3.19 \(\int (b \tan ^n(e+f x))^{5/2} \, dx\)

Optimal. Leaf size=71 \[ \frac{2 b^2 \tan ^{2 n+1}(e+f x) \sqrt{b \tan ^n(e+f x)} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (5 n+2),\frac{1}{4} (5 n+6),-\tan ^2(e+f x)\right )}{f (5 n+2)} \]

[Out]

(2*b^2*Hypergeometric2F1[1, (2 + 5*n)/4, (6 + 5*n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 + 2*n)*Sqrt[b*Tan[e + f
*x]^n])/(f*(2 + 5*n))

________________________________________________________________________________________

Rubi [A]  time = 0.0439371, antiderivative size = 71, 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} \[ \frac{2 b^2 \tan ^{2 n+1}(e+f x) \sqrt{b \tan ^n(e+f x)} \, _2F_1\left (1,\frac{1}{4} (5 n+2);\frac{1}{4} (5 n+6);-\tan ^2(e+f x)\right )}{f (5 n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^(5/2),x]

[Out]

(2*b^2*Hypergeometric2F1[1, (2 + 5*n)/4, (6 + 5*n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]^(1 + 2*n)*Sqrt[b*Tan[e + f
*x]^n])/(f*(2 + 5*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 \left (b \tan ^n(e+f x)\right )^{5/2} \, dx &=\left (b^2 \tan ^{-\frac{n}{2}}(e+f x) \sqrt{b \tan ^n(e+f x)}\right ) \int \tan ^{\frac{5 n}{2}}(e+f x) \, dx\\ &=\frac{\left (b^2 \tan ^{-\frac{n}{2}}(e+f x) \sqrt{b \tan ^n(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^{5 n/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 b^2 \, _2F_1\left (1,\frac{1}{4} (2+5 n);\frac{1}{4} (6+5 n);-\tan ^2(e+f x)\right ) \tan ^{1+2 n}(e+f x) \sqrt{b \tan ^n(e+f x)}}{f (2+5 n)}\\ \end{align*}

Mathematica [A]  time = 0.108413, size = 62, normalized size = 0.87 \[ \frac{2 \tan (e+f x) \left (b \tan ^n(e+f x)\right )^{5/2} \text{Hypergeometric2F1}\left (1,\frac{1}{4} (5 n+2),\frac{1}{4} (5 n+6),-\tan ^2(e+f x)\right )}{f (5 n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^n)^(5/2),x]

[Out]

(2*Hypergeometric2F1[1, (2 + 5*n)/4, (6 + 5*n)/4, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^n)^(5/2))/(f*(
2 + 5*n))

________________________________________________________________________________________

Maple [F]  time = 0.615, size = 0, normalized size = 0. \begin{align*} \int \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{n} \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^n)^(5/2),x)

[Out]

int((b*tan(f*x+e)^n)^(5/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{n}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^n)^(5/2), x)

________________________________________________________________________________________

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)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**n)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{n}\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^n)^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^n)^(5/2), x)