Integrand size = 14, antiderivative size = 119 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=\frac {\cot (e+f x)}{3 b f \sqrt {b \tan ^4(e+f x)}}-\frac {\cot ^3(e+f x)}{5 b f \sqrt {b \tan ^4(e+f x)}}-\frac {\tan (e+f x)}{b f \sqrt {b \tan ^4(e+f x)}}-\frac {x \tan ^2(e+f x)}{b \sqrt {b \tan ^4(e+f x)}} \]
1/3*cot(f*x+e)/b/f/(b*tan(f*x+e)^4)^(1/2)-1/5*cot(f*x+e)^3/b/f/(b*tan(f*x+ e)^4)^(1/2)-tan(f*x+e)/b/f/(b*tan(f*x+e)^4)^(1/2)-x*tan(f*x+e)^2/b/(b*tan( f*x+e)^4)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{5 f \left (b \tan ^4(e+f x)\right )^{3/2}} \]
-1/5*(Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(f*( b*Tan[e + f*x]^4)^(3/2))
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.60, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4141, 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.
\(\displaystyle \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \tan (e+f x)^4\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\tan ^2(e+f x) \int \cot ^6(e+f x)dx}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(e+f x) \int \tan \left (e+f x+\frac {\pi }{2}\right )^6dx}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \cot ^4(e+f x)dx-\frac {\cot ^5(e+f x)}{5 f}\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (-\int \tan \left (e+f x+\frac {\pi }{2}\right )^4dx-\frac {\cot ^5(e+f x)}{5 f}\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (\int \cot ^2(e+f x)dx-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (\int \tan \left (e+f x+\frac {\pi }{2}\right )^2dx-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (-\int 1dx-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}-\frac {\cot (e+f x)}{f}\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\tan ^2(e+f x) \left (-\frac {\cot ^5(e+f x)}{5 f}+\frac {\cot ^3(e+f x)}{3 f}-\frac {\cot (e+f x)}{f}-x\right )}{b \sqrt {b \tan ^4(e+f x)}}\) |
((-x - Cot[e + f*x]/f + Cot[e + f*x]^3/(3*f) - Cot[e + f*x]^5/(5*f))*Tan[e + f*x]^2)/(b*Sqrt[b*Tan[e + f*x]^4])
3.1.17.3.1 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 = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.53
method | result | size |
derivativedivides | \(-\frac {\tan \left (f x +e \right ) \left (15 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{5}+15 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+3\right )}{15 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {3}{2}}}\) | \(63\) |
default | \(-\frac {\tan \left (f x +e \right ) \left (15 \arctan \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{5}+15 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+3\right )}{15 f \left (b \tan \left (f x +e \right )^{4}\right )^{\frac {3}{2}}}\) | \(63\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} x}{b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}}+\frac {2 i \left (45 \,{\mathrm e}^{8 i \left (f x +e \right )}-90 \,{\mathrm e}^{6 i \left (f x +e \right )}+140 \,{\mathrm e}^{4 i \left (f x +e \right )}-70 \,{\mathrm e}^{2 i \left (f x +e \right )}+23\right )}{15 b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} \sqrt {\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}}\, f}\) | \(174\) |
-1/15/f*tan(f*x+e)*(15*arctan(tan(f*x+e))*tan(f*x+e)^5+15*tan(f*x+e)^4-5*t an(f*x+e)^2+3)/(b*tan(f*x+e)^4)^(3/2)
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (15 \, f x \tan \left (f x + e\right )^{5} + 15 \, \tan \left (f x + e\right )^{4} - 5 \, \tan \left (f x + e\right )^{2} + 3\right )} \sqrt {b \tan \left (f x + e\right )^{4}}}{15 \, b^{2} f \tan \left (f x + e\right )^{7}} \]
-1/15*(15*f*x*tan(f*x + e)^5 + 15*tan(f*x + e)^4 - 5*tan(f*x + e)^2 + 3)*s qrt(b*tan(f*x + e)^4)/(b^2*f*tan(f*x + e)^7)
\[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, {\left (f x + e\right )}}{b^{\frac {3}{2}}} + \frac {15 \, \tan \left (f x + e\right )^{4} - 5 \, \tan \left (f x + e\right )^{2} + 3}{b^{\frac {3}{2}} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
-1/15*(15*(f*x + e)/b^(3/2) + (15*tan(f*x + e)^4 - 5*tan(f*x + e)^2 + 3)/( b^(3/2)*tan(f*x + e)^5))/f
Time = 0.51 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {480 \, {\left (f x + e\right )}}{\sqrt {b}} - \frac {3 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 330 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{b^{\frac {5}{2}}} + \frac {330 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 35 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3}{\sqrt {b} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{480 \, b f} \]
-1/480*(480*(f*x + e)/sqrt(b) - (3*b^2*tan(1/2*f*x + 1/2*e)^5 - 35*b^2*tan (1/2*f*x + 1/2*e)^3 + 330*b^2*tan(1/2*f*x + 1/2*e))/b^(5/2) + (330*tan(1/2 *f*x + 1/2*e)^4 - 35*tan(1/2*f*x + 1/2*e)^2 + 3)/(sqrt(b)*tan(1/2*f*x + 1/ 2*e)^5))/(b*f)
Timed out. \[ \int \frac {1}{\left (b \tan ^4(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}^{3/2}} \,d x \]