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

________________________________________________________________________________________

Rubi [A]  time = 0.0189031, antiderivative size = 25, normalized size of antiderivative = 1., 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 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]])

Rule 2638

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

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*}

Mathematica [A]  time = 0.0337482, size = 25, normalized size = 1. \[ -\frac{\cot (a+b x) \left (c \sin ^n(a+b x)\right )^{\frac{1}{n}}}{b} \]

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)

________________________________________________________________________________________

Maple [F]  time = 0.292, size = 0, normalized size = 0. \begin{align*} \int \sqrt [n]{c \left ( \sin \left ( bx+a \right ) \right ) ^{n}}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sin \left (b x + a\right )^{n}\right )^{\left (\frac{1}{n}\right )}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 1.643, size = 34, normalized size = 1.36 \begin{align*} -\frac{c^{\left (\frac{1}{n}\right )} \cos \left (b x + a\right )}{b} \end{align*}

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

________________________________________________________________________________________

Sympy [A]  time = 6.30427, size = 61, normalized size = 2.44 \begin{align*} \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{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} \end{align*}

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))

________________________________________________________________________________________

Giac [B]  time = 8.92025, size = 518, normalized size = 20.72 \begin{align*} \text{result too large to display} \end{align*}

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)