Integrand size = 25, antiderivative size = 108 \[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2},1-\frac {m}{2},-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{f} \]
AppellF1(1/2,1-1/2*m,-p,3/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*(d*sec(f*x+e) )^m*tan(f*x+e)*(a+b*tan(f*x+e)^2)^p/f/((sec(f*x+e)^2)^(1/2*m))/((1+b*tan(f *x+e)^2/a)^p)
Leaf count is larger than twice the leaf count of optimal. \(2033\) vs. \(2(108)=216\).
Time = 16.97 (sec) , antiderivative size = 2033, normalized size of antiderivative = 18.82 \[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \]
(3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2) /a)]*(d*Sec[e + f*x])^m*(Sec[e + f*x]^2)^(-1 + m/2)*Tan[e + f*x]*(a + b*Ta n[e + f*x]^2)^(2*p))/(f*(3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x] ^2, -((b*Tan[e + f*x]^2)/a)] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, - Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] + a*(-2 + m)*AppellF1[3/2, 2 - m/ 2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2)*((6 *a*b*p*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^ 2)/a)]*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p))/(3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x ]^2)/a)] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b *Tan[e + f*x]^2)/a)] + a*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2) + (3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*(Sec[e + f*x]^2) ^(m/2)*(a + b*Tan[e + f*x]^2)^p)/(3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan [e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] + a*(-2 + m)*AppellF1[3/ 2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f* x]^2) + (6*a*(-1 + m/2)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, - ((b*Tan[e + f*x]^2)/a)]*(Sec[e + f*x]^2)^(-1 + m/2)*Tan[e + f*x]^2*(a + b* Tan[e + f*x]^2)^p)/(3*a*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2...
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4162, 334, 333}
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.
\(\displaystyle \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (e+f x))^m \left (a+b \tan (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4162 |
\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \int \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}} \left (b \tan ^2(e+f x)+a\right )^pd\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \int \left (\tan ^2(e+f x)+1\right )^{\frac {m-2}{2}} \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^pd\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},\frac {2-m}{2},-p,\frac {3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{f}\) |
(AppellF1[1/2, (2 - m)/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a )]*(d*Sec[e + f*x])^m*Tan[e + f*x]*(a + b*Tan[e + f*x]^2)^p)/(f*(Sec[e + f *x]^2)^(m/2)*(1 + (b*Tan[e + f*x]^2)/a)^p)
3.5.75.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff *((d*Sec[e + f*x])^m/(f*(Sec[e + f*x]^2)^(m/2))) Subst[Int[(1 + ff^2*x^2) ^(m/2 - 1)*(a + b*ff^2*x^2)^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && !IntegerQ[m]
\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p}\, dx \]
\[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]