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

Optimal. Leaf size=121 \[ -\frac {b^{3/2} (5 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{5/2}}+\frac {x}{a^2}-\frac {(2 a-b) \coth (c+d x)}{2 a d (a+b)^2}-\frac {b \coth (c+d x)}{2 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

x/a^2-1/2*b^(3/2)*(5*a+2*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2/(a+b)^(5/2)/d-1/2*(2*a-b)*coth(d*x+c)
/a/(a+b)^2/d-1/2*b*coth(d*x+c)/a/(a+b)/d/(a+b-b*tanh(d*x+c)^2)

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Rubi [A]  time = 0.28, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4141, 1975, 472, 583, 522, 206, 208} \[ -\frac {b^{3/2} (5 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{5/2}}+\frac {x}{a^2}-\frac {(2 a-b) \coth (c+d x)}{2 a d (a+b)^2}-\frac {b \coth (c+d x)}{2 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a^2 - (b^(3/2)*(5*a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(5/2)*d) - ((2*a - b
)*Coth[c + d*x])/(2*a*(a + b)^2*d) - (b*Coth[c + d*x])/(2*a*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

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 208

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

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[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 583

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

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

\begin {align*} \int \frac {\coth ^2(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \left (a+b \left (1-x^2\right )\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 a+b-3 b x^2}{x^2 \left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 a (a+b) d}\\ &=-\frac {(2 a-b) \coth (c+d x)}{2 a (a+b)^2 d}-\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 a^2+6 a b+b^2-(2 a-b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 a (a+b)^2 d}\\ &=-\frac {(2 a-b) \coth (c+d x)}{2 a (a+b)^2 d}-\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d}-\frac {\left (b^2 (5 a+2 b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 a^2 (a+b)^2 d}\\ &=\frac {x}{a^2}-\frac {b^{3/2} (5 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{5/2} d}-\frac {(2 a-b) \coth (c+d x)}{2 a (a+b)^2 d}-\frac {b \coth (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 2.82, size = 268, normalized size = 2.21 \[ \frac {\text {sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (\frac {b^2 \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{a^2 d (a+b)^2}-\frac {b^2 (5 a+2 b) (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{a^2 d (a+b)^{5/2} \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {2 x (a \cosh (2 (c+d x))+a+2 b)}{a^2}+\frac {2 \text {csch}(c) \sinh (d x) \text {csch}(c+d x) (a \cosh (2 (c+d x))+a+2 b)}{d (a+b)^2}\right )}{8 \left (a+b \text {sech}^2(c+d x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^4*((2*x*(a + 2*b + a*Cosh[2*(c + d*x)]))/a^2 - (b^2*(5*a + 2*b)
*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(
Cosh[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])*(Cosh[2*c] - Sinh[2*c]))/(a^2*(a + b)^(5/2)*d*Sqrt[b*(
Cosh[c] - Sinh[c])^4]) + (2*(a + 2*b + a*Cosh[2*(c + d*x)])*Csch[c]*Csch[c + d*x]*Sinh[d*x])/((a + b)^2*d) + (
b^2*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/(a^2*(a + b)^2*d)))/(8*(a + b*Sech[c + d*x]^2)^2)

