3.2.50 \(\int \frac {\coth ^2(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\) [150]

Optimal. Leaf size=665 \[ \frac {4 a \coth (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{(a-b) (a+b)^{3/2} d}-\frac {2 \coth (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{a \sqrt {a+b} d}-\frac {(3 a-b) \coth (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{(a-b) (a+b)^{3/2} d}+\frac {2 \coth (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{a \sqrt {a+b} d}+\frac {2 \sqrt {a+b} \coth (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\text {sech}(c+d x))}{a+b}} \sqrt {-\frac {b (1+\text {sech}(c+d x))}{a-b}}}{a^2 d}-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^{3/2}}-\frac {4 a b^2 \tanh (c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}} \]

[Out]

-coth(d*x+c)/d/(a+b*sech(d*x+c))^(3/2)+4*a*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)/(a+b)^(3/2)/d-(3*a-b)*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)/(a+b)^(3/2)/d-2*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/d/(a+b)^(1/2)+2*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/d/(a+b)^(1/2)+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))*(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^2/d-b^2*
tanh(d*x+c)/(a^2-b^2)/d/(a+b*sech(d*x+c))^(3/2)-4*a*b^2*tanh(d*x+c)/(a^2-b^2)^2/d/(a+b*sech(d*x+c))^(1/2)+2*b^
2*tanh(d*x+c)/a/(a^2-b^2)/d/(a+b*sech(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.68, antiderivative size = 665, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3981, 3870, 4143, 4006, 3869, 3917, 4089, 3960, 3918, 4088, 4090} \begin {gather*} \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}} \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^2 d}+\frac {2 b^2 \tanh (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \text {sech}(c+d x)}}-\frac {4 a b^2 \tanh (c+d x)}{d \left (a^2-b^2\right )^2 \sqrt {a+b \text {sech}(c+d x)}}-\frac {b^2 \tanh (c+d x)}{d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^{3/2}}+\frac {2 \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}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}-\frac {(3 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}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d (a-b) (a+b)^{3/2}}-\frac {2 \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 (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d \sqrt {a+b}}+\frac {4 a \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 (\text {ArcSin}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d (a-b) (a+b)^{3/2}}-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^2/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

(4*a*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))])/((a - b)*(a + b)^(3/2)*d) - (2*Coth[c + d*x]*El
lipticE[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))])/(a*Sqrt[a + b]*d) - ((3*a - b)*Coth[c + d*x]*EllipticF[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))])/((a - b)*(a + b)^(3/2)*d) + (2*Coth[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sech[c + d*x]]/S
qrt[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*Sqrt[a + b]*d) + (2*Sqrt[a + b]*Coth[c + d*x]*EllipticPi[(a + b)/a, 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))])/(a^2*
d) - Coth[c + d*x]/(d*(a + b*Sech[c + d*x])^(3/2)) - (b^2*Tanh[c + d*x])/((a^2 - b^2)*d*(a + b*Sech[c + d*x])^
(3/2)) - (4*a*b^2*Tanh[c + d*x])/((a^2 - b^2)^2*d*Sqrt[a + b*Sech[c + d*x]]) + (2*b^2*Tanh[c + d*x])/(a*(a^2 -
b^2)*d*Sqrt[a + b*Sech[c + d*x]])

Rule 3869

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b
*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b)*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b
*Csc[c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3870

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[c + d*x]*((a + b*Csc[c + d*x])^(n +
1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3918

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-b)*Cot[e + f*x]*(
(a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 -
b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3960

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)/cos[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Simp[Tan[e + f*x]*((a
+ b*Csc[e + f*x])^m/f), x] + Dist[b*m, Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e
, f, m}, x]

Rule 3981

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]

Rule 4006

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 4088

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]

Rule 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 4143

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[Csc[e + f*x
]*((1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]

