Integrand size = 23, antiderivative size = 81 \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right ) \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{3 f} \]
1/3*AppellF1(3/2,2,-p,5/2,sec(f*x+e)^2,-b*sec(f*x+e)^2/a)*sec(f*x+e)^3*(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. \(266\) vs. \(2(81)=162\).
Time = 3.22 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.28 \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1}{2}-p,-\frac {1}{2},-p,\frac {3}{2}-p,-\cot ^2(e+f x),-\frac {(a+b) \cot ^2(e+f x)}{b}\right ) \csc ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^p}{f (-1+2 p) \left (-\frac {\left (2 (a+b) p \operatorname {AppellF1}\left (\frac {3}{2}-p,-\frac {1}{2},1-p,\frac {5}{2}-p,-\cot ^2(e+f x),-\frac {(a+b) \cot ^2(e+f x)}{b}\right )+b \operatorname {AppellF1}\left (\frac {3}{2}-p,\frac {1}{2},-p,\frac {5}{2}-p,-\cot ^2(e+f x),-\frac {(a+b) \cot ^2(e+f x)}{b}\right )\right ) \cot (e+f x) \csc (e+f x)}{b (-3+2 p)}+\operatorname {AppellF1}\left (\frac {1}{2}-p,-\frac {1}{2},-p,\frac {3}{2}-p,-\cot ^2(e+f x),-\frac {(a+b) \cot ^2(e+f x)}{b}\right ) \sec (e+f x)\right )} \]
(AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f*x]^2, -(((a + b)*Cot[e + f*x]^2)/b)]*Csc[e + f*x]^2*(a + b*Sec[e + f*x]^2)^p)/(f*(-1 + 2*p)*(-(((2* (a + b)*p*AppellF1[3/2 - p, -1/2, 1 - p, 5/2 - p, -Cot[e + f*x]^2, -(((a + b)*Cot[e + f*x]^2)/b)] + b*AppellF1[3/2 - p, 1/2, -p, 5/2 - p, -Cot[e + f *x]^2, -(((a + b)*Cot[e + f*x]^2)/b)])*Cot[e + f*x]*Csc[e + f*x])/(b*(-3 + 2*p))) + AppellF1[1/2 - p, -1/2, -p, 3/2 - p, -Cot[e + f*x]^2, -(((a + b) *Cot[e + f*x]^2)/b)]*Sec[e + f*x]))
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4622, 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.
\(\displaystyle \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^p}{\sin (e+f x)^3}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int \frac {\sec ^2(e+f x) \left (b \sec ^2(e+f x)+a\right )^p}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 395 |
\(\displaystyle \frac {\left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \int \frac {\sec ^2(e+f x) \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^p}{\left (1-\sec ^2(e+f x)\right )^2}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 394 |
\(\displaystyle \frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},2,-p,\frac {5}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right )}{3 f}\) |
(AppellF1[3/2, 2, -p, 5/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/a)]*Sec[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p)/(3*f*(1 + (b*Sec[e + f*x]^2)/a)^p)
3.2.37.3.1 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[((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])
\[\int \csc \left (f x +e \right )^{3} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \]
\[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \csc ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \]