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

Optimal. Leaf size=59 \[ \frac {\tan (e+f x) \left (b \tan ^n(e+f x)\right )^p \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right )}{f (n p+1)} \]

[Out]

hypergeom([1, 1/2*n*p+1/2],[1/2*n*p+3/2],-tan(f*x+e)^2)*tan(f*x+e)*(b*tan(f*x+e)^n)^p/f/(n*p+1)

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Rubi [A]  time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3659, 3476, 364} \[ \frac {\tan (e+f x) \left (b \tan ^n(e+f x)\right )^p \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right )}{f (n p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^n)^p,x]

[Out]

(Hypergeometric2F1[1, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^n)^p)/(f*(1 + n*
p))

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])

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 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 )^p \, dx &=\left (\tan ^{-n p}(e+f x) \left (b \tan ^n(e+f x)\right )^p\right ) \int \tan ^{n p}(e+f x) \, dx\\ &=\frac {\left (\tan ^{-n p}(e+f x) \left (b \tan ^n(e+f x)\right )^p\right ) \operatorname {Subst}\left (\int \frac {x^{n p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\, _2F_1\left (1,\frac {1}{2} (1+n p);\frac {1}{2} (3+n p);-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^n(e+f x)\right )^p}{f (1+n p)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.97 \[ \frac {\tan (e+f x) \left (b \tan ^n(e+f x)\right )^p \, _2F_1\left (1,\frac {1}{2} (n p+1);\frac {1}{2} (n p+3);-\tan ^2(e+f x)\right )}{f n p+f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^n)^p,x]

[Out]

(Hypergeometric2F1[1, (1 + n*p)/2, (3 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^n)^p)/(f + f*n*p
)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (b \tan \left (f x + e\right )^{n}\right )^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(f*x + e)^n)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e)^n)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e)^n)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^n\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^n)^p,x)

[Out]

int((b*tan(e + f*x)^n)^p, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(e + f*x)**n)**p, x)

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