3.1.9 \(\int x^2 \cos ^2(a+b x^2) \, dx\) [9]

Optimal. Leaf size=91 \[ \frac {x^3}{6}-\frac {\sqrt {\pi } \cos (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a)}{16 b^{3/2}}+\frac {x \sin \left (2 a+2 b x^2\right )}{8 b} \]

[Out]

1/6*x^3+1/8*x*sin(2*b*x^2+2*a)/b-1/16*cos(2*a)*FresnelS(2*x*b^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(3/2)-1/16*FresnelC(2
*x*b^(1/2)/Pi^(1/2))*sin(2*a)*Pi^(1/2)/b^(3/2)

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Rubi [A]
time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3485, 3467, 3434, 3433, 3432} \begin {gather*} -\frac {\sqrt {\pi } \sin (2 a) \text {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {\pi } \cos (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}+\frac {x \sin \left (2 a+2 b x^2\right )}{8 b}+\frac {x^3}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x^2]^2,x]

[Out]

x^3/6 - (Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]])/(16*b^(3/2)) - (Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/S
qrt[Pi]]*Sin[2*a])/(16*b^(3/2)) + (x*Sin[2*a + 2*b*x^2])/(8*b)

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 3434

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

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 3485

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^2 \cos ^2\left (a+b x^2\right ) \, dx &=\int \left (\frac {x^2}{2}+\frac {1}{2} x^2 \cos \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac {x^3}{6}+\frac {1}{2} \int x^2 \cos \left (2 a+2 b x^2\right ) \, dx\\ &=\frac {x^3}{6}+\frac {x \sin \left (2 a+2 b x^2\right )}{8 b}-\frac {\int \sin \left (2 a+2 b x^2\right ) \, dx}{8 b}\\ &=\frac {x^3}{6}+\frac {x \sin \left (2 a+2 b x^2\right )}{8 b}-\frac {\cos (2 a) \int \sin \left (2 b x^2\right ) \, dx}{8 b}-\frac {\sin (2 a) \int \cos \left (2 b x^2\right ) \, dx}{8 b}\\ &=\frac {x^3}{6}-\frac {\sqrt {\pi } \cos (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a)}{16 b^{3/2}}+\frac {x \sin \left (2 a+2 b x^2\right )}{8 b}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 87, normalized size = 0.96 \begin {gather*} \frac {-3 \sqrt {\pi } \cos (2 a) S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )-3 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right ) \sin (2 a)+2 \sqrt {b} x \left (4 b x^2+3 \sin \left (2 \left (a+b x^2\right )\right )\right )}{48 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x^2]^2,x]

[Out]

(-3*Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x)/Sqrt[Pi]] - 3*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x)/Sqrt[Pi]]*Sin[2*a]
+ 2*Sqrt[b]*x*(4*b*x^2 + 3*Sin[2*(a + b*x^2)]))/(48*b^(3/2))

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Maple [A]
time = 0.12, size = 63, normalized size = 0.69

method	result	size
default	\(\frac {x^{3}}{6}+\frac {x \sin \left (2 b \,x^{2}+2 a \right )}{8 b}-\frac {\sqrt {\pi }\, \left (\cos \left (2 a \right ) \mathrm {S}\left (\frac {2 x \sqrt {b}}{\sqrt {\pi }}\right )+\sin \left (2 a \right ) \FresnelC \left (\frac {2 x \sqrt {b}}{\sqrt {\pi }}\right )\right )}{16 b^{\frac {3}{2}}}\)	\(63\)
risch	\(\frac {x^{3}}{6}-\frac {i {\mathrm e}^{-2 i a} \sqrt {\pi }\, \sqrt {2}\, \erf \left (\sqrt {2}\, \sqrt {i b}\, x \right )}{64 b \sqrt {i b}}+\frac {i {\mathrm e}^{2 i a} \sqrt {\pi }\, \erf \left (\sqrt {-2 i b}\, x \right )}{32 b \sqrt {-2 i b}}+\frac {x \sin \left (2 b \,x^{2}+2 a \right )}{8 b}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3+1/8*x*sin(2*b*x^2+2*a)/b-1/16/b^(3/2)*Pi^(1/2)*(cos(2*a)*FresnelS(2*x*b^(1/2)/Pi^(1/2))+sin(2*a)*Fresn
elC(2*x*b^(1/2)/Pi^(1/2)))

