Integrand size = 25, antiderivative size = 105 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},1,-p,\frac {3+m}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) (d \tan (e+f x))^{1+m} \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{d f (1+m)} \]
AppellF1(1/2+1/2*m,1,-p,3/2+1/2*m,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*(d* tan(f*x+e))^(1+m)*(a+b+b*tan(f*x+e)^2)^p/d/f/(1+m)/((1+b*tan(f*x+e)^2/(a+b ))^p)
Leaf count is larger than twice the leaf count of optimal. \(259\) vs. \(2(105)=210\).
Time = 5.25 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.47 \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin (e+f x) (d \tan (e+f x))^m}{f (1+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )+\frac {2 \left (b p \operatorname {AppellF1}\left (\frac {3+m}{2},1-p,1,\frac {5+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )-(a+b) \operatorname {AppellF1}\left (\frac {3+m}{2},-p,2,\frac {5+m}{2},-\frac {b \tan ^2(e+f x)}{a+b},-\tan ^2(e+f x)\right )\right ) \tan ^2(e+f x)}{(a+b) (3+m)}\right )} \]
(AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan [e + f*x]^2]*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]*(d*Tan[e + f*x])^m)/(f*(1 + m)*(AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*Tan[e + f *x]^2)/(a + b)), -Tan[e + f*x]^2] + (2*(b*p*AppellF1[(3 + m)/2, 1 - p, 1, (5 + m)/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*Appel lF1[(3 + m)/2, -p, 2, (5 + m)/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f *x]^2])*Tan[e + f*x]^2)/((a + b)*(3 + m))))
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4629, 2075, 395, 394}
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.
\(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sec (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4629 |
\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^m \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2075 |
\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^m \left (b \tan ^2(e+f x)+a+b\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \frac {(d \tan (e+f x))^m \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {(d \tan (e+f x))^{m+1} \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {m+1}{2},1,-p,\frac {m+3}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{d f (m+1)}\) |
(AppellF1[(1 + m)/2, 1, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^ 2)/(a + b))]*(d*Tan[e + f*x])^(1 + m)*(a + b + b*Tan[e + f*x]^2)^p)/(d*f*( 1 + m)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)
3.5.41.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 , -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ FracPart[p]) Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ [{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && ! BinomialMatchQ[{u, v}, x]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f _.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/f Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 )), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \left (d \tan \left (f x +e \right )\right )^{m}d x\]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\text {Timed out} \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \tan (e+f x))^m \, dx=\int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]