3.1.26 \(\int x^2 S(a+b x) \, dx\) [26]

Optimal. Leaf size=147 \[ \frac {a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {a \text {FresnelC}(a+b x)}{b^3 \pi }+\frac {a^3 S(a+b x)}{3 b^3}+\frac {1}{3} x^3 S(a+b x)-\frac {2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi ^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {6563, 3514, 3432, 3460, 2718, 3466, 3433, 3377, 2717} \begin {gather*} \frac {a^3 S(a+b x)}{3 b^3}+\frac {a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {a \text {FresnelC}(a+b x)}{\pi b^3}-\frac {2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}+\frac {1}{3} x^3 S(a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*FresnelS[a + b*x],x]

[Out]

(a^2*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) - (a*(a + b*x)*Cos[(Pi*(a + b*x)^2)/2])/(b^3*Pi) + ((a + b*x)^2*Cos[(Pi
*(a + b*x)^2)/2])/(3*b^3*Pi) + (a*FresnelC[a + b*x])/(b^3*Pi) + (a^3*FresnelS[a + b*x])/(3*b^3) + (x^3*Fresnel
S[a + b*x])/3 - (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 x^2 S(a+b x) \, dx &=\frac {1}{3} x^3 S(a+b x)-\frac {1}{3} b \int x^3 \sin \left (\frac {1}{2} \pi (a+b x)^2\right ) \, dx\\ &=\frac {1}{3} x^3 S(a+b x)-\frac {\text {Subst}\left (\int \left (-a^3 \sin \left (\frac {\pi x^2}{2}\right )+3 a^2 x \sin \left (\frac {\pi x^2}{2}\right )-3 a x^2 \sin \left (\frac {\pi x^2}{2}\right )+x^3 \sin \left (\frac {\pi x^2}{2}\right )\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=\frac {1}{3} x^3 S(a+b x)-\frac {\text {Subst}\left (\int x^3 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}+\frac {a \text {Subst}\left (\int x^2 \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}-\frac {a^2 \text {Subst}\left (\int x \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3}+\frac {a^3 \text {Subst}\left (\int \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3}\\ &=-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {a^3 S(a+b x)}{3 b^3}+\frac {1}{3} x^3 S(a+b x)-\frac {\text {Subst}\left (\int x \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{6 b^3}-\frac {a^2 \text {Subst}\left (\int \sin \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{2 b^3}+\frac {a \text {Subst}\left (\int \cos \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=\frac {a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {a C(a+b x)}{b^3 \pi }+\frac {a^3 S(a+b x)}{3 b^3}+\frac {1}{3} x^3 S(a+b x)-\frac {\text {Subst}\left (\int \cos \left (\frac {\pi x}{2}\right ) \, dx,x,(a+b x)^2\right )}{3 b^3 \pi }\\ &=\frac {a^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {a (a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }+\frac {(a+b x)^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {a C(a+b x)}{b^3 \pi }+\frac {a^3 S(a+b x)}{3 b^3}+\frac {1}{3} x^3 S(a+b x)-\frac {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 = 0.17, size = 115, normalized size = 0.78 \begin {gather*} \frac {a^2 \pi \cos \left (\frac {1}{2} \pi (a+b x)^2\right )-a b \pi x \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+b^2 \pi x^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )+3 a \pi \text {FresnelC}(a+b x)+\pi ^2 \left (a^3+b^3 x^3\right ) S(a+b x)-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[x^2*FresnelS[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.40, size = 121, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \mathrm {S}\left (b x +a \right )}{3}+\frac {a^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \FresnelC \left (b x +a \right )}{\pi }+\frac {\left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(121\)
default	\(\frac {\frac {\mathrm {S}\left (b x +a \right ) b^{3} x^{3}}{3}+\frac {a^{3} \mathrm {S}\left (b x +a \right )}{3}+\frac {a^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \left (b x +a \right ) \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a \FresnelC \left (b x +a \right )}{\pi }+\frac {\left (b x +a \right )^{2} \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}}{b^{3}}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.91, size = 424, normalized size = 2.88 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {S}\left (b x + a\right ) + \frac {{\left (12 \, {\left (\pi e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{3} + 4 \, {\left (3 \, {\left (\pi e^{\left (\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right )} + \pi e^{\left (-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )}\right )} a^{2} - 2 i \, \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + 2 i \, \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} b x + 8 \, a {\left (-i \, \Gamma \left (2, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) + i \, \Gamma \left (2, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} - \sqrt {2 \, \pi b^{2} x^{2} + 4 \, \pi a b x + 2 \, \pi a^{2}} {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \pi ^{\frac {3}{2}} {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \pi ^{\frac {3}{2}} {\left (\operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}}\right ) - 1\right )}\right )} a^{3} - 6 \, {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2} + i \, \pi a b x + \frac {1}{2} i \, \pi a^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2} - i \, \pi a b x - \frac {1}{2} i \, \pi a^{2}\right )\right )} a\right )}\right )} b}{24 \, {\left (\pi ^{2} b^{5} x + \pi ^{2} a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*fresnel_sin(b*x + a) + 1/24*(12*(pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi*e^(-1/2*I*pi
*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^3 + 4*(3*(pi*e^(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) + pi*e^
(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a^2 - 2*I*gamma(2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a
^2) + 2*I*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2))*b*x + 8*a*(-I*gamma(2, 1/2*I*pi*b^2*x^2 + I
*pi*a*b*x + 1/2*I*pi*a^2) + I*gamma(2, -1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - sqrt(2*pi*b^2*x^2 + 4
*pi*a*b*x + 2*pi*a^2)*((-(I + 1)*sqrt(2)*pi^(3/2)*(erf(sqrt(1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2)) - 1
) + (I - 1)*sqrt(2)*pi^(3/2)*(erf(sqrt(-1/2*I*pi*b^2*x^2 - I*pi*a*b*x - 1/2*I*pi*a^2)) - 1))*a^3 - 6*((I - 1)*
sqrt(2)*gamma(3/2, 1/2*I*pi*b^2*x^2 + I*pi*a*b*x + 1/2*I*pi*a^2) - (I + 1)*sqrt(2)*gamma(3/2, -1/2*I*pi*b^2*x^
2 - I*pi*a*b*x - 1/2*I*pi*a^2))*a))*b/(pi^2*b^5*x + pi^2*a*b^4)

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Fricas [A]
time = 0.35, size = 147, normalized size = 1.00 \begin {gather*} \frac {\pi ^{2} b^{4} x^{3} \operatorname {S}\left (b x + a\right ) + \pi ^{2} a^{3} \sqrt {b^{2}} \operatorname {S}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + 3 \, \pi a \sqrt {b^{2}} \operatorname {C}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right ) + {\left (\pi b^{3} x^{2} - \pi a b^{2} x + \pi a^{2} b\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right ) - 2 \, b \sin \left (\frac {1}{2} \, \pi b^{2} x^{2} + \pi a b x + \frac {1}{2} \, \pi a^{2}\right )}{3 \, \pi ^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(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(x^2*fresnel_sin(b*x+a),x, algorithm="giac")

[Out]

integrate(x^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 x^2\,\mathrm {FresnelS}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelS(a + b*x),x)

[Out]

int(x^2*FresnelS(a + b*x), x)

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