3.29 \(\int (b \tan ^n(e+f x))^{\frac {1}{n}} \, dx\)

Optimal. Leaf size=32 \[ -\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f} \]

[Out]

-cot(f*x+e)*ln(cos(f*x+e))*(b*tan(f*x+e)^n)^(1/n)/f

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Rubi [A]  time = 0.02, antiderivative size = 32, 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 = {3659, 3475} \[ -\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^n^(-1),x]

[Out]

-((Cot[e + f*x]*Log[Cos[e + f*x]]*(b*Tan[e + f*x]^n)^n^(-1))/f)

Rule 3475

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

Rule 3659

Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(b*(c*Tan[e + f*x
])^n)^FracPart[p])/(c*Tan[e + f*x])^(n*FracPart[p]), Int[ActivateTrig[u]*(c*Tan[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 (b \tan ^n(e+f x)\right )^{\frac {1}{n}} \, dx &=\left (\cot (e+f x) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}\right ) \int \tan (e+f x) \, dx\\ &=-\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 32, normalized size = 1.00 \[ -\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^n)^n^(-1),x]

[Out]

-((Cot[e + f*x]*Log[Cos[e + f*x]]*(b*Tan[e + f*x]^n)^n^(-1))/f)

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fricas [A]  time = 0.47, size = 23, normalized size = 0.72 \[ -\frac {b^{\left (\frac {1}{n}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/n),x, algorithm="fricas")

[Out]

-1/2*b^(1/n)*log(1/(tan(f*x + e)^2 + 1))/f

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan \left (f x + e\right )^{n}\right )^{\left (\frac {1}{n}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/n),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^n)^(1/n), x)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (b \left (\tan ^{n}\left (f x +e \right )\right )\right )^{\frac {1}{n}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^n)^(1/n),x)

[Out]

int((b*tan(f*x+e)^n)^(1/n),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/n),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^n)^(1/n), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^n\right )}^{1/n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^n)^(1/n),x)

[Out]

int((b*tan(e + f*x)^n)^(1/n), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{n}{\left (e + f x \right )}\right )^{\frac {1}{n}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**n)**(1/n),x)

[Out]

Integral((b*tan(e + f*x)**n)**(1/n), x)

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