3.20 \(\int \frac{\cos ^3(a+b x^2)}{x^2} \, dx\)

Optimal. Leaf size=168 \[ -\frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \sin (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \cos (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-\frac{\cos ^3\left (a+b x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.143579, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3394, 4574, 3353, 3352, 3351} \[ -\frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \sin (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \cos (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-\frac{\cos ^3\left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^3/x^2,x]

[Out]

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

Rule 3394

Int[Cos[(a_.) + (b_.)*(x_)^(n_)]^(p_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*Cos[a + b*x^n]^p)/(m + 1), x] +
Dist[(b*n*p)/(m + 1), Int[Cos[a + b*x^n]^(p - 1)*Sin[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4574

Int[Cos[w_]^(q_.)*Sin[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sin[v]^p*Cos[w]^q, x], x] /; IGtQ[p, 0] &&
IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w],
x]))

Rule 3353

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 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 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 \frac{\cos ^3\left (a+b x^2\right )}{x^2} \, dx &=-\frac{\cos ^3\left (a+b x^2\right )}{x}-(6 b) \int \cos ^2\left (a+b x^2\right ) \sin \left (a+b x^2\right ) \, dx\\ &=-\frac{\cos ^3\left (a+b x^2\right )}{x}-(6 b) \int \left (\frac{1}{4} \sin \left (a+b x^2\right )+\frac{1}{4} \sin \left (3 a+3 b x^2\right )\right ) \, dx\\ &=-\frac{\cos ^3\left (a+b x^2\right )}{x}-\frac{1}{2} (3 b) \int \sin \left (a+b x^2\right ) \, dx-\frac{1}{2} (3 b) \int \sin \left (3 a+3 b x^2\right ) \, dx\\ &=-\frac{\cos ^3\left (a+b x^2\right )}{x}-\frac{1}{2} (3 b \cos (a)) \int \sin \left (b x^2\right ) \, dx-\frac{1}{2} (3 b \cos (3 a)) \int \sin \left (3 b x^2\right ) \, dx-\frac{1}{2} (3 b \sin (a)) \int \cos \left (b x^2\right ) \, dx-\frac{1}{2} (3 b \sin (3 a)) \int \cos \left (3 b x^2\right ) \, dx\\ &=-\frac{\cos ^3\left (a+b x^2\right )}{x}-\frac{3}{2} \sqrt{b} \sqrt{\frac{\pi }{2}} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{1}{2} \sqrt{b} \sqrt{\frac{3 \pi }{2}} \cos (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-\frac{3}{2} \sqrt{b} \sqrt{\frac{\pi }{2}} C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a)-\frac{1}{2} \sqrt{b} \sqrt{\frac{3 \pi }{2}} C\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right ) \sin (3 a)\\ \end{align*}

Mathematica [A]  time = 0.643549, size = 166, normalized size = 0.99 \[ -\frac{3 \sqrt{2 \pi } \sqrt{b} x \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )+\sqrt{6 \pi } \sqrt{b} x \sin (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )+3 \sqrt{2 \pi } \sqrt{b} x \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )+\sqrt{6 \pi } \sqrt{b} x \cos (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )+3 \cos \left (a+b x^2\right )+\cos \left (3 \left (a+b x^2\right )\right )}{4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^3/x^2,x]

[Out]

-(3*Cos[a + b*x^2] + Cos[3*(a + b*x^2)] + 3*Sqrt[b]*Sqrt[2*Pi]*x*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x] + Sqrt[
b]*Sqrt[6*Pi]*x*Cos[3*a]*FresnelS[Sqrt[b]*Sqrt[6/Pi]*x] + 3*Sqrt[b]*Sqrt[2*Pi]*x*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x
]*Sin[a] + Sqrt[b]*Sqrt[6*Pi]*x*FresnelC[Sqrt[b]*Sqrt[6/Pi]*x]*Sin[3*a])/(4*x)

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Maple [A]  time = 0.033, size = 128, normalized size = 0.8 \begin{align*} -{\frac{3\,\cos \left ( b{x}^{2}+a \right ) }{4\,x}}-{\frac{3\,\sqrt{2}\sqrt{\pi }}{4}\sqrt{b} \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) \right ) }-{\frac{\cos \left ( 3\,b{x}^{2}+3\,a \right ) }{4\,x}}-{\frac{\sqrt{2}\sqrt{\pi }\sqrt{3}}{4}\sqrt{b} \left ( \cos \left ( 3\,a \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{3}x}{\sqrt{\pi }}\sqrt{b}} \right ) +\sin \left ( 3\,a \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}x}{\sqrt{\pi }}\sqrt{b}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x^2+a)^3/x^2,x)

