\(\int (c \sin ^n(a+b x))^{\frac {1}{n}} \, dx\) [29]

Optimal result

Integrand size = 14, antiderivative size = 25 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=-\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

[Out]

-cot(b*x+a)*(c*sin(b*x+a)^n)^(1/n)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3287, 2718} \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=-\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

[In]

Int[(c*Sin[a + b*x]^n)^n^(-1),x]

[Out]

-((Cot[a + b*x]*(c*Sin[a + b*x]^n)^n^(-1))/b)

Rule 2718

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

Rule 3287

Int[(u_.)*((b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((b*(c*Sin[e + f*x
])^n)^FracPart[p]/(c*Sin[e + f*x])^(n*FracPart[p])), Int[ActivateTrig[u]*(c*Sin[e + f*x])^(n*p), x], x] /; Fre
eQ[{b, c, e, f, n, p}, x] &&  !IntegerQ[p] &&  !IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x]
)^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])

Rubi steps \begin{align*} \text {integral}& = \left (\csc (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}\right ) \int \sin (a+b x) \, dx \\ & = -\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=-\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

[In]

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

[Out]

-((Cot[a + b*x]*(c*Sin[a + b*x]^n)^n^(-1))/b)

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16

method	result	size
parallelrisch	\(-\frac {{\left (c \left (\sin ^{n}\left (b x +a \right )\right )\right )}^{\frac {1}{n}} \cot \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}\)	\(29\)

[In]

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

[Out]

-1/b*(c*sin(b*x+a)^n)^(1/n)*cot(1/2*b*x+1/2*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=-\frac {c^{\left (\frac {1}{n}\right )} \cos \left (b x + a\right )}{b} \]

[In]

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

[Out]

-c^(1/n)*cos(b*x + a)/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (22) = 44\).

Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=\begin {cases} x \left (c \sin ^{n}{\left (a \right )}\right )^{\frac {1}{n}} & \text {for}\: b = 0 \\x \left (0^{n} c\right )^{\frac {1}{n}} & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {\left (c \sin ^{n}{\left (a + b x \right )}\right )^{\frac {1}{n}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*sin(b*x+a)**n)**(1/n),x)

[Out]

Piecewise((x*(c*sin(a)**n)**(1/n), Eq(b, 0)), (x*(0**n*c)**(1/n), Eq(a, -b*x) | Eq(a, -b*x + pi)), (-(c*sin(a
+ b*x)**n)**(1/n)*cos(a + b*x)/(b*sin(a + b*x)), True))

Maxima [F]

\[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=\int { \left (c \sin \left (b x + a\right )^{n}\right )^{\left (\frac {1}{n}\right )} \,d x } \]

[In]

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

[Out]

integrate((c*sin(b*x + a)^n)^(1/n), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (25) = 50\).

Time = 0.74 (sec) , antiderivative size = 384, normalized size of antiderivative = 15.36 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=\frac {{\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} - 2 \, {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + 4 \, {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} - {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} - 4 \, {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 2 \, {\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - {\left | c \right |}^{\left (\frac {1}{n}\right )}}{b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\left (c\right )}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b} \]

[In]

integrate((c*sin(b*x+a)^n)^(1/n),x, algorithm="giac")

[Out]

(abs(c)^(1/n)*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/n)^2*tan(1/2*b*x + 1/2*a)^4 - 2*abs(c)^(1/n)*tan(
1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/n)^2*tan(1/2*b*x + 1/2*a)^2 + 4*abs(c)^(1/n)*tan(1/2*b*x + 1/2*a +
1/4*pi*sgn(c)/n - 1/4*pi/n)*tan(1/2*b*x + 1/2*a)^3 - abs(c)^(1/n)*tan(1/2*b*x + 1/2*a)^4 + abs(c)^(1/n)*tan(1/
2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/n)^2 - 4*abs(c)^(1/n)*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/
n)*tan(1/2*b*x + 1/2*a) + 2*abs(c)^(1/n)*tan(1/2*b*x + 1/2*a)^2 - abs(c)^(1/n))/(b*tan(1/2*b*x + 1/2*a + 1/4*p
i*sgn(c)/n - 1/4*pi/n)^2*tan(1/2*b*x + 1/2*a)^4 + 2*b*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/n)^2*tan(
1/2*b*x + 1/2*a)^2 + b*tan(1/2*b*x + 1/2*a)^4 + b*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/n - 1/4*pi/n)^2 + 2*b*ta
n(1/2*b*x + 1/2*a)^2 + b)

Mupad [B] (verification not implemented)

Time = 14.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}} \, dx=-\frac {\sin \left (2\,a+2\,b\,x\right )\,{\left (c\,{\sin \left (a+b\,x\right )}^n\right )}^{1/n}}{2\,b\,{\sin \left (a+b\,x\right )}^2} \]

[In]

int((c*sin(a + b*x)^n)^(1/n),x)

[Out]

-(sin(2*a + 2*b*x)*(c*sin(a + b*x)^n)^(1/n))/(2*b*sin(a + b*x)^2)