Integrand size = 15, antiderivative size = 59 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=-\frac {(2 a+3 b) x}{2 b^2}+\frac {(a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{\sqrt {a} b^2}+\frac {\cosh (x) \sinh (x)}{2 b} \]
-1/2*(2*a+3*b)*x/b^2+1/2*cosh(x)*sinh(x)/b+(a+b)^(3/2)*arctanh(a^(1/2)*tan h(x)/(a+b)^(1/2))/b^2/a^(1/2)
Time = 0.10 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=\frac {-4 a x-6 b x+\frac {4 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \tanh (x)}{\sqrt {a+b}}\right )}{\sqrt {a}}+b \sinh (2 x)}{4 b^2} \]
(-4*a*x - 6*b*x + (4*(a + b)^(3/2)*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]]) /Sqrt[a] + b*Sinh[2*x])/(4*b^2)
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3670, 316, 25, 397, 219, 221}
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.
| \(\displaystyle \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (\frac {\pi }{2}+i x\right )^4}{a+b \sin \left (\frac {\pi }{2}+i x\right )^2}dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \int \frac {1}{\left (1-\coth ^2(x)\right )^2 \left (a-(a+b) \coth ^2(x)\right )}d\coth (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle -\frac {\int -\frac {(a+b) \coth ^2(x)+a+2 b}{\left (1-\coth ^2(x)\right ) \left (a-(a+b) \coth ^2(x)\right )}d\coth (x)}{2 b}-\frac {\coth (x)}{2 b \left (1-\coth ^2(x)\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(a+b) \coth ^2(x)+a+2 b}{\left (1-\coth ^2(x)\right ) \left (a-(a+b) \coth ^2(x)\right )}d\coth (x)}{2 b}-\frac {\coth (x)}{2 b \left (1-\coth ^2(x)\right )}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {2 (a+b)^2 \int \frac {1}{a-(a+b) \coth ^2(x)}d\coth (x)}{b}-\frac {(2 a+3 b) \int \frac {1}{1-\coth ^2(x)}d\coth (x)}{b}}{2 b}-\frac {\coth (x)}{2 b \left (1-\coth ^2(x)\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {2 (a+b)^2 \int \frac {1}{a-(a+b) \coth ^2(x)}d\coth (x)}{b}-\frac {(2 a+3 b) \text {arctanh}(\coth (x))}{b}}{2 b}-\frac {\coth (x)}{2 b \left (1-\coth ^2(x)\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b} \coth (x)}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {(2 a+3 b) \text {arctanh}(\coth (x))}{b}}{2 b}-\frac {\coth (x)}{2 b \left (1-\coth ^2(x)\right )}\) |
(-(((2*a + 3*b)*ArcTanh[Coth[x]])/b) + (2*(a + b)^(3/2)*ArcTanh[(Sqrt[a + b]*Coth[x])/Sqrt[a]])/(Sqrt[a]*b))/(2*b) - Coth[x]/(2*b*(1 - Coth[x]^2))
3.1.14.3.1 Defintions of rubi rules used
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(184\) vs. \(2(47)=94\).
Time = 26.57 (sec) , antiderivative size = 185, normalized size of antiderivative = 3.14
| method | result | size |
| default | \(\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \left (\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {x}{2}\right )^{2}+2 \tanh \left (\frac {x}{2}\right ) \sqrt {a}+\sqrt {a +b}\right )}{4 \sqrt {a}\, \sqrt {a +b}}-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {x}{2}\right )^{2}-2 \tanh \left (\frac {x}{2}\right ) \sqrt {a}+\sqrt {a +b}\right )}{4 \sqrt {a}\, \sqrt {a +b}}\right )}{b^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (2 a +3 b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}\) | \(185\) |
| risch | \(-\frac {a x}{b^{2}}-\frac {3 x}{2 b}+\frac {{\mathrm e}^{2 x}}{8 b}-\frac {{\mathrm e}^{-2 x}}{8 b}+\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 x}-\frac {2 \sqrt {a \left (a +b \right )}-2 a -b}{b}\right )}{2 b^{2}}+\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 x}-\frac {2 \sqrt {a \left (a +b \right )}-2 a -b}{b}\right )}{2 a b}-\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 x}+\frac {2 \sqrt {a \left (a +b \right )}+2 a +b}{b}\right )}{2 b^{2}}-\frac {\sqrt {a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 x}+\frac {2 \sqrt {a \left (a +b \right )}+2 a +b}{b}\right )}{2 a b}\) | \(189\) |
2/b^2*(a^2+2*a*b+b^2)*(1/4/a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^ 2+2*tanh(1/2*x)*a^(1/2)+(a+b)^(1/2))-1/4/a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2 )*tanh(1/2*x)^2-2*tanh(1/2*x)*a^(1/2)+(a+b)^(1/2)))-1/2/b/(tanh(1/2*x)+1)^ 2+1/2/b/(tanh(1/2*x)+1)+1/2/b^2*(-2*a-3*b)*ln(tanh(1/2*x)+1)+1/2/b/(tanh(1 /2*x)-1)^2+1/2/b/(tanh(1/2*x)-1)+1/2*(2*a+3*b)/b^2*ln(tanh(1/2*x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (47) = 94\).
