Integrand size = 14, antiderivative size = 66 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\cot (c+d x)}{2 b d \sqrt {b \tan ^2(c+d x)}}-\frac {\log (\sin (c+d x)) \tan (c+d x)}{b d \sqrt {b \tan ^2(c+d x)}} \] Output:
-1/2*cot(d*x+c)/b/d/(b*tan(d*x+c)^2)^(1/2)-ln(sin(d*x+c))*tan(d*x+c)/b/d/( b*tan(d*x+c)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\left (\csc ^2(c+d x)+2 \log (\sin (c+d x))\right ) \tan ^3(c+d x)}{2 d \left (b \tan ^2(c+d x)\right )^{3/2}} \] Input:
Integrate[(b*Tan[c + d*x]^2)^(-3/2),x]
Output:
-1/2*((Csc[c + d*x]^2 + 2*Log[Sin[c + d*x]])*Tan[c + d*x]^3)/(d*(b*Tan[c + d*x]^2)^(3/2))
Time = 0.31 (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.
\(\displaystyle \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (b \tan (c+d x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle \frac {\tan (c+d x) \int \cot ^3(c+d x)dx}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x) \int -\tan \left (c+d x+\frac {\pi }{2}\right )^3dx}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (c+d x) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )^3dx}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}-\int -\cot (c+d x)dx\right )}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (c+d x) \left (\int \cot (c+d x)dx+\frac {\cot ^2(c+d x)}{2 d}\right )}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tan (c+d x) \left (\int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\cot ^2(c+d x)}{2 d}\right )}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}-\int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx\right )}{b \sqrt {b \tan ^2(c+d x)}}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {\tan (c+d x) \left (\frac {\cot ^2(c+d x)}{2 d}+\frac {\log (-\sin (c+d x))}{d}\right )}{b \sqrt {b \tan ^2(c+d x)}}\) |
Input:
Int[(b*Tan[c + d*x]^2)^(-3/2),x]
Output:
-(((Cot[c + d*x]^2/(2*d) + Log[-Sin[c + d*x]]/d)*Tan[c + d*x])/(b*Sqrt[b*T an[c + d*x]^2]))
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 = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(-\frac {\tan \left (d x +c \right ) \left (2 \ln \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2}-\ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2}+1\right )}{2 d \left (b \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}\) | \(64\) |
default | \(-\frac {\tan \left (d x +c \right ) \left (2 \ln \left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2}-\ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2}+1\right )}{2 d \left (b \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}\) | \(64\) |
risch | \(\frac {i {\mathrm e}^{4 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )+{\mathrm e}^{4 i \left (d x +c \right )} d x +2 \,{\mathrm e}^{4 i \left (d x +c \right )} c -2 i {\mathrm e}^{2 i \left (d x +c \right )} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )-2 \,{\mathrm e}^{2 i \left (d x +c \right )} d x -2 i {\mathrm e}^{2 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )} c +i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )+d x +2 c}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {-\frac {b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\) | \(196\) |
Input:
int(1/(b*tan(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/d*tan(d*x+c)*(2*ln(tan(d*x+c))*tan(d*x+c)^2-ln(1+tan(d*x+c)^2)*tan(d* x+c)^2+1)/(b*tan(d*x+c)^2)^(3/2)
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\sqrt {b \tan \left (d x + c\right )^{2}} {\left (\log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + \tan \left (d x + c\right )^{2} + 1\right )}}{2 \, b^{2} d \tan \left (d x + c\right )^{3}} \] Input:
integrate(1/(b*tan(d*x+c)^2)^(3/2),x, algorithm="fricas")
Output:
-1/2*sqrt(b*tan(d*x + c)^2)*(log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan( d*x + c)^2 + tan(d*x + c)^2 + 1)/(b^2*d*tan(d*x + c)^3)
\[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \tan ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b*tan(d*x+c)**2)**(3/2),x)
Output:
Integral((b*tan(c + d*x)**2)**(-3/2), x)
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=\frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{b^{\frac {3}{2}}} - \frac {2 \, \log \left (\tan \left (d x + c\right )\right )}{b^{\frac {3}{2}}} - \frac {1}{b^{\frac {3}{2}} \tan \left (d x + c\right )^{2}}}{2 \, d} \] Input:
integrate(1/(b*tan(d*x+c)^2)^(3/2),x, algorithm="maxima")
Output:
1/2*(log(tan(d*x + c)^2 + 1)/b^(3/2) - 2*log(tan(d*x + c))/b^(3/2) - 1/(b^ (3/2)*tan(d*x + c)^2))/d
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=\frac {\frac {\log \left (\tan \left (d x + c\right )^{2} + 1\right )}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {\log \left (\tan \left (d x + c\right )^{2}\right )}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} + \frac {\tan \left (d x + c\right )^{2} - 1}{b d \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \tan \left (d x + c\right )^{2}}}{2 \, \sqrt {b}} \] Input:
integrate(1/(b*tan(d*x+c)^2)^(3/2),x, algorithm="giac")
Output:
1/2*(log(tan(d*x + c)^2 + 1)/(b*d*sgn(tan(d*x + c))) - log(tan(d*x + c)^2) /(b*d*sgn(tan(d*x + c))) + (tan(d*x + c)^2 - 1)/(b*d*sgn(tan(d*x + c))*tan (d*x + c)^2))/sqrt(b)
Timed out. \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}^{3/2}} \,d x \] Input:
int(1/(b*tan(c + d*x)^2)^(3/2),x)
Output:
int(1/(b*tan(c + d*x)^2)^(3/2), x)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (b \tan ^2(c+d x)\right )^{3/2}} \, dx=\frac {\sqrt {b}\, \left (\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right )^{2}-2 \,\mathrm {log}\left (\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2}-1\right )}{2 \tan \left (d x +c \right )^{2} b^{2} d} \] Input:
int(1/(b*tan(d*x+c)^2)^(3/2),x)
Output:
(sqrt(b)*(log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2 - 2*log(tan(c + d*x))*t an(c + d*x)**2 - 1))/(2*tan(c + d*x)**2*b**2*d)