3.64 \(\int (c \cos ^m(a+b x))^{\frac{1}{m}} \, dx\)

Optimal. Leaf size=24 \[ \frac{\tan (a+b x) \left (c \cos ^m(a+b x)\right )^{\frac{1}{m}}}{b} \]

[Out]

((c*Cos[a + b*x]^m)^m^(-1)*Tan[a + b*x])/b

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Rubi [A]  time = 0.0198843, antiderivative size = 24, 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, 2637} \[ \frac{\tan (a+b x) \left (c \cos ^m(a+b x)\right )^{\frac{1}{m}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cos[a + b*x]^m)^m^(-1),x]

[Out]

((c*Cos[a + b*x]^m)^m^(-1)*Tan[a + b*x])/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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \left (c \cos ^m(a+b x)\right )^{\frac{1}{m}} \, dx &=\left (\left (c \cos ^m(a+b x)\right )^{\frac{1}{m}} \sec (a+b x)\right ) \int \cos (a+b x) \, dx\\ &=\frac{\left (c \cos ^m(a+b x)\right )^{\frac{1}{m}} \tan (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.029611, size = 24, normalized size = 1. \[ \frac{\tan (a+b x) \left (c \cos ^m(a+b x)\right )^{\frac{1}{m}}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cos[a + b*x]^m)^m^(-1),x]

[Out]

((c*Cos[a + b*x]^m)^m^(-1)*Tan[a + b*x])/b

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Maple [F]  time = 0.26, size = 0, normalized size = 0. \begin{align*} \int \sqrt [m]{c \left ( \cos \left ( bx+a \right ) \right ) ^{m}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*cos(b*x+a)^m)^(1/m),x)

[Out]

int((c*cos(b*x+a)^m)^(1/m),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*cos(b*x + a)^m)^(1/m), x)

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Fricas [A]  time = 1.0493, size = 32, normalized size = 1.33 \begin{align*} \frac{c^{\left (\frac{1}{m}\right )} \sin \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^(1/m)*sin(b*x + a)/b

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Sympy [A]  time = 1.54197, size = 65, normalized size = 2.71 \begin{align*} \begin{cases} x \left (c \cos ^{m}{\left (a \right )}\right )^{\frac{1}{m}} & \text{for}\: b = 0 \\x \left (0^{m} c\right )^{\frac{1}{m}} & \text{for}\: a = - b x + \frac{\pi }{2} \vee a = - b x + \frac{3 \pi }{2} \\\frac{c^{\frac{1}{m}} \left (\cos ^{m}{\left (a + b x \right )}\right )^{\frac{1}{m}} \sin{\left (a + b x \right )}}{b \cos{\left (a + b x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*cos(b*x+a)**m)**(1/m),x)

[Out]

Piecewise((x*(c*cos(a)**m)**(1/m), Eq(b, 0)), (x*(0**m*c)**(1/m), Eq(a, -b*x + pi/2) | Eq(a, -b*x + 3*pi/2)),
(c**(1/m)*(cos(a + b*x)**m)**(1/m)*sin(a + b*x)/(b*cos(a + b*x)), True))

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Giac [B]  time = 8.07252, size = 405, normalized size = 16.88 \begin{align*} \frac{2 \,{\left ({\left | c \right |}^{\left (\frac{1}{m}\right )} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{\pi \mathrm{sgn}\left (c\right )}{4 \, m} - \frac{\pi }{4 \, m}\right )^{2} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} -{\left | c \right |}^{\left (\frac{1}{m}\right )} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{\pi \mathrm{sgn}\left (c\right )}{4 \, m} - \frac{\pi }{4 \, m}\right )^{2} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) + 4 \,{\left | c \right |}^{\left (\frac{1}{m}\right )} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{\pi \mathrm{sgn}\left (c\right )}{4 \, m} - \frac{\pi }{4 \, m}\right ) \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} -{\left | c \right |}^{\left (\frac{1}{m}\right )} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{3} +{\left | c \right |}^{\left (\frac{1}{m}\right )} \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )\right )}}{b \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a + \frac{\pi \mathrm{sgn}\left (c\right )}{4 \, m} - \frac{\pi }{4 \, m}\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 \, m} - \frac{\pi }{4 \, m}\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 \, m} - \frac{\pi }{4 \, m}\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )^{2} + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(abs(c)^(1/m)*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/m - 1/4*pi/m)^2*tan(1/2*b*x + 1/2*a)^3 - abs(c)^(1/m)*tan(
1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/m - 1/4*pi/m)^2*tan(1/2*b*x + 1/2*a) + 4*abs(c)^(1/m)*tan(1/2*b*x + 1/2*a + 1/
4*pi*sgn(c)/m - 1/4*pi/m)*tan(1/2*b*x + 1/2*a)^2 - abs(c)^(1/m)*tan(1/2*b*x + 1/2*a)^3 + abs(c)^(1/m)*tan(1/2*
b*x + 1/2*a))/(b*tan(1/2*b*x + 1/2*a + 1/4*pi*sgn(c)/m - 1/4*pi/m)^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)/m - 1/4*pi/m)^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)/m - 1/4*pi/m)^2 + 2*b*tan(1/2*b*x + 1/2*a)^2 + b)