Integrand size = 25, antiderivative size = 97 \[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\cos ^2(e+f x)^{\frac {1}{2} (1+m+n p)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+n p),\frac {1}{2} (1+m+n p),\frac {1}{2} (3+n p),\sin ^2(e+f x)\right ) (d \sec (e+f x))^m \tan (e+f x) \left (b (c \tan (e+f x))^n\right )^p}{f (1+n p)} \]
(cos(f*x+e)^2)^(1/2*n*p+1/2*m+1/2)*hypergeom([1/2*n*p+1/2, 1/2*n*p+1/2*m+1 /2],[1/2*n*p+3/2],sin(f*x+e)^2)*(d*sec(f*x+e))^m*tan(f*x+e)*(b*(c*tan(f*x+ e))^n)^p/f/(n*p+1)
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.92 \[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\frac {\cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {m}{2},\frac {1}{2} (1-n p),\frac {2+m}{2},\sec ^2(e+f x)\right ) (d \sec (e+f x))^m \left (-\tan ^2(e+f x)\right )^{\frac {1}{2} (1-n p)} \left (b (c \tan (e+f x))^n\right )^p}{f m} \]
(Cot[e + f*x]*Hypergeometric2F1[m/2, (1 - n*p)/2, (2 + m)/2, Sec[e + f*x]^ 2]*(d*Sec[e + f*x])^m*(-Tan[e + f*x]^2)^((1 - n*p)/2)*(b*(c*Tan[e + f*x])^ n)^p)/(f*m)
Time = 0.37 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4142, 3042, 3097}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^pdx\) |
\(\Big \downarrow \) 4142 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int (d \sec (e+f x))^m (c \tan (e+f x))^{n p}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (c \tan (e+f x))^{-n p} \left (b (c \tan (e+f x))^n\right )^p \int (d \sec (e+f x))^m (c \tan (e+f x))^{n p}dx\) |
\(\Big \downarrow \) 3097 |
\(\displaystyle \frac {\tan (e+f x) (d \sec (e+f x))^m \cos ^2(e+f x)^{\frac {1}{2} (m+n p+1)} \left (b (c \tan (e+f x))^n\right )^p \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (n p+1),\frac {1}{2} (m+n p+1),\frac {1}{2} (n p+3),\sin ^2(e+f x)\right )}{f (n p+1)}\) |
((Cos[e + f*x]^2)^((1 + m + n*p)/2)*Hypergeometric2F1[(1 + n*p)/2, (1 + m + n*p)/2, (3 + n*p)/2, Sin[e + f*x]^2]*(d*Sec[e + f*x])^m*Tan[e + f*x]*(b* (c*Tan[e + f*x])^n)^p)/(f*(1 + n*p))
3.5.76.3.1 Defintions of rubi rules used
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] && !IntegerQ[m/2]
Int[(u_.)*((b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> S imp[b^IntPart[p]*((b*(c*Tan[e + f*x])^n)^FracPart[p]/(c*Tan[e + f*x])^(n*Fr acPart[p])) Int[ActivateTrig[u]*(c*Tan[e + f*x])^(n*p), x], x] /; FreeQ[{ b, c, e, f, n, p}, x] && !IntegerQ[p] && !IntegerQ[n] && (EqQ[u, 1] || Ma tchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (b \left (c \tan \left (f x +e \right )\right )^{n}\right )^{p}d x\]
\[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\int \left (b \left (c \tan {\left (e + f x \right )}\right )^{n}\right )^{p} \left (d \sec {\left (e + f x \right )}\right )^{m}\, dx \]
\[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\int { \left (\left (c \tan \left (f x + e\right )\right )^{n} b\right )^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^m \left (b (c \tan (e+f x))^n\right )^p \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (b\,{\left (c\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\right )}^p \,d x \]