3.177 \(\int \frac {\coth ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx\)

Optimal. Leaf size=60 \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2} d (a+b)}+\frac {x}{a+b}-\frac {\coth (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3670, 480, 522, 206, 205} \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2} d (a+b)}+\frac {x}{a+b}-\frac {\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(a + b) - (b^(3/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^(3/2)*(a + b)*d) - Coth[c + d*x]/(a*d)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\coth ^2(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {\coth (c+d x)}{a d}+\frac {\operatorname {Subst}\left (\int \frac {a-b+b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{a d}\\ &=-\frac {\coth (c+d x)}{a d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b) d}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{a (a+b) d}\\ &=\frac {x}{a+b}-\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2} (a+b) d}-\frac {\coth (c+d x)}{a d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 67, normalized size = 1.12 \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2} d (a+b)}+\frac {c+d x}{d (a+b)}-\frac {\coth (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)/((a + b)*d) - (b^(3/2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(a^(3/2)*(a + b)*d) - Coth[c + d*x]/
(a*d)

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fricas [B]  time = 0.45, size = 784, normalized size = 13.07 \[ \left [\frac {2 \, a d x \cosh \left (d x + c\right )^{2} + 4 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + 2 \, a d x \sinh \left (d x + c\right )^{2} - 2 \, a d x + {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} - b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} - 6 \, a b + b^{2} + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (a^{2} - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 4 \, {\left ({\left (a^{2} + a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} + a b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} - a b\right )} \sqrt {-\frac {b}{a}}}{{\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + {\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b}\right ) - 4 \, a - 4 \, b}{2 \, {\left ({\left (a^{2} + a b\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} + a b\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + a b\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{2} + a b\right )} d\right )}}, \frac {a d x \cosh \left (d x + c\right )^{2} + 2 \, a d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a d x \sinh \left (d x + c\right )^{2} - a d x - {\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )} \sqrt {\frac {b}{a}}}{2 \, b}\right ) - 2 \, a - 2 \, b}{{\left (a^{2} + a b\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{2} + a b\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + a b\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{2} + a b\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*d*x*cosh(d*x + c)^2 + 4*a*d*x*cosh(d*x + c)*sinh(d*x + c) + 2*a*d*x*sinh(d*x + c)^2 - 2*a*d*x + (b*c
osh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 - b)*sqrt(-b/a)*log(((a^2 + 2*a*b + b^2)*
cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 +
2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 - b^2)*sinh(d*x + c)^2 + a^2 -
6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a^2 + a*
b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 - a*b)*sqrt
(-b/a))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a -
b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^3 + (a -
b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - 4*a - 4*b)/((a^2 + a*b)*d*cosh(d*x + c)^2 + 2*(a^2 + a*b)*d*cosh(
d*x + c)*sinh(d*x + c) + (a^2 + a*b)*d*sinh(d*x + c)^2 - (a^2 + a*b)*d), (a*d*x*cosh(d*x + c)^2 + 2*a*d*x*cosh
(d*x + c)*sinh(d*x + c) + a*d*x*sinh(d*x + c)^2 - a*d*x - (b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c)
+ b*sinh(d*x + c)^2 - b)*sqrt(b/a)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c
) + (a + b)*sinh(d*x + c)^2 + a - b)*sqrt(b/a)/b) - 2*a - 2*b)/((a^2 + a*b)*d*cosh(d*x + c)^2 + 2*(a^2 + a*b)*
d*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a*b)*d*sinh(d*x + c)^2 - (a^2 + a*b)*d)]

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giac [A]  time = 0.21, size = 89, normalized size = 1.48 \[ -\frac {\frac {b^{2} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{2} + a b\right )} \sqrt {a b}} - \frac {d x + c}{a + b} + \frac {2}{a {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2 + a*b)*sqrt(a*b)) - (d*x + c
)/(a + b) + 2/(a*(e^(2*d*x + 2*c) - 1)))/d

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maple [B]  time = 0.40, size = 494, normalized size = 8.23 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a +b \right )}-\frac {1}{2 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b^{2} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right ) \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}-\frac {b^{2} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right ) a \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {b^{3} \arctanh \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}\right )}{d \left (a +b \right ) \sqrt {b \left (a +b \right )}\, a \sqrt {\left (2 \sqrt {b \left (a +b \right )}-a -2 b \right ) a}}+\frac {b^{2} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right ) \sqrt {b \left (a +b \right )}\, \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}+\frac {b^{2} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right ) a \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}+\frac {b^{3} \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}}\right )}{d \left (a +b \right ) \sqrt {b \left (a +b \right )}\, a \sqrt {\left (2 \sqrt {b \left (a +b \right )}+a +2 b \right ) a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/a*tanh(1/2*d*x+1/2*c)-1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/(a+b)*ln(tanh(1/2*d*x+1/2*c)+1)-1/2/d/a/t
anh(1/2*d*x+1/2*c)+1/d*b^2/(a+b)/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/
2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d*b^2/(a+b)/a/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1
/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*b^3/(a+b)/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)-a-2*b)*
a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*b^2/(a+b)/(b*(a+b))^(1/2)/((2*
(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+1/d*b^2/(a+b
)/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+1/d*
b^3/(a+b)/(b*(a+b))^(1/2)/a/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/
2)+a+2*b)*a)^(1/2))

