\(\int \frac {1}{(-\tanh ^2(c+d x))^{5/2}} \, dx\) [32]

Optimal result

Integrand size = 14, antiderivative size = 88 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\coth (c+d x)}{2 d \sqrt {-\tanh ^2(c+d x)}}-\frac {\coth ^3(c+d x)}{4 d \sqrt {-\tanh ^2(c+d x)}}+\frac {\log (\sinh (c+d x)) \tanh (c+d x)}{d \sqrt {-\tanh ^2(c+d x)}} \] Output:

-1/2*coth(d*x+c)/d/(-tanh(d*x+c)^2)^(1/2)-1/4*coth(d*x+c)^3/d/(-tanh(d*x+c 
)^2)^(1/2)+ln(sinh(d*x+c))*tanh(d*x+c)/d/(-tanh(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\left (4 \text {csch}^2(c+d x)+\text {csch}^4(c+d x)-4 \log (\sinh (c+d x))\right ) \tanh ^5(c+d x)}{4 d \left (-\tanh ^2(c+d x)\right )^{5/2}} \] Input:

Integrate[(-Tanh[c + d*x]^2)^(-5/2),x]
 

Output:

-1/4*((4*Csch[c + d*x]^2 + Csch[c + d*x]^4 - 4*Log[Sinh[c + d*x]])*Tanh[c 
+ d*x]^5)/(d*(-Tanh[c + d*x]^2)^(5/2))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88, 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, 26, 3954, 26, 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.

\(\displaystyle \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (\tan (i c+i d x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle \frac {\tanh (c+d x) \int \coth ^5(c+d x)dx}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tanh (c+d x) \int -i \tan \left (i c+i d x+\frac {\pi }{2}\right )^5dx}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \tanh (c+d x) \int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^5dx}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3954

\(\displaystyle -\frac {i \tanh (c+d x) \left (-\int -i \coth ^3(c+d x)dx-\frac {i \coth ^4(c+d x)}{4 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \tanh (c+d x) \left (i \int \coth ^3(c+d x)dx-\frac {i \coth ^4(c+d x)}{4 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {i \tanh (c+d x) \left (i \int i \tan \left (i c+i d x+\frac {\pi }{2}\right )^3dx-\frac {i \coth ^4(c+d x)}{4 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \tanh (c+d x) \left (-\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^3dx-\frac {i \coth ^4(c+d x)}{4 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3954

\(\displaystyle -\frac {i \tanh (c+d x) \left (\int i \coth (c+d x)dx-\frac {i \coth ^4(c+d x)}{4 d}-\frac {i \coth ^2(c+d x)}{2 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \tanh (c+d x) \left (i \int \coth (c+d x)dx-\frac {i \coth ^4(c+d x)}{4 d}-\frac {i \coth ^2(c+d x)}{2 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {i \tanh (c+d x) \left (i \int -i \tan \left (i c+i d x+\frac {\pi }{2}\right )dx-\frac {i \coth ^4(c+d x)}{4 d}-\frac {i \coth ^2(c+d x)}{2 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {i \tanh (c+d x) \left (\int \tan \left (\frac {1}{2} (2 i c+\pi )+i d x\right )dx-\frac {i \coth ^4(c+d x)}{4 d}-\frac {i \coth ^2(c+d x)}{2 d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

\(\Big \downarrow \) 3956

\(\displaystyle -\frac {i \tanh (c+d x) \left (-\frac {i \coth ^4(c+d x)}{4 d}-\frac {i \coth ^2(c+d x)}{2 d}+\frac {i \log (-i \sinh (c+d x))}{d}\right )}{\sqrt {-\tanh ^2(c+d x)}}\)

Input:

Int[(-Tanh[c + d*x]^2)^(-5/2),x]
 

Output:

((-I)*(((-1/2*I)*Coth[c + d*x]^2)/d - ((I/4)*Coth[c + d*x]^4)/d + (I*Log[( 
-I)*Sinh[c + d*x]])/d)*Tanh[c + d*x])/Sqrt[-Tanh[c + d*x]^2]
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3954

