Integrand size = 18, antiderivative size = 89 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=-\frac {8 a \text {arctanh}\left (\frac {b-2 a \tanh (c+d x)}{\sqrt {4 a^2+b^2}}\right )}{\left (4 a^2+b^2\right )^{3/2} d}-\frac {2 b \cosh (2 c+2 d x)}{\left (4 a^2+b^2\right ) d (2 a+b \sinh (2 c+2 d x))} \]
-8*a*arctanh((b-2*a*tanh(d*x+c))/(4*a^2+b^2)^(1/2))/(4*a^2+b^2)^(3/2)/d-2* b*cosh(2*d*x+2*c)/(4*a^2+b^2)/d/(2*a+b*sinh(2*d*x+2*c))
Time = 0.50 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=\frac {2 \left (-\frac {4 a \arctan \left (\frac {b-2 a \tanh (c+d x)}{\sqrt {-4 a^2-b^2}}\right )}{\left (-4 a^2-b^2\right )^{3/2}}-\frac {b \cosh (2 (c+d x))}{\left (4 a^2+b^2\right ) (2 a+b \sinh (2 (c+d x)))}\right )}{d} \]
(2*((-4*a*ArcTan[(b - 2*a*Tanh[c + d*x])/Sqrt[-4*a^2 - b^2]])/(-4*a^2 - b^ 2)^(3/2) - (b*Cosh[2*(c + d*x)])/((4*a^2 + b^2)*(2*a + b*Sinh[2*(c + d*x)] ))))/d
Time = 0.41 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3145, 3042, 3143, 27, 3042, 3139, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{(a+b \sinh (c+d x) \cosh (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a-i b \sin (i c+i d x) \cos (i c+i d x))^2}dx\) |
| \(\Big \downarrow \) 3145 |
| \(\displaystyle \int \frac {1}{\left (a+\frac {1}{2} b \sinh (2 c+2 d x)\right )^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-\frac {1}{2} i b \sin (2 i c+2 i d x)\right )^2}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle -\frac {4 \int -\frac {2 a}{2 a+b \sinh (2 c+2 d x)}dx}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {8 a \int \frac {1}{2 a+b \sinh (2 c+2 d x)}dx}{4 a^2+b^2}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}+\frac {8 a \int \frac {1}{2 a-i b \sin (2 i c+2 i d x)}dx}{4 a^2+b^2}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}-\frac {8 i a \int \frac {1}{-2 a \tanh ^2(c+d x)+2 b \tanh (c+d x)+2 a}d(i \tanh (c+d x))}{d \left (4 a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle -\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}+\frac {16 i a \int \frac {1}{\tanh ^2(c+d x)-4 \left (4 a^2+b^2\right )}d(4 i a \tanh (c+d x)-2 i b)}{d \left (4 a^2+b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {8 a \text {arctanh}\left (\frac {\tanh (c+d x)}{2 \sqrt {4 a^2+b^2}}\right )}{d \left (4 a^2+b^2\right )^{3/2}}-\frac {2 b \cosh (2 c+2 d x)}{d \left (4 a^2+b^2\right ) (2 a+b \sinh (2 c+2 d x))}\) |
(8*a*ArcTanh[Tanh[c + d*x]/(2*Sqrt[4*a^2 + b^2])])/((4*a^2 + b^2)^(3/2)*d) - (2*b*Cosh[2*c + 2*d*x])/((4*a^2 + b^2)*d*(2*a + b*Sinh[2*c + 2*d*x]))
3.9.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_ Symbol] :> Int[(a + b*(Sin[2*c + 2*d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(212\) vs. \(2(85)=170\).