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maxima [B]  time = 0.47, size = 329, normalized size = 5.48 \[ -\frac {b \log \left ({\left (a + b\right )} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a - b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + a b\right )} d} + \frac {b \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{4 \, {\left (a^{2} + a b\right )} d} + \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + a b\right )} \sqrt {a b} d} - \frac {{\left (a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{4 \, {\left (a^{2} + a b\right )} \sqrt {a b} d} + \frac {b \arctan \left (\frac {{\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{2 \, \sqrt {a b} a d} + \frac {\log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{2 \, a d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}{2 \, a d} - \frac {1}{2 \, {\left (a e^{\left (2 \, d x + 2 \, c\right )} - a\right )} d} + \frac {3}{2 \, {\left (a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*log((a + b)*e^(4*d*x + 4*c) + 2*(a - b)*e^(2*d*x + 2*c) + a + b)/((a^2 + a*b)*d) + 1/4*b*log(2*(a - b)*
e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2 + a*b)*d) + 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(
2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^2 + a*b)*sqrt(a*b)*d) - 1/4*(a*b - b^2)*arctan(1/2*((a + b)*e^(-2*d*x - 2
*c) + a - b)/sqrt(a*b))/((a^2 + a*b)*sqrt(a*b)*d) + 1/2*b*arctan(1/2*((a + b)*e^(-2*d*x - 2*c) + a - b)/sqrt(a
*b))/(sqrt(a*b)*a*d) + 1/2*log(e^(2*d*x + 2*c) - 1)/(a*d) - 1/2*log(e^(-2*d*x - 2*c) - 1)/(a*d) - 1/2/((a*e^(2
*d*x + 2*c) - a)*d) + 3/2/((a*e^(-2*d*x - 2*c) - a)*d)

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mupad [B]  time = 1.62, size = 402, normalized size = 6.70 \[ \frac {x}{a+b}-\frac {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {4\,b^2}{a\,d\,{\left (a+b\right )}^3\,\left (a^2+b\,a\right )\,\sqrt {b^3}}+\frac {\left (a^3\,d\,\sqrt {b^3}-a\,b^2\,d\,\sqrt {b^3}\right )\,\left (a-b\right )}{b^2\,{\left (a+b\right )}^2\,\left (a^2+b\,a\right )\,\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}\,\sqrt {a^3\,d^2\,{\left (a+b\right )}^2}}\right )+\frac {\left (a-b\right )\,\left (a^3\,d\,\sqrt {b^3}+a\,b^2\,d\,\sqrt {b^3}+2\,a^2\,b\,d\,\sqrt {b^3}\right )}{b^2\,{\left (a+b\right )}^2\,\left (a^2+b\,a\right )\,\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}\,\sqrt {a^3\,d^2\,{\left (a+b\right )}^2}}\right )\,\left (\frac {a^3\,\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}}{2}+\frac {a\,b^2\,\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}}{2}+a^2\,b\,\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}\right )\right )\,\sqrt {b^3}}{\sqrt {a^5\,d^2+2\,a^4\,b\,d^2+a^3\,b^2\,d^2}}-\frac {2}{a\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a + b) - (atan((exp(2*c)*exp(2*d*x)*((4*b^2)/(a*d*(a + b)^3*(a*b + a^2)*(b^3)^(1/2)) + ((a^3*d*(b^3)^(1/2)
- a*b^2*d*(b^3)^(1/2))*(a - b))/(b^2*(a + b)^2*(a*b + a^2)*(a^5*d^2 + 2*a^4*b*d^2 + a^3*b^2*d^2)^(1/2)*(a^3*d^
2*(a + b)^2)^(1/2))) + ((a - b)*(a^3*d*(b^3)^(1/2) + a*b^2*d*(b^3)^(1/2) + 2*a^2*b*d*(b^3)^(1/2)))/(b^2*(a + b
)^2*(a*b + a^2)*(a^5*d^2 + 2*a^4*b*d^2 + a^3*b^2*d^2)^(1/2)*(a^3*d^2*(a + b)^2)^(1/2)))*((a^3*(a^5*d^2 + 2*a^4
*b*d^2 + a^3*b^2*d^2)^(1/2))/2 + (a*b^2*(a^5*d^2 + 2*a^4*b*d^2 + a^3*b^2*d^2)^(1/2))/2 + a^2*b*(a^5*d^2 + 2*a^
4*b*d^2 + a^3*b^2*d^2)^(1/2)))*(b^3)^(1/2))/(a^5*d^2 + 2*a^4*b*d^2 + a^3*b^2*d^2)^(1/2) - 2/(a*d*(exp(2*c + 2*
d*x) - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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