Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cot (e+f x)}{2 b f \sqrt {b \tan ^2(e+f x)}}-\frac {\log (\sin (e+f x)) \tan (e+f x)}{b f \sqrt {b \tan ^2(e+f x)}} \]
-1/2*cot(f*x+e)/b/f/(b*tan(f*x+e)^2)^(1/2)-ln(sin(f*x+e))*tan(f*x+e)/b/f/( b*tan(f*x+e)^2)^(1/2)
Time = 0.41 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\left (\cot ^2(e+f x)+2 \log (\cos (e+f x))+2 \log (\tan (e+f x))\right ) \tan ^3(e+f x)}{2 f \left (b \tan ^2(e+f x)\right )^{3/2}} \]
-1/2*((Cot[e + f*x]^2 + 2*Log[Cos[e + f*x]] + 2*Log[Tan[e + f*x]])*Tan[e + f*x]^3)/(f*(b*Tan[e + f*x]^2)^(3/2))
Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82, 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, 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.
\(\displaystyle \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \tan (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\tan (e+f x) \int \cot ^3(e+f x)dx}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (e+f x) \int -\tan \left (e+f x+\frac {\pi }{2}\right )^3dx}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (e+f x) \int \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^3dx}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^2(e+f x)}{2 f}-\int -\cot (e+f x)dx\right )}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (e+f x) \left (\int \cot (e+f x)dx+\frac {\cot ^2(e+f x)}{2 f}\right )}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tan (e+f x) \left (\int -\tan \left (e+f x+\frac {\pi }{2}\right )dx+\frac {\cot ^2(e+f x)}{2 f}\right )}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^2(e+f x)}{2 f}-\int \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )dx\right )}{b \sqrt {b \tan ^2(e+f x)}}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {\tan (e+f x) \left (\frac {\cot ^2(e+f x)}{2 f}+\frac {\log (-\sin (e+f x))}{f}\right )}{b \sqrt {b \tan ^2(e+f x)}}\) |
3.1.5.3.1 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 = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(-\frac {\tan \left (f x +e \right ) \left (2 \ln \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{2}-\ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2}+1\right )}{2 f \left (b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(64\) |
default | \(-\frac {\tan \left (f x +e \right ) \left (2 \ln \left (\tan \left (f x +e \right )\right ) \tan \left (f x +e \right )^{2}-\ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2}+1\right )}{2 f \left (b \tan \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(64\) |
risch | \(\frac {i {\mathrm e}^{4 i \left (f x +e \right )} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )+{\mathrm e}^{4 i \left (f x +e \right )} f x +2 \,{\mathrm e}^{4 i \left (f x +e \right )} e -2 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )-2 \,{\mathrm e}^{2 i \left (f x +e \right )} f x -2 i {\mathrm e}^{2 i \left (f x +e \right )}-4 \,{\mathrm e}^{2 i \left (f x +e \right )} e +i \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )+f x +2 e}{b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}}\, f}\) | \(196\) |
-1/2/f*tan(f*x+e)*(2*ln(tan(f*x+e))*tan(f*x+e)^2-ln(1+tan(f*x+e)^2)*tan(f* x+e)^2+1)/(b*tan(f*x+e)^2)^(3/2)
Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\sqrt {b \tan \left (f x + e\right )^{2}} {\left (\log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + \tan \left (f x + e\right )^{2} + 1\right )}}{2 \, b^{2} f \tan \left (f x + e\right )^{3}} \]
-1/2*sqrt(b*tan(f*x + e)^2)*(log(tan(f*x + e)^2/(tan(f*x + e)^2 + 1))*tan( f*x + e)^2 + tan(f*x + e)^2 + 1)/(b^2*f*tan(f*x + e)^3)
\[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\log \left (\tan \left (f x + e\right )^{2} + 1\right )}{b^{\frac {3}{2}}} - \frac {2 \, \log \left (\tan \left (f x + e\right )\right )}{b^{\frac {3}{2}}} - \frac {1}{b^{\frac {3}{2}} \tan \left (f x + e\right )^{2}}}{2 \, f} \]
1/2*(log(tan(f*x + e)^2 + 1)/b^(3/2) - 2*log(tan(f*x + e))/b^(3/2) - 1/(b^ (3/2)*tan(f*x + e)^2))/f
Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (60) = 120\).
Time = 0.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.56 \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {{\left (\frac {4 \, {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}{\sqrt {b} {\left (\cos \left (f x + e\right ) - 1\right )} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} - \frac {4 \, \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{\sqrt {b} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {8 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{\sqrt {b} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )} + \frac {\cos \left (f x + e\right ) - 1}{\sqrt {b} {\left (\cos \left (f x + e\right ) + 1\right )} \mathrm {sgn}\left (\tan \left (f x + e\right )\right )}}{8 \, b f} \]
1/8*((4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1)*(cos(f*x + e) + 1)/(sqr t(b)*(cos(f*x + e) - 1)*sgn(tan(f*x + e))) - 4*log(abs(-cos(f*x + e) + 1)/ abs(cos(f*x + e) + 1))/(sqrt(b)*sgn(tan(f*x + e))) + 8*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/(sqrt(b)*sgn(tan(f*x + e))) + (cos(f*x + e) - 1)/(sqrt(b)*(cos(f*x + e) + 1)*sgn(tan(f*x + e))))/(b*f)
Timed out. \[ \int \frac {1}{\left (b \tan ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )}^{3/2}} \,d x \]