3.1.20 \(\int (c+d x)^2 S(a+b x) \, dx\) [20]

Optimal. Leaf size=193 \[ \frac {(b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {d (b c-a d) (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }-\frac {d (b c-a d) \text {FresnelC}(a+b x)}{b^3 \pi }-\frac {(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac {(c+d x)^3 S(a+b x)}{3 d}-\frac {2 d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 193, normalized size of antiderivative = 1.00, 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 = {6563, 3514, 3432, 3460, 2718, 3466, 3433, 3377, 2717} \begin {gather*} -\frac {d (b c-a d) \text {FresnelC}(a+b x)}{\pi b^3}-\frac {(b c-a d)^3 S(a+b x)}{3 b^3 d}+\frac {(b c-a d)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {d (a+b x) (b c-a d) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {2 d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {d^2 (a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac {(c+d x)^3 S(a+b x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*FresnelS[a + b*x],x]

[Out]

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

Rule 2717

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

Rule 2718

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

Rule 3377

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 3432

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

Rule 3433

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

Rule 3460

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 3466

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 3514

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Module[{k = If[FractionQ[n], Denominator[n], 1]}, Dist[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 6563

Int[FresnelS[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(FresnelS[a +
b*x]/(d*(m + 1))), x] - Dist[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]

Rubi steps

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

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Mathematica [A]
time = 0.30, size = 236, normalized size = 1.22 \begin {gather*} \frac {3 b^2 c^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-3 a b c d \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+a^2 d^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+3 b^2 c d \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b d^2 \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^2 d^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+3 d (-b c+a d) \pi \text {FresnelC}(a+b x)+\pi ^2 \left (3 a b^2 c^2-3 a^2 b c d+a^3 d^2+b^3 x \left (3 c^2+3 c d x+d^2 x^2\right )\right ) S(a+b x)-2 d^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.51, size = 251, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {-\frac {\mathrm {S}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{3}}{3 b^{2} d}+\frac {\frac {d^{3} \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {2 d^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (3 a \,d^{3}-3 b c \,d^{2}\right ) \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (3 a \,d^{3}-3 b c \,d^{2}\right ) \FresnelC \left (b x +a \right )}{\pi }-\frac {\left (-3 a^{2} d^{3}+6 a b c \,d^{2}-3 b^{2} c^{2} d \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{3} d^{3} \mathrm {S}\left (b x +a \right )-3 a^{2} b c \,d^{2} \mathrm {S}\left (b x +a \right )+3 a \,b^{2} c^{2} d \,\mathrm {S}\left (b x +a \right )-b^{3} c^{3} \mathrm {S}\left (b x +a \right )}{3 b^{2} d}}{b}\)	\(251\)
default	\(\frac {-\frac {\mathrm {S}\left (b x +a \right ) \left (a d -c b -d \left (b x +a \right )\right )^{3}}{3 b^{2} d}+\frac {\frac {d^{3} \left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {2 d^{3} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi ^{2}}-\frac {\left (3 a \,d^{3}-3 b c \,d^{2}\right ) \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {\left (3 a \,d^{3}-3 b c \,d^{2}\right ) \FresnelC \left (b x +a \right )}{\pi }-\frac {\left (-3 a^{2} d^{3}+6 a b c \,d^{2}-3 b^{2} c^{2} d \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+a^{3} d^{3} \mathrm {S}\left (b x +a \right )-3 a^{2} b c \,d^{2} \mathrm {S}\left (b x +a \right )+3 a \,b^{2} c^{2} d \,\mathrm {S}\left (b x +a \right )-b^{3} c^{3} \mathrm {S}\left (b x +a \right )}{3 b^{2} d}}{b}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/3*FresnelS(b*x+a)*(a*d-c*b-d*(b*x+a))^3/b^2/d+1/3/b^2/d*(d^3/Pi*(b*x+a)^2*cos(1/2*Pi*(b*x+a)^2)-2*d^3/
Pi^2*sin(1/2*Pi*(b*x+a)^2)-(3*a*d^3-3*b*c*d^2)/Pi*(b*x+a)*cos(1/2*Pi*(b*x+a)^2)+(3*a*d^3-3*b*c*d^2)/Pi*Fresnel
C(b*x+a)-(-3*a^2*d^3+6*a*b*c*d^2-3*b^2*c^2*d)/Pi*cos(1/2*Pi*(b*x+a)^2)+a^3*d^3*FresnelS(b*x+a)-3*a^2*b*c*d^2*F
resnelS(b*x+a)+3*a*b^2*c^2*d*FresnelS(b*x+a)-b^3*c^3*FresnelS(b*x+a)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 248, normalized size = 1.28 \begin {gather*} \frac {\pi ^{2} {\left (3 \, a b^{2} c^{2} - 3 \, a^{2} b c d + a^{3} d^{2}\right )} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) - 2 \, b d^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - 3 \, \pi {\left (b c d - a d^{2}\right )} \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{3} d^{2} x^{2} + \pi {\left (3 \, b^{3} c d - a b^{2} d^{2}\right )} x + \pi {\left (3 \, b^{3} c^{2} - 3 \, a b^{2} c d + a^{2} b d^{2}\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^{4} d^{2} x^{3} + 3 \, \pi ^{2} b^{4} c d x^{2} + 3 \, \pi ^{2} b^{4} c^{2} x\right )} \operatorname {S}\left (b x + a\right )}{3 \, \pi ^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(pi^2*(3*a*b^2*c^2 - 3*a^2*b*c*d + a^3*d^2)*sqrt(b^2)*fresnel_sin(sqrt(b^2)*(b*x + a)/b) - 2*b*d^2*sin(1/2
*pi*b^2*x^2 + pi*a*b*x + 1/2*pi*a^2) - 3*pi*(b*c*d - a*d^2)*sqrt(b^2)*fresnel_cos(sqrt(b^2)*(b*x + a)/b) + (pi
*b^3*d^2*x^2 + pi*(3*b^3*c*d - a*b^2*d^2)*x + pi*(3*b^3*c^2 - 3*a*b^2*c*d + a^2*b*d^2))*cos(1/2*pi*b^2*x^2 + p
i*a*b*x + 1/2*pi*a^2) + (pi^2*b^4*d^2*x^3 + 3*pi^2*b^4*c*d*x^2 + 3*pi^2*b^4*c^2*x)*fresnel_sin(b*x + a))/(pi^2
*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {FresnelS}\left (a+b\,x\right )\,{\left (c+d\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(a + b*x)*(c + d*x)^2,x)

[Out]

int(FresnelS(a + b*x)*(c + d*x)^2, x)

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