3.135 \(\int x^2 C(a+b x) \, dx\)

Optimal. Leaf size=148 \[ \frac {a^3 C(a+b x)}{3 b^3}-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {a S(a+b x)}{\pi b^3}+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 C(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 148, 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 = {6429, 3434, 3352, 3380, 2637, 3386, 3351, 3296, 2638} \[ \frac {a^3 \text {FresnelC}(a+b x)}{3 b^3}-\frac {a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {a S(a+b x)}{\pi b^3}+\frac {a (a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {(a+b x)^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}+\frac {1}{3} x^3 \text {FresnelC}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*FresnelC[a + b*x],x]

[Out]

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

Rule 2637

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

Rule 2638

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

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 116, normalized size = 0.78 \[ -\frac {-\pi ^2 \left (a^3+b^3 x^3\right ) C(a+b x)+\pi a^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+\pi b^2 x^2 \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+3 \pi a S(a+b x)-\pi a b x \sin \left (\frac {1}{2} \pi (a+b x)^2\right )+2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*FresnelC[a + b*x],x]

[Out]

-1/3*(2*Cos[(Pi*(a + b*x)^2)/2] - Pi^2*(a^3 + b^3*x^3)*FresnelC[a + b*x] + 3*a*Pi*FresnelS[a + b*x] + a^2*Pi*S
in[(Pi*(a + b*x)^2)/2] - a*b*Pi*x*Sin[(Pi*(a + b*x)^2)/2] + b^2*Pi*x^2*Sin[(Pi*(a + b*x)^2)/2])/(b^3*Pi^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*fresnelc(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnelc(b*x + a), x)

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maple [A]  time = 0.02, size = 122, normalized size = 0.82 \[ \frac {\frac {b^{3} x^{3} \FresnelC \left (b x +a \right )}{3}-\frac {\left (b x +a \right )^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi }-\frac {2 \cos \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{3 \pi ^{2}}+\frac {a \left (b x +a \right ) \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }-\frac {a \,\mathrm {S}\left (b x +a \right )}{\pi }-\frac {a^{2} \sin \left (\frac {\pi \left (b x +a \right )^{2}}{2}\right )}{\pi }+\frac {a^{3} \FresnelC \left (b x +a \right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*fresnelc(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {FresnelC}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*FresnelC(a + b*x),x)

[Out]

int(x^2*FresnelC(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*fresnelc(a + b*x), x)

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