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

Optimal. Leaf size=91 \[ \frac {3}{8} b \cos (a) \text {Ci}\left (b x^2\right )-\frac {3}{8} b \cos (3 a) \text {Ci}\left (3 b x^2\right )-\frac {3}{8} b \sin (a) \text {Si}\left (b x^2\right )+\frac {3}{8} b \sin (3 a) \text {Si}\left (3 b x^2\right )-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

[Out]

3/8*b*Ci(b*x^2)*cos(a)-3/8*b*Ci(3*b*x^2)*cos(3*a)-3/8*b*Si(b*x^2)*sin(a)+3/8*b*Si(3*b*x^2)*sin(3*a)-3/8*sin(b*
x^2+a)/x^2+1/8*sin(3*b*x^2+3*a)/x^2

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Rubi [A]  time = 0.22, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3403, 3379, 3297, 3303, 3299, 3302} \[ \frac {3}{8} b \cos (a) \text {CosIntegral}\left (b x^2\right )-\frac {3}{8} b \cos (3 a) \text {CosIntegral}\left (3 b x^2\right )-\frac {3}{8} b \sin (a) \text {Si}\left (b x^2\right )+\frac {3}{8} b \sin (3 a) \text {Si}\left (3 b x^2\right )-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x^2]^3/x^3,x]

[Out]

(3*b*Cos[a]*CosIntegral[b*x^2])/8 - (3*b*Cos[3*a]*CosIntegral[3*b*x^2])/8 - (3*Sin[a + b*x^2])/(8*x^2) + Sin[3
*(a + b*x^2)]/(8*x^2) - (3*b*Sin[a]*SinIntegral[b*x^2])/8 + (3*b*Sin[3*a]*SinIntegral[3*b*x^2])/8

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^3\left (a+b x^2\right )}{x^3} \, dx &=\int \left (\frac {3 \sin \left (a+b x^2\right )}{4 x^3}-\frac {\sin \left (3 a+3 b x^2\right )}{4 x^3}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {\sin \left (3 a+3 b x^2\right )}{x^3} \, dx\right )+\frac {3}{4} \int \frac {\sin \left (a+b x^2\right )}{x^3} \, dx\\ &=-\left (\frac {1}{8} \operatorname {Subst}\left (\int \frac {\sin (3 a+3 b x)}{x^2} \, dx,x,x^2\right )\right )+\frac {3}{8} \operatorname {Subst}\left (\int \frac {\sin (a+b x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2}+\frac {1}{8} (3 b) \operatorname {Subst}\left (\int \frac {\cos (a+b x)}{x} \, dx,x,x^2\right )-\frac {1}{8} (3 b) \operatorname {Subst}\left (\int \frac {\cos (3 a+3 b x)}{x} \, dx,x,x^2\right )\\ &=-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2}+\frac {1}{8} (3 b \cos (a)) \operatorname {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,x^2\right )-\frac {1}{8} (3 b \cos (3 a)) \operatorname {Subst}\left (\int \frac {\cos (3 b x)}{x} \, dx,x,x^2\right )-\frac {1}{8} (3 b \sin (a)) \operatorname {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,x^2\right )+\frac {1}{8} (3 b \sin (3 a)) \operatorname {Subst}\left (\int \frac {\sin (3 b x)}{x} \, dx,x,x^2\right )\\ &=\frac {3}{8} b \cos (a) \text {Ci}\left (b x^2\right )-\frac {3}{8} b \cos (3 a) \text {Ci}\left (3 b x^2\right )-\frac {3 \sin \left (a+b x^2\right )}{8 x^2}+\frac {\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac {3}{8} b \sin (a) \text {Si}\left (b x^2\right )+\frac {3}{8} b \sin (3 a) \text {Si}\left (3 b x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 90, normalized size = 0.99 \[ \frac {3 b x^2 \cos (a) \text {Ci}\left (b x^2\right )-3 b x^2 \cos (3 a) \text {Ci}\left (3 b x^2\right )-3 b x^2 \sin (a) \text {Si}\left (b x^2\right )+3 b x^2 \sin (3 a) \text {Si}\left (3 b x^2\right )-3 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x^2]^3/x^3,x]

[Out]

(3*b*x^2*Cos[a]*CosIntegral[b*x^2] - 3*b*x^2*Cos[3*a]*CosIntegral[3*b*x^2] - 3*Sin[a + b*x^2] + Sin[3*(a + b*x
^2)] - 3*b*x^2*Sin[a]*SinIntegral[b*x^2] + 3*b*x^2*Sin[3*a]*SinIntegral[3*b*x^2])/(8*x^2)

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fricas [A]  time = 0.79, size = 118, normalized size = 1.30 \[ \frac {6 \, b x^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (3 \, b x^{2}\right ) - 6 \, b x^{2} \sin \relax (a) \operatorname {Si}\left (b x^{2}\right ) - 3 \, {\left (b x^{2} \operatorname {Ci}\left (3 \, b x^{2}\right ) + b x^{2} \operatorname {Ci}\left (-3 \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) + 3 \, {\left (b x^{2} \operatorname {Ci}\left (b x^{2}\right ) + b x^{2} \operatorname {Ci}\left (-b x^{2}\right )\right )} \cos \relax (a) + 8 \, {\left (\cos \left (b x^{2} + a\right )^{2} - 1\right )} \sin \left (b x^{2} + a\right )}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*b*x^2*sin(3*a)*sin_integral(3*b*x^2) - 6*b*x^2*sin(a)*sin_integral(b*x^2) - 3*(b*x^2*cos_integral(3*b*
x^2) + b*x^2*cos_integral(-3*b*x^2))*cos(3*a) + 3*(b*x^2*cos_integral(b*x^2) + b*x^2*cos_integral(-b*x^2))*cos
(a) + 8*(cos(b*x^2 + a)^2 - 1)*sin(b*x^2 + a))/x^2

