3.417 \(\int \tan (e+f x) (b (c \tan (e+f x))^n)^p \, dx\)

Optimal. Leaf size=63 \[ \frac{\tan ^2(e+f x) \text{Hypergeometric2F1}\left (1,\frac{1}{2} (n p+2),\frac{1}{2} (n p+4),-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \]

[Out]

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

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Rubi [A]  time = 0.068729, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3659, 16, 3476, 364} \[ \frac{\tan ^2(e+f x) \, _2F_1\left (1,\frac{1}{2} (n p+2);\frac{1}{2} (n p+4);-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

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

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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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

Mathematica [A]  time = 0.0574892, size = 61, normalized size = 0.97 \[ \frac{\tan ^2(e+f x) \text{Hypergeometric2F1}\left (1,\frac{n p}{2}+1,\frac{n p}{2}+2,-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Hypergeometric2F1[1, 1 + (n*p)/2, 2 + (n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p)/(f*(
2 + n*p))

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Maple [F]  time = 5.706, size = 0, normalized size = 0. \begin{align*} \int \tan \left ( fx+e \right ) \left ( b \left ( c\tan \left ( fx+e \right ) \right ) ^{n} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(tan(f*x+e)*(b*(c*tan(f*x+e))^n)^p,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*tan(f*x + e), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \tan \left (f x + e\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(((c*tan(f*x + e))^n*b)^p*tan(f*x + e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*tan(e + f*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(((c*tan(f*x + e))^n*b)^p*tan(f*x + e), x)