\(\int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx\) [249]

Optimal result

Integrand size = 15, antiderivative size = 89 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=\frac {b B \arctan (\sinh (x))}{a^2+b^2}-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {a B \log (\cosh (x))}{a^2+b^2}-\frac {a B \log (a+b \sinh (x))}{a^2+b^2} \]

[Out]

b*B*arctan(sinh(x))/(a^2+b^2)+a*B*ln(cosh(x))/(a^2+b^2)-a*B*ln(a+b*sinh(x))/(a^2+b^2)-2*A*arctanh((b-a*tanh(1/
2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4486, 2739, 632, 212, 2800, 815, 649, 209, 266} \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {b B \arctan (\sinh (x))}{a^2+b^2}-\frac {a B \log (a+b \sinh (x))}{a^2+b^2}+\frac {a B \log (\cosh (x))}{a^2+b^2} \]

[In]

Int[(A + B*Tanh[x])/(a + b*Sinh[x]),x]

[Out]

(b*B*ArcTan[Sinh[x]])/(a^2 + b^2) - (2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] + (a*B*Lo
g[Cosh[x]])/(a^2 + b^2) - (a*B*Log[a + b*Sinh[x]])/(a^2 + b^2)

Rule 209

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

Rule 212

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 632

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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2739

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 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {A}{a+b \sinh (x)}+\frac {B \tanh (x)}{a+b \sinh (x)}\right ) \, dx \\ & = A \int \frac {1}{a+b \sinh (x)} \, dx+B \int \frac {\tanh (x)}{a+b \sinh (x)} \, dx \\ & = (2 A) \text {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )-B \text {Subst}\left (\int \frac {x}{(a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (x)\right ) \\ & = -\left ((4 A) \text {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )\right )-B \text {Subst}\left (\int \left (\frac {a}{\left (a^2+b^2\right ) (a+x)}+\frac {-b^2-a x}{\left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (x)\right ) \\ & = -\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {a B \log (a+b \sinh (x))}{a^2+b^2}-\frac {B \text {Subst}\left (\int \frac {-b^2-a x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2} \\ & = -\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {a B \log (a+b \sinh (x))}{a^2+b^2}+\frac {(a B) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2}+\frac {\left (b^2 B\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \sinh (x)\right )}{a^2+b^2} \\ & = \frac {b B \arctan (\sinh (x))}{a^2+b^2}-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {a B \log (\cosh (x))}{a^2+b^2}-\frac {a B \log (a+b \sinh (x))}{a^2+b^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=\frac {\cosh (x) \left (2 b \sqrt {-a^2-b^2} B \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+2 A \left (a^2+b^2\right ) \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )+a \sqrt {-a^2-b^2} B (\log (\cosh (x))-\log (a+b \sinh (x)))\right ) (A+B \tanh (x))}{(a-i b) (a+i b) \sqrt {-a^2-b^2} (A \cosh (x)+B \sinh (x))} \]

[In]

Integrate[(A + B*Tanh[x])/(a + b*Sinh[x]),x]

[Out]

(Cosh[x]*(2*b*Sqrt[-a^2 - b^2]*B*ArcTan[Tanh[x/2]] + 2*A*(a^2 + b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]
] + a*Sqrt[-a^2 - b^2]*B*(Log[Cosh[x]] - Log[a + b*Sinh[x]]))*(A + B*Tanh[x]))/((a - I*b)*(a + I*b)*Sqrt[-a^2
- b^2]*(A*Cosh[x] + B*Sinh[x]))

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {-B a \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )-\frac {2 \left (-a^{2} A -A \,b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{a^{2}+b^{2}}+\frac {2 B \left (\frac {a \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )}{2}+b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )\right )}{a^{2}+b^{2}}\)	\(117\)
risch	\(-\frac {2 x B a}{a^{2}+b^{2}}-\frac {2 x \,a^{3} B}{-a^{4}-2 a^{2} b^{2}-b^{4}}-\frac {2 x B a \,b^{2}}{-a^{4}-2 a^{2} b^{2}-b^{4}}+\frac {i B \ln \left ({\mathrm e}^{x}+i\right ) b}{a^{2}+b^{2}}+\frac {B \ln \left ({\mathrm e}^{x}+i\right ) a}{a^{2}+b^{2}}-\frac {i B \ln \left ({\mathrm e}^{x}-i\right ) b}{a^{2}+b^{2}}+\frac {B \ln \left ({\mathrm e}^{x}-i\right ) a}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {A a -\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) a B}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a -\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {A^{2} a^{2}+A^{2} b^{2}}}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {A a +\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) a B}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {A a +\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {A^{2} a^{2}+A^{2} b^{2}}}{a^{2}+b^{2}}\)	\(362\)

[In]

int((A+B*tanh(x))/(a+b*sinh(x)),x,method=_RETURNVERBOSE)

[Out]

2/(a^2+b^2)*(-1/2*B*a*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)-(-A*a^2-A*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*ta
nh(1/2*x)-2*b)/(a^2+b^2)^(1/2)))+2*B/(a^2+b^2)*(1/2*a*ln(1+tanh(1/2*x)^2)+b*arctan(tanh(1/2*x)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (85) = 170\).

