Integrand size = 21, antiderivative size = 89 \[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\frac {\cos ^2(e+f x)^{\frac {1+n}{2}} (a \csc (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1-m+n),\frac {1}{2} (3-m+n),\sin ^2(e+f x)\right ) (b \tan (e+f x))^{1+n}}{b f (1-m+n)} \]
(cos(f*x+e)^2)^(1/2+1/2*n)*(a*csc(f*x+e))^m*hypergeom([1/2+1/2*n, 1/2-1/2* m+1/2*n],[3/2-1/2*m+1/2*n],sin(f*x+e)^2)*(b*tan(f*x+e))^(1+n)/b/f/(1-m+n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 2.00 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.22 \[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=-\frac {a (-3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1-m+n),n,1-m,\frac {1}{2} (3-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (a \csc (e+f x))^{-1+m} (b \tan (e+f x))^n}{f (-1+m-n) \left ((-3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1-m+n),n,1-m,\frac {1}{2} (3-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((-1+m) \operatorname {AppellF1}\left (\frac {1}{2} (3-m+n),n,2-m,\frac {1}{2} (5-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n \operatorname {AppellF1}\left (\frac {1}{2} (3-m+n),1+n,1-m,\frac {1}{2} (5-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]
-((a*(-3 + m - n)*AppellF1[(1 - m + n)/2, n, 1 - m, (3 - m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(a*Csc[e + f*x])^(-1 + m)*(b*Tan[e + f*x ])^n)/(f*(-1 + m - n)*((-3 + m - n)*AppellF1[(1 - m + n)/2, n, 1 - m, (3 - m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((-1 + m)*AppellF1 [(3 - m + n)/2, n, 2 - m, (5 - m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x )/2]^2] + n*AppellF1[(3 - m + n)/2, 1 + n, 1 - m, (5 - m + n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))
Time = 0.49 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3098, 3042, 3082, 3042, 3057}
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.
\(\displaystyle \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \csc (e+f x))^m (b \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 3098 |
\(\displaystyle \left (\frac {\sin (e+f x)}{a}\right )^m (a \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{a}\right )^{-m} (b \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (\frac {\sin (e+f x)}{a}\right )^m (a \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{a}\right )^{-m} (b \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 3082 |
\(\displaystyle \frac {\cos ^{n+1}(e+f x) (a \csc (e+f x))^{m+1} (b \tan (e+f x))^{n+1} \left (\frac {\sin (e+f x)}{a}\right )^{m-n} \int \cos ^{-n}(e+f x) \left (\frac {\sin (e+f x)}{a}\right )^{n-m}dx}{a b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cos ^{n+1}(e+f x) (a \csc (e+f x))^{m+1} (b \tan (e+f x))^{n+1} \left (\frac {\sin (e+f x)}{a}\right )^{m-n} \int \cos (e+f x)^{-n} \left (\frac {\sin (e+f x)}{a}\right )^{n-m}dx}{a b}\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\cos ^2(e+f x)^{\frac {n+1}{2}} (a \csc (e+f x))^m (b \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1),\frac {1}{2} (-m+n+3),\sin ^2(e+f x)\right )}{b f (-m+n+1)}\) |
((Cos[e + f*x]^2)^((1 + n)/2)*(a*Csc[e + f*x])^m*Hypergeometric2F1[(1 + n) /2, (1 - m + n)/2, (3 - m + n)/2, Sin[e + f*x]^2]*(b*Tan[e + f*x])^(1 + n) )/(b*f*(1 - m + n))
3.4.87.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* (a*Sin[e + f*x])^(n + 1))) 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]
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(a*Csc[e + f*x])^FracPart[m]*(Sin[e + f*x]/a)^FracPar t[m] Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
\[\int \left (a \csc \left (f x +e \right )\right )^{m} \left (b \tan \left (f x +e \right )\right )^{n}d x\]
\[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \csc \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\int \left (a \csc {\left (e + f x \right )}\right )^{m} \left (b \tan {\left (e + f x \right )}\right )^{n}\, dx \]
\[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \csc \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\int { \left (a \csc \left (f x + e\right )\right )^{m} \left (b \tan \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=\int {\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (\frac {a}{\sin \left (e+f\,x\right )}\right )}^m \,d x \]