3.2.48 \(\int x^{-1-n} \sin ^2(a+b x^n) \, dx\) [148]

Optimal. Leaf size=67 \[ -\frac {x^{-n}}{2 n}+\frac {x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}+\frac {b \text {Ci}\left (2 b x^n\right ) \sin (2 a)}{n}+\frac {b \cos (2 a) \text {Si}\left (2 b x^n\right )}{n} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3506, 3461, 3378, 3384, 3380, 3383} \begin {gather*} \frac {b \sin (2 a) \text {CosIntegral}\left (2 b x^n\right )}{n}+\frac {b \cos (2 a) \text {Si}\left (2 b x^n\right )}{n}+\frac {x^{-n} \cos \left (2 \left (a+b x^n\right )\right )}{2 n}-\frac {x^{-n}}{2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n)*Sin[a + b*x^n]^2,x]

[Out]

-1/2*1/(n*x^n) + Cos[2*(a + b*x^n)]/(2*n*x^n) + (b*CosIntegral[2*b*x^n]*Sin[2*a])/n + (b*Cos[2*a]*SinIntegral[
2*b*x^n])/n

Rule 3378

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 3380

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 3383

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 3384

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 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 3506

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, n}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 58, normalized size = 0.87 \begin {gather*} \frac {x^{-n} \left (-1+\cos \left (2 \left (a+b x^n\right )\right )+2 b x^n \text {Ci}\left (2 b x^n\right ) \sin (2 a)+2 b x^n \cos (2 a) \text {Si}\left (2 b x^n\right )\right )}{2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n)*Sin[a + b*x^n]^2,x]

[Out]

(-1 + Cos[2*(a + b*x^n)] + 2*b*x^n*CosIntegral[2*b*x^n]*Sin[2*a] + 2*b*x^n*Cos[2*a]*SinIntegral[2*b*x^n])/(2*n
*x^n)

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Maple [A]
time = 0.07, size = 66, normalized size = 0.99

method	result	size
default	\(-\frac {x^{-n}}{2 n}-\frac {b \left (-\frac {\cos \left (2 a +2 b \,x^{n}\right ) x^{-n}}{2 b}-\sinIntegral \left (2 b \,x^{n}\right ) \cos \left (2 a \right )-\cosineIntegral \left (2 b \,x^{n}\right ) \sin \left (2 a \right )\right )}{n}\)	\(66\)
risch	\(-\frac {b \,{\mathrm e}^{-2 i a} \pi \,\mathrm {csgn}\left (b \,x^{n}\right )}{2 n}+\frac {b \,{\mathrm e}^{-2 i a} \sinIntegral \left (2 b \,x^{n}\right )}{n}-\frac {i b \,{\mathrm e}^{-2 i a} \expIntegral \left (1, -2 i b \,x^{n}\right )}{2 n}+\frac {i b \,{\mathrm e}^{2 i a} \expIntegral \left (1, -2 i b \,x^{n}\right )}{2 n}-\frac {x^{-n}}{2 n}+\frac {\cos \left (2 a +2 b \,x^{n}\right ) x^{-n}}{2 n}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-n)*sin(a+b*x^n)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*sin(a+b*x^n)^2,x, algorithm="maxima")

[Out]

-1/2*(n*x^n*integrate(cos(2*b*x^n + 2*a)/(x*x^n), x) + 1)/(n*x^n)

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Fricas [A]
time = 0.52, size = 73, normalized size = 1.09 \begin {gather*} \frac {b x^{n} \operatorname {Ci}\left (2 \, b x^{n}\right ) \sin \left (2 \, a\right ) + b x^{n} \operatorname {Ci}\left (-2 \, b x^{n}\right ) \sin \left (2 \, a\right ) + 2 \, b x^{n} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{n}\right ) + 2 \, \cos \left (b x^{n} + a\right )^{2} - 2}{2 \, n x^{n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-n)*sin(a+b*x^n)^2,x, algorithm="fricas")

[Out]

1/2*(b*x^n*cos_integral(2*b*x^n)*sin(2*a) + b*x^n*cos_integral(-2*b*x^n)*sin(2*a) + 2*b*x^n*cos(2*a)*sin_integ
ral(2*b*x^n) + 2*cos(b*x^n + a)^2 - 2)/(n*x^n)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{- n - 1} \sin ^{2}{\left (a + b x^{n} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-n)*sin(a+b*x**n)**2,x)

[Out]

Integral(x**(-n - 1)*sin(a + b*x**n)**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(x^(-1-n)*sin(a+b*x^n)^2,x, algorithm="giac")

[Out]

integrate(x^(-n - 1)*sin(b*x^n + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^n)^2/x^(n + 1),x)

[Out]

int(sin(a + b*x^n)^2/x^(n + 1), x)

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