3.1.13 \(\int x^{-1+n} \sin (x^n) \, dx\) [13]

Optimal. Leaf size=9 \[ -\frac {\cos \left (x^n\right )}{n} \]

[Out]

-cos(x^n)/n

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Rubi [A]
time = 0.01, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3460, 2718} \begin {gather*} -\frac {\cos \left (x^n\right )}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + n)*Sin[x^n],x]

[Out]

-(Cos[x^n]/n)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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+n} \sin \left (x^n\right ) \, dx &=\frac {\text {Subst}\left (\int \sin (x) \, dx,x,x^n\right )}{n}\\ &=-\frac {\cos \left (x^n\right )}{n}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 9, normalized size = 1.00 \begin {gather*} -\frac {\cos \left (x^n\right )}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + n)*Sin[x^n],x]

[Out]

-(Cos[x^n]/n)

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Mathics [A]
time = 5.10, size = 9, normalized size = 1.00 \begin {gather*} -\frac {\text {Cos}\left [x^n\right ]}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^(n - 1)*Sin[x^n],x]')

[Out]

-Cos[x ^ n] / n

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Maple [A]
time = 0.06, size = 10, normalized size = 1.11

method	result	size
default	\(-\frac {\cos \left (x^{n}\right )}{n}\)	\(10\)
risch	\(-\frac {\cos \left (x^{n}\right )}{n}\)	\(10\)
norman	\(\frac {2 \left (\tan ^{2}\left (\frac {{\mathrm e}^{n \ln \left (x \right )}}{2}\right )\right )}{n \left (1+\tan ^{2}\left (\frac {{\mathrm e}^{n \ln \left (x \right )}}{2}\right )\right )}\)	\(30\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {2^{1-\frac {-1+n}{n}-\frac {1}{n}} \left (-1\right )^{\frac {1}{2}-\frac {-1+n}{2 n}-\frac {1}{2 n}}}{\sqrt {\pi }\, \Gamma \left (3-\frac {-1+n}{n}-\frac {1}{n}\right )}-\frac {\left (-1\right )^{\frac {1}{2}-\frac {-1+n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1+n}{n}-\frac {1}{n}} \cos \left (x^{n}\right )}{\sqrt {\pi }\, \Gamma \left (3-\frac {-1+n}{n}-\frac {1}{n}\right )}\right )}{n}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x^n)/n

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Maxima [A]
time = 0.25, size = 9, normalized size = 1.00 \begin {gather*} -\frac {\cos \left (x^{n}\right )}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x^n)/n

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Fricas [A]
time = 0.34, size = 9, normalized size = 1.00 \begin {gather*} -\frac {\cos \left (x^{n}\right )}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x^n)/n

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Sympy [A]
time = 3.80, size = 7, normalized size = 0.78 \begin {gather*} - \frac {\cos {\left (x^{n} \right )}}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+n)*sin(x**n),x)

[Out]

-cos(x**n)/n

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Giac [A]
time = 0.00, size = 8, normalized size = 0.89 \begin {gather*} -\frac {\cos \left (x^{n}\right )}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x^n)/n

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Mupad [B]
time = 0.21, size = 9, normalized size = 1.00 \begin {gather*} -\frac {\cos \left (x^n\right )}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n - 1)*sin(x^n),x)

[Out]

-cos(x^n)/n

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