3.158 \(\int (c+d x)^2 \text{FresnelC}(a+b x)^2 \, dx\)

Optimal. Leaf size=495 \[ \frac{i d (a+b x)^2 (b c-a d) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac{i d (a+b x)^2 (b c-a d) \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}+\frac{d (b c-a d) \text{FresnelC}(a+b x) S(a+b x)}{\pi b^3}+\frac{d (a+b x)^2 (b c-a d) \text{FresnelC}(a+b x)^2}{b^3}+\frac{(a+b x) (b c-a d)^2 \text{FresnelC}(a+b x)^2}{b^3}-\frac{2 d (a+b x) (b c-a d) \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{2 (b c-a d)^2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{(b c-a d)^2 S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b^3}-\frac{d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}+\frac{d^2 (a+b x)^3 \text{FresnelC}(a+b x)^2}{3 b^3}+\frac{5 d^2 \text{FresnelC}\left (\sqrt{2} (a+b x)\right )}{6 \sqrt{2} \pi ^2 b^3}-\frac{2 d^2 (a+b x)^2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{4 d^2 \text{FresnelC}(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac{d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac{2 d^2 x}{3 \pi ^2 b^2} \]

[Out]

(2*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) - (4*d^2*Cos[(Pi*(a + b*x)^2)/2]*FresnelC[a + b*x])/(3*b^3*Pi^2) + ((b*c - a*d)^2*(a + b*x)*Fre
snelC[a + b*x]^2)/b^3 + (d*(b*c - a*d)*(a + b*x)^2*FresnelC[a + b*x]^2)/b^3 + (d^2*(a + b*x)^3*FresnelC[a + b*
x]^2)/(3*b^3) + (5*d^2*FresnelC[Sqrt[2]*(a + b*x)])/(6*Sqrt[2]*b^3*Pi^2) + (d*(b*c - a*d)*FresnelC[a + b*x]*Fr
esnelS[a + b*x])/(b^3*Pi) + ((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)*Pi*(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) - (2*(b*c - a*d)^2*FresnelC[a + b*
x]*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (2*d*(b*c - a*d)*(a + b*x)*FresnelC[a + b*x]*Sin[(Pi*(a + b*x)^2)/2])/(
b^3*Pi) - (2*d^2*(a + b*x)^2*FresnelC[a + b*x]*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi)

________________________________________________________________________________________

Rubi [A]  time = 0.396293, antiderivative size = 495, normalized size of antiderivative = 1., 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 = {6433, 6421, 6453, 3351, 6431, 6455, 6447, 3379, 2638, 6461, 3358, 3352, 3385} \[ \frac{i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac{i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}+\frac{d (b c-a d) \text{FresnelC}(a+b x) S(a+b x)}{\pi b^3}+\frac{d (a+b x)^2 (b c-a d) \text{FresnelC}(a+b x)^2}{b^3}+\frac{(a+b x) (b c-a d)^2 \text{FresnelC}(a+b x)^2}{b^3}-\frac{2 d (a+b x) (b c-a d) \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{2 (b c-a d)^2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac{(b c-a d)^2 S\left (\sqrt{2} (a+b x)\right )}{\sqrt{2} \pi b^3}-\frac{d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}+\frac{d^2 (a+b x)^3 \text{FresnelC}(a+b x)^2}{3 b^3}+\frac{5 d^2 \text{FresnelC}\left (\sqrt{2} (a+b x)\right )}{6 \sqrt{2} \pi ^2 b^3}-\frac{2 d^2 (a+b x)^2 \text{FresnelC}(a+b x) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{4 d^2 \text{FresnelC}(a+b x) \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac{d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac{2 d^2 x}{3 \pi ^2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*FresnelC[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) - (4*d^2*Cos[(Pi*(a + b*x)^2)/2]*FresnelC[a + b*x])/(3*b^3*Pi^2) + ((b*c - a*d)^2*(a + b*x)*Fre
snelC[a + b*x]^2)/b^3 + (d*(b*c - a*d)*(a + b*x)^2*FresnelC[a + b*x]^2)/b^3 + (d^2*(a + b*x)^3*FresnelC[a + b*
x]^2)/(3*b^3) + (5*d^2*FresnelC[Sqrt[2]*(a + b*x)])/(6*Sqrt[2]*b^3*Pi^2) + (d*(b*c - a*d)*FresnelC[a + b*x]*Fr
esnelS[a + b*x])/(b^3*Pi) + ((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)*Pi*(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) - (2*(b*c - a*d)^2*FresnelC[a + b*
x]*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (2*d*(b*c - a*d)*(a + b*x)*FresnelC[a + b*x]*Sin[(Pi*(a + b*x)^2)/2])/(
b^3*Pi) - (2*d^2*(a + b*x)^2*FresnelC[a + b*x]*Sin[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi)

Rule 6433

Int[FresnelC[(a_) + (b_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/b^(m + 1), Subst[Int[ExpandI
ntegrand[FresnelC[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 6421

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

Rule 6453

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

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 6431

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

Rule 6455

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelC[b*x])/(
2*d), x] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(4*d), 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 6447

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

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 2638

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

Rule 6461

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

Rule 3358

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

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 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]

Rubi steps

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

Mathematica [F]  time = 0.6492, size = 0, normalized size = 0. \[ \int (c+d x)^2 \text{FresnelC}(a+b x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.227, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{2} \left ({\it FresnelC} \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2}{\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )}{\rm fresnelc}\left (b x + a\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{2} C^{2}\left (a + b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{2}{\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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