Time = 15.76 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.39
| method | result | size |
| risch | \(\frac {8 a \,{\mathrm e}^{2 d x +2 c}-4 b}{\left (4 a^{2}+b^{2}\right ) d \left (4 a \,{\mathrm e}^{2 d x +2 c}+{\mathrm e}^{4 d x +4 c} b -b \right )}+\frac {4 a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a -16 a^{4}-8 a^{2} b^{2}-b^{4}}{b \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {4 a \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a +16 a^{4}+8 a^{2} b^{2}+b^{4}}{b \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} d}\) | \(213\) |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a \left (4 a^{2}+b^{2}\right )}+\frac {4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a^{2}+b^{2}}-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (4 a^{2}+b^{2}\right )}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a}-\frac {8 a^{2} \left (\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}+\frac {\ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}\right )}{4 a^{2}+b^{2}}}{d}\) | \(309\) |
| default | \(\frac {-\frac {2 \left (-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a \left (4 a^{2}+b^{2}\right )}+\frac {4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{4 a^{2}+b^{2}}-\frac {b^{2} \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \left (4 a^{2}+b^{2}\right )}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a}-\frac {8 a^{2} \left (\frac {\left (-4 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{2 \left (4 a^{2}+b^{2}\right )^{\frac {3}{2}} a}+\frac {\ln \left (-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +\sqrt {4 a^{2}+b^{2}}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{2 \sqrt {4 a^{2}+b^{2}}\, a}\right )}{4 a^{2}+b^{2}}}{d}\) | \(309\) |
4*(2*a*exp(2*d*x+2*c)-b)/(4*a^2+b^2)/d/(4*a*exp(2*d*x+2*c)+exp(4*d*x+4*c)* b-b)+4/(4*a^2+b^2)^(3/2)*a/d*ln(exp(2*d*x+2*c)+(2*(4*a^2+b^2)^(3/2)*a-16*a ^4-8*a^2*b^2-b^4)/b/(4*a^2+b^2)^(3/2))-4/(4*a^2+b^2)^(3/2)*a/d*ln(exp(2*d* x+2*c)+(2*(4*a^2+b^2)^(3/2)*a+16*a^4+8*a^2*b^2+b^4)/b/(4*a^2+b^2)^(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 765 vs. \(2 (88) = 176\).
Time = 0.26 (sec) , antiderivative size = 765, normalized size of antiderivative = 8.60 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=-\frac {4 \, {\left (4 \, a^{2} b + b^{3} - 2 \, {\left (4 \, a^{3} + a b^{2}\right )} \cosh \left (d x + c\right )^{2} - 4 \, {\left (4 \, a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - 2 \, {\left (4 \, a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )^{2} - {\left (a b \cosh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a b \sinh \left (d x + c\right )^{4} + 4 \, a^{2} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a b \cosh \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sinh \left (d x + c\right )^{2} - a b + 4 \, {\left (a b \cosh \left (d x + c\right )^{3} + 2 \, a^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \sqrt {4 \, a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, a b\right )} \sinh \left (d x + c\right )^{2} + 8 \, a^{2} + b^{2} + 4 \, {\left (b^{2} \cosh \left (d x + c\right )^{3} + 2 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 2 \, {\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} + 2 \, a\right )} \sqrt {4 \, a^{2} + b^{2}}}{b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} + 4 \, a \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, b \cosh \left (d x + c\right )^{2} + 2 \, a\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (b \cosh \left (d x + c\right )^{3} + 2 \, a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - b}\right )\right )}}{{\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{4} + 4 \, {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \sinh \left (d x + c\right )^{4} + 4 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d\right )} \sinh \left (d x + c\right )^{2} - {\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d + 4 \, {\left ({\left (16 \, a^{4} b + 8 \, a^{2} b^{3} + b^{5}\right )} d \cosh \left (d x + c\right )^{3} + 2 \, {\left (16 \, a^{5} + 8 \, a^{3} b^{2} + a b^{4}\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )} \]
