Integrand size = 23, antiderivative size = 124 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {\coth ^2(c+d x)}{2 a^2 d}+\frac {\log (\cosh (c+d x))}{(a+b)^2 d}+\frac {(a-2 b) \log (\tanh (c+d x))}{a^3 d}+\frac {b^2 (3 a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{2 a^3 (a+b)^2 d}-\frac {b^2}{2 a^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \]
-1/2*coth(d*x+c)^2/a^2/d+ln(cosh(d*x+c))/(a+b)^2/d+(a-2*b)*ln(tanh(d*x+c)) /a^3/d+1/2*b^2*(3*a+2*b)*ln(a+b*tanh(d*x+c)^2)/a^3/(a+b)^2/d-1/2*b^2/a^2/( a+b)/d/(a+b*tanh(d*x+c)^2)
Time = 1.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.75 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {-\frac {\coth ^2(c+d x)}{a^2}+\frac {b^3}{a^3 (a+b) \left (b+a \coth ^2(c+d x)\right )}+\frac {b^2 (3 a+2 b) \log \left (b+a \coth ^2(c+d x)\right )}{a^3 (a+b)^2}+\frac {2 \log (\sinh (c+d x))}{(a+b)^2}}{2 d} \]
(-(Coth[c + d*x]^2/a^2) + b^3/(a^3*(a + b)*(b + a*Coth[c + d*x]^2)) + (b^2 *(3*a + 2*b)*Log[b + a*Coth[c + d*x]^2])/(a^3*(a + b)^2) + (2*Log[Sinh[c + d*x]])/(a + b)^2)/(2*d)
Time = 0.36 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 26, 4153, 26, 354, 99, 2009}
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.
| \(\displaystyle \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i}{\tan (i c+i d x)^3 \left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {1}{\tan (i c+i d x)^3 \left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {i \int \frac {i \coth ^3(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\coth ^3(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\coth ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {(3 a+2 b) b^3}{a^3 (a+b)^2 \left (b \tanh ^2(c+d x)+a\right )}+\frac {b^3}{a^2 (a+b) \left (b \tanh ^2(c+d x)+a\right )^2}+\frac {\coth ^2(c+d x)}{a^2}+\frac {(a-2 b) \coth (c+d x)}{a^3}-\frac {1}{(a+b)^2 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^2 (3 a+2 b) \log \left (a+b \tanh ^2(c+d x)\right )}{a^3 (a+b)^2}+\frac {(a-2 b) \log \left (\tanh ^2(c+d x)\right )}{a^3}-\frac {b^2}{a^2 (a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\coth (c+d x)}{a^2}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\) |
(-(Coth[c + d*x]/a^2) + ((a - 2*b)*Log[Tanh[c + d*x]^2])/a^3 - Log[1 - Tan h[c + d*x]^2]/(a + b)^2 + (b^2*(3*a + 2*b)*Log[a + b*Tanh[c + d*x]^2])/(a^ 3*(a + b)^2) - b^2/(a^2*(a + b)*(a + b*Tanh[c + d*x]^2)))/(2*d)
3.2.88.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(-\frac {\frac {\left (-a +2 b \right ) \ln \left (\tanh \left (d x +c \right )\right )}{a^{3}}+\frac {1}{2 a^{2} \tanh \left (d x +c \right )^{2}}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {b^{3} \left (-\frac {a \left (a +b \right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (3 a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}\right )}{2 \left (a +b \right )^{2} a^{3}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(131\) |
| default | \(-\frac {\frac {\left (-a +2 b \right ) \ln \left (\tanh \left (d x +c \right )\right )}{a^{3}}+\frac {1}{2 a^{2} \tanh \left (d x +c \right )^{2}}+\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {b^{3} \left (-\frac {a \left (a +b \right )}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\left (3 a +2 b \right ) \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}\right )}{2 \left (a +b \right )^{2} a^{3}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(131\) |
| parallelrisch | \(\frac {3 \left (a +\frac {2 b}{3}\right ) \left (a +b \tanh \left (d x +c \right )^{2}\right ) b^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )+\left (-2 \tanh \left (d x +c \right )^{2} a^{3} b -2 a^{4}\right ) \ln \left (1-\tanh \left (d x +c \right )\right )+2 \left (a -2 b \right ) \left (a +b \right )^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right ) \ln \left (\tanh \left (d x +c \right )\right )+\left (-2 a^{3} b d x +a^{2} b^{2}+3 a \,b^{3}+2 b^{4}\right ) \tanh \left (d x +c \right )^{2}-2 \left (\frac {\left (a +b \right )^{2} \coth \left (d x +c \right )^{2}}{2}+a^{2} d x \right ) a^{2}}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} a^{3}}\) | \(190\) |
| risch | \(\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 x}{a^{2}}-\frac {2 c}{a^{2} d}+\frac {4 b x}{a^{3}}+\frac {4 b c}{a^{3} d}-\frac {6 b^{2} x}{a^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {6 b^{2} c}{d \,a^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 b^{3} x}{a^{3} \left (a^{2}+2 a b +b^{2}\right )}-\frac {4 b^{3} c}{a^{3} d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \,{\mathrm e}^{2 d x +2 c} \left (a^{3} {\mathrm e}^{4 d x +4 c}+3 a^{2} b \,{\mathrm e}^{4 d x +4 c}+3 a \,b^{2} {\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{4 d x +4 c} b^{3}+2 a^{3} {\mathrm e}^{2 d x +2 c}+2 a^{2} b \,{\mathrm e}^{2 d x +2 c}-2 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}-4 \,{\mathrm e}^{2 d x +2 c} b^{3}+a^{3}+3 a^{2} b +3 a \,b^{2}+2 b^{3}\right )}{a^{2} d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}-1\right )}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{2 d x +2 c}-1\right ) b}{a^{3} d}+\frac {3 b^{2} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \,a^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {b^{3} \ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{a^{3} d \left (a^{2}+2 a b +b^{2}\right )}\) | \(518\) |
-1/d*((-a+2*b)/a^3*ln(tanh(d*x+c))+1/2/a^2/tanh(d*x+c)^2+1/2/(a+b)^2*ln(ta nh(d*x+c)-1)-1/2*b^3/(a+b)^2/a^3*(-a*(a+b)/b/(a+b*tanh(d*x+c)^2)+(3*a+2*b) /b*ln(a+b*tanh(d*x+c)^2))+1/2/(a+b)^2*ln(tanh(d*x+c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 3468 vs. \(2 (118) = 236\).