Time = 1.19 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=\frac {2 \, B b \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - B a \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + B a \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \sqrt {a^{2} + b^{2}} A \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right )}{a^{2} + b^{2}} \]

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x, algorithm="fricas")

[Out]

(2*B*b*arctan(cosh(x) + sinh(x)) - B*a*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) + B*a*log(2*cosh(x)/(cosh(x)
- sinh(x))) + sqrt(a^2 + b^2)*A*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cos
h(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x)
+ 2*(b*cosh(x) + a)*sinh(x) - b)))/(a^2 + b^2)

Sympy [F]

\[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=\int \frac {A + B \tanh {\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \]

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x)

[Out]

Integral((A + B*tanh(x))/(a + b*sinh(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.40 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=-B {\left (\frac {2 \, b \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} + \frac {a \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} - \frac {a \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}}\right )} + \frac {A \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} \]

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x, algorithm="maxima")

[Out]

-B*(2*b*arctan(e^(-x))/(a^2 + b^2) + a*log(-2*a*e^(-x) + b*e^(-2*x) - b)/(a^2 + b^2) - a*log(e^(-2*x) + 1)/(a^
2 + b^2)) + A*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=\frac {2 \, B b \arctan \left (e^{x}\right )}{a^{2} + b^{2}} + \frac {B a \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{2} + b^{2}} - \frac {B a \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{a^{2} + b^{2}} + \frac {A \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} \]

[In]

integrate((A+B*tanh(x))/(a+b*sinh(x)),x, algorithm="giac")

[Out]

2*B*b*arctan(e^x)/(a^2 + b^2) + B*a*log(e^(2*x) + 1)/(a^2 + b^2) - B*a*log(abs(b*e^(2*x) + 2*a*e^x - b))/(a^2
+ b^2) + A*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)

Mupad [B] (verification not implemented)

Time = 9.69 (sec) , antiderivative size = 914, normalized size of antiderivative = 10.27 \[ \int \frac {A+B \tanh (x)}{a+b \sinh (x)} \, dx=-\frac {\ln \left (\frac {32\,B\,\left (A^2\,a\,b+{\mathrm {e}}^x\,A^2\,b^2-4\,{\mathrm {e}}^x\,A\,B\,a^2+2\,A\,B\,a\,b-{\mathrm {e}}^x\,A\,B\,b^2-4\,{\mathrm {e}}^x\,B^2\,a^2+B^2\,a\,b\right )}{b^5}-\frac {\left (\frac {32\,\left (-A^2\,a^2\,b-2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3+8\,{\mathrm {e}}^x\,A\,B\,a^3-4\,A\,B\,a^2\,b+2\,{\mathrm {e}}^x\,A\,B\,a\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^3-3\,B^2\,a^2\,b-5\,{\mathrm {e}}^x\,B^2\,a\,b^2+B^2\,b^3\right )}{b^5}-\frac {\left (A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a^3+B\,a\,b^2\right )\,\left (a\,b^5\,\left (64\,A-128\,B\right )+a^5\,b\,\left (64\,A-128\,B\right )+96\,b^6\,{\mathrm {e}}^x\,\left (A-3\,B\right )+a^3\,b^3\,\left (128\,A-256\,B\right )-128\,{\mathrm {e}}^x\,\left (A-2\,B\right )\,{\left (a^2+b^2\right )}^3+192\,a^2\,b^4\,{\mathrm {e}}^x\,\left (A-3\,B\right )+96\,a^4\,b^2\,{\mathrm {e}}^x\,\left (A-3\,B\right )-96\,A\,a^2\,b\,\sqrt {{\left (a^2+b^2\right )}^3}+128\,A\,a^3\,{\mathrm {e}}^x\,\sqrt {{\left (a^2+b^2\right )}^3}+32\,A\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {{\left (a^2+b^2\right )}^3}\right )}{b^5\,{\left (a^2+b^2\right )}^3}\right )\,\left (A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a^3+B\,a\,b^2\right )}{{\left (a^2+b^2\right )}^2}\right )\,\left (A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a^3+B\,a\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {\ln \left (\frac {32\,B\,\left (A^2\,a\,b+{\mathrm {e}}^x\,A^2\,b^2-4\,{\mathrm {e}}^x\,A\,B\,a^2+2\,A\,B\,a\,b-{\mathrm {e}}^x\,A\,B\,b^2-4\,{\mathrm {e}}^x\,B^2\,a^2+B^2\,a\,b\right )}{b^5}-\frac {\left (\frac {32\,\left (-A^2\,a^2\,b-2\,{\mathrm {e}}^x\,A^2\,a\,b^2+A^2\,b^3+8\,{\mathrm {e}}^x\,A\,B\,a^3-4\,A\,B\,a^2\,b+2\,{\mathrm {e}}^x\,A\,B\,a\,b^2+4\,{\mathrm {e}}^x\,B^2\,a^3-3\,B^2\,a^2\,b-5\,{\mathrm {e}}^x\,B^2\,a\,b^2+B^2\,b^3\right )}{b^5}-\frac {\left (B\,a^3-A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a\,b^2\right )\,\left (a\,b^5\,\left (64\,A-128\,B\right )+a^5\,b\,\left (64\,A-128\,B\right )+96\,b^6\,{\mathrm {e}}^x\,\left (A-3\,B\right )+a^3\,b^3\,\left (128\,A-256\,B\right )-128\,{\mathrm {e}}^x\,\left (A-2\,B\right )\,{\left (a^2+b^2\right )}^3+192\,a^2\,b^4\,{\mathrm {e}}^x\,\left (A-3\,B\right )+96\,a^4\,b^2\,{\mathrm {e}}^x\,\left (A-3\,B\right )+96\,A\,a^2\,b\,\sqrt {{\left (a^2+b^2\right )}^3}-128\,A\,a^3\,{\mathrm {e}}^x\,\sqrt {{\left (a^2+b^2\right )}^3}-32\,A\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {{\left (a^2+b^2\right )}^3}\right )}{b^5\,{\left (a^2+b^2\right )}^3}\right )\,\left (B\,a^3-A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a\,b^2\right )}{{\left (a^2+b^2\right )}^2}\right )\,\left (B\,a^3-A\,\sqrt {{\left (a^2+b^2\right )}^3}+B\,a\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {B\,\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{a-b\,1{}\mathrm {i}}+\frac {B\,\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}}{-b+a\,1{}\mathrm {i}} \]

