Integrand size = 23, antiderivative size = 86 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\frac {\left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b f (1+p)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)} \]
1/2*(a+b*sec(f*x+e)^2)^(p+1)/b/f/(p+1)+1/2*hypergeom([1, p+1],[2+p],1+b*se c(f*x+e)^2/a)*(a+b*sec(f*x+e)^2)^(p+1)/a/f/(p+1)
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.71 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\frac {\left (a+b \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right )\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a b f (1+p)} \]
((a + b*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sec[e + f*x]^2)/a])*(a + b*Sec[e + f*x]^2)^(1 + p))/(2*a*b*f*(1 + p))
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4627, 25, 354, 90, 75}
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.
\(\displaystyle \int \tan ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4627 |
\(\displaystyle \frac {\int -\cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (1-\sec ^2(e+f x)\right ) \left (b \sec ^2(e+f x)+a\right )^pd\sec ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle -\frac {\int \cos (e+f x) \left (b \sec ^2(e+f x)+a\right )^pd\sec ^2(e+f x)-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1}}{b (p+1)}}{2 f}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle -\frac {-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sec ^2(e+f x)}{a}+1\right )}{a (p+1)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1}}{b (p+1)}}{2 f}\) |
-1/2*(-((a + b*Sec[e + f*x]^2)^(1 + p)/(b*(1 + p))) - (Hypergeometric2F1[1 , 1 + p, 2 + p, 1 + (b*Sec[e + f*x]^2)/a]*(a + b*Sec[e + f*x]^2)^(1 + p))/ (a*(1 + p)))/f
3.5.43.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Si mp[1/f Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x), x] , x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[( m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4] || IGtQ[p, 0] || Integers Q[2*n, p])
\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \tan \left (f x +e \right )^{3}d x\]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{p} \tan ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^3(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]