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giac [B]  time = 0.47, size = 186, normalized size = 2.04 \[ -\frac {3 \, {\left (b x^{2} + a\right )} b^{2} \cos \left (3 \, a\right ) \operatorname {Ci}\left (3 \, b x^{2}\right ) - 3 \, a b^{2} \cos \left (3 \, a\right ) \operatorname {Ci}\left (3 \, b x^{2}\right ) - 3 \, {\left (b x^{2} + a\right )} b^{2} \cos \relax (a) \operatorname {Ci}\left (b x^{2}\right ) + 3 \, a b^{2} \cos \relax (a) \operatorname {Ci}\left (b x^{2}\right ) + 3 \, {\left (b x^{2} + a\right )} b^{2} \sin \relax (a) \operatorname {Si}\left (b x^{2}\right ) - 3 \, a b^{2} \sin \relax (a) \operatorname {Si}\left (b x^{2}\right ) + 3 \, {\left (b x^{2} + a\right )} b^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (-3 \, b x^{2}\right ) - 3 \, a b^{2} \sin \left (3 \, a\right ) \operatorname {Si}\left (-3 \, b x^{2}\right ) - b^{2} \sin \left (3 \, b x^{2} + 3 \, a\right ) + 3 \, b^{2} \sin \left (b x^{2} + a\right )}{8 \, b^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*(b*x^2 + a)*b^2*cos(3*a)*cos_integral(3*b*x^2) - 3*a*b^2*cos(3*a)*cos_integral(3*b*x^2) - 3*(b*x^2 + a
)*b^2*cos(a)*cos_integral(b*x^2) + 3*a*b^2*cos(a)*cos_integral(b*x^2) + 3*(b*x^2 + a)*b^2*sin(a)*sin_integral(
b*x^2) - 3*a*b^2*sin(a)*sin_integral(b*x^2) + 3*(b*x^2 + a)*b^2*sin(3*a)*sin_integral(-3*b*x^2) - 3*a*b^2*sin(
3*a)*sin_integral(-3*b*x^2) - b^2*sin(3*b*x^2 + 3*a) + 3*b^2*sin(b*x^2 + a))/(b^2*x^2)

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maple [C]  time = 0.54, size = 162, normalized size = 1.78 \[ -\frac {3 i \pi \,{\mathrm e}^{-3 i a} \mathrm {csgn}\left (b \,x^{2}\right ) b}{16}+\frac {3 i {\mathrm e}^{-3 i a} \Si \left (3 b \,x^{2}\right ) b}{8}+\frac {3 \,{\mathrm e}^{-3 i a} \Ei \left (1, -3 i b \,x^{2}\right ) b}{16}+\frac {3 \,{\mathrm e}^{3 i a} b \Ei \left (1, -3 i b \,x^{2}\right )}{16}-\frac {3 \,{\mathrm e}^{i a} b \Ei \left (1, -i b \,x^{2}\right )}{16}+\frac {3 i \pi \,\mathrm {csgn}\left (b \,x^{2}\right ) {\mathrm e}^{-i a} b}{16}-\frac {3 i {\mathrm e}^{-i a} \Si \left (b \,x^{2}\right ) b}{8}-\frac {3 \Ei \left (1, -i b \,x^{2}\right ) {\mathrm e}^{-i a} b}{16}-\frac {3 \sin \left (b \,x^{2}+a \right )}{8 x^{2}}+\frac {\sin \left (3 b \,x^{2}+3 a \right )}{8 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*I*Pi*exp(-3*I*a)*csgn(b*x^2)*b+3/8*I*exp(-3*I*a)*Si(3*b*x^2)*b+3/16*exp(-3*I*a)*Ei(1,-3*I*b*x^2)*b+3/16*
exp(3*I*a)*b*Ei(1,-3*I*b*x^2)-3/16*exp(I*a)*b*Ei(1,-I*b*x^2)+3/16*I*Pi*csgn(b*x^2)*exp(-I*a)*b-3/8*I*exp(-I*a)
*Si(b*x^2)*b-3/16*Ei(1,-I*b*x^2)*exp(-I*a)*b-3/8*sin(b*x^2+a)/x^2+1/8*sin(3*b*x^2+3*a)/x^2

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maxima [C]  time = 0.60, size = 100, normalized size = 1.10 \[ -\frac {1}{16} \, {\left (3 \, {\left (\Gamma \left (-1, 3 i \, b x^{2}\right ) + \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) - 3 \, {\left (\Gamma \left (-1, i \, b x^{2}\right ) + \Gamma \left (-1, -i \, b x^{2}\right )\right )} \cos \relax (a) - {\left (3 i \, \Gamma \left (-1, 3 i \, b x^{2}\right ) - 3 i \, \Gamma \left (-1, -3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) - {\left (-3 i \, \Gamma \left (-1, i \, b x^{2}\right ) + 3 i \, \Gamma \left (-1, -i \, b x^{2}\right )\right )} \sin \relax (a)\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(3*(gamma(-1, 3*I*b*x^2) + gamma(-1, -3*I*b*x^2))*cos(3*a) - 3*(gamma(-1, I*b*x^2) + gamma(-1, -I*b*x^2)
)*cos(a) - (3*I*gamma(-1, 3*I*b*x^2) - 3*I*gamma(-1, -3*I*b*x^2))*sin(3*a) - (-3*I*gamma(-1, I*b*x^2) + 3*I*ga
mma(-1, -I*b*x^2))*sin(a))*b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^2)^3/x^3,x)

[Out]

int(sin(a + b*x^2)^3/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{3}{\left (a + b x^{2} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x**2)**3/x**3, x)

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