3.1.26 \(\int \frac {\sin ^3(a+b x^2)}{x} \, dx\) [26]

Optimal. Leaf size=55 \[ \frac {3}{8} \text {Ci}\left (b x^2\right ) \sin (a)-\frac {1}{8} \text {Ci}\left (3 b x^2\right ) \sin (3 a)+\frac {3}{8} \cos (a) \text {Si}\left (b x^2\right )-\frac {1}{8} \cos (3 a) \text {Si}\left (3 b x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3484, 3458, 3457, 3456} \begin {gather*} \frac {3}{8} \sin (a) \text {CosIntegral}\left (b x^2\right )-\frac {1}{8} \sin (3 a) \text {CosIntegral}\left (3 b x^2\right )+\frac {3}{8} \cos (a) \text {Si}\left (b x^2\right )-\frac {1}{8} \cos (3 a) \text {Si}\left (3 b x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x^2]^3/x,x]

[Out]

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

Rule 3456

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3457

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 3458

Int[Sin[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Sin[c], Int[Cos[d*x^n]/x, x], x] + Dist[Cos[c], Int[Si
n[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 3484

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

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Mathematica [A]
time = 0.05, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{8} \left (3 \text {Ci}\left (b x^2\right ) \sin (a)-\text {Ci}\left (3 b x^2\right ) \sin (3 a)+3 \cos (a) \text {Si}\left (b x^2\right )-\cos (3 a) \text {Si}\left (3 b x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x^2]^3/x,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.31, size = 125, normalized size = 2.27

method	result	size
risch	\(-\frac {i {\mathrm e}^{3 i a} \expIntegral \left (1, -3 i x^{2} b \right )}{16}+\frac {\pi \,{\mathrm e}^{-3 i a} \mathrm {csgn}\left (b \,x^{2}\right )}{16}-\frac {{\mathrm e}^{-3 i a} \sinIntegral \left (3 b \,x^{2}\right )}{8}+\frac {i \expIntegral \left (1, -3 i x^{2} b \right ) {\mathrm e}^{-3 i a}}{16}-\frac {3 \pi \,\mathrm {csgn}\left (b \,x^{2}\right ) {\mathrm e}^{-i a}}{16}+\frac {3 \,{\mathrm e}^{-i a} \sinIntegral \left (b \,x^{2}\right )}{8}-\frac {3 i {\mathrm e}^{-i a} \expIntegral \left (1, -i x^{2} b \right )}{16}+\frac {3 i {\mathrm e}^{i a} \expIntegral \left (1, -i x^{2} b \right )}{16}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.37, size = 89, normalized size = 1.62 \begin {gather*} \frac {1}{16} \, {\left (i \, {\rm Ei}\left (3 i \, b x^{2}\right ) - i \, {\rm Ei}\left (-3 i \, b x^{2}\right )\right )} \cos \left (3 \, a\right ) - \frac {3}{16} \, {\left (i \, {\rm Ei}\left (i \, b x^{2}\right ) - i \, {\rm Ei}\left (-i \, b x^{2}\right )\right )} \cos \left (a\right ) - \frac {1}{16} \, {\left ({\rm Ei}\left (3 i \, b x^{2}\right ) + {\rm Ei}\left (-3 i \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) + \frac {3}{16} \, {\left ({\rm Ei}\left (i \, b x^{2}\right ) + {\rm Ei}\left (-i \, b x^{2}\right )\right )} \sin \left (a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(I*Ei(3*I*b*x^2) - I*Ei(-3*I*b*x^2))*cos(3*a) - 3/16*(I*Ei(I*b*x^2) - I*Ei(-I*b*x^2))*cos(a) - 1/16*(Ei(3
*I*b*x^2) + Ei(-3*I*b*x^2))*sin(3*a) + 3/16*(Ei(I*b*x^2) + Ei(-I*b*x^2))*sin(a)

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Fricas [A]
time = 0.36, size = 63, normalized size = 1.15 \begin {gather*} -\frac {1}{16} \, {\left (\operatorname {Ci}\left (3 \, b x^{2}\right ) + \operatorname {Ci}\left (-3 \, b x^{2}\right )\right )} \sin \left (3 \, a\right ) + \frac {3}{16} \, {\left (\operatorname {Ci}\left (b x^{2}\right ) + \operatorname {Ci}\left (-b x^{2}\right )\right )} \sin \left (a\right ) - \frac {1}{8} \, \cos \left (3 \, a\right ) \operatorname {Si}\left (3 \, b x^{2}\right ) + \frac {3}{8} \, \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x**2)**3/x, x)

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Giac [A]
time = 3.87, size = 47, normalized size = 0.85 \begin {gather*} -\frac {1}{8} \, \operatorname {Ci}\left (3 \, b x^{2}\right ) \sin \left (3 \, a\right ) + \frac {3}{8} \, \operatorname {Ci}\left (b x^{2}\right ) \sin \left (a\right ) + \frac {3}{8} \, \cos \left (a\right ) \operatorname {Si}\left (b x^{2}\right ) + \frac {1}{8} \, \cos \left (3 \, a\right ) \operatorname {Si}\left (-3 \, b x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\sin \left (b\,x^2+a\right )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^2)^3/x,x)

[Out]

int(sin(a + b*x^2)^3/x, x)

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