3.484 \(\int \frac {\coth ^2(c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=111 \[ \frac {b \coth (c+d x)}{a^2 d}+\frac {2 b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d}-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2889, 3056, 3055, 3001, 3770, 2660, 618, 204} \[ \frac {2 b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d}-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3001

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

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\coth ^2(c+d x) \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx &=\int \frac {\text {csch}^3(c+d x) \left (1+\sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx\\ &=-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {i \int \frac {\text {csch}^2(c+d x) \left (2 i b-i a \sinh (c+d x)+i b \sinh ^2(c+d x)\right )}{a+b \sinh (c+d x)} \, dx}{2 a}\\ &=\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}-\frac {\int \frac {\text {csch}(c+d x) \left (-a^2-2 b^2+a b \sinh (c+d x)\right )}{a+b \sinh (c+d x)} \, dx}{2 a^2}\\ &=\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}-\frac {\left (b \left (a^2+b^2\right )\right ) \int \frac {1}{a+b \sinh (c+d x)} \, dx}{a^3}+\frac {\left (a^2+2 b^2\right ) \int \text {csch}(c+d x) \, dx}{2 a^3}\\ &=-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}+\frac {\left (2 i b \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^3 d}\\ &=-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}-\frac {\left (4 i b \left (a^2+b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{a^3 d}\\ &=-\frac {\left (a^2+2 b^2\right ) \tanh ^{-1}(\cosh (c+d x))}{2 a^3 d}+\frac {2 b \sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{a^3 d}+\frac {b \coth (c+d x)}{a^2 d}-\frac {\coth (c+d x) \text {csch}(c+d x)}{2 a d}\\ \end {align*}

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Mathematica [A]  time = 1.22, size = 145, normalized size = 1.31 \[ \frac {4 \left (a^2+2 b^2\right ) \log \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+16 b \sqrt {-a^2-b^2} \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-a^2 \text {csch}^2\left (\frac {1}{2} (c+d x)\right )-a^2 \text {sech}^2\left (\frac {1}{2} (c+d x)\right )+4 a b \tanh \left (\frac {1}{2} (c+d x)\right )+4 a b \coth \left (\frac {1}{2} (c+d x)\right )}{8 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*b*Sqrt[-a^2 - b^2]*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]] + 4*a*b*Coth[(c + d*x)/2] - a^2*Csch
[(c + d*x)/2]^2 + 4*(a^2 + 2*b^2)*Log[Tanh[(c + d*x)/2]] - a^2*Sech[(c + d*x)/2]^2 + 4*a*b*Tanh[(c + d*x)/2])/
(8*a^3*d)

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fricas [B]  time = 0.55, size = 892, normalized size = 8.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*cosh(d*x + c)^3 + 2*a^2*sinh(d*x + c)^3 - 4*a*b*cosh(d*x + c)^2 + 2*a^2*cosh(d*x + c) + 2*(3*a^2*c
osh(d*x + c) - 2*a*b)*sinh(d*x + c)^2 - 2*(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x
+ c)^4 - 2*b*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 - b*cosh(d*x
+ c))*sinh(d*x + c) + b)*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c)
+ 2*a^2 + b^2 + 2*(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x +
c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) -
b)) + 4*a*b + ((a^2 + 2*b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*b^2)*
sinh(d*x + c)^4 - 2*(a^2 + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*b^2)*cosh(d*x + c)^2 - a^2 - 2*b^2)*sinh(d*x
+ c)^2 + a^2 + 2*b^2 + 4*((a^2 + 2*b^2)*cosh(d*x + c)^3 - (a^2 + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))*log(cos
h(d*x + c) + sinh(d*x + c) + 1) - ((a^2 + 2*b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)
^3 + (a^2 + 2*b^2)*sinh(d*x + c)^4 - 2*(a^2 + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*b^2)*cosh(d*x + c)^2 - a^
2 - 2*b^2)*sinh(d*x + c)^2 + a^2 + 2*b^2 + 4*((a^2 + 2*b^2)*cosh(d*x + c)^3 - (a^2 + 2*b^2)*cosh(d*x + c))*sin
h(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 2*(3*a^2*cosh(d*x + c)^2 - 4*a*b*cosh(d*x + c) + a^2)*sin
h(d*x + c))/(a^3*d*cosh(d*x + c)^4 + 4*a^3*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^3*d*sinh(d*x + c)^4 - 2*a^3*d*c
osh(d*x + c)^2 + a^3*d + 2*(3*a^3*d*cosh(d*x + c)^2 - a^3*d)*sinh(d*x + c)^2 + 4*(a^3*d*cosh(d*x + c)^3 - a^3*
d*cosh(d*x + c))*sinh(d*x + c))