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Maxima [C] Result contains complex when optimal does not.
time = 0.52, size = 90, normalized size = 0.99 \begin {gather*} \frac {64 \, b^{3} x^{3} + 48 \, b^{2} x \sin \left (2 \, b x^{2} + 2 \, a\right ) - 3 \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, \cos \left (2 \, a\right ) - \left (i - 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {2 i \, b} x\right ) + {\left (-\left (i - 1\right ) \, \cos \left (2 \, a\right ) + \left (i + 1\right ) \, \sin \left (2 \, a\right )\right )} \operatorname {erf}\left (\sqrt {-2 i \, b} x\right )\right )} b^{\frac {3}{2}}}{384 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(64*b^3*x^3 + 48*b^2*x*sin(2*b*x^2 + 2*a) - 3*4^(1/4)*sqrt(2)*sqrt(pi)*(((I + 1)*cos(2*a) - (I - 1)*sin(
2*a))*erf(sqrt(2*I*b)*x) + (-(I - 1)*cos(2*a) + (I + 1)*sin(2*a))*erf(sqrt(-2*I*b)*x))*b^(3/2))/b^3

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Fricas [A]
time = 0.39, size = 84, normalized size = 0.92 \begin {gather*} \frac {8 \, b^{2} x^{3} + 12 \, b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right ) - 3 \, \pi \sqrt {\frac {b}{\pi }} \cos \left (2 \, a\right ) \operatorname {S}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) - 3 \, \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (2 \, x \sqrt {\frac {b}{\pi }}\right ) \sin \left (2 \, a\right )}{48 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(8*b^2*x^3 + 12*b*x*cos(b*x^2 + a)*sin(b*x^2 + a) - 3*pi*sqrt(b/pi)*cos(2*a)*fresnel_sin(2*x*sqrt(b/pi))
- 3*pi*sqrt(b/pi)*fresnel_cos(2*x*sqrt(b/pi))*sin(2*a))/b^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (85) = 170\).
time = 1.61, size = 201, normalized size = 2.21 \begin {gather*} \frac {b^{\frac {3}{2}} x^{5} \sqrt {\frac {1}{b}} \sin {\left (2 a \right )} \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {9}{4} \end {matrix}\middle | {- b^{2} x^{4}} \right )}}{8 \Gamma \left (\frac {7}{4}\right ) \Gamma \left (\frac {9}{4}\right )} - \frac {\sqrt {b} x^{3} \sqrt {\frac {1}{b}} \cos {\left (2 a \right )} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {7}{4} \end {matrix}\middle | {- b^{2} x^{4}} \right )}}{16 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} + \frac {x^{3}}{6} - \frac {\sqrt {\pi } x^{2} \sqrt {\frac {1}{b}} \sin {\left (2 a \right )} S\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{4} + \frac {\sqrt {\pi } x^{2} \sqrt {\frac {1}{b}} \cos {\left (2 a \right )} C\left (\frac {2 \sqrt {b} x}{\sqrt {\pi }}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**(3/2)*x**5*sqrt(1/b)*sin(2*a)*gamma(3/4)*gamma(5/4)*hyper((3/4, 5/4), (3/2, 7/4, 9/4), -b**2*x**4)/(8*gamma
(7/4)*gamma(9/4)) - sqrt(b)*x**3*sqrt(1/b)*cos(2*a)*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -
b**2*x**4)/(16*gamma(5/4)*gamma(7/4)) + x**3/6 - sqrt(pi)*x**2*sqrt(1/b)*sin(2*a)*fresnels(2*sqrt(b)*x/sqrt(pi
))/4 + sqrt(pi)*x**2*sqrt(1/b)*cos(2*a)*fresnelc(2*sqrt(b)*x/sqrt(pi))/4

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Giac [C] Result contains complex when optimal does not.
time = 0.49, size = 118, normalized size = 1.30 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {i \, x e^{\left (2 i \, b x^{2} + 2 i \, a\right )}}{16 \, b} + \frac {i \, x e^{\left (-2 i \, b x^{2} - 2 i \, a\right )}}{16 \, b} - \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\sqrt {b} x {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (2 i \, a\right )}}{32 \, b^{\frac {3}{2}} {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}} + \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\sqrt {b} x {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-2 i \, a\right )}}{32 \, b^{\frac {3}{2}} {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/16*I*x*e^(2*I*b*x^2 + 2*I*a)/b + 1/16*I*x*e^(-2*I*b*x^2 - 2*I*a)/b - 1/32*I*sqrt(pi)*erf(-sqrt(b)*
x*(-I*b/abs(b) + 1))*e^(2*I*a)/(b^(3/2)*(-I*b/abs(b) + 1)) + 1/32*I*sqrt(pi)*erf(-sqrt(b)*x*(I*b/abs(b) + 1))*
e^(-2*I*a)/(b^(3/2)*(I*b/abs(b) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*x^2)^2,x)

[Out]

int(x^2*cos(a + b*x^2)^2, x)

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