3.29 \(\int \sin ^3(a+b x^2) \, dx\)

Optimal. Leaf size=153 \[ \frac {3 \sqrt {\frac {\pi }{2}} \sin (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{6}} \sin (3 a) C\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )}{4 \sqrt {b}}+\frac {3 \sqrt {\frac {\pi }{2}} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{6}} \cos (3 a) S\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )}{4 \sqrt {b}} \]

[Out]

-1/24*cos(3*a)*FresnelS(x*b^(1/2)*6^(1/2)/Pi^(1/2))*6^(1/2)*Pi^(1/2)/b^(1/2)-1/24*FresnelC(x*b^(1/2)*6^(1/2)/P
i^(1/2))*sin(3*a)*6^(1/2)*Pi^(1/2)/b^(1/2)+3/8*cos(a)*FresnelS(x*b^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)/b^
(1/2)+3/8*FresnelC(x*b^(1/2)*2^(1/2)/Pi^(1/2))*sin(a)*2^(1/2)*Pi^(1/2)/b^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3357, 3353, 3352, 3351} \[ \frac {3 \sqrt {\frac {\pi }{2}} \sin (a) \text {FresnelC}\left (\sqrt {\frac {2}{\pi }} \sqrt {b} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{6}} \sin (3 a) \text {FresnelC}\left (\sqrt {\frac {6}{\pi }} \sqrt {b} x\right )}{4 \sqrt {b}}+\frac {3 \sqrt {\frac {\pi }{2}} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )}{4 \sqrt {b}}-\frac {\sqrt {\frac {\pi }{6}} \cos (3 a) S\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )}{4 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x^2]^3,x]

[Out]

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

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]

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 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 3357

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 117, normalized size = 0.76 \[ \frac {\sqrt {\frac {\pi }{6}} \left (3 \sqrt {3} \sin (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\sin (3 a) C\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )+3 \sqrt {3} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} x\right )-\cos (3 a) S\left (\sqrt {b} \sqrt {\frac {6}{\pi }} x\right )\right )}{4 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x^2]^3,x]

[Out]

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

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fricas [A]  time = 0.72, size = 120, normalized size = 0.78 \[ -\frac {\sqrt {6} \pi \sqrt {\frac {b}{\pi }} \cos \left (3 \, a\right ) \operatorname {S}\left (\sqrt {6} x \sqrt {\frac {b}{\pi }}\right ) - 9 \, \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {S}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) + \sqrt {6} \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {6} x \sqrt {\frac {b}{\pi }}\right ) \sin \left (3 \, a\right ) - 9 \, \sqrt {2} \pi \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} x \sqrt {\frac {b}{\pi }}\right ) \sin \relax (a)}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(sqrt(6)*pi*sqrt(b/pi)*cos(3*a)*fresnel_sin(sqrt(6)*x*sqrt(b/pi)) - 9*sqrt(2)*pi*sqrt(b/pi)*cos(a)*fresn
el_sin(sqrt(2)*x*sqrt(b/pi)) + sqrt(6)*pi*sqrt(b/pi)*fresnel_cos(sqrt(6)*x*sqrt(b/pi))*sin(3*a) - 9*sqrt(2)*pi
*sqrt(b/pi)*fresnel_cos(sqrt(2)*x*sqrt(b/pi))*sin(a))/b

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giac [C]  time = 0.93, size = 185, normalized size = 1.21 \[ -\frac {i \, \sqrt {6} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {6} \sqrt {b} x {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (3 i \, a\right )}}{48 \, \sqrt {b} {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )}} + \frac {3 i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}\right ) e^{\left (i \, a\right )}}{16 \, {\left (-\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}} - \frac {3 i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {2} x {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}\right ) e^{\left (-i \, a\right )}}{16 \, {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )} \sqrt {{\left | b \right |}}} + \frac {i \, \sqrt {6} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {6} \sqrt {b} x {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-3 i \, a\right )}}{48 \, \sqrt {b} {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*I*sqrt(6)*sqrt(pi)*erf(-1/2*sqrt(6)*sqrt(b)*x*(-I*b/abs(b) + 1))*e^(3*I*a)/(sqrt(b)*(-I*b/abs(b) + 1)) +
3/16*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(-I*b/abs(b) + 1)*sqrt(abs(b)))*e^(I*a)/((-I*b/abs(b) + 1)*sqrt(ab
s(b))) - 3/16*I*sqrt(2)*sqrt(pi)*erf(-1/2*sqrt(2)*x*(I*b/abs(b) + 1)*sqrt(abs(b)))*e^(-I*a)/((I*b/abs(b) + 1)*
sqrt(abs(b))) + 1/48*I*sqrt(6)*sqrt(pi)*erf(-1/2*sqrt(6)*sqrt(b)*x*(I*b/abs(b) + 1))*e^(-3*I*a)/(sqrt(b)*(I*b/
abs(b) + 1))