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]
 

rule 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4141

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]])
 

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(-\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )+1\right ) \tanh \left (d x +c \right )^{4}-4 \ln \left (\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{4}+2 \ln \left (\tanh \left (d x +c \right )-1\right ) \tanh \left (d x +c \right )^{4}+2 \tanh \left (d x +c \right )^{2}+1\right )}{4 d \left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}\)	\(91\)
default	\(-\frac {\tanh \left (d x +c \right ) \left (2 \ln \left (\tanh \left (d x +c \right )+1\right ) \tanh \left (d x +c \right )^{4}-4 \ln \left (\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{4}+2 \ln \left (\tanh \left (d x +c \right )-1\right ) \tanh \left (d x +c \right )^{4}+2 \tanh \left (d x +c \right )^{2}+1\right )}{4 d \left (-\tanh \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}\)	\(91\)
risch	\(\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) x}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}+1\right )}-\frac {2 \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left (d x +c \right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}+1\right ) d}-\frac {4 \,{\mathrm e}^{2 d x +2 c} \left ({\mathrm e}^{4 d x +4 c}-{\mathrm e}^{2 d x +2 c}+1\right )}{\left ({\mathrm e}^{2 d x +2 c}-1\right )^{3} \left ({\mathrm e}^{2 d x +2 c}+1\right ) \sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, d}+\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right ) \ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 d x +2 c}-1\right )^{2}}{\left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}}\, \left ({\mathrm e}^{2 d x +2 c}+1\right ) d}\)	\(284\)

Input:

int(1/(-tanh(d*x+c)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/d*tanh(d*x+c)*(2*ln(tanh(d*x+c)+1)*tanh(d*x+c)^4-4*ln(tanh(d*x+c))*ta 
nh(d*x+c)^4+2*ln(tanh(d*x+c)-1)*tanh(d*x+c)^4+2*tanh(d*x+c)^2+1)/(-tanh(d* 
x+c)^2)^(5/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.05 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=\frac {i \, d x e^{\left (8 \, d x + 8 \, c\right )} + i \, d x - 4 \, {\left (i \, d x - i\right )} e^{\left (6 \, d x + 6 \, c\right )} - 2 \, {\left (-3 i \, d x + 2 i\right )} e^{\left (4 \, d x + 4 \, c\right )} - 4 \, {\left (i \, d x - i\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-i \, e^{\left (8 \, d x + 8 \, c\right )} + 4 i \, e^{\left (6 \, d x + 6 \, c\right )} - 6 i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 i \, e^{\left (2 \, d x + 2 \, c\right )} - i\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{d e^{\left (8 \, d x + 8 \, c\right )} - 4 \, d e^{\left (6 \, d x + 6 \, c\right )} + 6 \, d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, d e^{\left (2 \, d x + 2 \, c\right )} + d} \] Input:

integrate(1/(-tanh(d*x+c)^2)^(5/2),x, algorithm="fricas")
 

Output:

(I*d*x*e^(8*d*x + 8*c) + I*d*x - 4*(I*d*x - I)*e^(6*d*x + 6*c) - 2*(-3*I*d 
*x + 2*I)*e^(4*d*x + 4*c) - 4*(I*d*x - I)*e^(2*d*x + 2*c) + (-I*e^(8*d*x + 
 8*c) + 4*I*e^(6*d*x + 6*c) - 6*I*e^(4*d*x + 4*c) + 4*I*e^(2*d*x + 2*c) - 
I)*log(e^(2*d*x + 2*c) - 1))/(d*e^(8*d*x + 8*c) - 4*d*e^(6*d*x + 6*c) + 6* 
d*e^(4*d*x + 4*c) - 4*d*e^(2*d*x + 2*c) + d)
 

Sympy [F]

\[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{\left (- \tanh ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(-tanh(d*x+c)**2)**(5/2),x)
 

Output:

