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

Optimal. Leaf size=92 \[ \frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}}-\frac {3 \coth (c+d x)}{2 d (a+b)^2}+\frac {\coth (c+d x)}{2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4132, 290, 325, 208} \[ \frac {3 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 d (a+b)^{5/2}}-\frac {3 \coth (c+d x)}{2 d (a+b)^2}+\frac {\coth (c+d x)}{2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

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

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Mathematica [B]  time = 2.62, size = 220, normalized size = 2.39 \[ \frac {\text {sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (2 \text {csch}(c) \sinh (d x) \text {csch}(c+d x) (a \cosh (2 (c+d x))+a+2 b)+\frac {3 b (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}-\frac {b (a+2 b) \tanh (2 c)}{a}+b \text {sech}(2 c) \sinh (2 d x)\right )}{8 d (a+b)^2 \left (a+b \text {sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[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*((3*b*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*S
inh[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]))/(Sqrt[a + b]*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] + b*Sech[2*c]*Sinh[2*d*x] - (b*(a + 2*b)*Tanh[2*c])/a))/(8*(a + b)^2*d*(a + b*Sech[
c + d*x]^2)^2)

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fricas [B]  time = 0.62, size = 2407, normalized size = 26.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^4 + 16*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(2
*a^2 + a*b + 2*b^2)*sinh(d*x + c)^4 + 8*(2*a^2 + 4*a*b - b^2)*cosh(d*x + c)^2 + 8*(3*(2*a^2 + a*b + 2*b^2)*cos
h(d*x + c)^2 + 2*a^2 + 4*a*b - b^2)*sinh(d*x + c)^2 - 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x +
c)^5 + a^2*sinh(d*x + c)^6 + (a^2 + 4*a*b)*cosh(d*x + c)^4 + (15*a^2*cosh(d*x + c)^2 + a^2 + 4*a*b)*sinh(d*x +
c)^4 + 4*(5*a^2*cosh(d*x + c)^3 + (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c)^
2 + (15*a^2*cosh(d*x + c)^4 + 6*(a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - 4*a*b)*sinh(d*x + c)^2 - a^2 + 2*(3*a^2*
cosh(d*x + c)^5 + 2*(a^2 + 4*a*b)*cosh(d*x + c)^3 - (a^2 + 4*a*b)*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*cosh(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)*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 + 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*a^2 - 4*a*b
+ 16*((2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^3 + (2*a^2 + 4*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + 2*a
^3*b + a^2*b^2)*d*cosh(d*x + c)^6 + 6*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^4 + 2*a^3
*b + a^2*b^2)*d*sinh(d*x + c)^6 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^4 + (15*(a^4 + 2*a^3*b
+ a^2*b^2)*d*cosh(d*x + c)^2 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^4 - (a^4 + 6*a^3*b + 9*
a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^3 + (a^4 + 6*a^3*b + 9*a
^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^4 + 6*(a^4
+ 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^2
- (a^4 + 2*a^3*b + a^2*b^2)*d + 2*(3*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^5 + 2*(a^4 + 6*a^3*b + 9*a^2*b^
2 + 4*a*b^3)*d*cosh(d*x + c)^3 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), -1/2*(
2*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^4 + 8*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(2*a^2 + a
*b + 2*b^2)*sinh(d*x + c)^4 + 4*(2*a^2 + 4*a*b - b^2)*cosh(d*x + c)^2 + 4*(3*(2*a^2 + a*b + 2*b^2)*cosh(d*x +
c)^2 + 2*a^2 + 4*a*b - b^2)*sinh(d*x + c)^2 - 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a
^2*sinh(d*x + c)^6 + (a^2 + 4*a*b)*cosh(d*x + c)^4 + (15*a^2*cosh(d*x + c)^2 + a^2 + 4*a*b)*sinh(d*x + c)^4 +
4*(5*a^2*cosh(d*x + c)^3 + (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c)^2 + (15*
a^2*cosh(d*x + c)^4 + 6*(a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - 4*a*b)*sinh(d*x + c)^2 - a^2 + 2*(3*a^2*cosh(d*x
+ c)^5 + 2*(a^2 + 4*a*b)*cosh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/(a + b))*arcta
n(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*a^2 - 2*a*b + 8*((2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^3 + (2*a^2 + 4*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c
))/((a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^6 + 6*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5
+ (a^4 + 2*a^3*b + a^2*b^2)*d*sinh(d*x + c)^6 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^4 + (15*
(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^2 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^4 - (a^4
+ 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^3 + (a^4 +
6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x +
c)^4 + 6*(a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*si
nh(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d + 2*(3*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^5 + 2*(a^4 + 6*a^
3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^3 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x
+ c))]

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giac [B]  time = 0.73, size = 239, normalized size = 2.60 \[ \frac {\frac {3 \, b \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (2 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + a b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} - a b\right )}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a e^{\left (6 \, d x + 6 \, c\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} - a\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*b*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^2 + 2*a*b + b^2)*sqrt(-a*b - b^2)) - 2
*(2*a^2*e^(4*d*x + 4*c) + a*b*e^(4*d*x + 4*c) + 2*b^2*e^(4*d*x + 4*c) + 4*a^2*e^(2*d*x + 2*c) + 8*a*b*e^(2*d*x
+ 2*c) - 2*b^2*e^(2*d*x + 2*c) + 2*a^2 - a*b)/((a^3 + 2*a^2*b + a*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 = 0.39, size = 313, normalized size = 3.40 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a +b \right )^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a +b \right )^{2} \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {3 \sqrt {b}\, \ln \left (-\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {a +b}\right )}{4 d \left (a +b \right )^{\frac {5}{2}}}+\frac {3 \sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d \left (a +b \right )^{\frac {5}{2}}}-\frac {1}{2 d \left (a +b \right )^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(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*b/(a+b)^2/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*ta
nh(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/(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)-3/4/d
*b^(1/2)/(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))+3/4/d*b^
(1/2)/(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 = 0.46, size = 262, normalized size = 2.85 \[ -\frac {3 \, b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {2 \, a^{2} - a b + 2 \, {\left (2 \, a^{2} + 4 \, a b - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (2 \, a^{2} + a b + 2 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} + 6 \, a^{3} b + 9 \, a^{2} b^{2} + 4 \, a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (a^{4} + 6 \, a^{3} b + 9 \, a^{2} b^{2} + 4 \, a b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*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) - (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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cosh(c + d*x)^4/(sinh(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 \[ \int \frac {\operatorname {csch}^{2}{\left (c + d x \right )}}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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