\(\int (d \sec (e+f x))^m (b \tan ^2(e+f x))^p \, dx\) [474]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int (d \sec (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+m+2 p)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 p),\frac {1}{2} (1+m+2 p),\frac {1}{2} (3+2 p),\sin ^2(e+f x)\right ) (d \sec (e+f x))^m \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+2 p)} \]

[Out]

(cos(f*x+e)^2)^(1/2+1/2*m+p)*hypergeom([1/2+p, 1/2+1/2*m+p],[3/2+p],sin(f*x+e)^2)*(d*sec(f*x+e))^m*tan(f*x+e)*
(b*tan(f*x+e)^2)^p/f/(1+2*p)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3739, 2697} \[ \int (d \sec (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\frac {\tan (e+f x) \left (b \tan ^2(e+f x)\right )^p (d \sec (e+f x))^m \cos ^2(e+f x)^{\frac {1}{2} (m+2 p+1)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 p+1),\frac {1}{2} (m+2 p+1),\frac {1}{2} (2 p+3),\sin ^2(e+f x)\right )}{f (2 p+1)} \]

[In]

Int[(d*Sec[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

((Cos[e + f*x]^2)^((1 + m + 2*p)/2)*Hypergeometric2F1[(1 + 2*p)/2, (1 + m + 2*p)/2, (3 + 2*p)/2, Sin[e + f*x]^
2]*(d*Sec[e + f*x])^m*Tan[e + f*x]*(b*Tan[e + f*x]^2)^p)/(f*(1 + 2*p))

Rule 2697

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*Sec[e + f
*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

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 ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int (d \sec (e+f x))^m \tan ^{2 p}(e+f x) \, dx \\ & = \frac {\cos ^2(e+f x)^{\frac {1}{2} (1+m+2 p)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 p),\frac {1}{2} (1+m+2 p),\frac {1}{2} (3+2 p),\sin ^2(e+f x)\right ) (d \sec (e+f x))^m \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{f (1+2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int (d \sec (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx=\frac {\cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {m}{2},\frac {1}{2}-p,\frac {2+m}{2},\sec ^2(e+f x)\right ) (d \sec (e+f x))^m \left (-\tan ^2(e+f x)\right )^{\frac {1}{2}-p} \left (b \tan ^2(e+f x)\right )^p}{f m} \]

[In]

Integrate[(d*Sec[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

(Cot[e + f*x]*Hypergeometric2F1[m/2, 1/2 - p, (2 + m)/2, Sec[e + f*x]^2]*(d*Sec[e + f*x])^m*(-Tan[e + f*x]^2)^
(1/2 - p)*(b*Tan[e + f*x]^2)^p)/(f*m)

Maple [F]

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

[In]

int((d*sec(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

[Out]

int((d*sec(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

Fricas [F]

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

[In]

integrate((d*sec(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2)^p*(d*sec(f*x + e))^m, x)

Sympy [F]

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

[In]

integrate((d*sec(f*x+e))**m*(b*tan(f*x+e)**2)**p,x)

[Out]

Integral((b*tan(e + f*x)**2)**p*(d*sec(e + f*x))**m, x)

Maxima [F]

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

[In]

integrate((d*sec(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2)^p*(d*sec(f*x + e))^m, x)

Giac [F]

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

[In]

integrate((d*sec(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2)^p*(d*sec(f*x + e))^m, x)

Mupad [F(-1)]

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

[In]

int((d/cos(e + f*x))^m*(b*tan(e + f*x)^2)^p,x)

[Out]

int((d/cos(e + f*x))^m*(b*tan(e + f*x)^2)^p, x)