[Out]

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

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Maxima [C]  time = 1.38955, size = 721, normalized size = 4.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^3/x^2,x, algorithm="maxima")

[Out]

-1/32*(sqrt(3)*sqrt(x^2*abs(b))*(((gamma(-1/2, 3*I*b*x^2) + gamma(-1/2, -3*I*b*x^2))*cos(1/4*pi + 1/2*arctan2(
0, b)) + (gamma(-1/2, 3*I*b*x^2) + gamma(-1/2, -3*I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (I*gamma(-1/2,
3*I*b*x^2) - I*gamma(-1/2, -3*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) + (-I*gamma(-1/2, 3*I*b*x^2) + I*gamma
(-1/2, -3*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(3*a) + ((-I*gamma(-1/2, 3*I*b*x^2) + I*gamma(-1/2, -
3*I*b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) + (-I*gamma(-1/2, 3*I*b*x^2) + I*gamma(-1/2, -3*I*b*x^2))*cos(-1/4
*pi + 1/2*arctan2(0, b)) + (gamma(-1/2, 3*I*b*x^2) + gamma(-1/2, -3*I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b))
- (gamma(-1/2, 3*I*b*x^2) + gamma(-1/2, -3*I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(3*a)) + sqrt(x^2*ab
s(b))*((3*(gamma(-1/2, I*b*x^2) + gamma(-1/2, -I*b*x^2))*cos(1/4*pi + 1/2*arctan2(0, b)) + 3*(gamma(-1/2, I*b*
x^2) + gamma(-1/2, -I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (3*I*gamma(-1/2, I*b*x^2) - 3*I*gamma(-1/2, -
I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) + (-3*I*gamma(-1/2, I*b*x^2) + 3*I*gamma(-1/2, -I*b*x^2))*sin(-1/4*p
i + 1/2*arctan2(0, b)))*cos(a) + ((-3*I*gamma(-1/2, I*b*x^2) + 3*I*gamma(-1/2, -I*b*x^2))*cos(1/4*pi + 1/2*arc
tan2(0, b)) + (-3*I*gamma(-1/2, I*b*x^2) + 3*I*gamma(-1/2, -I*b*x^2))*cos(-1/4*pi + 1/2*arctan2(0, b)) + 3*(ga
mma(-1/2, I*b*x^2) + gamma(-1/2, -I*b*x^2))*sin(1/4*pi + 1/2*arctan2(0, b)) - 3*(gamma(-1/2, I*b*x^2) + gamma(
-1/2, -I*b*x^2))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a)))/x

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Fricas [A]  time = 1.82808, size = 412, normalized size = 2.45 \begin{align*} -\frac{\sqrt{6} \pi x \sqrt{\frac{b}{\pi }} \cos \left (3 \, a\right ) \operatorname{S}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) + 3 \, \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) + \sqrt{6} \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (3 \, a\right ) + 3 \, \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) + 4 \, \cos \left (b x^{2} + a\right )^{3}}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^3/x^2,x, algorithm="fricas")

[Out]

-1/4*(sqrt(6)*pi*x*sqrt(b/pi)*cos(3*a)*fresnel_sin(sqrt(6)*x*sqrt(b/pi)) + 3*sqrt(2)*pi*x*sqrt(b/pi)*cos(a)*fr
esnel_sin(sqrt(2)*x*sqrt(b/pi)) + sqrt(6)*pi*x*sqrt(b/pi)*fresnel_cos(sqrt(6)*x*sqrt(b/pi))*sin(3*a) + 3*sqrt(
2)*pi*x*sqrt(b/pi)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(a) + 4*cos(b*x^2 + a)^3)/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{3}{\left (a + b x^{2} \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)**3/x**2,x)

[Out]

Integral(cos(a + b*x**2)**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x^{2} + a\right )^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^3/x^2,x, algorithm="giac")

[Out]

integrate(cos(b*x^2 + a)^3/x^2, x)