Time = 0.28 (sec) , antiderivative size = 568, normalized size of antiderivative = 9.63 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=\left [\frac {b \cosh \left (x\right )^{4} + 4 \, b \cosh \left (x\right ) \sinh \left (x\right )^{3} + b \sinh \left (x\right )^{4} - 4 \, {\left (2 \, a + 3 \, b\right )} x \cosh \left (x\right )^{2} + 2 \, {\left (3 \, b \cosh \left (x\right )^{2} - 2 \, {\left (2 \, a + 3 \, b\right )} x\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2}\right )} \sqrt {\frac {a + b}{a}} \log \left (\frac {b^{2} \cosh \left (x\right )^{4} + 4 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + b^{2} \sinh \left (x\right )^{4} + 2 \, {\left (2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, b^{2} \cosh \left (x\right )^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{2} + 8 \, a^{2} + 8 \, a b + b^{2} + 4 \, {\left (b^{2} \cosh \left (x\right )^{3} + {\left (2 \, a b + b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) - 4 \, {\left (a b \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) \sinh \left (x\right ) + a b \sinh \left (x\right )^{2} + 2 \, a^{2} + a b\right )} \sqrt {\frac {a + b}{a}}}{b \cosh \left (x\right )^{4} + 4 \, b \cosh \left (x\right ) \sinh \left (x\right )^{3} + b \sinh \left (x\right )^{4} + 2 \, {\left (2 \, a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, b \cosh \left (x\right )^{2} + 2 \, a + b\right )} \sinh \left (x\right )^{2} + 4 \, {\left (b \cosh \left (x\right )^{3} + {\left (2 \, a + b\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + b}\right ) + 4 \, {\left (b \cosh \left (x\right )^{3} - 2 \, {\left (2 \, a + 3 \, b\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right ) - b}{8 \, {\left (b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2}\right )}}, \frac {b \cosh \left (x\right )^{4} + 4 \, b \cosh \left (x\right ) \sinh \left (x\right )^{3} + b \sinh \left (x\right )^{4} - 4 \, {\left (2 \, a + 3 \, b\right )} x \cosh \left (x\right )^{2} + 2 \, {\left (3 \, b \cosh \left (x\right )^{2} - 2 \, {\left (2 \, a + 3 \, b\right )} x\right )} \sinh \left (x\right )^{2} + 8 \, {\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2}\right )} \sqrt {-\frac {a + b}{a}} \arctan \left (\frac {{\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2} + 2 \, a + b\right )} \sqrt {-\frac {a + b}{a}}}{2 \, {\left (a + b\right )}}\right ) + 4 \, {\left (b \cosh \left (x\right )^{3} - 2 \, {\left (2 \, a + 3 \, b\right )} x \cosh \left (x\right )\right )} \sinh \left (x\right ) - b}{8 \, {\left (b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2}\right )}}\right ] \]
[1/8*(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 4*(2*a + 3*b)*x* cosh(x)^2 + 2*(3*b*cosh(x)^2 - 2*(2*a + 3*b)*x)*sinh(x)^2 + 4*((a + b)*cos h(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2)*sqrt((a + b)/a)*lo g((b^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*(2*a*b + b^ 2)*cosh(x)^2 + 2*(3*b^2*cosh(x)^2 + 2*a*b + b^2)*sinh(x)^2 + 8*a^2 + 8*a*b + b^2 + 4*(b^2*cosh(x)^3 + (2*a*b + b^2)*cosh(x))*sinh(x) - 4*(a*b*cosh(x )^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 + 2*a^2 + a*b)*sqrt((a + b)/a) )/(b*cosh(x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 + 2*(2*a + b)*cosh(x) ^2 + 2*(3*b*cosh(x)^2 + 2*a + b)*sinh(x)^2 + 4*(b*cosh(x)^3 + (2*a + b)*co sh(x))*sinh(x) + b)) + 4*(b*cosh(x)^3 - 2*(2*a + 3*b)*x*cosh(x))*sinh(x) - b)/(b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2), 1/8*(b*cosh(x )^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 - 4*(2*a + 3*b)*x*cosh(x)^2 + 2* (3*b*cosh(x)^2 - 2*(2*a + 3*b)*x)*sinh(x)^2 + 8*((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2)*sqrt(-(a + b)/a)*arctan(1/2*(b*c osh(x)^2 + 2*b*cosh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(-(a + b)/a)/( a + b)) + 4*(b*cosh(x)^3 - 2*(2*a + 3*b)*x*cosh(x))*sinh(x) - b)/(b^2*cosh (x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2)]
Timed out. \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (47) = 94\).
