Integrand size = 14, antiderivative size = 183 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=\frac {\cot (c+d x)}{3 b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b^2 d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot ^5(c+d x)}{7 b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^7(c+d x)}{9 b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b^2 d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b^2 \sqrt {b \tan ^4(c+d x)}} \]
1/3*cot(d*x+c)/b^2/d/(tan(d*x+c)^4*b)^(1/2)-1/5*cot(d*x+c)^3/b^2/d/(tan(d* x+c)^4*b)^(1/2)+1/7*cot(d*x+c)^5/b^2/d/(tan(d*x+c)^4*b)^(1/2)-1/9*cot(d*x+ c)^7/b^2/d/(tan(d*x+c)^4*b)^(1/2)-tan(d*x+c)/b^2/d/(tan(d*x+c)^4*b)^(1/2)- x*tan(d*x+c)^2/b^2/(tan(d*x+c)^4*b)^(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 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {9}{2},1,-\frac {7}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{9 d \left (b \tan ^4(c+d x)\right )^{5/2}} \]
-1/9*(Hypergeometric2F1[-9/2, 1, -7/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(d*( b*Tan[c + d*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.
| \(\displaystyle \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (b \tan (c+d x)^4\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4141 |
| \(\displaystyle \frac {\tan ^2(c+d x) \int \cot ^{10}(c+d x)dx}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(c+d x) \int \tan \left (c+d x+\frac {\pi }{2}\right )^{10}dx}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \cot ^8(c+d x)dx-\frac {\cot ^9(c+d x)}{9 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \tan \left (c+d x+\frac {\pi }{2}\right )^8dx-\frac {\cot ^9(c+d x)}{9 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (\int \cot ^6(c+d x)dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (\int \tan \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \cot ^4(c+d x)dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\int \tan \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (\int \cot ^2(c+d x)dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (\int \tan \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\int 1dx-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}-\frac {\cot (c+d x)}{d}\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\tan ^2(c+d x) \left (-\frac {\cot ^9(c+d x)}{9 d}+\frac {\cot ^7(c+d x)}{7 d}-\frac {\cot ^5(c+d x)}{5 d}+\frac {\cot ^3(c+d x)}{3 d}-\frac {\cot (c+d x)}{d}-x\right )}{b^2 \sqrt {b \tan ^4(c+d x)}}\) |
((-x - Cot[c + d*x]/d + Cot[c + d*x]^3/(3*d) - Cot[c + d*x]^5/(5*d) + Cot[ c + d*x]^7/(7*d) - Cot[c + d*x]^9/(9*d))*Tan[c + d*x]^2)/(b^2*Sqrt[b*Tan[c + d*x]^4])
3.1.41.3.1 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 = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45
| method | result | size |
| derivativedivides | \(-\frac {\tan \left (d x +c \right ) \left (315 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{9}\left (d x +c \right )\right )+315 \left (\tan ^{8}\left (d x +c \right )\right )-105 \left (\tan ^{6}\left (d x +c \right )\right )+63 \left (\tan ^{4}\left (d x +c \right )\right )-45 \left (\tan ^{2}\left (d x +c \right )\right )+35\right )}{315 d {\left (\left (\tan ^{4}\left (d x +c \right )\right ) b \right )}^{\frac {5}{2}}}\) | \(83\) |
| default | \(-\frac {\tan \left (d x +c \right ) \left (315 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{9}\left (d x +c \right )\right )+315 \left (\tan ^{8}\left (d x +c \right )\right )-105 \left (\tan ^{6}\left (d x +c \right )\right )+63 \left (\tan ^{4}\left (d x +c \right )\right )-45 \left (\tan ^{2}\left (d x +c \right )\right )+35\right )}{315 d {\left (\left (\tan ^{4}\left (d x +c \right )\right ) b \right )}^{\frac {5}{2}}}\) | \(83\) |
| risch | \(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} x}{b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} b}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}}+\frac {2 i \left (1575 \,{\mathrm e}^{16 i \left (d x +c \right )}-6300 \,{\mathrm e}^{14 i \left (d x +c \right )}+21000 \,{\mathrm e}^{12 i \left (d x +c \right )}-31500 \,{\mathrm e}^{10 i \left (d x +c \right )}+39438 \,{\mathrm e}^{8 i \left (d x +c \right )}-26292 \,{\mathrm e}^{6 i \left (d x +c \right )}+13968 \,{\mathrm e}^{4 i \left (d x +c \right )}-3492 \,{\mathrm e}^{2 i \left (d x +c \right )}+563\right )}{315 b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} b}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, d}\) | \(218\) |
-1/315/d*tan(d*x+c)*(315*arctan(tan(d*x+c))*tan(d*x+c)^9+315*tan(d*x+c)^8- 105*tan(d*x+c)^6+63*tan(d*x+c)^4-45*tan(d*x+c)^2+35)/(tan(d*x+c)^4*b)^(5/2 )
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=-\frac {{\left (315 \, d x \tan \left (d x + c\right )^{9} + 315 \, \tan \left (d x + c\right )^{8} - 105 \, \tan \left (d x + c\right )^{6} + 63 \, \tan \left (d x + c\right )^{4} - 45 \, \tan \left (d x + c\right )^{2} + 35\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{315 \, b^{3} d \tan \left (d x + c\right )^{11}} \]
-1/315*(315*d*x*tan(d*x + c)^9 + 315*tan(d*x + c)^8 - 105*tan(d*x + c)^6 + 63*tan(d*x + c)^4 - 45*tan(d*x + c)^2 + 35)*sqrt(b*tan(d*x + c)^4)/(b^3*d *tan(d*x + c)^11)
\[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.46 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {315 \, {\left (d x + c\right )}}{b^{\frac {5}{2}}} + \frac {315 \, \tan \left (d x + c\right )^{8} - 105 \, \tan \left (d x + c\right )^{6} + 63 \, \tan \left (d x + c\right )^{4} - 45 \, \tan \left (d x + c\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (d x + c\right )^{9}}}{315 \, d} \]
-1/315*(315*(d*x + c)/b^(5/2) + (315*tan(d*x + c)^8 - 105*tan(d*x + c)^6 + 63*tan(d*x + c)^4 - 45*tan(d*x + c)^2 + 35)/(b^(5/2)*tan(d*x + c)^9))/d
Time = 1.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {161280 \, {\left (d x + c\right )}}{b^{\frac {5}{2}}} + \frac {121590 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 18480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3528 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 495 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35}{b^{\frac {5}{2}} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}} - \frac {35 \, b^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 495 \, b^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3528 \, b^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 18480 \, b^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 121590 \, b^{20} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{b^{\frac {45}{2}}}}{161280 \, d} \]
-1/161280*(161280*(d*x + c)/b^(5/2) + (121590*tan(1/2*d*x + 1/2*c)^8 - 184 80*tan(1/2*d*x + 1/2*c)^6 + 3528*tan(1/2*d*x + 1/2*c)^4 - 495*tan(1/2*d*x + 1/2*c)^2 + 35)/(b^(5/2)*tan(1/2*d*x + 1/2*c)^9) - (35*b^20*tan(1/2*d*x + 1/2*c)^9 - 495*b^20*tan(1/2*d*x + 1/2*c)^7 + 3528*b^20*tan(1/2*d*x + 1/2* c)^5 - 18480*b^20*tan(1/2*d*x + 1/2*c)^3 + 121590*b^20*tan(1/2*d*x + 1/2*c ))/b^(45/2))/d
Timed out. \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{5/2}} \,d x \]