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fricas [B]  time = 0.51, size = 3624, normalized size = 29.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^6 + 24*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x +
c)^5 + 4*(a^3 + 2*a^2*b + a*b^2)*d*x*sinh(d*x + c)^6 - 4*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4
*b^3)*d*x)*cosh(d*x + c)^4 + 4*(15*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 - 2*a^3 + a*b^2 + 2*b^3 + (a^3
+ 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*sinh(d*x + c)^4 + 16*(5*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 - (2*a^3
- a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 - 8*a^3 - 4*a*b^2 - 4
*(a^3 + 2*a^2*b + a*b^2)*d*x - 4*(4*a^3 + 8*a^2*b + 2*b^3 + (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x +
c)^2 + 4*(15*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 - 4*a^3 - 8*a^2*b - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2
+ 4*b^3)*d*x - 6*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c
)^2 + ((5*a^2*b + 2*a*b^2)*cosh(d*x + c)^6 + 6*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (5*a^2*b +
2*a*b^2)*sinh(d*x + c)^6 + (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + (5*a^2*b + 22*a*b^2 + 8*b^3 + 15*(5*
a^2*b + 2*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^3 + (5*a^2*b + 22*a
*b^2 + 8*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - 5*a^2*b - 2*a*b^2 - (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c)^
2 + (15*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^4 - 5*a^2*b - 22*a*b^2 - 8*b^3 + 6*(5*a^2*b + 22*a*b^2 + 8*b^3)*cosh
(d*x + c)^2)*sinh(d*x + c)^2 + 2*(3*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^5 + 2*(5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(
d*x + c)^3 - (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt(b/(a + b))*log((a^2*cosh(d*x + c)
^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*co
sh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x + c)^3 + (a^2 + 2*a*b)*co
sh(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 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*
x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2
+ 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 8*(3*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh
(d*x + c)^5 - 2*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c)^3 - (4*a^3 + 8*a
^2*b + 2*b^3 + (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^5 + 2*a^4*b + a^3*b^2)
*d*cosh(d*x + c)^6 + 6*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^5 + 2*a^4*b + a^3*b^2)*d
*sinh(d*x + c)^6 + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^4 + (15*(a^5 + 2*a^4*b + a^3*b^2)*d
*cosh(d*x + c)^2 + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d)*sinh(d*x + c)^4 - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4
*a^2*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^5 + 6*a^4*b + 9*a^3*b^2 +
4*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^4 + 6*(a^5 + 6*a^4
*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^2 - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d)*sinh(d*x + c)^2 - (
a^5 + 2*a^4*b + a^3*b^2)*d + 2*(3*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^5 + 2*(a^5 + 6*a^4*b + 9*a^3*b^2 +
4*a^2*b^3)*d*cosh(d*x + c)^3 - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(
2*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^6 + 12*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^5 +
2*(a^3 + 2*a^2*b + a*b^2)*d*x*sinh(d*x + c)^6 - 2*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*
d*x)*cosh(d*x + c)^4 + 2*(15*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^2 - 2*a^3 + a*b^2 + 2*b^3 + (a^3 + 6*a^
2*b + 9*a*b^2 + 4*b^3)*d*x)*sinh(d*x + c)^4 + 8*(5*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^3 - (2*a^3 - a*b^
2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 - 4*a^3 - 2*a*b^2 - 2*(a^3 +
2*a^2*b + a*b^2)*d*x - 2*(4*a^3 + 8*a^2*b + 2*b^3 + (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c)^2 +
2*(15*(a^3 + 2*a^2*b + a*b^2)*d*x*cosh(d*x + c)^4 - 4*a^3 - 8*a^2*b - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3
)*d*x - 6*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - (
(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^6 + 6*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (5*a^2*b + 2*a*b^2
)*sinh(d*x + c)^6 + (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c)^4 + (5*a^2*b + 22*a*b^2 + 8*b^3 + 15*(5*a^2*b +
2*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(5*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^3 + (5*a^2*b + 22*a*b^2 +
8*b^3)*cosh(d*x + c))*sinh(d*x + c)^3 - 5*a^2*b - 2*a*b^2 - (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c)^2 + (15
*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^4 - 5*a^2*b - 22*a*b^2 - 8*b^3 + 6*(5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x +
c)^2)*sinh(d*x + c)^2 + 2*(3*(5*a^2*b + 2*a*b^2)*cosh(d*x + c)^5 + 2*(5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c
)^3 - (5*a^2*b + 22*a*b^2 + 8*b^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^
2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) + 4*(3*(a^3 + 2*a^2*b +
a*b^2)*d*x*cosh(d*x + c)^5 - 2*(2*a^3 - a*b^2 - 2*b^3 - (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c)^
3 - (4*a^3 + 8*a^2*b + 2*b^3 + (a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^5 + 2*
a^4*b + a^3*b^2)*d*cosh(d*x + c)^6 + 6*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^5 + 2*a^
4*b + a^3*b^2)*d*sinh(d*x + c)^6 + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^4 + (15*(a^5 + 2*a^
4*b + a^3*b^2)*d*cosh(d*x + c)^2 + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d)*sinh(d*x + c)^4 - (a^5 + 6*a^4*b
+ 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^3 + (a^5 + 6*a^4*
b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^4
+ 6*(a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^2 - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d)*sin
h(d*x + c)^2 - (a^5 + 2*a^4*b + a^3*b^2)*d + 2*(3*(a^5 + 2*a^4*b + a^3*b^2)*d*cosh(d*x + c)^5 + 2*(a^5 + 6*a^4
*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c)^3 - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*d*cosh(d*x + c))*sinh(
d*x + c))]

