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

Optimal result

Integrand size = 10, antiderivative size = 9 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^n\right )}{n} \]

[Out]

-cos(x^n)/n

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3460, 2718} \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^n\right )}{n} \]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^n\right )}{n} \]

[In]

Integrate[x^(-1 + n)*Sin[x^n],x]

[Out]

-(Cos[x^n]/n)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 10, normalized size of antiderivative = 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	\(\frac {\sqrt {\pi }\, \left (\frac {2^{1-\frac {-1+n}{n}-\frac {1}{n}} \left (-1\right )^{\frac {1}{2}-\frac {-1+n}{2 n}-\frac {1}{2 n}}}{\sqrt {\pi }\, \Gamma \left (3-\frac {-1+n}{n}-\frac {1}{n}\right )}-\frac {\left (-1\right )^{\frac {1}{2}-\frac {-1+n}{2 n}-\frac {1}{2 n}} 2^{1-\frac {-1+n}{n}-\frac {1}{n}} \cos \left (x^{n}\right )}{\sqrt {\pi }\, \Gamma \left (3-\frac {-1+n}{n}-\frac {1}{n}\right )}\right )}{n}\)	\(126\)

[In]

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

[Out]

-cos(x^n)/n

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^{n}\right )}{n} \]

[In]

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

[Out]

-cos(x^n)/n

Sympy [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=- \frac {\cos {\left (x^{n} \right )}}{n} \]

[In]

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

[Out]

-cos(x**n)/n

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^{n}\right )}{n} \]

[In]

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

[Out]

-cos(x^n)/n

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^{n}\right )}{n} \]

[In]

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

[Out]

-cos(x^n)/n

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int x^{-1+n} \sin \left (x^n\right ) \, dx=-\frac {\cos \left (x^n\right )}{n} \]

[In]

int(x^(n - 1)*sin(x^n),x)

[Out]

-cos(x^n)/n