Integrand size = 23, antiderivative size = 192 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {(10 a+b (7+2 p)) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{15 (a+b)^2 f}-\frac {\cot ^5(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{5 (a+b) f}-\frac {\left (15 a^2+20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 (a+b)^2 f} \]
-1/15*(10*a+b*(7+2*p))*cot(f*x+e)^3*(a+b+b*tan(f*x+e)^2)^(p+1)/(a+b)^2/f-1 /5*cot(f*x+e)^5*(a+b+b*tan(f*x+e)^2)^(p+1)/(a+b)/f-1/15*(15*a^2+20*a*b*(p+ 1)+4*b^2*(p^2+3*p+2))*cot(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*tan(f*x+e)^ 2/(a+b))*(a+b+b*tan(f*x+e)^2)^p/(a+b)^2/f/((1+b*tan(f*x+e)^2/(a+b))^p)
Time = 1.44 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.78 \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\cot (e+f x) \left (3 \cot ^4(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+10 \cot ^2(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+15 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{15 f} \]
-1/15*(Cot[e + f*x]*(3*Cot[e + f*x]^4*Hypergeometric2F1[-5/2, -p, -3/2, -( (b*Tan[e + f*x]^2)/(a + b))] + 10*Cot[e + f*x]^2*Hypergeometric2F1[-3/2, - p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))] + 15*Hypergeometric2F1[-1/2, -p, 1 /2, -((b*Tan[e + f*x]^2)/(a + b))])*(a + b*Sec[e + f*x]^2)^p)/(f*(1 + (b*T an[e + f*x]^2)/(a + b))^p)
Time = 0.34 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4620, 365, 359, 279, 278}
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.
| \(\displaystyle \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^p}{\sin (e+f x)^6}dx\) |
| \(\Big \downarrow \) 4620 |
| \(\displaystyle \frac {\int \cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a+b\right )^pd\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int \cot ^4(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^p \left (5 (a+b) \tan ^2(e+f x)+10 a+b (2 p+7)\right )d\tan (e+f x)}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {\frac {\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \int \cot ^2(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^pd\tan (e+f x)}{3 (a+b)}-\frac {(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 279 |
| \(\displaystyle \frac {\frac {\frac {\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \cot ^2(e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^pd\tan (e+f x)}{3 (a+b)}-\frac {(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 (a+b)}}{f}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {\frac {-\frac {\left (15 a^2+20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cot (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{3 (a+b)}-\frac {(10 a+b (2 p+7)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{3 (a+b)}}{5 (a+b)}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{5 (a+b)}}{f}\) |
(-1/5*(Cot[e + f*x]^5*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(a + b) + (-1/3* ((10*a + b*(7 + 2*p))*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^(1 + p))/( a + b) - ((15*a^2 + 20*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cot[e + f*x]*H ypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + b + b* Tan[e + f*x]^2)^p)/(3*(a + b)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/(5*(a + b)))/f
3.2.43.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
\[\int \csc \left (f x +e \right )^{6} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]
\[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \csc \left (f x + e\right )^{6} \,d x } \]
Timed out. \[ \int \csc ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\sin \left (e+f\,x\right )}^6} \,d x \]