d)^2 S(a+b x)^2}{b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b^3}+\frac {2 d (a+b x) (b c-a d) S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {2 (b c-a d)^2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}-\frac {5 d^2 \text {FresnelC}\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} \pi ^2 b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {2 d^2 (a+b x)^2 S(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac {2 d^2 x}{3 \pi ^2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*FresnelS[a + b*x]^2,x]

[Out]

(2*d^2*x)/(3*b^2*Pi^2) + (d*(b*c - a*d)*Cos[Pi*(a + b*x)^2])/(2*b^3*Pi^2) + (d^2*(a + b*x)*Cos[Pi*(a + b*x)^2]
)/(6*b^3*Pi^2) - (5*d^2*FresnelC[Sqrt[2]*(a + b*x)])/(6*Sqrt[2]*b^3*Pi^2) + (2*(b*c - a*d)^2*Cos[(Pi*(a + b*x)
^2)/2]*FresnelS[a + b*x])/(b^3*Pi) + (2*d*(b*c - a*d)*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2]*FresnelS[a + b*x])/(b^
3*Pi) + (2*d^2*(a + b*x)^2*Cos[(Pi*(a + b*x)^2)/2]*FresnelS[a + b*x])/(3*b^3*Pi) - (d*(b*c - a*d)*FresnelC[a +
b*x]*FresnelS[a + b*x])/(b^3*Pi) + ((b*c - a*d)^2*(a + b*x)*FresnelS[a + b*x]^2)/b^3 + (d*(b*c - a*d)*(a + b*
x)^2*FresnelS[a + b*x]^2)/b^3 + (d^2*(a + b*x)^3*FresnelS[a + b*x]^2)/(3*b^3) - ((b*c - a*d)^2*FresnelS[Sqrt[2
]*(a + b*x)])/(Sqrt[2]*b^3*Pi) + ((I/4)*d*(b*c - a*d)*(a + b*x)^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*P
i*(a + b*x)^2])/(b^3*Pi) - ((I/4)*d*(b*c - a*d)*(a + b*x)^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*Pi*(a +
b*x)^2])/(b^3*Pi) - (4*d^2*FresnelS[a + b*x]*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2)

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3357

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Sin[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3385

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] + Dist[(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 6420

Int[FresnelS[(a_.) + (b_.)*(x_)]^2, x_Symbol] :> Simp[((a + b*x)*FresnelS[a + b*x]^2)/b, x] - Dist[2, Int[(a +
b*x)*Sin[(Pi*(a + b*x)^2)/2]*FresnelS[a + b*x], x], x] /; FreeQ[{a, b}, x]

Rule 6430

Int[FresnelS[(b_.)*(x_)]^2*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*FresnelS[b*x]^2)/(m + 1), x] - Dist[(2*b)/
(m + 1), Int[x^(m + 1)*Sin[(Pi*b^2*x^2)/2]*FresnelS[b*x], x], x] /; FreeQ[b, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6432

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

Rule 6446

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)], x_Symbol] :> Simp[(FresnelC[b*x]*FresnelS[b*x])/(2*b), x] + (-Simp
[(1*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, -((I*b^2*Pi*x^2)/2)])/8, x] + Simp[(1*I*b*x^2*HypergeometricPF
Q[{1, 1}, {3/2, 2}, (1*I*b^2*Pi*x^2)/2])/8, x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6452

Int[FresnelS[(b_.)*(x_)]*(x_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(Cos[d*x^2]*FresnelS[b*x])/(2*d), x] + Dis
t[1/(2*b*Pi), Int[Sin[2*d*x^2], x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6454

Int[FresnelS[(b_.)*(x_)]*(x_)^(m_)*Sin[(d_.)*(x_)^2], x_Symbol] :> -Simp[(x^(m - 1)*Cos[d*x^2]*FresnelS[b*x])/
(2*d), x] + (Dist[(m - 1)/(2*d), Int[x^(m - 2)*Cos[d*x^2]*FresnelS[b*x], x], x] + Dist[1/(2*b*Pi), Int[x^(m -
1)*Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 6460

