Integrand size = 14, antiderivative size = 60 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\log (\sinh (c+d x)) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}} \] Output:
1/2*coth(d*x+c)/d/(-tanh(d*x+c)^2)^(1/2)-ln(sinh(d*x+c))*tanh(d*x+c)/d/(-t anh(d*x+c)^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=-\frac {\left (\text {csch}^2(c+d x)-2 \log (\sinh (c+d x))\right ) \tanh ^3(c+d x)}{2 d \left (-\tanh ^2(c+d x)\right )^{3/2}} \] Input:
Integrate[(-Tanh[c + d*x]^2)^(-3/2),x]
Output:
-1/2*((Csch[c + d*x]^2 - 2*Log[Sinh[c + d*x]])*Tanh[c + d*x]^3)/(d*(-Tanh[ c + d*x]^2)^(3/2))
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, 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, 26, 3954, 26, 3042, 26, 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.
\(\displaystyle \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\tan (i c+i d x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4141 |
\(\displaystyle -\frac {\tanh (c+d x) \int \coth ^3(c+d x)dx}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh (c+d x) \int i \tan \left (i c+i d x+\frac {\pi }{2}\right )^3dx}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {i \tanh (c+d x) \int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3dx}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 3954 |
\(\displaystyle -\frac {i \tanh (c+d x) \left (\frac {i \coth ^2(c+d x)}{2 d}-\int i \coth (c+d x)dx\right )}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {i \tanh (c+d x) \left (\frac {i \coth ^2(c+d x)}{2 d}-i \int \coth (c+d x)dx\right )}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \tanh (c+d x) \left (\frac {i \coth ^2(c+d x)}{2 d}-i \int -i \tan \left (i c+i d x+\frac {\pi }{2}\right )dx\right )}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -\frac {i \tanh (c+d x) \left (\frac {i \coth ^2(c+d x)}{2 d}-\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )dx\right )}{\sqrt {-\tanh ^2(c+d x)}}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {i \tanh (c+d x) \left (\frac {i \coth ^2(c+d x)}{2 d}-\frac {i \log (-i \sinh (c+d x))}{d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\) |
Input:
Int[(-Tanh[c + d*x]^2)^(-3/2),x]
Output:
((-I)*(((I/2)*Coth[c + d*x]^2)/d - (I*Log[(-I)*Sinh[c + d*x]])/d)*Tanh[c + d*x])/Sqrt[-Tanh[c + d*x]^2]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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 = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2}-\ln \left (\tanh \left (d x +c \right )-1\right ) \tanh \left (d x +c \right )^{2}-\ln \left (\tanh \left (d x +c \right )+1\right ) \tanh \left (d x +c \right )^{2}-1\right )}{2 d \left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}\) | \(81\) |
default | \(\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2}-\ln \left (\tanh \left (d x +c \right )-1\right ) \tanh \left (d x +c \right )^{2}-\ln \left (\tanh \left (d x +c \right )+1\right ) \tanh \left (d x +c \right )^{2}-1\right )}{2 d \left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}\) | \(81\) |
risch | \(-\frac {-{\mathrm e}^{4 d x +4 c} d x +{\mathrm e}^{4 d x +4 c} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )-2 \,{\mathrm e}^{4 d x +4 c} c +2 \,{\mathrm e}^{2 d x +2 c} d x -2 \,{\mathrm e}^{2 d x +2 c} \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )+4 \,{\mathrm e}^{2 d x +2 c} c -d x -2 \,{\mathrm e}^{2 d x +2 c}+\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )-2 c}{\left ({\mathrm e}^{2 d x +2 c}-1\right ) \left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, d}\) | \(188\) |
Input:
int(1/(-tanh(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/d*tanh(d*x+c)*(2*ln(tanh(d*x+c))*tanh(d*x+c)^2-ln(tanh(d*x+c)-1)*tanh( d*x+c)^2-ln(tanh(d*x+c)+1)*tanh(d*x+c)^2-1)/(-tanh(d*x+c)^2)^(3/2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.67 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\frac {-i \, d x e^{\left (4 \, d x + 4 \, c\right )} - i \, d x - 2 \, {\left (-i \, d x + i\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (i \, e^{\left (4 \, d x + 4 \, c\right )} - 2 i \, e^{\left (2 \, d x + 2 \, c\right )} + i\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{d e^{\left (4 \, d x + 4 \, c\right )} - 2 \, d e^{\left (2 \, d x + 2 \, c\right )} + d} \] Input:
integrate(1/(-tanh(d*x+c)^2)^(3/2),x, algorithm="fricas")
Output:
(-I*d*x*e^(4*d*x + 4*c) - I*d*x - 2*(-I*d*x + I)*e^(2*d*x + 2*c) + (I*e^(4 *d*x + 4*c) - 2*I*e^(2*d*x + 2*c) + I)*log(e^(2*d*x + 2*c) - 1))/(d*e^(4*d *x + 4*c) - 2*d*e^(2*d*x + 2*c) + d)
\[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(-tanh(d*x+c)**2)**(3/2),x)
Output:
Integral((-tanh(c + d*x)**2)**(-3/2), x)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=-\frac {i \, {\left (d x + c\right )}}{d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {2 i \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \] Input:
integrate(1/(-tanh(d*x+c)^2)^(3/2),x, algorithm="maxima")
Output:
-I*(d*x + c)/d - I*log(e^(-d*x - c) + 1)/d - I*log(e^(-d*x - c) - 1)/d - 2 *I*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) - e^(-4*d*x - 4*c) - 1))
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.67 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\frac {\frac {i \, d x + i \, c}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} - \frac {i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} + \frac {2 i \, e^{\left (2 \, d x + 2 \, c\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}}{d} \] Input:
integrate(1/(-tanh(d*x+c)^2)^(3/2),x, algorithm="giac")
Output:
((I*d*x + I*c)/sgn(-e^(4*d*x + 4*c) + 1) - I*log(e^(2*d*x + 2*c) - 1)/sgn( -e^(4*d*x + 4*c) + 1) + 2*I*e^(2*d*x + 2*c)/((e^(2*d*x + 2*c) - 1)^2*sgn(- e^(4*d*x + 4*c) + 1)))/d
Timed out. \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (-{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}^{3/2}} \,d x \] Input:
int(1/(-tanh(c + d*x)^2)^(3/2),x)
Output:
int(1/(-tanh(c + d*x)^2)^(3/2), x)
Time = 0.25 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.05 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{3/2}} \, dx=\frac {i \left (e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right )+e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right )-e^{4 d x +4 c} d x -e^{4 d x +4 c}-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right )-2 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right )+2 e^{2 d x +2 c} d x +\mathrm {log}\left (e^{d x +c}-1\right )+\mathrm {log}\left (e^{d x +c}+1\right )-d x -1\right )}{d \left (e^{4 d x +4 c}-2 e^{2 d x +2 c}+1\right )} \] Input:
int(1/(-tanh(d*x+c)^2)^(3/2),x)
Output:
(i*(e**(4*c + 4*d*x)*log(e**(c + d*x) - 1) + e**(4*c + 4*d*x)*log(e**(c + d*x) + 1) - e**(4*c + 4*d*x)*d*x - e**(4*c + 4*d*x) - 2*e**(2*c + 2*d*x)*l og(e**(c + d*x) - 1) - 2*e**(2*c + 2*d*x)*log(e**(c + d*x) + 1) + 2*e**(2* c + 2*d*x)*d*x + log(e**(c + d*x) - 1) + log(e**(c + d*x) + 1) - d*x - 1)) /(d*(e**(4*c + 4*d*x) - 2*e**(2*c + 2*d*x) + 1))