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(b*x+a)^m)^(1/m)*tan(b*x+a)/b

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Rubi [A]  time = 0.02, antiderivative size = 24, 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, 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 \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*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 1.00 \[ \frac {\tan (a+b x) \left (c \cos ^m(a+b x)\right )^{\frac {1}{m}}}{b} \]

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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fricas [A]  time = 0.54, size = 15, normalized size = 0.62 \[ \frac {c^{\left (\frac {1}{m}\right )} \sin \left (b x + a\right )}{b} \]

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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giac [B]  time = 5.14, size = 300, normalized size = 12.50 \[ \frac {2 \, {\left ({\left | c \right |}^{\left (\frac {1}{m}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} - {\left | c \right |}^{\left (\frac {1}{m}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) + 4 \, {\left | c \right |}^{\left (\frac {1}{m}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right ) \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - {\left | c \right |}^{\left (\frac {1}{m}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} + {\left | c \right |}^{\left (\frac {1}{m}\right )} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )}}{b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right )^{2} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a + \frac {\pi \mathrm {sgn}\relax (c)}{4 \, m} - \frac {\pi }{4 \, m}\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*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)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (c \left (\cos ^{m}\left (b x +a \right )\right )\right )^{\frac {1}{m}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \cos \left (b x + a\right )^{m}\right )^{\left (\frac {1}{m}\right )}\,{d x} \]

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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mupad [B]  time = 0.40, size = 40, normalized size = 1.67 \[ \frac {\sin \left (2\,a+2\,b\,x\right )\,{\left (c\,{\cos \left (a+b\,x\right )}^m\right )}^{1/m}}{b\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*cos(a + b*x)^m)^(1/m),x)

[Out]

(sin(2*a + 2*b*x)*(c*cos(a + b*x)^m)^(1/m))/(b*(cos(2*a + 2*b*x) + 1))

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sympy [A]  time = 1.58, size = 65, normalized size = 2.71 \[ \begin {cases} x \left (c \cos ^{m}{\relax (a )}\right )^{\frac {1}{m}} & \text {for}\: b = 0 \\x \left (0^{m} c\right )^{\frac {1}{m}} & \text {for}\: a = - b x + \frac {\pi }{2} \vee a = - b x + \frac {3 \pi }{2} \\\frac {c^{\frac {1}{m}} \left (\cos ^{m}{\left (a + b x \right )}\right )^{\frac {1}{m}} \sin {\left (a + b x \right )}}{b \cos {\left (a + b x \right )}} & \text {otherwise} \end {cases} \]

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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