Integrand size = 17, antiderivative size = 67 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=\frac {\cot (c+d x)}{2 a d \sqrt {-a \tan ^2(c+d x)}}+\frac {\log (\sin (c+d x)) \tan (c+d x)}{a d \sqrt {-a \tan ^2(c+d x)}} \]
1/2*cot(d*x+c)/a/d/(-a*tan(d*x+c)^2)^(1/2)+ln(sin(d*x+c))*tan(d*x+c)/a/d/( -a*tan(d*x+c)^2)^(1/2)
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right ) \tan ^3(c+d x)}{2 d \left (-a \tan ^2(c+d x)\right )^{3/2}} \]
-1/2*((Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]])*Tan[c + d*x]^3)/(d*(-(a*Tan[c + d*x]^2))^(3/2))
Time = 0.37 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.647, Rules used = {3042, 4609, 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.
\(\displaystyle \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a-a \sec (c+d x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4609 |
\(\displaystyle \int \frac {1}{\left (-a \tan ^2(c+d x)\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (-a \tan (c+d x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle -\frac {\tan (c+d x) \int \cot ^3(c+d x)dx}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tan (c+d x) \int -\tan \left (c+d x+\frac {\pi }{2}\right )^3dx}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle \frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}-\int -\cot (c+d x)dx\right )}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \left (\int \cot (c+d x)dx+\frac {\cot ^2(c+d x)}{2 d}\right )}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \left (\int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\cot ^2(c+d x)}{2 d}\right )}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}-\int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\right )}{a \sqrt {-a \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}+\frac {\log (-\sin (c+d x))}{d}\right )}{a \sqrt {-a \tan ^2(c+d x)}}\) |
3.3.25.3.1 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]])
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(b*tan[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Time = 0.83 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.30
method | result | size |
default | \(\frac {4 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \tan \left (d x +c \right )-4 \ln \left (\frac {2}{\cos \left (d x +c \right )+1}\right ) \tan \left (d x +c \right )+\cot \left (d x +c \right )+\sec \left (d x +c \right ) \csc \left (d x +c \right )}{4 d \sqrt {-a \tan \left (d x +c \right )^{2}}\, a}\) | \(87\) |
risch | \(-\frac {i {\mathrm e}^{4 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )+{\mathrm e}^{4 i \left (d x +c \right )} d x -2 i {\mathrm e}^{2 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )+2 \,{\mathrm e}^{4 i \left (d x +c \right )} c -2 \,{\mathrm e}^{2 i \left (d x +c \right )} d x -2 i {\mathrm e}^{2 i \left (d x +c \right )}+i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )-4 \,{\mathrm e}^{2 i \left (d x +c \right )} c +d x +2 c}{a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\) | \(196\) |
1/4/d/(-a*tan(d*x+c)^2)^(1/2)/a*(4*ln(-cot(d*x+c)+csc(d*x+c))*tan(d*x+c)-4 *ln(2/(cos(d*x+c)+1))*tan(d*x+c)+cot(d*x+c)+sec(d*x+c)*csc(d*x+c))
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=-\frac {{\left (2 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right )^{2} - a}{\cos \left (d x + c\right )^{2}}}}{2 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )} \]
-1/2*(2*(cos(d*x + c)^3 - cos(d*x + c))*log(1/2*sin(d*x + c)) - cos(d*x + c))*sqrt((a*cos(d*x + c)^2 - a)/cos(d*x + c)^2)/((a^2*d*cos(d*x + c)^2 - a ^2*d)*sin(d*x + c))
\[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (- a \sec ^{2}{\left (c + d x \right )} + a\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{\sqrt {-a} a} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{\sqrt {-a} a} + \frac {\sqrt {-a}}{a^{2} \tan \left (d x + c\right )^{2}}}{2 \, d} \]
-1/2*(log(tan(d*x + c)^2 + 1)/(sqrt(-a)*a) - 2*log(tan(d*x + c))/(sqrt(-a) *a) + sqrt(-a)/(a^2*tan(d*x + c)^2))/d
Time = 0.43 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\sqrt {-a} a} - \frac {8 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}{\sqrt {-a} a} + \frac {4 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )}{\sqrt {-a} a} - \frac {4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}{\sqrt {-a} a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]
-1/8*(tan(1/2*d*x + 1/2*c)^2/(sqrt(-a)*a) - 8*log(tan(1/2*d*x + 1/2*c)^2 + 1)/(sqrt(-a)*a) + 4*log(tan(1/2*d*x + 1/2*c)^2)/(sqrt(-a)*a) - (4*tan(1/2 *d*x + 1/2*c)^2 - 1)/(sqrt(-a)*a*tan(1/2*d*x + 1/2*c)^2))/d
Timed out. \[ \int \frac {1}{\left (a-a \sec ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (a-\frac {a}{{\cos \left (c+d\,x\right )}^2}\right )}^{3/2}} \,d x \]