3.85 \(\int x^2 \cos ((a+b x)^2) \, dx\)

Optimal. Leaf size=99 \[ \frac{\sqrt{\frac{\pi }{2}} a^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )}{b^3}-\frac{\sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )}{2 b^3}-\frac{a \sin \left ((a+b x)^2\right )}{b^3}+\frac{(a+b x) \sin \left ((a+b x)^2\right )}{2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0675234, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3434, 3352, 3380, 2637, 3386, 3351} \[ \frac{\sqrt{\frac{\pi }{2}} a^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )}{b^3}-\frac{\sqrt{\frac{\pi }{2}} S\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )}{2 b^3}-\frac{a \sin \left ((a+b x)^2\right )}{b^3}+\frac{(a+b x) \sin \left ((a+b x)^2\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[(a + b*x)^2],x]

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.265072, size = 76, normalized size = 0.77 \[ -\frac{-2 \sqrt{2 \pi } a^2 \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )+\sqrt{2 \pi } S\left (\sqrt{\frac{2}{\pi }} (a+b x)\right )+2 (a-b x) \sin \left ((a+b x)^2\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[(a + b*x)^2],x]

[Out]

-(-2*a^2*Sqrt[2*Pi]*FresnelC[Sqrt[2/Pi]*(a + b*x)] + Sqrt[2*Pi]*FresnelS[Sqrt[2/Pi]*(a + b*x)] + 2*(a - b*x)*S
in[(a + b*x)^2])/(4*b^3)

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Maple [A]  time = 0.029, size = 131, normalized size = 1.3 \begin{align*}{\frac{x\sin \left ({x}^{2}{b}^{2}+2\,abx+{a}^{2} \right ) }{2\,{b}^{2}}}-{\frac{a}{b} \left ({\frac{\sin \left ({x}^{2}{b}^{2}+2\,abx+{a}^{2} \right ) }{2\,{b}^{2}}}-{\frac{\sqrt{2}a\sqrt{\pi }}{2\,b}{\it FresnelC} \left ({\frac{\sqrt{2} \left ({b}^{2}x+ab \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{b}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \right ) }-{\frac{\sqrt{2}\sqrt{\pi }}{4\,{b}^{2}}{\it FresnelS} \left ({\frac{\sqrt{2} \left ({b}^{2}x+ab \right ) }{\sqrt{\pi }}{\frac{1}{\sqrt{{b}^{2}}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b^2*x*sin(b^2*x^2+2*a*b*x+a^2)-a/b*(1/2/b^2*sin(b^2*x^2+2*a*b*x+a^2)-1/2*a/b*2^(1/2)*Pi^(1/2)/(b^2)^(1/2)*
FresnelC(2^(1/2)/Pi^(1/2)/(b^2)^(1/2)*(b^2*x+a*b)))-1/4/b^2*2^(1/2)*Pi^(1/2)/(b^2)^(1/2)*FresnelS(2^(1/2)/Pi^(
1/2)/(b^2)^(1/2)*(b^2*x+a*b))

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Maxima [C]  time = 2.78108, size = 346, normalized size = 3.49 \begin{align*} \frac{a b x{\left (8 i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} - 8 i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )} + a^{2}{\left (8 i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} - 8 i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )} + 2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}{\left ({\left (-\left (i - 1\right ) \, \sqrt{2} \sqrt{\pi }{\left (\operatorname{erf}\left (\sqrt{i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}}\right ) - 1\right )} + \left (i + 1\right ) \, \sqrt{2} \sqrt{\pi }{\left (\operatorname{erf}\left (\sqrt{-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}}\right ) - 1\right )}\right )} a^{2} + \left (i + 1\right ) \, \sqrt{2} \Gamma \left (\frac{3}{2}, i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right ) - \left (i - 1\right ) \, \sqrt{2} \Gamma \left (\frac{3}{2}, -i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )\right )}}{16 \,{\left (b^{4} x + a b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a*b*x*(8*I*e^(I*b^2*x^2 + 2*I*a*b*x + I*a^2) - 8*I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)) + a^2*(8*I*e^(I*b
^2*x^2 + 2*I*a*b*x + I*a^2) - 8*I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*((-(I
- 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(I*b^2*x^2 + 2*I*a*b*x + I*a^2)) - 1) + (I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*b
^2*x^2 - 2*I*a*b*x - I*a^2)) - 1))*a^2 + (I + 1)*sqrt(2)*gamma(3/2, I*b^2*x^2 + 2*I*a*b*x + I*a^2) - (I - 1)*s
qrt(2)*gamma(3/2, -I*b^2*x^2 - 2*I*a*b*x - I*a^2)))/(b^4*x + a*b^3)

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Fricas [A]  time = 1.66701, size = 278, normalized size = 2.81 \begin{align*} \frac{2 \, \sqrt{2} \pi a^{2} \sqrt{\frac{b^{2}}{\pi }} \operatorname{C}\left (\frac{\sqrt{2}{\left (b x + a\right )} \sqrt{\frac{b^{2}}{\pi }}}{b}\right ) - \sqrt{2} \pi \sqrt{\frac{b^{2}}{\pi }} \operatorname{S}\left (\frac{\sqrt{2}{\left (b x + a\right )} \sqrt{\frac{b^{2}}{\pi }}}{b}\right ) + 2 \,{\left (b^{2} x - a b\right )} \sin \left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{4 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(2)*pi*a^2*sqrt(b^2/pi)*fresnel_cos(sqrt(2)*(b*x + a)*sqrt(b^2/pi)/b) - sqrt(2)*pi*sqrt(b^2/pi)*fre
snel_sin(sqrt(2)*(b*x + a)*sqrt(b^2/pi)/b) + 2*(b^2*x - a*b)*sin(b^2*x^2 + 2*a*b*x + a^2))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cos(a**2 + 2*a*b*x + b**2*x**2), x)

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Giac [C]  time = 1.13045, size = 215, normalized size = 2.17 \begin{align*} -\frac{\frac{\left (i + 1\right ) \, \sqrt{2} \sqrt{\pi }{\left (2 \, a^{2} + i\right )} \operatorname{erf}\left (\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2}{\left (x + \frac{a}{b}\right )}{\left | b \right |}\right )}{{\left | b \right |}} + \frac{4 \,{\left (i \, b{\left (x + \frac{a}{b}\right )} - 2 i \, a\right )} e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )}}{b}}{16 \, b^{2}} - \frac{-\frac{\left (i - 1\right ) \, \sqrt{2} \sqrt{\pi }{\left (2 \, a^{2} - i\right )} \operatorname{erf}\left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2}{\left (x + \frac{a}{b}\right )}{\left | b \right |}\right )}{{\left | b \right |}} + \frac{4 \,{\left (-i \, b{\left (x + \frac{a}{b}\right )} + 2 i \, a\right )} e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}}{b}}{16 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((I + 1)*sqrt(2)*sqrt(pi)*(2*a^2 + I)*erf((1/2*I - 1/2)*sqrt(2)*(x + a/b)*abs(b))/abs(b) + 4*(I*b*(x + a
/b) - 2*I*a)*e^(I*b^2*x^2 + 2*I*a*b*x + I*a^2)/b)/b^2 - 1/16*(-(I - 1)*sqrt(2)*sqrt(pi)*(2*a^2 - I)*erf(-(1/2*
I + 1/2)*sqrt(2)*(x + a/b)*abs(b))/abs(b) + 4*(-I*b*(x + a/b) + 2*I*a)*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)/b)/b
^2