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giac [B]  time = 0.96, size = 282, normalized size = 2.33 \[ \frac {\frac {{\left (5 \, a b^{2} e^{\left (2 \, c\right )} + 2 \, b^{3} e^{\left (2 \, c\right )}\right )} \arctan \left (-\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right ) e^{\left (-2 \, c\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \sqrt {-a b - b^{2}}} + \frac {2 \, d x}{a^{2}} - \frac {2 \, {\left (2 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 2 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} + a b^{2}\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (a e^{\left (6 \, d x + 6 \, c\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} - a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2*e^(2*c) + 2*b^3*e^(2*c))*arctan(-1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))*e^(-2*c)/((
a^4 + 2*a^3*b + a^2*b^2)*sqrt(-a*b - b^2)) + 2*d*x/a^2 - 2*(2*a^3*e^(4*d*x + 4*c) - a*b^2*e^(4*d*x + 4*c) - 2*
b^3*e^(4*d*x + 4*c) + 4*a^3*e^(2*d*x + 2*c) + 8*a^2*b*e^(2*d*x + 2*c) + 2*b^3*e^(2*d*x + 2*c) + 2*a^3 + a*b^2)
/((a^4 + 2*a^3*b + a^2*b^2)*(a*e^(6*d*x + 6*c) + a*e^(4*d*x + 4*c) + 4*b*e^(4*d*x + 4*c) - a*e^(2*d*x + 2*c) -
4*b*e^(2*d*x + 2*c) - a)))/d

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maple [B]  time = 0.46, size = 481, normalized size = 3.98 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {b^{2} \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a \left (a +b \right )^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a +b \right )^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {5 b^{\frac {3}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{4 d a \left (a +b \right )^{\frac {5}{2}}}-\frac {5 b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d a \left (a +b \right )^{\frac {5}{2}}}+\frac {b^{\frac {5}{2}} \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{2 d \,a^{2} \left (a +b \right )^{\frac {5}{2}}}-\frac {b^{\frac {5}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{2 d \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.92, size = 1070, normalized size = 8.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b + b^2)*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/((a^4 + 2*a^3*b + a^2*b^2)*d) - 1/4
*(2*a*b + b^2)*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/((a^4 + 2*a^3*b + a^2*b^2)*d) - 1/16
*(3*a^2*b + 10*a*b^2 + 4*b^3)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2
*b + 2*sqrt((a + b)*b)))/((a^4 + 2*a^3*b + a^2*b^2)*sqrt((a + b)*b)*d) + 1/16*(3*a^2*b + 10*a*b^2 + 4*b^3)*log
((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^4
+ 2*a^3*b + a^2*b^2)*sqrt((a + b)*b)*d) - 3/8*b*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b))/(a*e^(-
2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + 2*a*b + b^2)*sqrt((a + b)*b)*d) + 1/4*(2*a^3 + a^2*b + 2*
a*b^2 + (2*a^3 - a^2*b - 8*a*b^2 - 8*b^3)*e^(4*d*x + 4*c) + 2*(2*a^3 + 4*a^2*b + 3*a*b^2 + 4*b^3)*e^(2*d*x + 2
*c))/((a^5 + 2*a^4*b + a^3*b^2 - (a^5 + 2*a^4*b + a^3*b^2)*e^(6*d*x + 6*c) - (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^
2*b^3)*e^(4*d*x + 4*c) + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*e^(2*d*x + 2*c))*d) - 1/4*(2*a^3 + a^2*b + 2*
a*b^2 + 2*(2*a^3 + 4*a^2*b + 3*a*b^2 + 4*b^3)*e^(-2*d*x - 2*c) + (2*a^3 - a^2*b - 8*a*b^2 - 8*b^3)*e^(-4*d*x -
4*c))/((a^5 + 2*a^4*b + a^3*b^2 + (a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*e^(-2*d*x - 2*c) - (a^5 + 6*a^4*b +
9*a^3*b^2 + 4*a^2*b^3)*e^(-4*d*x - 4*c) - (a^5 + 2*a^4*b + a^3*b^2)*e^(-6*d*x - 6*c))*d) - 1/2*(2*a^2 - a*b +
2*(2*a^2 + 4*a*b - b^2)*e^(-2*d*x - 2*c) + (2*a^2 + a*b + 2*b^2)*e^(-4*d*x - 4*c))/((a^4 + 2*a^3*b + a^2*b^2
+ (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*e^(-2*d*x - 2*c) - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*e^(-4*d*x - 4
*c) - (a^4 + 2*a^3*b + a^2*b^2)*e^(-6*d*x - 6*c))*d) + 1/2*log(e^(2*d*x + 2*c) - 1)/((a^2 + 2*a*b + b^2)*d) -
1/2*log(e^(-2*d*x - 2*c) - 1)/((a^2 + 2*a*b + b^2)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^4\,{\mathrm {coth}\left (c+d\,x\right )}^2}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((cosh(c + d*x)^4*coth(c + d*x)^2)/(b + a*cosh(c + d*x)^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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