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

Optimal. Leaf size=25 \[ -\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3208, 2638} \[ -\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

Antiderivative was successfully verified.

[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 2638

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

Rule 3208

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

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Mathematica [A]  time = 0.03, size = 25, normalized size = 1.00 \[ -\frac {\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac {1}{n}}}{b} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.44, size = 16, normalized size = 0.64 \[ -\frac {c^{\left (\frac {1}{n}\right )} \cos \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^(1/n)*cos(b*x + a)/b

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giac [B]  time = 3.43, size = 384, normalized size = 15.36 \[ \frac {{\left | c \right |}^{\left (\frac {1}{n}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{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}\relax (c)}{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}\relax (c)}{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}\relax (c)}{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}\relax (c)}{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}\relax (c)}{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}\relax (c)}{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}\relax (c)}{4 \, n} - \frac {\pi }{4 \, n}\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.85, size = 0, normalized size = 0.00 \[ \int \left (c \left (\sin ^{n}\left (b x +a \right )\right )\right )^{\frac {1}{n}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin \left (b x + a\right )^{n}\right )^{\left (\frac {1}{n}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.75, size = 36, normalized size = 1.44 \[ -\frac {\sin \left (2\,a+2\,b\,x\right )\,{\left (c\,{\sin \left (a+b\,x\right )}^n\right )}^{1/n}}{2\,b\,{\sin \left (a+b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 2.11, size = 61, normalized size = 2.44 \[ \begin {cases} x \left (c \sin ^{n}{\relax (a )}\right )^{\frac {1}{n}} & \text {for}\: b = 0 \\x \left (0^{n} c\right )^{\frac {1}{n}} & \text {for}\: a = - b x \vee a = - b x + \pi \\- \frac {c^{\frac {1}{n}} \left (\sin ^{n}{\left (a + b x \right )}\right )^{\frac {1}{n}} \cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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**(1/n)*
(sin(a + b*x)**n)**(1/n)*cos(a + b*x)/(b*sin(a + b*x)), True))

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