[In]

int((A + B*tanh(x))/(a + b*sinh(x)),x)

[Out]

(B*log(exp(x) + 1i))/(a - b*1i) - (log((32*B*(A^2*b^2*exp(x) - 4*B^2*a^2*exp(x) + A^2*a*b + B^2*a*b - 4*A*B*a^
2*exp(x) - A*B*b^2*exp(x) + 2*A*B*a*b))/b^5 - (((32*(A^2*b^3 + B^2*b^3 - A^2*a^2*b - 3*B^2*a^2*b + 4*B^2*a^3*e
xp(x) - 5*B^2*a*b^2*exp(x) - 4*A*B*a^2*b + 8*A*B*a^3*exp(x) - 2*A^2*a*b^2*exp(x) + 2*A*B*a*b^2*exp(x)))/b^5 -
((B*a^3 - A*((a^2 + b^2)^3)^(1/2) + B*a*b^2)*(a*b^5*(64*A - 128*B) + a^5*b*(64*A - 128*B) + 96*b^6*exp(x)*(A -
3*B) + a^3*b^3*(128*A - 256*B) - 128*exp(x)*(A - 2*B)*(a^2 + b^2)^3 + 192*a^2*b^4*exp(x)*(A - 3*B) + 96*a^4*b
^2*exp(x)*(A - 3*B) + 96*A*a^2*b*((a^2 + b^2)^3)^(1/2) - 128*A*a^3*exp(x)*((a^2 + b^2)^3)^(1/2) - 32*A*a*b^2*e
xp(x)*((a^2 + b^2)^3)^(1/2)))/(b^5*(a^2 + b^2)^3))*(B*a^3 - A*((a^2 + b^2)^3)^(1/2) + B*a*b^2))/(a^2 + b^2)^2)
*(B*a^3 - A*((a^2 + b^2)^3)^(1/2) + B*a*b^2))/(a^4 + b^4 + 2*a^2*b^2) - (log((32*B*(A^2*b^2*exp(x) - 4*B^2*a^2
*exp(x) + A^2*a*b + B^2*a*b - 4*A*B*a^2*exp(x) - A*B*b^2*exp(x) + 2*A*B*a*b))/b^5 - (((32*(A^2*b^3 + B^2*b^3 -
A^2*a^2*b - 3*B^2*a^2*b + 4*B^2*a^3*exp(x) - 5*B^2*a*b^2*exp(x) - 4*A*B*a^2*b + 8*A*B*a^3*exp(x) - 2*A^2*a*b^
2*exp(x) + 2*A*B*a*b^2*exp(x)))/b^5 - ((A*((a^2 + b^2)^3)^(1/2) + B*a^3 + B*a*b^2)*(a*b^5*(64*A - 128*B) + a^5
*b*(64*A - 128*B) + 96*b^6*exp(x)*(A - 3*B) + a^3*b^3*(128*A - 256*B) - 128*exp(x)*(A - 2*B)*(a^2 + b^2)^3 + 1
92*a^2*b^4*exp(x)*(A - 3*B) + 96*a^4*b^2*exp(x)*(A - 3*B) - 96*A*a^2*b*((a^2 + b^2)^3)^(1/2) + 128*A*a^3*exp(x
)*((a^2 + b^2)^3)^(1/2) + 32*A*a*b^2*exp(x)*((a^2 + b^2)^3)^(1/2)))/(b^5*(a^2 + b^2)^3))*(A*((a^2 + b^2)^3)^(1
/2) + B*a^3 + B*a*b^2))/(a^2 + b^2)^2)*(A*((a^2 + b^2)^3)^(1/2) + B*a^3 + B*a*b^2))/(a^4 + b^4 + 2*a^2*b^2) +
(B*log(exp(x) - 1i)*1i)/(a*1i - b)