Integrand size = 19, antiderivative size = 78 \[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {4-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n}{f (1-n)} \]
-cos(f*x+e)*hypergeom([2-1/2*n, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*ta n(f*x+e))^n/f/(1-n)/((sin(f*x+e)^2)^(1/2*n))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 13.22 (sec) , antiderivative size = 743, normalized size of antiderivative = 9.53 \[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\frac {\cot ^2\left (\frac {1}{2} (e+f x)\right ) \operatorname {Hypergeometric2F1}\left (-1+\frac {n}{2},n,\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )^n (b \tan (e+f x))^n}{f (-8+4 n)}+\frac {(4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,1,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin ^2(e+f x) (b \tan (e+f x))^n}{4 f (2+n) \left (2 \left (\operatorname {AppellF1}\left (2+\frac {n}{2},n,2,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (2+\frac {n}{2},1+n,1,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (-1+\cos (e+f x))+(4+n) \operatorname {AppellF1}\left (1+\frac {n}{2},n,1,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))\right )}+\frac {\operatorname {Hypergeometric2F1}\left (1+\frac {n}{2},n,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )^n \tan ^2\left (\frac {1}{2} (e+f x)\right ) (b \tan (e+f x))^n}{f (8+4 n)}+\frac {4 \cos ^2\left (\frac {1}{2} (e+f x)\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \left ((2+n) \operatorname {Hypergeometric2F1}\left (\frac {n}{2},n,1+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (1+\frac {n}{2},n,1,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (b \tan (e+f x))^n}{f n (2+n) \left (-8 \operatorname {AppellF1}\left (1+\frac {n}{2},n,1,2+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \tan \left (\frac {1}{2} (e+f x)\right )+\frac {8 \left (2 \operatorname {AppellF1}\left (2+\frac {n}{2},n,2,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 n \operatorname {AppellF1}\left (2+\frac {n}{2},1+n,1,3+\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(4+n) \cot ^4\left (\frac {1}{2} (e+f x)\right ) \left (\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )\right )^{-n}\right ) \tan ^3\left (\frac {1}{2} (e+f x)\right )}{4+n}\right )} \]
(Cot[(e + f*x)/2]^2*Hypergeometric2F1[-1 + n/2, n, n/2, Tan[(e + f*x)/2]^2 ]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*(b*Tan[e + f*x])^n)/(f*(-8 + 4*n)) + ((4 + n)*AppellF1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f *x)/2]^2]*Sin[e + f*x]^2*(b*Tan[e + f*x])^n)/(4*f*(2 + n)*(2*(AppellF1[2 + n/2, n, 2, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1 [2 + n/2, 1 + n, 1, 3 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*(-1 + Cos[e + f*x]) + (4 + n)*AppellF1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/ 2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]))) + (Hypergeometric2F1[1 + n /2, n, 2 + n/2, Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^n*Ta n[(e + f*x)/2]^2*(b*Tan[e + f*x])^n)/(f*(8 + 4*n)) + (4*Cos[(e + f*x)/2]^2 *Cot[(e + f*x)/2]*((2 + n)*Hypergeometric2F1[n/2, n, 1 + n/2, Tan[(e + f*x )/2]^2] - n*AppellF1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)*(b*Tan[e + f*x])^n)/(f*n*(2 + n)*(-8*Appel lF1[1 + n/2, n, 1, 2 + n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[( e + f*x)/2] + (8*(2*AppellF1[2 + n/2, n, 2, 3 + n/2, Tan[(e + f*x)/2]^2, - Tan[(e + f*x)/2]^2] - 2*n*AppellF1[2 + n/2, 1 + n, 1, 3 + n/2, Tan[(e + f* x)/2]^2, -Tan[(e + f*x)/2]^2] + ((4 + n)*Cot[(e + f*x)/2]^4)/(Cos[e + f*x] *Sec[(e + f*x)/2]^2)^n)*Tan[(e + f*x)/2]^3)/(4 + n)))
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3081, 3042, 3056}
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.
| \(\displaystyle \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(b \tan (e+f x))^n}{\sin (e+f x)^3}dx\) |
| \(\Big \downarrow \) 3081 |
| \(\displaystyle \sin ^{-n}(e+f x) \cos ^n(e+f x) (b \tan (e+f x))^n \int \cos ^{-n}(e+f x) \sin ^{n-3}(e+f x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sin ^{-n}(e+f x) \cos ^n(e+f x) (b \tan (e+f x))^n \int \cos (e+f x)^{-n} \sin (e+f x)^{n-3}dx\) |
| \(\Big \downarrow \) 3056 |
| \(\displaystyle -\frac {\cos (e+f x) \sin ^2(e+f x)^{-n/2} (b \tan (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {4-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n)}\) |
-((Cos[e + f*x]*Hypergeometric2F1[(1 - n)/2, (4 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Tan[e + f*x])^n)/(f*(1 - n)*(Sin[e + f*x]^2)^(n/2)))
3.2.83.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F racPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*x]^2) ^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, C os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[Cos[e + f*x]^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^ n) Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(- 1)]) || IntegersQ[m - 1/2, n - 1/2])
\[\int \left (\csc ^{3}\left (f x +e \right )\right ) \left (b \tan \left (f x +e \right )\right )^{n}d x\]
\[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{3} \,d x } \]
\[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\int \left (b \tan {\left (e + f x \right )}\right )^{n} \csc ^{3}{\left (e + f x \right )}\, dx \]
\[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{3} \,d x } \]
\[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\int { \left (b \tan \left (f x + e\right )\right )^{n} \csc \left (f x + e\right )^{3} \,d x } \]
Timed out. \[ \int \csc ^3(e+f x) (b \tan (e+f x))^n \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{{\sin \left (e+f\,x\right )}^3} \,d x \]