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

Optimal. Leaf size=62 \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a d (a+b)^{3/2}}-\frac {\coth (c+d x)}{d (a+b)}+\frac {x}{a} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4141, 1975, 480, 522, 206, 208} \[ -\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a d (a+b)^{3/2}}-\frac {\coth (c+d x)}{d (a+b)}+\frac {x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 1.16, size = 193, normalized size = 3.11 \[ \frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (b^2 (\sinh (2 c)-\cosh (2 c)) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )+\sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4} (d x (a+b)+a \text {csch}(c) \sinh (d x) \text {csch}(c+d x))\right )}{2 a d (a+b)^{3/2} \sqrt {b (\cosh (c)-\sinh (c))^4} \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*(b^2*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*Si
nh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(-Cosh[2*c] + Sinh[2*c]) + Sqrt[a
+ b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]*((a + b)*d*x + a*Csch[c]*Csch[c + d*x]*Sinh[d*x])))/(2*a*(a + b)^(3/2)*d*(
a + b*Sech[c + d*x]^2)*Sqrt[b*(Cosh[c] - Sinh[c])^4])

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fricas [B]  time = 0.46, size = 749, normalized size = 12.08 \[ \left [\frac {2 \, {\left (a + b\right )} d x \cosh \left (d x + c\right )^{2} + 4 \, {\left (a + b\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + 2 \, {\left (a + b\right )} d x \sinh \left (d x + c\right )^{2} - 2 \, {\left (a + b\right )} 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 + b}} \log \left (\frac {a^{2} \cosh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2} + 4 \, {\left (a^{2} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b\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} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a + b}}}{a \cosh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a \cosh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (a \cosh \left (d x + c\right )^{3} + {\left (a + 2 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a}\right ) - 4 \, a}{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 {{\left (a + b\right )} d x \cosh \left (d x + c\right )^{2} + 2 \, {\left (a + b\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a + b\right )} d x \sinh \left (d x + c\right )^{2} - {\left (a + b\right )} 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 + b}} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a \sinh \left (d x + c\right )^{2} + a + 2 \, b\right )} \sqrt {-\frac {b}{a + b}}}{2 \, b}\right ) - 2 \, a}{{\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*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/2*(2*(a + b)*d*x*cosh(d*x + c)^2 + 4*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c) + 2*(a + b)*d*x*sinh(d*x + c)^
2 - 2*(a + b)*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 +
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)*cos
h(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)) - 4*a)/((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 + b)*d*x*cosh(d*x + c)^2 + 2*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c) + (a + b)*d*x*sinh(d*x + c
)^2 - (a + b)*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 +
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) - 2*a)/((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 [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueError index.cc
index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueDone

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maple [B]  time = 0.46, size = 189, normalized size = 3.05 \[ -\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \left (a +b \right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}-\frac {1}{2 d \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \left (a +b \right )^{\frac {3}{2}}}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d a \left (a +b \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.48, size = 429, normalized size = 6.92 \[ \frac {b \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, {\left (a + 2 \, b\right )} e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{4 \, {\left (a^{2} + a b\right )} d} - \frac {b \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{4 \, {\left (a^{2} + a b\right )} d} - \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{2} + a b\right )} \sqrt {{\left (a + b\right )} b} d} + \frac {{\left (a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{8 \, {\left (a^{2} + a b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {b \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, \sqrt {{\left (a + b\right )} b} {\left (a + b\right )} d} + \frac {\log \left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}{2 \, {\left (a + b\right )} d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}{2 \, {\left (a + b\right )} d} - \frac {1}{2 \, {\left ({\left (a + b\right )} e^{\left (2 \, d x + 2 \, c\right )} - a - b\right )} d} + \frac {3}{2 \, {\left ({\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - a - b\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*log(a*e^(4*d*x + 4*c) + 2*(a + 2*b)*e^(2*d*x + 2*c) + a)/((a^2 + a*b)*d) - 1/4*b*log(2*(a + 2*b)*e^(-2*d
*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/((a^2 + a*b)*d) - 1/8*(a*b + 2*b^2)*log((a*e^(2*d*x + 2*c) + a + 2*b - 2*s
qrt((a + b)*b))/(a*e^(2*d*x + 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^2 + a*b)*sqrt((a + b)*b)*d) + 1/8*(a*b
+ 2*b^2)*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 + a*b)*sqrt((a + b)*b)*d) - 1/4*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)))/(sqrt((a + b)*b)*(a + b)*d) + 1/2*log(e^(2*d*x + 2*c) - 1)/((a + b)
*d) - 1/2*log(e^(-2*d*x - 2*c) - 1)/((a + b)*d) - 1/2/(((a + b)*e^(2*d*x + 2*c) - a - b)*d) + 3/2/(((a + b)*e^
(-2*d*x - 2*c) - a - b)*d)