Time = 0.29 (sec) , antiderivative size = 348, normalized size of antiderivative = 5.90 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=\frac {{\left (2 \, a + b\right )} \log \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{4 \, \sqrt {{\left (a + b\right )} a} b} - \frac {3 \, \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{16 \, \sqrt {{\left (a + b\right )} a}} - \frac {{\left (2 \, a + b\right )} x}{b^{2}} - \frac {x}{b} + \frac {e^{\left (2 \, x\right )}}{8 \, b} - \frac {e^{\left (-2 \, x\right )}}{8 \, b} + \frac {{\left (2 \, a + b\right )} \log \left (b e^{\left (4 \, x\right )} + 2 \, {\left (2 \, a + b\right )} e^{\left (2 \, x\right )} + b\right )}{8 \, b^{2}} - \frac {{\left (2 \, a + b\right )} \log \left (2 \, {\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} + b e^{\left (-4 \, x\right )} + b\right )}{8 \, b^{2}} + \frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \log \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{32 \, \sqrt {{\left (a + b\right )} a} b^{2}} - \frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \log \left (\frac {b e^{\left (-2 \, x\right )} + 2 \, a + b - 2 \, \sqrt {{\left (a + b\right )} a}}{b e^{\left (-2 \, x\right )} + 2 \, a + b + 2 \, \sqrt {{\left (a + b\right )} a}}\right )}{32 \, \sqrt {{\left (a + b\right )} a} b^{2}} \]
1/4*(2*a + b)*log((b*e^(2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(2*x) + 2 *a + b + 2*sqrt((a + b)*a)))/(sqrt((a + b)*a)*b) - 3/16*log((b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2*sqrt((a + b)*a)))/s qrt((a + b)*a) - (2*a + b)*x/b^2 - x/b + 1/8*e^(2*x)/b - 1/8*e^(-2*x)/b + 1/8*(2*a + b)*log(b*e^(4*x) + 2*(2*a + b)*e^(2*x) + b)/b^2 - 1/8*(2*a + b) *log(2*(2*a + b)*e^(-2*x) + b*e^(-4*x) + b)/b^2 + 1/32*(8*a^2 + 8*a*b + b^ 2)*log((b*e^(2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(2*x) + 2*a + b + 2* sqrt((a + b)*a)))/(sqrt((a + b)*a)*b^2) - 1/32*(8*a^2 + 8*a*b + b^2)*log(( b*e^(-2*x) + 2*a + b - 2*sqrt((a + b)*a))/(b*e^(-2*x) + 2*a + b + 2*sqrt(( a + b)*a)))/(sqrt((a + b)*a)*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (47) = 94\).
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.75 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=-\frac {{\left (2 \, a + 3 \, b\right )} x}{2 \, b^{2}} + \frac {e^{\left (2 \, x\right )}}{8 \, b} + \frac {{\left (4 \, a e^{\left (2 \, x\right )} + 6 \, b e^{\left (2 \, x\right )} - b\right )} e^{\left (-2 \, x\right )}}{8 \, b^{2}} + \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac {b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} b^{2}} \]
-1/2*(2*a + 3*b)*x/b^2 + 1/8*e^(2*x)/b + 1/8*(4*a*e^(2*x) + 6*b*e^(2*x) - b)*e^(-2*x)/b^2 + (a^2 + 2*a*b + b^2)*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqr t(-a^2 - a*b))/(sqrt(-a^2 - a*b)*b^2)
Time = 2.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.47 \[ \int \frac {\sinh ^4(x)}{a+b \cosh ^2(x)} \, dx=\frac {{\mathrm {e}}^{2\,x}}{8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b}-\frac {x\,\left (2\,a+3\,b\right )}{2\,b^2}+\frac {\ln \left (-\frac {4\,{\mathrm {e}}^{2\,x}\,{\left (a+b\right )}^2}{b^3}-\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,b^3}\right )\,{\left (a+b\right )}^{3/2}}{2\,\sqrt {a}\,b^2}-\frac {\ln \left (\frac {2\,{\left (a+b\right )}^{3/2}\,\left (b+2\,a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{\sqrt {a}\,b^3}-\frac {4\,{\mathrm {e}}^{2\,x}\,{\left (a+b\right )}^2}{b^3}\right )\,{\left (a+b\right )}^{3/2}}{2\,\sqrt {a}\,b^2} \]
exp(2*x)/(8*b) - exp(-2*x)/(8*b) - (x*(2*a + 3*b))/(2*b^2) + (log(- (4*exp (2*x)*(a + b)^2)/b^3 - (2*(a + b)^(3/2)*(b + 2*a*exp(2*x) + b*exp(2*x)))/( a^(1/2)*b^3))*(a + b)^(3/2))/(2*a^(1/2)*b^2) - (log((2*(a + b)^(3/2)*(b + 2*a*exp(2*x) + b*exp(2*x)))/(a^(1/2)*b^3) - (4*exp(2*x)*(a + b)^2)/b^3)*(a + b)^(3/2))/(2*a^(1/2)*b^2)