\(\int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\) [141]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 362 \[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\frac {\coth (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{\sqrt {a+b} d}-\frac {\coth (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{\sqrt {a+b} d}+\frac {2 \sqrt {a+b} \coth (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{a d}-\frac {\coth (c+d x)}{d \sqrt {a+b \text {sech}(c+d x)}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}} \] Output:

coth(d*x+c)*EllipticE((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1 
/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/(a+b 
)^(1/2)/d-coth(d*x+c)*EllipticF((a+b*sech(d*x+c))^(1/2)/(a+b)^(1/2),((a+b) 
/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^(1/2)*(-b*(1+sech(d*x+c))/(a-b))^ 
(1/2)/(a+b)^(1/2)/d+2*(a+b)^(1/2)*coth(d*x+c)*EllipticPi((a+b*sech(d*x+c)) 
^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b*(1-sech(d*x+c))/(a+b))^ 
(1/2)*(-b*(1+sech(d*x+c))/(a-b))^(1/2)/a/d-coth(d*x+c)/d/(a+b*sech(d*x+c)) 
^(1/2)-b^2*tanh(d*x+c)/(a^2-b^2)/d/(a+b*sech(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx \] Input:

Integrate[Coth[c + d*x]^2/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

Integrate[Coth[c + d*x]^2/Sqrt[a + b*Sech[c + d*x]], x]
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 25, 4384, 2009}

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.

\(\displaystyle \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {1}{\cot \left (i c+i d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 i c+\pi )+i d x\right )^2 \sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}dx\)

\(\Big \downarrow \) 4384

\(\displaystyle -\int \left (-\frac {\text {csch}^2(c+d x)}{\sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}-\frac {1}{\sqrt {a+b \csc \left (\frac {1}{2} (2 i c+\pi )+i d x\right )}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {b^2 \tanh (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}-\frac {\coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d \sqrt {a+b}}+\frac {\coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d \sqrt {a+b}}+\frac {2 \sqrt {a+b} \coth (c+d x) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (\text {sech}(c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {\coth (c+d x)}{d \sqrt {a+b \text {sech}(c+d x)}}\)

Input:

Int[Coth[c + d*x]^2/Sqrt[a + b*Sech[c + d*x]],x]
 

Output:

(Coth[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a 
 + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c 
 + d*x]))/(a - b))])/(Sqrt[a + b]*d) - (Coth[c + d*x]*EllipticF[ArcSin[Sqr 
t[a + b*Sech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c 
+ d*x]))/(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(Sqrt[a + b]*d 
) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*S 
ech[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sech[c + d*x]))/ 
(a + b)]*Sqrt[-((b*(1 + Sech[c + d*x]))/(a - b))])/(a*d) - Coth[c + d*x]/( 
d*Sqrt[a + b*Sech[c + d*x]]) - (b^2*Tanh[c + d*x])/((a^2 - b^2)*d*Sqrt[a + 
 b*Sech[c + d*x]])
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4384
Int[cot[(c_.) + (d_.)*(x_)]^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_ 
), x_Symbol] :> Int[ExpandIntegrand[(a + b*Csc[c + d*x])^n, (-1 + Sec[c + d 
*x]^2)^(-m/2), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && 
ILtQ[m/2, 0] && IntegerQ[n - 1/2] && EqQ[m, -2]
 
Maple [F]

\[\int \frac {\coth \left (d x +c \right )^{2}}{\sqrt {a +b \,\operatorname {sech}\left (d x +c \right )}}d x\]

Input:

int(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x)
 

Fricas [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )^{2}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

integral(coth(d*x + c)^2/sqrt(b*sech(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\coth ^{2}{\left (c + d x \right )}}{\sqrt {a + b \operatorname {sech}{\left (c + d x \right )}}}\, dx \] Input:

integrate(coth(d*x+c)**2/(a+b*sech(d*x+c))**(1/2),x)
 

Output:

Integral(coth(c + d*x)**2/sqrt(a + b*sech(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )^{2}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(coth(d*x + c)^2/sqrt(b*sech(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int { \frac {\coth \left (d x + c\right )^{2}}{\sqrt {b \operatorname {sech}\left (d x + c\right ) + a}} \,d x } \] Input:

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(coth(d*x + c)^2/sqrt(b*sech(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^2}{\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}} \,d x \] Input:

int(coth(c + d*x)^2/(a + b/cosh(c + d*x))^(1/2),x)
 

Output:

int(coth(c + d*x)^2/(a + b/cosh(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\coth ^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx=\int \frac {\sqrt {\mathrm {sech}\left (d x +c \right ) b +a}\, \coth \left (d x +c \right )^{2}}{\mathrm {sech}\left (d x +c \right ) b +a}d x \] Input:

int(coth(d*x+c)^2/(a+b*sech(d*x+c))^(1/2),x)
 

Output:

int((sqrt(sech(c + d*x)*b + a)*coth(c + d*x)**2)/(sech(c + d*x)*b + a),x)