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mupad [B]  time = 3.39, size = 977, normalized size = 15.76 \[ \frac {x}{a}-\frac {2}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a\,d+b\,d\right )}+\frac {\mathrm {atan}\left (\frac {\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {8\,\left (a+2\,b\right )\,\left (4\,d\,a^4\,b^2+16\,d\,a^3\,b^3+20\,d\,a^2\,b^4+8\,d\,a\,b^5\right )}{a^6\,\left (a+b\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}+\frac {2\,\sqrt {b^3}\,\left (a^2+8\,a\,b+8\,b^2\right )\,\left (a^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+8\,b^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+8\,a\,b\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )}{a^7\,b^2\,d\,{\left (a+b\right )}^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}\right )+\frac {8\,\left (a+2\,b\right )\,\left (2\,d\,a^4\,b^2+4\,d\,a^3\,b^3+2\,d\,a^2\,b^4\right )}{a^6\,\left (a+b\right )\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^2\,d^2\,{\left (a+b\right )}^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}+\frac {2\,\left (a^2\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+2\,a\,b\,\sqrt {b^3}\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )\,\sqrt {b^3}\,\left (a^2+8\,a\,b+8\,b^2\right )}{a^7\,b^2\,d\,{\left (a+b\right )}^3\,\left (a^3+2\,a^2\,b+a\,b^2\right )\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}}\right )\,\left (a^7\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+a^4\,b^3\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+3\,a^5\,b^2\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}+3\,a^6\,b\,\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}\right )}{4\,\sqrt {b^3}}\right )\,\sqrt {b^3}}{\sqrt {-a^5\,d^2-3\,a^4\,b\,d^2-3\,a^3\,b^2\,d^2-a^2\,b^3\,d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a - 2/((exp(2*c + 2*d*x) - 1)*(a*d + b*d)) + (atan(((exp(2*c)*exp(2*d*x)*((8*(a + 2*b)*(20*a^2*b^4*d + 16*a^
3*b^3*d + 4*a^4*b^2*d + 8*a*b^5*d))/(a^6*(a + b)*(a*b^2 + 2*a^2*b + a^3)*(-a^2*d^2*(a + b)^3)^(1/2)*(- a^5*d^2
- 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)) + (2*(b^3)^(1/2)*(8*a*b + a^2 + 8*b^2)*(a^2*(b^3)^(1/2)*(
- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 8*b^2*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^
2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 8*a*b*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(
1/2)))/(a^7*b^2*d*(a + b)^3*(a*b^2 + 2*a^2*b + a^3)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1
/2))) + (8*(a + 2*b)*(2*a^2*b^4*d + 4*a^3*b^3*d + 2*a^4*b^2*d))/(a^6*(a + b)*(a*b^2 + 2*a^2*b + a^3)*(-a^2*d^2
*(a + b)^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)) + (2*(a^2*(b^3)^(1/2)*(- a^5*
d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 2*a*b*(b^3)^(1/2)*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*
d^2 - 3*a^3*b^2*d^2)^(1/2))*(b^3)^(1/2)*(8*a*b + a^2 + 8*b^2))/(a^7*b^2*d*(a + b)^3*(a*b^2 + 2*a^2*b + a^3)*(-
a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)))*(a^7*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*
a^3*b^2*d^2)^(1/2) + a^4*b^3*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 3*a^5*b^2*(- a^5*
d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2) + 3*a^6*b*(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^
3*b^2*d^2)^(1/2)))/(4*(b^3)^(1/2)))*(b^3)^(1/2))/(- a^5*d^2 - 3*a^4*b*d^2 - a^2*b^3*d^2 - 3*a^3*b^2*d^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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