3.1.29 \(\int (b \tan ^n(e+f x))^{\frac {1}{n}} \, dx\) [29]

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 = 0.01, 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 = {3740, 3556} \begin {gather*} -\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f} \end {gather*}

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 3556

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

Rule 3740

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 \begin {gather*} -\frac {\cot (e+f x) \log (\cos (e+f x)) \left (b \tan ^n(e+f x)\right )^{\frac {1}{n}}}{f} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 4.99, size = 12884, normalized size = 402.62

method	result	size
risch	\(\text {Expression too large to display}\)	\(12884\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 1.81, size = 24, normalized size = 0.75 \begin {gather*} -\frac {b^{\left (\frac {1}{n}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f} \end {gather*}

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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