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giac [B]  time = 1.85, size = 221, normalized size = 1.99 \[ -\frac {\frac {{\left (a^{2} e^{c} + 2 \, b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left (e^{\left (d x + c\right )} + 1\right )}{a^{3}} - \frac {{\left (a^{2} e^{c} + 2 \, b^{2} e^{c}\right )} e^{\left (-c\right )} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right )}{a^{3}} + \frac {2 \, {\left (a^{2} b e^{c} + b^{3} e^{c}\right )} e^{\left (-c\right )} \log \left (\frac {{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} - 2 \, \sqrt {a^{2} + b^{2}} e^{c} \right |}}{{\left | 2 \, b e^{\left (d x + 2 \, c\right )} + 2 \, a e^{c} + 2 \, \sqrt {a^{2} + b^{2}} e^{c} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {2 \, {\left (a e^{\left (3 \, d x + 3 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a e^{\left (d x + c\right )} + 2 \, b\right )}}{a^{2} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 162, normalized size = 1.46 \[ \frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{2 d \,a^{2}}-\frac {2 b \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{d \,a^{3}}-\frac {1}{8 d a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{3}}+\frac {b}{2 d \,a^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d/a*tanh(1/2*d*x+1/2*c)^2+1/2/d/a^2*tanh(1/2*d*x+1/2*c)*b-2/d*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(2*a*tanh(
1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-1/8/d/a/tanh(1/2*d*x+1/2*c)^2+1/2/d/a*ln(tanh(1/2*d*x+1/2*c))+1/d/a^3*ln(
tanh(1/2*d*x+1/2*c))*b^2+1/2/d*b/a^2/tanh(1/2*d*x+1/2*c)

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maxima [B]  time = 0.40, size = 217, normalized size = 1.95 \[ \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-3 \, d x - 3 \, c\right )} - 2 \, b}{{\left (2 \, a^{2} e^{\left (-2 \, d x - 2 \, c\right )} - a^{2} e^{\left (-4 \, d x - 4 \, c\right )} - a^{2}\right )} d} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{2 \, a^{3} d} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{2 \, a^{3} d} - \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.69, size = 628, normalized size = 5.66 \[ \frac {{\mathrm {e}}^{c+d\,x}}{a\,d-a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}-\frac {2\,{\mathrm {e}}^{c+d\,x}}{a\,d-2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+a\,d\,{\mathrm {e}}^{4\,c+4\,d\,x}}-\frac {2\,b}{a^2\,d-a^2\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {\ln \left (4\,a^4+8\,b^4+12\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-12\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{2\,a\,d}-\frac {\ln \left (4\,a^4+8\,b^4+12\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+12\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{2\,a\,d}+\frac {b^2\,\ln \left (4\,a^4+8\,b^4+12\,a^2\,b^2-4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-12\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^3\,d}-\frac {b^2\,\ln \left (4\,a^4+8\,b^4+12\,a^2\,b^2+4\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+12\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{a^3\,d}-\frac {b\,\ln \left (32\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,a\,b^3-16\,a^3\,b-8\,b^3\,\sqrt {a^2+b^2}+8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,a^2\,b\,\sqrt {a^2+b^2}+40\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}+24\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^3\,d}+\frac {b\,\ln \left (8\,b^3\,\sqrt {a^2+b^2}-16\,a\,b^3-16\,a^3\,b+32\,a^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+8\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+16\,a^2\,b\,\sqrt {a^2+b^2}+40\,a^2\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,a^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}-24\,a\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2+b^2}\right )\,\sqrt {a^2+b^2}}{a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(c + d*x)/(a*d - a*d*exp(2*c + 2*d*x)) - (2*exp(c + d*x))/(a*d - 2*a*d*exp(2*c + 2*d*x) + a*d*exp(4*c + 4*d
*x)) - (2*b)/(a^2*d - a^2*d*exp(2*c + 2*d*x)) + log(4*a^4 + 8*b^4 + 12*a^2*b^2 - 4*a^4*exp(d*x)*exp(c) - 8*b^4
*exp(d*x)*exp(c) - 12*a^2*b^2*exp(d*x)*exp(c))/(2*a*d) - log(4*a^4 + 8*b^4 + 12*a^2*b^2 + 4*a^4*exp(d*x)*exp(c
) + 8*b^4*exp(d*x)*exp(c) + 12*a^2*b^2*exp(d*x)*exp(c))/(2*a*d) + (b^2*log(4*a^4 + 8*b^4 + 12*a^2*b^2 - 4*a^4*
exp(d*x)*exp(c) - 8*b^4*exp(d*x)*exp(c) - 12*a^2*b^2*exp(d*x)*exp(c)))/(a^3*d) - (b^2*log(4*a^4 + 8*b^4 + 12*a
^2*b^2 + 4*a^4*exp(d*x)*exp(c) + 8*b^4*exp(d*x)*exp(c) + 12*a^2*b^2*exp(d*x)*exp(c)))/(a^3*d) - (b*log(32*a^4*
exp(d*x)*exp(c) - 16*a*b^3 - 16*a^3*b - 8*b^3*(a^2 + b^2)^(1/2) + 8*b^4*exp(d*x)*exp(c) - 16*a^2*b*(a^2 + b^2)
^(1/2) + 40*a^2*b^2*exp(d*x)*exp(c) + 32*a^3*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2) + 24*a*b^2*exp(d*x)*exp(c)*(a^2
+ b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^3*d) + (b*log(8*b^3*(a^2 + b^2)^(1/2) - 16*a*b^3 - 16*a^3*b + 32*a^4*exp(
d*x)*exp(c) + 8*b^4*exp(d*x)*exp(c) + 16*a^2*b*(a^2 + b^2)^(1/2) + 40*a^2*b^2*exp(d*x)*exp(c) - 32*a^3*exp(d*x
)*exp(c)*(a^2 + b^2)^(1/2) - 24*a*b^2*exp(d*x)*exp(c)*(a^2 + b^2)^(1/2))*(a^2 + b^2)^(1/2))/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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