Integrand size = 21, antiderivative size = 77 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right ) \sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{f} \]
-AppellF1(1/2,1,-p,3/2,sec(f*x+e)^2,-b*sec(f*x+e)^2/a)*sec(f*x+e)*(a+b*sec (f*x+e)^2)^p/f/((1+b*sec(f*x+e)^2/a)^p)
Leaf count is larger than twice the leaf count of optimal. \(1532\) vs. \(2(77)=154\).
Time = 16.05 (sec) , antiderivative size = 1532, normalized size of antiderivative = 19.90 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx =\text {Too large to display} \]
((a + 2*b + a*Cos[2*(e + f*x)])^p*Csc[e + f*x]*(Sec[e + f*x]^2)^p*(a + b*S ec[e + f*x]^2)^p*((2*AppellF1[-1/2 - p, -1/2, -p, 1/2 - p, -Cot[e + f*x]^2 , -(((a + b)*Cot[e + f*x]^2)/b)]*Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + ((a + b)*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x]^2]) - (AppellF1[1, 1/2, -p, 2 , -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/((a + b + b*Tan[e + f*x]^2)/(a + b))^p))/(2*f*(-(a*p*(a + 2*b + a*Cos[2*(e + f*x)] )^(-1 + p)*(Sec[e + f*x]^2)^p*Sin[2*(e + f*x)]*((2*AppellF1[-1/2 - p, -1/2 , -p, 1/2 - p, -Cot[e + f*x]^2, -(((a + b)*Cot[e + f*x]^2)/b)]*Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + ((a + b)*Cot[e + f*x]^2)/b)^p*Sqrt[Csc[e + f*x] ^2]) - (AppellF1[1, 1/2, -p, 2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/((a + b + b*Tan[e + f*x]^2)/(a + b))^p)) + p*(a + 2* b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*Tan[e + f*x]*((2*AppellF1[-1/ 2 - p, -1/2, -p, 1/2 - p, -Cot[e + f*x]^2, -(((a + b)*Cot[e + f*x]^2)/b)]* Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + ((a + b)*Cot[e + f*x]^2)/b)^p*Sqrt[C sc[e + f*x]^2]) - (AppellF1[1, 1/2, -p, 2, -Tan[e + f*x]^2, -((b*Tan[e + f *x]^2)/(a + b))]*Tan[e + f*x]^2)/((a + b + b*Tan[e + f*x]^2)/(a + b))^p) + ((a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^p*((2*AppellF1[-1/2 - p, -1/2, -p, 1/2 - p, -Cot[e + f*x]^2, -(((a + b)*Cot[e + f*x]^2)/b)]*Cot[ e + f*x]*Sqrt[Sec[e + f*x]^2])/((1 + 2*p)*(1 + ((a + b)*Cot[e + f*x]^2)/b) ^p*Sqrt[Csc[e + f*x]^2]) + (4*(a + b)*p*AppellF1[-1/2 - p, -1/2, -p, 1/...
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4622, 25, 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.
\(\displaystyle \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^p}{\sin (e+f x)}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int -\frac {\left (b \sec ^2(e+f x)+a\right )^p}{1-\sec ^2(e+f x)}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\left (b \sec ^2(e+f x)+a\right )^p}{1-\sec ^2(e+f x)}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle -\frac {\left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\left (\frac {b \sec ^2(e+f x)}{a}+1\right )^p}{1-\sec ^2(e+f x)}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle -\frac {\sec (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},1,-p,\frac {3}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right )}{f}\) |
-((AppellF1[1/2, 1, -p, 3/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/a)]*Sec[ e + f*x]*(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b*Sec[e + f*x]^2)/a)^p))
3.2.36.3.1 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[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), 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])
\[\int \csc \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right ) \,d x } \]
\[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{\sin \left (e+f\,x\right )} \,d x \]