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

Optimal. Leaf size=194 \[ -\frac{(b c-a d)^3 \text{FresnelC}(a+b x)}{3 b^3 d}+\frac{d (b c-a d) S(a+b x)}{\pi b^3}-\frac{(b c-a d)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{d (a+b x) (b c-a d) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{d^2 (a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{2 d^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{(c+d x)^3 \text{FresnelC}(a+b x)}{3 d} \]

[Out]

(-2*d^2*Cos[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2) - ((b*c - a*d)^3*FresnelC[a + b*x])/(3*b^3*d) + ((c + d*x)^3*Fre
snelC[a + b*x])/(3*d) + (d*(b*c - a*d)*FresnelS[a + b*x])/(b^3*Pi) - ((b*c - a*d)^2*Sin[(Pi*(a + b*x)^2)/2])/(
b^3*Pi) - (d*(b*c - a*d)*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (d^2*(a + b*x)^2*Sin[(Pi*(a + b*x)^2)/2
])/(3*b^3*Pi)

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Rubi [A]  time = 0.204614, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638} \[ -\frac{(b c-a d)^3 \text{FresnelC}(a+b x)}{3 b^3 d}+\frac{d (b c-a d) S(a+b x)}{\pi b^3}-\frac{(b c-a d)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{d (a+b x) (b c-a d) \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac{d^2 (a+b x)^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac{2 d^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac{(c+d x)^3 \text{FresnelC}(a+b x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*FresnelC[a + b*x],x]

[Out]

(-2*d^2*Cos[(Pi*(a + b*x)^2)/2])/(3*b^3*Pi^2) - ((b*c - a*d)^3*FresnelC[a + b*x])/(3*b^3*d) + ((c + d*x)^3*Fre
snelC[a + b*x])/(3*d) + (d*(b*c - a*d)*FresnelS[a + b*x])/(b^3*Pi) - ((b*c - a*d)^2*Sin[(Pi*(a + b*x)^2)/2])/(
b^3*Pi) - (d*(b*c - a*d)*(a + b*x)*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (d^2*(a + b*x)^2*Sin[(Pi*(a + b*x)^2)/2
])/(3*b^3*Pi)

Rule 6429

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

Rule 3434

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.)*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[k/f^(m + 1), Subst[Int[ExpandIntegrand[(a + b*Cos[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 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 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rubi steps

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

Mathematica [A]  time = 0.46643, size = 237, normalized size = 1.22 \[ \frac{\pi ^2 \text{FresnelC}(a+b x) \left (-3 a^2 b c d+a^3 d^2+3 a b^2 c^2+b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )\right )-\pi a^2 d^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-3 \pi b^2 c^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-3 \pi b^2 c d x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi b^2 \left (-d^2\right ) x^2 \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+3 \pi d (b c-a d) S(a+b x)+3 \pi a b c d \sin \left (\frac{1}{2} \pi (a+b x)^2\right )+\pi a b d^2 x \sin \left (\frac{1}{2} \pi (a+b x)^2\right )-2 d^2 \cos \left (\frac{1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 249, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ({\frac{{\it FresnelC} \left ( bx+a \right ) \left ( d \left ( bx+a \right ) -ad+bc \right ) ^{3}}{3\,d{b}^{2}}}-{\frac{1}{3\,d{b}^{2}} \left ({\frac{{d}^{3} \left ( bx+a \right ) ^{2}}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }+2\,{\frac{{d}^{3}\cos \left ( 1/2\,\pi \, \left ( bx+a \right ) ^{2} \right ) }{{\pi }^{2}}}+{\frac{ \left ( -3\,a{d}^{3}+3\,bc{d}^{2} \right ) \left ( bx+a \right ) }{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{\frac{ \left ( -3\,a{d}^{3}+3\,bc{d}^{2} \right ){\it FresnelS} \left ( bx+a \right ) }{\pi }}+{\frac{3\,{a}^{2}{d}^{3}-6\,abc{d}^{2}+3\,{b}^{2}{c}^{2}d}{\pi }\sin \left ({\frac{\pi \, \left ( bx+a \right ) ^{2}}{2}} \right ) }-{a}^{3}{d}^{3}{\it FresnelC} \left ( bx+a \right ) +3\,{a}^{2}bc{d}^{2}{\it FresnelC} \left ( bx+a \right ) -3\,a{b}^{2}{c}^{2}d{\it FresnelC} \left ( bx+a \right ) +{b}^{3}{c}^{3}{\it FresnelC} \left ( bx+a \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/3*FresnelC(b*x+a)*(d*(b*x+a)-a*d+b*c)^3/b^2/d-1/3/b^2/d*(d^3/Pi*(b*x+a)^2*sin(1/2*Pi*(b*x+a)^2)+2*d^3/P
i^2*cos(1/2*Pi*(b*x+a)^2)+(-3*a*d^3+3*b*c*d^2)/Pi*(b*x+a)*sin(1/2*Pi*(b*x+a)^2)-(-3*a*d^3+3*b*c*d^2)/Pi*Fresne
lS(b*x+a)+(3*a^2*d^3-6*a*b*c*d^2+3*b^2*c^2*d)/Pi*sin(1/2*Pi*(b*x+a)^2)-a^3*d^3*FresnelC(b*x+a)+3*a^2*b*c*d^2*F
resnelC(b*x+a)-3*a*b^2*c^2*d*FresnelC(b*x+a)+b^3*c^3*FresnelC(b*x+a)))

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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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2*fresnelc(b*x + a), 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 ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Integral((c + d*x)**2*fresnelc(a + b*x), 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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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