3.1.85 \(\int x^2 \cos ((a+b x)^2) \, dx\) [85]

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

[Out]

-a*sin((b*x+a)^2)/b^3+1/2*(b*x+a)*sin((b*x+a)^2)/b^3+1/2*a^2*FresnelC((b*x+a)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/
2)/b^3-1/4*FresnelS((b*x+a)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/b^3

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Rubi [A]
time = 0.05, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3515, 3433, 3461, 2717, 3467, 3432} \begin {gather*} \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} \end {gather*}

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 2717

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

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 3461

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 3467

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 3515

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]

Rubi steps

\begin {align*} \int x^2 \cos \left ((a+b x)^2\right ) \, dx &=\frac {\text {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 {\text {Subst}\left (\int x^2 \cos \left (x^2\right ) \, dx,x,a+b x\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int x \cos \left (x^2\right ) \, dx,x,a+b x\right )}{b^3}+\frac {a^2 \text {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 {\text {Subst}\left (\int \sin \left (x^2\right ) \, dx,x,a+b x\right )}{2 b^3}-\frac {a \text {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*}

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Mathematica [A]
time = 0.29, size = 76, normalized size = 0.77 \begin {gather*} -\frac {-2 a^2 \sqrt {2 \pi } \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[(a + b*x)^2],x]

[Out]

-1/4*(-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)*Sin[(a + b*x)^2])/b^3

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Maple [A]
time = 0.17, size = 131, normalized size = 1.32

method	result	size
default	\(\frac {x \sin \left (x^{2} b^{2}+2 a b x +a^{2}\right )}{2 b^{2}}-\frac {a \left (\frac {\sin \left (x^{2} b^{2}+2 a b x +a^{2}\right )}{2 b^{2}}-\frac {a \sqrt {2}\, \sqrt {\pi }\, \FresnelC \left (\frac {\sqrt {2}\, \left (b^{2} x +a b \right )}{\sqrt {\pi }\, \sqrt {b^{2}}}\right )}{2 b \sqrt {b^{2}}}\right )}{b}-\frac {\sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (b^{2} x +a b \right )}{\sqrt {\pi }\, \sqrt {b^{2}}}\right )}{4 b^{2} \sqrt {b^{2}}}\)	\(131\)
risch	\(-\frac {a^{2} \left (-1\right )^{\frac {3}{4}} \sqrt {\pi }\, \erf \left (b \left (-1\right )^{\frac {1}{4}} x +\left (-1\right )^{\frac {1}{4}} a \right )}{4 b^{3}}-\frac {\left (-1\right )^{\frac {1}{4}} \sqrt {\pi }\, \erf \left (b \left (-1\right )^{\frac {1}{4}} x +\left (-1\right )^{\frac {1}{4}} a \right )}{8 b^{3}}-\frac {a^{2} \sqrt {\pi }\, \erf \left (-b \sqrt {-i}\, x +\frac {i a}{\sqrt {-i}}\right )}{4 b^{3} \sqrt {-i}}-\frac {i \sqrt {\pi }\, \erf \left (-b \sqrt {-i}\, x +\frac {i a}{\sqrt {-i}}\right )}{8 b^{3} \sqrt {-i}}-2 i \left (\frac {i x}{4 b^{2}}-\frac {i a}{4 b^{3}}\right ) \sin \left (\left (b x +a \right )^{2}\right )\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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] Result contains complex when optimal does not.
time = 0.78, size = 258, normalized size = 2.61 \begin {gather*} -\frac {4 \, a b x {\left (-i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} + i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )} + 4 \, a^{2} {\left (-i \, e^{\left (i \, b^{2} x^{2} + 2 i \, a b x + i \, a^{2}\right )} + i \, e^{\left (-i \, b^{2} x^{2} - 2 i \, a b x - i \, a^{2}\right )}\right )} - \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 )}}{8 \, {\left (b^{4} x + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*a*b*x*(-I*e^(I*b^2*x^2 + 2*I*a*b*x + I*a^2) + I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^2)) + 4*a^2*(-I*e^(I*b
^2*x^2 + 2*I*a*b*x + I*a^2) + I*e^(-I*b^2*x^2 - 2*I*a*b*x - I*a^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)*sqrt(
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 = 0.36, size = 112, normalized size = 1.13 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \cos {\left (a^{2} + 2 a b x + b^{2} x^{2} \right )}\, dx \end {gather*}

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] Result contains complex when optimal does not.
time = 0.47, size = 159, normalized size = 1.61 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.15, size = 80, normalized size = 0.81 \begin {gather*} \frac {x\,\sin \left ({\left (a+b\,x\right )}^2\right )}{2\,b^2}-\frac {a\,\sin \left ({\left (a+b\,x\right )}^2\right )}{2\,b^3}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (a+b\,x\right )}{\sqrt {\pi }}\right )}{4\,b^3}+\frac {\sqrt {2}\,a^2\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (a+b\,x\right )}{\sqrt {\pi }}\right )}{2\,b^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos((a + b*x)^2),x)

[Out]

(x*sin((a + b*x)^2))/(2*b^2) - (a*sin((a + b*x)^2))/(2*b^3) - (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*(a + b*x))/p
i^(1/2)))/(4*b^3) + (2^(1/2)*a^2*pi^(1/2)*fresnelc((2^(1/2)*(a + b*x))/pi^(1/2)))/(2*b^3)

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