Rubi steps

\begin {align*} \int \frac {\coth ^2(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx &=-\int \left (-\frac {1}{(a+b \text {sech}(c+d x))^{3/2}}-\frac {\text {csch}^2(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}}\right ) \, dx\\ &=\int \frac {1}{(a+b \text {sech}(c+d x))^{3/2}} \, dx+\int \frac {\text {csch}^2(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx\\ &=-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}+\frac {1}{2} (3 b) \int \frac {\text {sech}(c+d x)}{(a+b \text {sech}(c+d x))^{5/2}} \, dx-\frac {2 \int \frac {\frac {1}{2} \left (-a^2+b^2\right )+\frac {1}{2} a b \text {sech}(c+d x)+\frac {1}{2} b^2 \text {sech}^2(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^{3/2}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2+b^2\right )+\left (\frac {a b}{2}-\frac {b^2}{2}\right ) \text {sech}(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}-\frac {b \int \frac {\text {sech}(c+d x) \left (-\frac {3 a}{2}+\frac {1}{2} b \text {sech}(c+d x)\right )}{(a+b \text {sech}(c+d x))^{3/2}} \, dx}{a^2-b^2}-\frac {b^2 \int \frac {\text {sech}(c+d x) (1+\text {sech}(c+d x))}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {2 \coth (c+d x) E\left (\sin ^{-1}\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 \sqrt {a+b} d}-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^{3/2}}-\frac {4 a b^2 \tanh (c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a}-\frac {b \int \frac {\text {sech}(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{a (a+b)}+\frac {(2 b) \int \frac {\text {sech}(c+d x) \left (\frac {1}{4} \left (3 a^2+b^2\right )+a b \text {sech}(c+d x)\right )}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {2 \coth (c+d x) E\left (\sin ^{-1}\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 \sqrt {a+b} d}+\frac {2 \coth (c+d x) F\left (\sin ^{-1}\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 \sqrt {a+b} d}+\frac {2 \sqrt {a+b} \coth (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\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^2 d}-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^{3/2}}-\frac {4 a b^2 \tanh (c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}+\frac {((3 a-b) b) \int \frac {\text {sech}(c+d x)}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{2 (a-b) (a+b)^2}+\frac {\left (2 a b^2\right ) \int \frac {\text {sech}(c+d x) (1+\text {sech}(c+d x))}{\sqrt {a+b \text {sech}(c+d x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {4 a \coth (c+d x) E\left (\sin ^{-1}\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-b) (a+b)^{3/2} d}-\frac {2 \coth (c+d x) E\left (\sin ^{-1}\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 \sqrt {a+b} d}-\frac {(3 a-b) \coth (c+d x) F\left (\sin ^{-1}\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-b) (a+b)^{3/2} d}+\frac {2 \coth (c+d x) F\left (\sin ^{-1}\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 \sqrt {a+b} d}+\frac {2 \sqrt {a+b} \coth (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\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^2 d}-\frac {\coth (c+d x)}{d (a+b \text {sech}(c+d x))^{3/2}}-\frac {b^2 \tanh (c+d x)}{\left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^{3/2}}-\frac {4 a b^2 \tanh (c+d x)}{\left (a^2-b^2\right )^2 d \sqrt {a+b \text {sech}(c+d x)}}+\frac {2 b^2 \tanh (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \text {sech}(c+d x)}}\\ \end {align*}

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Mathematica [F]
time = 150.72, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^2(c+d x)}{(a+b \text {sech}(c+d x))^{3/2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Coth[c + d*x]^2/(a + b*Sech[c + d*x])^(3/2),x]

[Out]

Integrate[Coth[c + d*x]^2/(a + b*Sech[c + d*x])^(3/2), x]

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Maple [F]
time = 2.84, size = 0, normalized size = 0.00 \[\int \frac {\coth ^{2}\left (d x +c \right )}{\left (a +b \,\mathrm {sech}\left (d x +c \right )\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^2/(a+b*sech(d*x+c))^(3/2),x)

[Out]

int(coth(d*x+c)^2/(a+b*sech(d*x+c))^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(coth(d*x + c)^2/(b*sech(d*x + c) + a)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\coth ^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**2/(a+b*sech(d*x+c))**(3/2),x)

[Out]

Integral(coth(c + d*x)**2/(a + b*sech(c + d*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^2/(a+b*sech(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(coth(d*x + c)^2/(b*sech(d*x + c) + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {coth}\left (c+d\,x\right )}^2}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^2/(a + b/cosh(c + d*x))^(3/2),x)

[Out]

int(coth(c + d*x)^2/(a + b/cosh(c + d*x))^(3/2), x)

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