Integrand size = 14, antiderivative size = 97 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {\cot (e+f x)}{2 b^2 f \sqrt {b \tan ^2(e+f x)}}-\frac {\cot ^3(e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}}+\frac {\log (\sin (e+f x)) \tan (e+f x)}{b^2 f \sqrt {b \tan ^2(e+f x)}} \]
1/2*cot(f*x+e)/b^2/f/(b*tan(f*x+e)^2)^(1/2)-1/4*cot(f*x+e)^3/b^2/f/(b*tan( f*x+e)^2)^(1/2)+ln(sin(f*x+e))*tan(f*x+e)/b^2/f/(b*tan(f*x+e)^2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {2 \cot (e+f x)-\cot ^3(e+f x)+4 (\log (\cos (e+f x))+\log (\tan (e+f x))) \tan (e+f x)}{4 b^2 f \sqrt {b \tan ^2(e+f x)}} \]
(2*Cot[e + f*x] - Cot[e + f*x]^3 + 4*(Log[Cos[e + f*x]] + Log[Tan[e + f*x] ])*Tan[e + f*x])/(4*b^2*f*Sqrt[b*Tan[e + f*x]^2])
Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72, 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, 25, 3954, 25, 3042, 25, 3954, 25, 3042, 25, 3956}
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.
| \(\displaystyle \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (b \tan (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4141 |
| \(\displaystyle \frac {\tan (e+f x) \int \cot ^5(e+f x)dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (e+f x) \int -\tan \left (e+f x+\frac {\pi }{2}\right )^5dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tan (e+f x) \int \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^5dx}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^4(e+f x)}{4 f}-\int -\cot ^3(e+f x)dx\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\int \cot ^3(e+f x)dx+\frac {\cot ^4(e+f x)}{4 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\int -\tan \left (e+f x+\frac {\pi }{2}\right )^3dx+\frac {\cot ^4(e+f x)}{4 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^4(e+f x)}{4 f}-\int \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^3dx\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\int -\cot (e+f x)dx+\frac {\cot ^4(e+f x)}{4 f}-\frac {\cot ^2(e+f x)}{2 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tan (e+f x) \left (-\int \cot (e+f x)dx+\frac {\cot ^4(e+f x)}{4 f}-\frac {\cot ^2(e+f x)}{2 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\tan (e+f x) \left (-\int -\tan \left (e+f x+\frac {\pi }{2}\right )dx+\frac {\cot ^4(e+f x)}{4 f}-\frac {\cot ^2(e+f x)}{2 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\int \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )dx+\frac {\cot ^4(e+f x)}{4 f}-\frac {\cot ^2(e+f x)}{2 f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^4(e+f x)}{4 f}-\frac {\cot ^2(e+f x)}{2 f}-\frac {\log (-\sin (e+f x))}{f}\right )}{b^2 \sqrt {b \tan ^2(e+f x)}}\) |
-(((-1/2*Cot[e + f*x]^2/f + Cot[e + f*x]^4/(4*f) - Log[-Sin[e + f*x]]/f)*T an[e + f*x])/(b^2*Sqrt[b*Tan[e + f*x]^2]))
3.1.6.3.1 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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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 = 74, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {\tan \left (f x +e \right ) \left (4 \ln \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{4}-2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}-1\right )}{4 f \left (b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(74\) |
| default | \(\frac {\tan \left (f x +e \right ) \left (4 \ln \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{4}-2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{2}-1\right )}{4 f \left (b \tan \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(74\) |
| risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) x}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}}-\frac {2 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, f}+\frac {4 i \left ({\mathrm e}^{6 i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, f}-\frac {i \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, f}\) | \(302\) |
1/4/f*tan(f*x+e)*(4*ln(tan(f*x+e))*tan(f*x+e)^4-2*ln(1+tan(f*x+e)^2)*tan(f *x+e)^4+2*tan(f*x+e)^2-1)/(b*tan(f*x+e)^2)^(5/2)
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left (2 \, \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} - 1\right )} \sqrt {b \tan \left (f x + e\right )^{2}}}{4 \, b^{3} f \tan \left (f x + e\right )^{5}} \]
1/4*(2*log(tan(f*x + e)^2/(tan(f*x + e)^2 + 1))*tan(f*x + e)^4 + 3*tan(f*x + e)^4 + 2*tan(f*x + e)^2 - 1)*sqrt(b*tan(f*x + e)^2)/(b^3*f*tan(f*x + e) ^5)
\[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{\left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {2 \, \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac {5}{2}}} - \frac {4 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sqrt {b} \tan \left (f x + e\right )^{2} - \sqrt {b}}{b^{3} \tan \left (f x + e\right )^{4}}}{4 \, f} \]
-1/4*(2*log(tan(f*x + e)^2 + 1)/b^(5/2) - 4*log(tan(f*x + e))/b^(5/2) - (2 *sqrt(b)*tan(f*x + e)^2 - sqrt(b))/(b^3*tan(f*x + e)^4))/f
Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (87) = 174\).
Time = 0.38 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.32 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {{\left (\frac {12 \, {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {48 \, {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{b^{\frac {5}{2}} {\left (\cos \left (f x + e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} - \frac {32 \, \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{b^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {64 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{b^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {\frac {12 \, b^{\frac {5}{2}} {\left (\cos \left (f x + e\right ) - 1\right )} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )}{\cos \left (f x + e\right ) + 1} + \frac {b^{\frac {5}{2}} {\left (\cos \left (f x + e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{b^{5}}}{64 \, f} \]
-1/64*((12*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 48*(cos(f*x + e) - 1)^2 /(cos(f*x + e) + 1)^2 + 1)*(cos(f*x + e) + 1)^2/(b^(5/2)*(cos(f*x + e) - 1 )^2*sgn(tan(f*x + e))) - 32*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/(b^(5/2)*sgn(tan(f*x + e))) + 64*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/(b^(5/2)*sgn(tan(f*x + e))) + (12*b^(5/2)*(cos(f*x + e) - 1)*sgn(tan(f*x + e))/(cos(f*x + e) + 1) + b^(5/2)*(cos(f*x + e) - 1)^2*sgn (tan(f*x + e))/(cos(f*x + e) + 1)^2)/b^5)/f
Timed out. \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{5/2}} \,d x \]