3.21 \(\int \sqrt{b \tan ^n(e+f x)} \, dx\)

Optimal. Leaf size=56 \[ \frac{2 \tan (e+f x) \sqrt{b \tan ^n(e+f x)} \text{Hypergeometric2F1}\left (1,\frac{n+2}{4},\frac{n+6}{4},-\tan ^2(e+f x)\right )}{f (n+2)} \]

[Out]

(2*Hypergeometric2F1[1, (2 + n)/4, (6 + n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^n])/(f*(2 + n)
)

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Rubi [A]  time = 0.0418333, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3659, 3476, 364} \[ \frac{2 \tan (e+f x) \sqrt{b \tan ^n(e+f x)} \, _2F_1\left (1,\frac{n+2}{4};\frac{n+6}{4};-\tan ^2(e+f x)\right )}{f (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Tan[e + f*x]^n],x]

[Out]

(2*Hypergeometric2F1[1, (2 + n)/4, (6 + n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^n])/(f*(2 + n)
)

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]])

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sqrt{b \tan ^n(e+f x)} \, dx &=\left (\tan ^{-\frac{n}{2}}(e+f x) \sqrt{b \tan ^n(e+f x)}\right ) \int \tan ^{\frac{n}{2}}(e+f x) \, dx\\ &=\frac{\left (\tan ^{-\frac{n}{2}}(e+f x) \sqrt{b \tan ^n(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^{n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 \, _2F_1\left (1,\frac{2+n}{4};\frac{6+n}{4};-\tan ^2(e+f x)\right ) \tan (e+f x) \sqrt{b \tan ^n(e+f x)}}{f (2+n)}\\ \end{align*}

Mathematica [A]  time = 0.037718, size = 56, normalized size = 1. \[ \frac{2 \tan (e+f x) \sqrt{b \tan ^n(e+f x)} \text{Hypergeometric2F1}\left (1,\frac{n+2}{4},\frac{n+6}{4},-\tan ^2(e+f x)\right )}{f (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Tan[e + f*x]^n],x]

[Out]

(2*Hypergeometric2F1[1, (2 + n)/4, (6 + n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^n])/(f*(2 + n)
)

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Maple [F]  time = 0.119, size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \left ( \tan \left ( fx+e \right ) \right ) ^{n}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^n)^(1/2),x)

[Out]

int((b*tan(f*x+e)^n)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^n), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**n)**(1/2),x)

[Out]

Integral(sqrt(b*tan(e + f*x)**n), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*tan(f*x + e)^n), x)