Integrand size = 14, antiderivative size = 183 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\frac {\cot (e+f x)}{3 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b^2 f \sqrt {b \tan ^4(e+f x)}}+\frac {\cot ^5(e+f x)}{7 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^7(e+f x)}{9 b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b^2 f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b^2 \sqrt {b \tan ^4(e+f x)}} \]
1/3*cot(f*x+e)/b^2/f/(b*tan(f*x+e)^4)^(1/2)-1/5*cot(f*x+e)^3/b^2/f/(b*tan( f*x+e)^4)^(1/2)+1/7*cot(f*x+e)^5/b^2/f/(b*tan(f*x+e)^4)^(1/2)-1/9*cot(f*x+ e)^7/b^2/f/(b*tan(f*x+e)^4)^(1/2)-tan(f*x+e)/b^2/f/(b*tan(f*x+e)^4)^(1/2)- x*tan(f*x+e)^2/b^2/(b*tan(f*x+e)^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.25 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {9}{2},1,-\frac {7}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{9 f \left (b \tan ^4(e+f x)\right )^{5/2}} \]
-1/9*(Hypergeometric2F1[-9/2, 1, -7/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*( b*Tan[e + f*x]^4)^(5/2))
Time = 0.56 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.55, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 3042, 3954, 3042, 3954, 3042, 3954, 24}
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.
| \(\displaystyle \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (b \tan (e+f x)^4\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4141 |
| \(\displaystyle \frac {\tan ^2(e+f x) \int \cot ^{10}(e+f x)dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(e+f x) \int \tan \left (e+f x+\frac {\pi }{2}\right )^{10}dx}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \cot ^8(e+f x)dx-\frac {\cot ^9(e+f x)}{9 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \tan \left (e+f x+\frac {\pi }{2}\right )^8dx-\frac {\cot ^9(e+f x)}{9 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (\int \cot ^6(e+f x)dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (\int \tan \left (e+f x+\frac {\pi }{2}\right )^6dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \cot ^4(e+f x)dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \tan \left (e+f x+\frac {\pi }{2}\right )^4dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (\int \cot ^2(e+f x)dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (\int \tan \left (e+f x+\frac {\pi }{2}\right )^2dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\int 1dx-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}-\frac {\cot (e+f x)}{f}\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\tan ^2(e+f x) \left (-\frac {\cot ^9(e+f x)}{9 f}+\frac {\cot ^7(e+f x)}{7 f}-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}-\frac {\cot (e+f x)}{f}-x\right )}{b^2 \sqrt {b \tan ^4(e+f x)}}\) |
((-x - Cot[e + f*x]/f + Cot[e + f*x]^3/(3*f) - Cot[e + f*x]^5/(5*f) + Cot[ e + f*x]^7/(7*f) - Cot[e + f*x]^9/(9*f))*Tan[e + f*x]^2)/(b^2*Sqrt[b*Tan[e + f*x]^4])
3.1.18.3.1 Defintions of rubi rules used
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Ta n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45
| method | result | size |
| derivativedivides | \(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{9}+315 \tan \left (f x +e \right )^{8}-105 \tan \left (f x +e \right )^{6}+63 \tan \left (f x +e \right )^{4}-45 \tan \left (f x +e \right )^{2}+35\right )}{315 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}}}\) | \(83\) |
| default | \(-\frac {\tan \left (f x +e \right ) \left (315 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{9}+315 \tan \left (f x +e \right )^{8}-105 \tan \left (f x +e \right )^{6}+63 \tan \left (f x +e \right )^{4}-45 \tan \left (f x +e \right )^{2}+35\right )}{315 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {5}{2}}}\) | \(83\) |
| risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}}+\frac {2 i \left (1575 \,{\mathrm e}^{16 i \left (f x +e \right )}-6300 \,{\mathrm e}^{14 i \left (f x +e \right )}+21000 \,{\mathrm e}^{12 i \left (f x +e \right )}-31500 \,{\mathrm e}^{10 i \left (f x +e \right )}+39438 \,{\mathrm e}^{8 i \left (f x +e \right )}-26292 \,{\mathrm e}^{6 i \left (f x +e \right )}+13968 \,{\mathrm e}^{4 i \left (f x +e \right )}-3492 \,{\mathrm e}^{2 i \left (f x +e \right )}+563\right )}{315 b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{7} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, f}\) | \(218\) |
-1/315/f*tan(f*x+e)*(315*arctan(tan(f*x+e))*tan(f*x+e)^9+315*tan(f*x+e)^8- 105*tan(f*x+e)^6+63*tan(f*x+e)^4-45*tan(f*x+e)^2+35)/(b*tan(f*x+e)^4)^(5/2 )
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (315 \, f x \tan \left (f x + e\right )^{9} + 315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{315 \, b^{3} f \tan \left (f x + e\right )^{11}} \]
-1/315*(315*f*x*tan(f*x + e)^9 + 315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)^2 + 35)*sqrt(b*tan(f*x + e)^4)/(b^3*f *tan(f*x + e)^11)
\[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {315 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (f x + e\right )^{8} - 105 \, \tan \left (f x + e\right )^{6} + 63 \, \tan \left (f x + e\right )^{4} - 45 \, \tan \left (f x + e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (f x + e\right )^{9}}}{315 \, f} \]
-1/315*(315*(f*x + e)/b^(5/2) + (315*tan(f*x + e)^8 - 105*tan(f*x + e)^6 + 63*tan(f*x + e)^4 - 45*tan(f*x + e)^2 + 35)/(b^(5/2)*tan(f*x + e)^9))/f
Time = 1.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {161280 \, {\left (f x + e\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 18480 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3528 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 495 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} - \frac {35 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 495 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3528 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18480 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 121590 \, b^{20} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{b^{\frac {45}{2}}}}{161280 \, f} \]
-1/161280*(161280*(f*x + e)/b^(5/2) + (121590*tan(1/2*f*x + 1/2*e)^8 - 184 80*tan(1/2*f*x + 1/2*e)^6 + 3528*tan(1/2*f*x + 1/2*e)^4 - 495*tan(1/2*f*x + 1/2*e)^2 + 35)/(b^(5/2)*tan(1/2*f*x + 1/2*e)^9) - (35*b^20*tan(1/2*f*x + 1/2*e)^9 - 495*b^20*tan(1/2*f*x + 1/2*e)^7 + 3528*b^20*tan(1/2*f*x + 1/2* e)^5 - 18480*b^20*tan(1/2*f*x + 1/2*e)^3 + 121590*b^20*tan(1/2*f*x + 1/2*e ))/b^(45/2))/f
Timed out. \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{5/2}} \,d x \]