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maple [A]  time = 0.04, size = 99, normalized size = 0.65 \[ \frac {3 \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \mathrm {S}\left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (a ) \FresnelC \left (\frac {x \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )}{8 \sqrt {b}}-\frac {\sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (3 a \right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )+\sin \left (3 a \right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {b}\, x}{\sqrt {\pi }}\right )\right )}{24 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*2^(1/2)*Pi^(1/2)/b^(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)))-1/24*2^(1/2)*Pi^(1/2)*3^(1/2)/b^(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 = 1.19, size = 112, normalized size = 0.73 \[ \frac {3 \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, \cos \left (3 \, a\right ) + \left (i - 1\right ) \, \sin \left (3 \, a\right )\right )} \operatorname {erf}\left (\sqrt {3 i \, b} x\right ) + {\left (\left (i - 1\right ) \, \cos \left (3 \, a\right ) - \left (i + 1\right ) \, \sin \left (3 \, a\right )\right )} \operatorname {erf}\left (\sqrt {-3 i \, b} x\right )\right )} b^{\frac {3}{2}} + \sqrt {2} \sqrt {\pi } {\left ({\left (\left (27 i + 27\right ) \, \cos \relax (a) - \left (27 i - 27\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\sqrt {i \, b} x\right ) + {\left (-\left (27 i - 27\right ) \, \cos \relax (a) + \left (27 i + 27\right ) \, \sin \relax (a)\right )} \operatorname {erf}\left (\sqrt {-i \, b} x\right )\right )} b^{\frac {3}{2}}}{288 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288*(3*9^(1/4)*sqrt(2)*sqrt(pi)*((-(I + 1)*cos(3*a) + (I - 1)*sin(3*a))*erf(sqrt(3*I*b)*x) + ((I - 1)*cos(3*
a) - (I + 1)*sin(3*a))*erf(sqrt(-3*I*b)*x))*b^(3/2) + sqrt(2)*sqrt(pi)*(((27*I + 27)*cos(a) - (27*I - 27)*sin(
a))*erf(sqrt(I*b)*x) + (-(27*I - 27)*cos(a) + (27*I + 27)*sin(a))*erf(sqrt(-I*b)*x))*b^(3/2))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^2)^3,x)

[Out]

int(sin(a + b*x^2)^3, x)

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sympy [A]  time = 1.36, size = 129, normalized size = 0.84 \[ \frac {3 \sqrt {2} \sqrt {\pi } \left (\sin {\relax (a )} C\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {\pi }}\right ) + \cos {\relax (a )} S\left (\frac {\sqrt {2} \sqrt {b} x}{\sqrt {\pi }}\right )\right ) \sqrt {\frac {1}{b}}}{8} - \frac {\sqrt {6} \sqrt {\pi } \left (\sin {\left (3 a \right )} C\left (\frac {\sqrt {6} \sqrt {b} x}{\sqrt {\pi }}\right ) + \cos {\left (3 a \right )} S\left (\frac {\sqrt {6} \sqrt {b} x}{\sqrt {\pi }}\right )\right ) \sqrt {\frac {1}{b}}}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(2)*sqrt(pi)*(sin(a)*fresnelc(sqrt(2)*sqrt(b)*x/sqrt(pi)) + cos(a)*fresnels(sqrt(2)*sqrt(b)*x/sqrt(pi)))
*sqrt(1/b)/8 - sqrt(6)*sqrt(pi)*(sin(3*a)*fresnelc(sqrt(6)*sqrt(b)*x/sqrt(pi)) + cos(3*a)*fresnels(sqrt(6)*sqr
t(b)*x/sqrt(pi)))*sqrt(1/b)/24

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