\(\int \tan ^2(e+f x) (b (c \tan (e+f x))^n)^p \, dx\) [411]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+n p),\frac {1}{2} (5+n p),-\tan ^2(e+f x)\right ) \tan ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3+n p)} \]

[Out]

hypergeom([1, 1/2*n*p+3/2],[1/2*n*p+5/2],-tan(f*x+e)^2)*tan(f*x+e)^3*(b*(c*tan(f*x+e))^n)^p/f/(n*p+3)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3740, 16, 3557, 371} \[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\tan ^3(e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (n p+3),\frac {1}{2} (n p+5),-\tan ^2(e+f x)\right ) \left (b (c \tan (e+f x))^n\right )^p}{f (n p+3)} \]

[In]

Int[Tan[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Hypergeometric2F1[1, (3 + n*p)/2, (5 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(
3 + n*p))

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

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 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*} \text {integral}& = \left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \int \tan ^2(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))^{2+n p} \, dx}{c^2} \\ & = \frac {\left ((c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p\right ) \text {Subst}\left (\int \frac {x^{2+n p}}{c^2+x^2} \, dx,x,c \tan (e+f x)\right )}{c f} \\ & = \frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+n p),\frac {1}{2} (5+n p),-\tan ^2(e+f x)\right ) \tan ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3+n p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+n p),\frac {1}{2} (5+n p),-\tan ^2(e+f x)\right ) \tan ^3(e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (3+n p)} \]

[In]

Integrate[Tan[e + f*x]^2*(b*(c*Tan[e + f*x])^n)^p,x]

[Out]

(Hypergeometric2F1[1, (3 + n*p)/2, (5 + n*p)/2, -Tan[e + f*x]^2]*Tan[e + f*x]^3*(b*(c*Tan[e + f*x])^n)^p)/(f*(
3 + n*p))

Maple [F]

\[\int \tan \left (f x +e \right )^{2} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]

[In]

int(tan(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)

[Out]

int(tan(f*x+e)^2*(b*(c*tan(f*x+e))^n)^p,x)

Fricas [F]

\[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

Sympy [F]

\[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \tan ^{2}{\left (e + f x \right )}\, dx \]

[In]

integrate(tan(f*x+e)**2*(b*(c*tan(f*x+e))**n)**p,x)

[Out]

Integral((b*(c*tan(e + f*x))**n)**p*tan(e + f*x)**2, x)

Maxima [F]

\[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

Giac [F]

\[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^2(e+f x) \left (b (c \tan (e+f x))^n\right )^p \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \]

[In]

int(tan(e + f*x)^2*(b*(c*tan(e + f*x))^n)^p,x)

[Out]

int(tan(e + f*x)^2*(b*(c*tan(e + f*x))^n)^p, x)