\(\int (b \tan ^3(e+f x))^p \, dx\) [27]

Optimal result

Integrand size = 12, antiderivative size = 57 \[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+3 p),\frac {3 (1+p)}{2},-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^3(e+f x)\right )^p}{f (1+3 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3739, 3557, 371} \[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\frac {\tan (e+f x) \left (b \tan ^3(e+f x)\right )^p \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3 p+1),\frac {3 (p+1)}{2},-\tan ^2(e+f x)\right )}{f (3 p+1)} \]

[In]

Int[(b*Tan[e + f*x]^3)^p,x]

[Out]

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

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 3739

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, 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 (\tan ^{-3 p}(e+f x) \left (b \tan ^3(e+f x)\right )^p\right ) \int \tan ^{3 p}(e+f x) \, dx \\ & = \frac {\left (\tan ^{-3 p}(e+f x) \left (b \tan ^3(e+f x)\right )^p\right ) \text {Subst}\left (\int \frac {x^{3 p}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+3 p),\frac {3 (1+p)}{2},-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^3(e+f x)\right )^p}{f (1+3 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (1+3 p),\frac {3 (1+p)}{2},-\tan ^2(e+f x)\right ) \tan (e+f x) \left (b \tan ^3(e+f x)\right )^p}{f (1+3 p)} \]

[In]

Integrate[(b*Tan[e + f*x]^3)^p,x]

[Out]

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

Maple [F]

\[\int \left (b \tan \left (f x +e \right )^{3}\right )^{p}d x\]

[In]

int((b*tan(f*x+e)^3)^p,x)

[Out]

int((b*tan(f*x+e)^3)^p,x)

Fricas [F]

\[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{3}\right )^{p} \,d x } \]

[In]

integrate((b*tan(f*x+e)^3)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^3)^p, x)

Sympy [F]

\[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\int \left (b \tan ^{3}{\left (e + f x \right )}\right )^{p}\, dx \]

[In]

integrate((b*tan(f*x+e)**3)**p,x)

[Out]

Integral((b*tan(e + f*x)**3)**p, x)

Maxima [F]

\[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{3}\right )^{p} \,d x } \]

[In]

integrate((b*tan(f*x+e)^3)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^3)^p, x)

Giac [F]

\[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\int { \left (b \tan \left (f x + e\right )^{3}\right )^{p} \,d x } \]

[In]

integrate((b*tan(f*x+e)^3)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^3)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (b \tan ^3(e+f x)\right )^p \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}^p \,d x \]

[In]

int((b*tan(e + f*x)^3)^p,x)

[Out]

int((b*tan(e + f*x)^3)^p, x)