-4*(4*a^2*b + b^3 - 2*(4*a^3 + a*b^2)*cosh(d*x + c)^2 - 4*(4*a^3 + a*b^2)* cosh(d*x + c)*sinh(d*x + c) - 2*(4*a^3 + a*b^2)*sinh(d*x + c)^2 - (a*b*cos h(d*x + c)^4 + 4*a*b*cosh(d*x + c)*sinh(d*x + c)^3 + a*b*sinh(d*x + c)^4 + 4*a^2*cosh(d*x + c)^2 + 2*(3*a*b*cosh(d*x + c)^2 + 2*a^2)*sinh(d*x + c)^2 - a*b + 4*(a*b*cosh(d*x + c)^3 + 2*a^2*cosh(d*x + c))*sinh(d*x + c))*sqrt (4*a^2 + b^2)*log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c) ^3 + b^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^ 2 + 2*a*b)*sinh(d*x + c)^2 + 8*a^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 + 2*a*b* cosh(d*x + c))*sinh(d*x + c) - 2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*si nh(d*x + c) + b*sinh(d*x + c)^2 + 2*a)*sqrt(4*a^2 + b^2))/(b*cosh(d*x + c) ^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 4*a*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c )^3 + 2*a*cosh(d*x + c))*sinh(d*x + c) - b)))/((16*a^4*b + 8*a^2*b^3 + b^5 )*d*cosh(d*x + c)^4 + 4*(16*a^4*b + 8*a^2*b^3 + b^5)*d*cosh(d*x + c)*sinh( d*x + c)^3 + (16*a^4*b + 8*a^2*b^3 + b^5)*d*sinh(d*x + c)^4 + 4*(16*a^5 + 8*a^3*b^2 + a*b^4)*d*cosh(d*x + c)^2 + 2*(3*(16*a^4*b + 8*a^2*b^3 + b^5)*d *cosh(d*x + c)^2 + 2*(16*a^5 + 8*a^3*b^2 + a*b^4)*d)*sinh(d*x + c)^2 - (16 *a^4*b + 8*a^2*b^3 + b^5)*d + 4*((16*a^4*b + 8*a^2*b^3 + b^5)*d*cosh(d*x + c)^3 + 2*(16*a^5 + 8*a^3*b^2 + a*b^4)*d*cosh(d*x + c))*sinh(d*x + c))
Timed out. \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.69 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=\frac {4 \, a \log \left (\frac {b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a - \sqrt {4 \, a^{2} + b^{2}}}{b e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, a + \sqrt {4 \, a^{2} + b^{2}}}\right )}{{\left (4 \, a^{2} + b^{2}\right )}^{\frac {3}{2}} d} - \frac {4 \, {\left (2 \, a e^{\left (-2 \, d x - 2 \, c\right )} + b\right )}}{{\left (4 \, a^{2} b + b^{3} + 4 \, {\left (4 \, a^{3} + a b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (4 \, a^{2} b + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]
4*a*log((b*e^(-2*d*x - 2*c) - 2*a - sqrt(4*a^2 + b^2))/(b*e^(-2*d*x - 2*c) - 2*a + sqrt(4*a^2 + b^2)))/((4*a^2 + b^2)^(3/2)*d) - 4*(2*a*e^(-2*d*x - 2*c) + b)/((4*a^2*b + b^3 + 4*(4*a^3 + a*b^2)*e^(-2*d*x - 2*c) - (4*a^2*b + b^3)*e^(-4*d*x - 4*c))*d)
Time = 0.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=\frac {4 \, {\left (\frac {a \log \left (\frac {{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a - 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 2 \, \sqrt {4 \, a^{2} + b^{2}} \right |}}\right )}{{\left (4 \, a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, a e^{\left (2 \, d x + 2 \, c\right )} - b}{{\left (4 \, a^{2} + b^{2}\right )} {\left (b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a e^{\left (2 \, d x + 2 \, c\right )} - b\right )}}\right )}}{d} \]
4*(a*log(abs(2*b*e^(2*d*x + 2*c) + 4*a - 2*sqrt(4*a^2 + b^2))/abs(2*b*e^(2 *d*x + 2*c) + 4*a + 2*sqrt(4*a^2 + b^2)))/(4*a^2 + b^2)^(3/2) + (2*a*e^(2* d*x + 2*c) - b)/((4*a^2 + b^2)*(b*e^(4*d*x + 4*c) + 4*a*e^(2*d*x + 2*c) - b)))/d
Time = 2.79 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.57 \[ \int \frac {1}{(a+b \cosh (c+d x) \sinh (c+d x))^2} \, dx=\frac {4\,a\,\ln \left (\frac {16\,a\,\left (b-2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {16\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{4\,a^2\,b+b^3}\right )}{d\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {4\,a\,\ln \left (-\frac {16\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}}{4\,a^2\,b+b^3}-\frac {16\,a\,\left (b-2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b\,{\left (4\,a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (4\,a^2+b^2\right )}^{3/2}}-\frac {\frac {4\,b^2}{d\,\left (4\,a^2\,b+b^3\right )}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{d\,\left (4\,a^2\,b+b^3\right )}}{4\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}-b+b\,{\mathrm {e}}^{4\,c+4\,d\,x}} \]
(4*a*log((16*a*(b - 2*a*exp(2*c + 2*d*x)))/(b*(4*a^2 + b^2)^(3/2)) - (16*a *exp(2*c + 2*d*x))/(4*a^2*b + b^3)))/(d*(4*a^2 + b^2)^(3/2)) - (4*a*log(- (16*a*exp(2*c + 2*d*x))/(4*a^2*b + b^3) - (16*a*(b - 2*a*exp(2*c + 2*d*x)) )/(b*(4*a^2 + b^2)^(3/2))))/(d*(4*a^2 + b^2)^(3/2)) - ((4*b^2)/(d*(4*a^2*b + b^3)) - (8*a*b*exp(2*c + 2*d*x))/(d*(4*a^2*b + b^3)))/(4*a*exp(2*c + 2* d*x) - b + b*exp(4*c + 4*d*x))