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

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3467, 3434, 3433, 3432} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} \sin (a) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} x\right )}{2 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{2 b^{3/2}}+\frac {x \sin \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*x^2],x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 82, normalized size = 0.90 \begin {gather*} \frac {-\sqrt {2 \pi } \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\sqrt {2 \pi } \text {FresnelC}\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)+2 \sqrt {b} x \sin \left (a+b x^2\right )}{4 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*x^2],x]

[Out]

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

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Maple [A]
time = 0.11, size = 58, normalized size = 0.64

method	result	size
default	\(\frac {x \sin \left (b \,x^{2}+a \right )}{2 b}-\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \mathrm {S}\left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \left (a \right ) \FresnelC \left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{4 b^{\frac {3}{2}}}\)	\(58\)
risch	\(-\frac {i {\mathrm e}^{-i a} \sqrt {\pi }\, \erf \left (\sqrt {i b}\, x \right )}{8 b \sqrt {i b}}+\frac {i {\mathrm e}^{i a} \sqrt {\pi }\, \erf \left (\sqrt {-i b}\, x \right )}{8 b \sqrt {-i b}}+\frac {x \sin \left (b \,x^{2}+a \right )}{2 b}\)	\(74\)
meijerg	\(\frac {\cos \left (a \right ) \sqrt {\pi }\, \sqrt {2}\, \left (\frac {x \sqrt {2}\, \left (b^{2}\right )^{\frac {3}{4}} \sin \left (b \,x^{2}\right )}{2 \sqrt {\pi }\, b}-\frac {\left (b^{2}\right )^{\frac {3}{4}} \mathrm {S}\left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )}{2 b^{\frac {3}{2}}}\right )}{2 \left (b^{2}\right )^{\frac {3}{4}}}-\frac {\sin \left (a \right ) \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {x \sqrt {2}\, \sqrt {b}\, \cos \left (b \,x^{2}\right )}{2 \sqrt {\pi }}+\frac {\FresnelC \left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )}{2}\right )}{2 b^{\frac {3}{2}}}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.30, size = 67, normalized size = 0.74 \begin {gather*} \frac {8 \, b^{2} x \sin \left (b x^{2} + a\right ) + \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \left (a\right ) + \left (i - 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {i \, b} x\right ) + {\left (\left (i - 1\right ) \, \cos \left (a\right ) - \left (i + 1\right ) \, \sin \left (a\right )\right )} \operatorname {erf}\left (\sqrt {-i \, b} x\right )\right )} b^{\frac {3}{2}}}{16 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(8*b^2*x*sin(b*x^2 + a) + sqrt(2)*sqrt(pi)*((-(I + 1)*cos(a) + (I - 1)*sin(a))*erf(sqrt(I*b)*x) + ((I - 1
)*cos(a) - (I + 1)*sin(a))*erf(sqrt(-I*b)*x))*b^(3/2))/b^3

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Fricas [A]
time = 0.37, size = 72, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \left (a\right ) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) + \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (a\right ) - 2 \, b x \sin \left (b x^{2} + a\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (90) = 180\).
time = 1.22, size = 209, normalized size = 2.30 \begin {gather*} \frac {b^{\frac {3}{2}} x^{5} \sqrt {\frac {1}{b}} \sin {\left (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 | {- \frac {b^{2} x^{4}}{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 (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 | {- \frac {b^{2} x^{4}}{4}} \right )}}{8 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} - \frac {\sqrt {2} \sqrt {\pi } x^{2} \sqrt {\frac {1}{b}} \sin {\left (a \right )} S\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {\pi }}\right )}{2} + \frac {\sqrt {2} \sqrt {\pi } x^{2} \sqrt {\frac {1}{b}} \cos {\left (a \right )} C\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {\pi }}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.43, size = 135, normalized size = 1.48 \begin {gather*} -\frac {i \, x e^{\left (i \, b x^{2} + i \, a\right )}}{4 \, b} + \frac {i \, x e^{\left (-i \, b x^{2} - i \, a\right )}}{4 \, b} - \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}\right ) e^{\left (i \, a\right )}}{8 \, b {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}} + \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}\right ) e^{\left (-i \, a\right )}}{8 \, b {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*x*e^(I*b*x^2 + I*a)/b + 1/4*I*x*e^(-I*b*x^2 - I*a)/b - 1/8*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(-I*b/
abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/(b*(-I*b/abs(b) + 1)*sqrt(abs(b))) + 1/8*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)
*x*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/(b*(I*b/abs(b) + 1)*sqrt(abs(b)))

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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 ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*x^2),x)

[Out]

int(x^2*cos(a + b*x^2), x)

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