Int[Cos[(d_.)*(x_)^2]*FresnelS[(b_.)*(x_)]*(x_), x_Symbol] :> Simp[(Sin[d*x^2]*FresnelS[b*x])/(2*d), x] - Dist
[1/(Pi*b), Int[Sin[d*x^2]^2, x], x] /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rubi steps

\begin {align*} \int (c+d x)^2 S(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (b^2 c^2 \left (1+\frac {a d (-2 b c+a d)}{b^2 c^2}\right ) S(x)^2+2 b c d \left (1-\frac {a d}{b c}\right ) x S(x)^2+d^2 x^2 S(x)^2\right ) \, dx,x,a+b x\right )}{b^3}\\ &=\frac {d^2 \operatorname {Subst}\left (\int x^2 S(x)^2 \, dx,x,a+b x\right )}{b^3}+\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int x S(x)^2 \, dx,x,a+b x\right )}{b^3}+\frac {(b c-a d)^2 \operatorname {Subst}\left (\int S(x)^2 \, dx,x,a+b x\right )}{b^3}\\ &=\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int x^3 S(x) \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}-\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int x^2 S(x) \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac {\left (2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int x S(x) \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}\\ &=\frac {2 (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{3 b^3 \pi }+\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {d^2 \operatorname {Subst}\left (\int x^2 \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{3 b^3 \pi }-\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) S(x) \, dx,x,a+b x\right )}{3 b^3 \pi }-\frac {(d (b c-a d)) \operatorname {Subst}\left (\int x \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b^3 \pi }-\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) S(x) \, dx,x,a+b x\right )}{b^3 \pi }-\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}+\frac {2 (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{3 b^3 \pi }-\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}-\frac {d^2 \operatorname {Subst}\left (\int \cos \left (\pi x^2\right ) \, dx,x,a+b x\right )}{6 b^3 \pi ^2}+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \sin ^2\left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}-\frac {(d (b c-a d)) \operatorname {Subst}\left (\int \sin (\pi x) \, dx,x,(a+b x)^2\right )}{2 b^3 \pi }\\ &=\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}+\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{3 b^3 \pi }-\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {1}{2} \cos \left (\pi x^2\right )\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}\\ &=\frac {2 d^2 x}{3 b^2 \pi ^2}+\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}+\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {2 (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{3 b^3 \pi }-\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \cos \left (\pi x^2\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}\\ &=\frac {2 d^2 x}{3 b^2 \pi ^2}+\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}+\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}-\frac {\sqrt {2} d^2 C\left (\sqrt {2} (a+b x)\right )}{3 b^3 \pi ^2}+\frac {2 (b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{b^3 \pi }+\frac {2 d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) S(a+b x)}{3 b^3 \pi }-\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 (a+b x) S(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 S(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 S(a+b x)^2}{3 b^3}-\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {4 d^2 S(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 0.88, size = 0, normalized size = 0.00 \[ \int (c+d x)^2 S(a+b x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c + d*x)^2*FresnelS[a + b*x]^2,x]

[Out]

Integrate[(c + d*x)^2*FresnelS[a + b*x]^2, x]

________________________________________________________________________________________

fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} {\rm fresnels}\left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnels(b*x+a)^2,x, algorithm="fricas")

[Out]

integral((d^2*x^2 + 2*c*d*x + c^2)*fresnels(b*x + a)^2, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnels(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*fresnels(b*x + a)^2, x)

________________________________________________________________________________________

maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{2} \mathrm {S}\left (b x +a \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*FresnelS(b*x+a)^2,x)

[Out]

int((d*x+c)^2*FresnelS(b*x+a)^2,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnels}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*fresnels(b*x+a)^2,x, algorithm="maxima")

[Out]

integrate((d*x + c)^2*fresnels(b*x + a)^2, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {FresnelS}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(a + b*x)^2*(c + d*x)^2,x)

[Out]

int(FresnelS(a + b*x)^2*(c + d*x)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} S^{2}\left (a + b x\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*fresnels(b*x+a)**2,x)

[Out]

Integral((c + d*x)**2*fresnels(a + b*x)**2, x)

________________________________________________________________________________________