Integral((-tanh(c + d*x)**2)**(-5/2), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=\frac {i \, {\left (d x + c\right )}}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {4 \, {\left (-i \, e^{\left (-2 \, d x - 2 \, c\right )} + i \, e^{\left (-4 \, d x - 4 \, c\right )} - i \, e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} - 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )} - 1\right )}} \] Input:

integrate(1/(-tanh(d*x+c)^2)^(5/2),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 - 4* 
(-I*e^(-2*d*x - 2*c) + I*e^(-4*d*x - 4*c) - I*e^(-6*d*x - 6*c))/(d*(4*e^(- 
2*d*x - 2*c) - 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) - e^(-8*d*x - 8*c) 
- 1))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=-\frac {\frac {i \, d x + i \, c}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} - \frac {i \, \log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{\mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} + \frac {4 \, {\left (i \, e^{\left (6 \, d x + 6 \, c\right )} - i \, e^{\left (4 \, d x + 4 \, c\right )} + i \, e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4} \mathrm {sgn}\left (-e^{\left (4 \, d x + 4 \, c\right )} + 1\right )}}{d} \] Input:

integrate(1/(-tanh(d*x+c)^2)^(5/2),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) + 4*(I*e^(6*d*x + 6*c) - I*e^(4*d*x + 4*c) + I*e^(2 
*d*x + 2*c))/((e^(2*d*x + 2*c) - 1)^4*sgn(-e^(4*d*x + 4*c) + 1)))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (-{\mathrm {tanh}\left (c+d\,x\right )}^2\right )}^{5/2}} \,d x \] Input:

int(1/(-tanh(c + d*x)^2)^(5/2),x)
 

Output:

int(1/(-tanh(c + d*x)^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.83 \[ \int \frac {1}{\left (-\tanh ^2(c+d x)\right )^{5/2}} \, dx=\frac {i \left (-e^{8 d x +8 c} \mathrm {log}\left (e^{d x +c}-1\right )-e^{8 d x +8 c} \mathrm {log}\left (e^{d x +c}+1\right )+e^{8 d x +8 c} d x +e^{8 d x +8 c}+4 e^{6 d x +6 c} \mathrm {log}\left (e^{d x +c}-1\right )+4 e^{6 d x +6 c} \mathrm {log}\left (e^{d x +c}+1\right )-4 e^{6 d x +6 c} d x -6 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}-1\right )-6 e^{4 d x +4 c} \mathrm {log}\left (e^{d x +c}+1\right )+6 e^{4 d x +4 c} d x +2 e^{4 d x +4 c}+4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}-1\right )+4 e^{2 d x +2 c} \mathrm {log}\left (e^{d x +c}+1\right )-4 e^{2 d x +2 c} d x -\mathrm {log}\left (e^{d x +c}-1\right )-\mathrm {log}\left (e^{d x +c}+1\right )+d x +1\right )}{d \left (e^{8 d x +8 c}-4 e^{6 d x +6 c}+6 e^{4 d x +4 c}-4 e^{2 d x +2 c}+1\right )} \] Input:

int(1/(-tanh(d*x+c)^2)^(5/2),x)
 

Output:

(i*( - e**(8*c + 8*d*x)*log(e**(c + d*x) - 1) - e**(8*c + 8*d*x)*log(e**(c 
 + d*x) + 1) + e**(8*c + 8*d*x)*d*x + e**(8*c + 8*d*x) + 4*e**(6*c + 6*d*x 
)*log(e**(c + d*x) - 1) + 4*e**(6*c + 6*d*x)*log(e**(c + d*x) + 1) - 4*e** 
(6*c + 6*d*x)*d*x - 6*e**(4*c + 4*d*x)*log(e**(c + d*x) - 1) - 6*e**(4*c + 
 4*d*x)*log(e**(c + d*x) + 1) + 6*e**(4*c + 4*d*x)*d*x + 2*e**(4*c + 4*d*x 
) + 4*e**(2*c + 2*d*x)*log(e**(c + d*x) - 1) + 4*e**(2*c + 2*d*x)*log(e**( 
c + d*x) + 1) - 4*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**(8*c + 8*d*x) - 4*e**(6*c + 6*d*x) + 6*e** 
(4*c + 4*d*x) - 4*e**(2*c + 2*d*x) + 1))