3.2.50 \(\int x^{-1-2 n} \sin (a+b x^n) \, dx\) [150]

Optimal. Leaf size=78 \[ -\frac {b x^{-n} \cos \left (a+b x^n\right )}{2 n}-\frac {b^2 \text {Ci}\left (b x^n\right ) \sin (a)}{2 n}-\frac {x^{-2 n} \sin \left (a+b x^n\right )}{2 n}-\frac {b^2 \cos (a) \text {Si}\left (b x^n\right )}{2 n} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3460, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {b^2 \sin (a) \text {CosIntegral}\left (b x^n\right )}{2 n}-\frac {b^2 \cos (a) \text {Si}\left (b x^n\right )}{2 n}-\frac {x^{-2 n} \sin \left (a+b x^n\right )}{2 n}-\frac {b x^{-n} \cos \left (a+b x^n\right )}{2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 2*n)*Sin[a + b*x^n],x]

[Out]

-1/2*(b*Cos[a + b*x^n])/(n*x^n) - (b^2*CosIntegral[b*x^n]*Sin[a])/(2*n) - Sin[a + b*x^n]/(2*n*x^(2*n)) - (b^2*
Cos[a]*SinIntegral[b*x^n])/(2*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 3460

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]))

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 68, normalized size = 0.87 \begin {gather*} -\frac {x^{-2 n} \left (b x^n \cos \left (a+b x^n\right )+b^2 x^{2 n} \text {Ci}\left (b x^n\right ) \sin (a)+\sin \left (a+b x^n\right )+b^2 x^{2 n} \cos (a) \text {Si}\left (b x^n\right )\right )}{2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 65, normalized size = 0.83

method	result	size
default	\(\frac {b^{2} \left (-\frac {x^{-2 n} \sin \left (a +b \,x^{n}\right )}{2 b^{2}}-\frac {\cos \left (a +b \,x^{n}\right ) x^{-n}}{2 b}-\frac {\sinIntegral \left (b \,x^{n}\right ) \cos \left (a \right )}{2}-\frac {\cosineIntegral \left (b \,x^{n}\right ) \sin \left (a \right )}{2}\right )}{n}\)	\(65\)
risch	\(\frac {b^{2} {\mathrm e}^{-i a} \pi \,\mathrm {csgn}\left (b \,x^{n}\right )}{4 n}-\frac {b^{2} {\mathrm e}^{-i a} \sinIntegral \left (b \,x^{n}\right )}{2 n}+\frac {i b^{2} {\mathrm e}^{-i a} \expIntegral \left (1, -i b \,x^{n}\right )}{4 n}-\frac {i b^{2} {\mathrm e}^{i a} \expIntegral \left (1, -i b \,x^{n}\right )}{4 n}-\frac {b \cos \left (a +b \,x^{n}\right ) x^{-n}}{2 n}-\frac {x^{-2 n} \sin \left (a +b \,x^{n}\right )}{2 n}\)	\(124\)
meijerg	\(\frac {b^{2} \sqrt {\pi }\, \left (-\frac {x^{2 \left (\frac {-1-2 n}{2 n}+\frac {1}{2 n}\right ) n} 2^{-\frac {-1-2 n}{n}-\frac {1}{n}}}{\sqrt {\pi }\, b^{2}}+\frac {\left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} \left (-\Psi \left (1-\frac {-1-2 n}{2 n}-\frac {1}{2 n}\right )-\Psi \left (\frac {1}{2}-\frac {-1-2 n}{2 n}-\frac {1}{2 n}\right )+2 n \ln \left (x \right )-2 \ln \left (2\right )+\ln \left (b^{2}\right )\right ) \sqrt {2}\, 2^{-\frac {-1-2 n}{n}-\frac {1}{n}-\frac {1}{2}}}{2 \sqrt {\pi }\, \Gamma \left (-\frac {-1-2 n}{n}-\frac {1}{n}\right )}-\frac {\left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} x^{-2 n} \left (-\frac {9 x^{2 n} b^{2}}{2}+3\right )}{\sqrt {\pi }\, b^{2} \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}-\frac {3 \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} \gamma }{\sqrt {\pi }\, \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}-\frac {3 \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} \ln \left (2\right )}{\sqrt {\pi }\, \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}-\frac {3 \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} \ln \left (\frac {b \,x^{n}}{2}\right )}{\sqrt {\pi }\, \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}+\frac {3 \,2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} x^{-2 n} \cos \left (b \,x^{n}\right )}{\sqrt {\pi }\, b^{2} \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}-\frac {3 \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} x^{-n} \sin \left (b \,x^{n}\right )}{\sqrt {\pi }\, b \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}+\frac {3 \,2^{1-\frac {-1-2 n}{n}-\frac {1}{n}} \left (-1\right )^{-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} \cosineIntegral \left (b \,x^{n}\right )}{\sqrt {\pi }\, \Gamma \left (2-\frac {-1-2 n}{n}-\frac {1}{n}\right )}\right ) \sin \left (a \right )}{8 n}+\frac {b^{2} \sqrt {\pi }\, \left (-\frac {4 x^{-n} \cos \left (b \,x^{n}\right )}{\sqrt {\pi }\, b}-\frac {4 x^{-2 n} \sin \left (b \,x^{n}\right )}{\sqrt {\pi }\, b^{2}}-\frac {4 \sinIntegral \left (b \,x^{n}\right )}{\sqrt {\pi }}\right ) \cos \left (a \right )}{8 n}\)	\(761\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/n*b^2*(-1/2/b^2/(x^n)^2*sin(a+b*x^n)-1/2*cos(a+b*x^n)/b/(x^n)-1/2*Si(b*x^n)*cos(a)-1/2*Ci(b*x^n)*sin(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^(-1-2*n)*sin(a+b*x^n),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.53, size = 90, normalized size = 1.15 \begin {gather*} -\frac {b^{2} x^{2 \, n} \operatorname {Ci}\left (b x^{n}\right ) \sin \left (a\right ) + b^{2} x^{2 \, n} \operatorname {Ci}\left (-b x^{n}\right ) \sin \left (a\right ) + 2 \, b^{2} x^{2 \, n} \cos \left (a\right ) \operatorname {Si}\left (b x^{n}\right ) + 2 \, b x^{n} \cos \left (b x^{n} + a\right ) + 2 \, \sin \left (b x^{n} + a\right )}{4 \, n x^{2 \, n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*x^(2*n)*cos_integral(b*x^n)*sin(a) + b^2*x^(2*n)*cos_integral(-b*x^n)*sin(a) + 2*b^2*x^(2*n)*cos(a)*
sin_integral(b*x^n) + 2*b*x^n*cos(b*x^n + a) + 2*sin(b*x^n + a))/(n*x^(2*n))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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

[Out]

integrate(x^(-2*n - 1)*sin(b*x^n + a), 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 )}{x^{2\,n+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x^n)/x^(2*n + 1),x)

[Out]

int(sin(a + b*x^n)/x^(2*n + 1), x)

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