3.1.30 \(\int \frac {\sin ^3(a+b x^2)}{x^2} \, dx\) [30]

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

[Out]

-sin(b*x^2+a)^3/x+3/4*cos(a)*FresnelC(x*b^(1/2)*2^(1/2)/Pi^(1/2))*b^(1/2)*2^(1/2)*Pi^(1/2)-3/4*FresnelS(x*b^(1
/2)*2^(1/2)/Pi^(1/2))*sin(a)*b^(1/2)*2^(1/2)*Pi^(1/2)-1/4*cos(3*a)*FresnelC(x*b^(1/2)*6^(1/2)/Pi^(1/2))*b^(1/2
)*6^(1/2)*Pi^(1/2)+1/4*FresnelS(x*b^(1/2)*6^(1/2)/Pi^(1/2))*sin(3*a)*b^(1/2)*6^(1/2)*Pi^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 168, normalized size of antiderivative = 1.00, 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 = {3474, 4670, 3435, 3433, 3432} \begin {gather*} \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} \end {gather*}

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 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 3435

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 3474

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 4670

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]))

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*}

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Mathematica [A]
time = 0.29, size = 167, normalized size = 0.99 \begin {gather*} \frac {3 \sqrt {b} \sqrt {2 \pi } x \cos (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\sqrt {b} \sqrt {6 \pi } x \cos (3 a) C\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )-3 \sqrt {b} \sqrt {2 \pi } x S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right ) \sin (a)+\sqrt {b} \sqrt {6 \pi } x S\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right ) \sin (3 a)-3 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )}{4 x} \end {gather*}

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 = 0.07, size = 130, normalized size = 0.77

method	result	size
default	\(-\frac {3 \sin \left (b \,x^{2}+a \right )}{4 x}+\frac {3 \sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (a \right ) \FresnelC \left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )-\sin \left (a \right ) \mathrm {S}\left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{4}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{4 x}-\frac {\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (3 a \right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )-\sin \left (3 a \right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )\right )}{4}\)	\(130\)
risch	\(-\frac {{\mathrm e}^{-3 i a} b \sqrt {\pi }\, \sqrt {3}\, \erf \left (\sqrt {3}\, \sqrt {i b}\, x \right )}{8 \sqrt {i b}}-\frac {3 \,{\mathrm e}^{3 i a} b \sqrt {\pi }\, \erf \left (\sqrt {-3 i b}\, x \right )}{8 \sqrt {-3 i b}}+\frac {3 \,{\mathrm e}^{i a} b \sqrt {\pi }\, \erf \left (\sqrt {-i b}\, x \right )}{8 \sqrt {-i b}}+\frac {3 \,{\mathrm e}^{-i a} b \sqrt {\pi }\, \erf \left (\sqrt {i b}\, x \right )}{8 \sqrt {i b}}-\frac {3 \sin \left (b \,x^{2}+a \right )}{4 x}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{4 x}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x^2+a)^3/x^2,x,method=_RETURNVERBOSE)

[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] Result contains complex when optimal does not.
time = 0.61, size = 152, normalized size = 0.90 \begin {gather*} \frac {\sqrt {3} \sqrt {b x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 3 i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, 3 i \, b x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right )\right )} - 3 \, \sqrt {b x^{2}} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \cos \left (a\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, i \, b x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -i \, b x^{2}\right )\right )} \sin \left (a\right )\right )}}{32 \, x} \end {gather*}

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(b*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, 3*I*b*x^2) - (I + 1)*sqrt(2)*gamma(-1/2, -3*I*b*x^2))*
cos(3*a) + ((I + 1)*sqrt(2)*gamma(-1/2, 3*I*b*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -3*I*b*x^2))*sin(3*a)) - 3*sq
rt(b*x^2)*(((I - 1)*sqrt(2)*gamma(-1/2, I*b*x^2) - (I + 1)*sqrt(2)*gamma(-1/2, -I*b*x^2))*cos(a) + ((I + 1)*sq
rt(2)*gamma(-1/2, I*b*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -I*b*x^2))*sin(a)))/x

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Fricas [A]
time = 0.37, size = 147, normalized size = 0.88 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{3}{\left (a + b x^{2} \right )}}{x^{2}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (b\,x^2+a\right )}^3}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^2)^3/x^2,x)

[Out]

int(sin(a + b*x^2)^3/x^2, x)

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