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

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

[Out]

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

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Rubi [A]  time = 0.145827, 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 = {3393, 4574, 3354, 3352, 3351} \[ \frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \cos (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )-\frac{3}{2} \sqrt{\frac{\pi }{2}} \sqrt{b} \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )+\frac{1}{2} \sqrt{\frac{3 \pi }{2}} \sqrt{b} \sin (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-\frac{\sin ^3\left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x^2]^3/x^2,x]

[Out]

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

Rule 3393

Int[(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[(x^(m + 1)*Sin[a + b*x^n]^p)/(m + 1), x] -
Dist[(b*n*p)/(m + 1), Int[Sin[a + b*x^n]^(p - 1)*Cos[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 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[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{\sin ^3\left (a+b x^2\right )}{x^2} \, dx &=-\frac{\sin ^3\left (a+b x^2\right )}{x}+(6 b) \int \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right ) \, dx\\ &=-\frac{\sin ^3\left (a+b x^2\right )}{x}+(6 b) \int \left (\frac{1}{4} \cos \left (a+b x^2\right )-\frac{1}{4} \cos \left (3 a+3 b x^2\right )\right ) \, dx\\ &=-\frac{\sin ^3\left (a+b x^2\right )}{x}+\frac{1}{2} (3 b) \int \cos \left (a+b x^2\right ) \, dx-\frac{1}{2} (3 b) \int \cos \left (3 a+3 b x^2\right ) \, dx\\ &=-\frac{\sin ^3\left (a+b x^2\right )}{x}+\frac{1}{2} (3 b \cos (a)) \int \cos \left (b x^2\right ) \, dx-\frac{1}{2} (3 b \cos (3 a)) \int \cos \left (3 b x^2\right ) \, dx-\frac{1}{2} (3 b \sin (a)) \int \sin \left (b x^2\right ) \, dx+\frac{1}{2} (3 b \sin (3 a)) \int \sin \left (3 b x^2\right ) \, dx\\ &=\frac{3}{2} \sqrt{b} \sqrt{\frac{\pi }{2}} \cos (a) C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )-\frac{1}{2} \sqrt{b} \sqrt{\frac{3 \pi }{2}} \cos (3 a) C\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-\frac{3}{2} \sqrt{b} \sqrt{\frac{\pi }{2}} S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right ) \sin (a)+\frac{1}{2} \sqrt{b} \sqrt{\frac{3 \pi }{2}} S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right ) \sin (3 a)-\frac{\sin ^3\left (a+b x^2\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.433062, size = 167, normalized size = 0.99 \[ \frac{3 \sqrt{2 \pi } \sqrt{b} x \cos (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} x\right )-\sqrt{6 \pi } \sqrt{b} x \cos (3 a) \text{FresnelC}\left (\sqrt{\frac{6}{\pi }} \sqrt{b} x\right )-3 \sqrt{2 \pi } \sqrt{b} x \sin (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} x\right )+\sqrt{6 \pi } \sqrt{b} x \sin (3 a) S\left (\sqrt{b} \sqrt{\frac{6}{\pi }} x\right )-3 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )}{4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x^2]^3/x^2,x]

[Out]

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

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Maple [A]  time = 0.01, size = 130, normalized size = 0.8 \begin{align*} -{\frac{3\,\sin \left ( b{x}^{2}+a \right ) }{4\,x}}+{\frac{3\,\sqrt{2}\sqrt{\pi }}{4}\sqrt{b} \left ( \cos \left ( a \right ){\it FresnelC} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) -\sin \left ( a \right ){\it FresnelS} \left ({\frac{x\sqrt{2}}{\sqrt{\pi }}\sqrt{b}} \right ) \right ) }+{\frac{\sin \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 FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}x}{\sqrt{\pi }}\sqrt{b}} \right ) -\sin \left ( 3\,a \right ){\it FresnelS} \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(sin(b*x^2+a)^3/x^2,x)

[Out]

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

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Maxima [C]  time = 1.2606, size = 725, normalized size = 4.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(sqrt(3)*sqrt(x^2*abs(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*arcta
n2(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)))*cos(3*a) + ((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*ar
ctan2(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*ga
mma(-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)))*sin(3*a)) + sqrt(x^2*abs(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*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*(gamma(-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)))*cos(a) - (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*pi + 1/2*arctan2(0, b)))*sin(a)))/x

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Fricas [A]  time = 2.48165, size = 440, normalized size = 2.62 \begin{align*} -\frac{\sqrt{6} \pi x \sqrt{\frac{b}{\pi }} \cos \left (3 \, a\right ) \operatorname{C}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) - 3 \, \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{C}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) - \sqrt{6} \pi x \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\sqrt{6} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (3 \, a\right ) + 3 \, \sqrt{2} \pi x \sqrt{\frac{b}{\pi }} \operatorname{S}\left (\sqrt{2} x \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) - 4 \,{\left (\cos \left (b x^{2} + a\right )^{2} - 1\right )} \sin \left (b x^{2} + a\right )}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(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{\sin \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(sin(b*x^2+a)^3/x^2,x, algorithm="giac")

[Out]

integrate(sin(b*x^2 + a)^3/x^2, x)