Time = 0.45 (sec) , antiderivative size = 3468, normalized size of antiderivative = 27.97 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \]
-1/2*(2*(a^4 + a^3*b)*d*x*cosh(d*x + c)^8 + 16*(a^4 + a^3*b)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 2*(a^4 + a^3*b)*d*x*sinh(d*x + c)^8 - 4*(2*a^3*b*d* x - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a*b^3)*cosh(d*x + c)^6 - 4*(2*a^3*b*d*x - 14*(a^4 + a^3*b)*d*x*cosh(d*x + c)^2 - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a*b ^3)*sinh(d*x + c)^6 + 8*(14*(a^4 + a^3*b)*d*x*cosh(d*x + c)^3 - 3*(2*a^3*b *d*x - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a*b^3)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*(2*a^4 + 2*a^3*b - 2*a^2*b^2 - 4*a*b^3 - (a^4 - 3*a^3*b)*d*x)*cosh(d* x + c)^4 + 4*(35*(a^4 + a^3*b)*d*x*cosh(d*x + c)^4 + 2*a^4 + 2*a^3*b - 2*a ^2*b^2 - 4*a*b^3 - (a^4 - 3*a^3*b)*d*x - 15*(2*a^3*b*d*x - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 16*(7*(a^4 + a^3* b)*d*x*cosh(d*x + c)^5 - 5*(2*a^3*b*d*x - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a* b^3)*cosh(d*x + c)^3 + (2*a^4 + 2*a^3*b - 2*a^2*b^2 - 4*a*b^3 - (a^4 - 3*a ^3*b)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(a^4 + a^3*b)*d*x - 4*(2*a^3 *b*d*x - a^4 - 3*a^3*b - 3*a^2*b^2 - 2*a*b^3)*cosh(d*x + c)^2 + 4*(14*(a^4 + a^3*b)*d*x*cosh(d*x + c)^6 - 2*a^3*b*d*x - 15*(2*a^3*b*d*x - a^4 - 3*a^ 3*b - 3*a^2*b^2 - 2*a*b^3)*cosh(d*x + c)^4 + a^4 + 3*a^3*b + 3*a^2*b^2 + 2 *a*b^3 + 6*(2*a^4 + 2*a^3*b - 2*a^2*b^2 - 4*a*b^3 - (a^4 - 3*a^3*b)*d*x)*c osh(d*x + c)^2)*sinh(d*x + c)^2 - ((3*a^2*b^2 + 5*a*b^3 + 2*b^4)*cosh(d*x + c)^8 + 8*(3*a^2*b^2 + 5*a*b^3 + 2*b^4)*cosh(d*x + c)*sinh(d*x + c)^7 + ( 3*a^2*b^2 + 5*a*b^3 + 2*b^4)*sinh(d*x + c)^8 - 4*(3*a*b^3 + 2*b^4)*cosh...
\[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {\coth ^{3}{\left (c + d x \right )}}{\left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (118) = 236\).
Time = 0.22 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.24 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {{\left (3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} - \frac {2 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - 2 \, b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3} - 4 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 2 \, {\left (a^{5} - a^{4} b - 5 \, a^{3} b^{2} - 3 \, a^{2} b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - 4 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{a^{3} d} + \frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{a^{3} d} \]
1/2*(3*a*b^2 + 2*b^3)*log(2*(a - b)*e^(-2*d*x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^5 + 2*a^4*b + a^3*b^2)*d) + (d*x + c)/((a^2 + 2*a*b + b ^2)*d) - 2*((a^3 + 3*a^2*b + 3*a*b^2 + 2*b^3)*e^(-2*d*x - 2*c) + 2*(a^3 + a^2*b - a*b^2 - 2*b^3)*e^(-4*d*x - 4*c) + (a^3 + 3*a^2*b + 3*a*b^2 + 2*b^3 )*e^(-6*d*x - 6*c))/((a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 - 4*(a^4*b + 2*a ^3*b^2 + a^2*b^3)*e^(-2*d*x - 2*c) - 2*(a^5 - a^4*b - 5*a^3*b^2 - 3*a^2*b^ 3)*e^(-4*d*x - 4*c) - 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*e^(-6*d*x - 6*c) + ( a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*e^(-8*d*x - 8*c))*d) + (a - 2*b)*log( e^(-d*x - c) + 1)/(a^3*d) + (a - 2*b)*log(e^(-d*x - c) - 1)/(a^3*d)
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (118) = 236\).
Time = 0.39 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.60 \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\left (3 \, a b^{2} + 2 \, b^{3}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}{a^{5} + 2 \, a^{4} b + a^{3} b^{2}} - \frac {2 \, {\left (d x + c\right )}}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a - 2 \, b\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (\frac {{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )}}{a + b} + \frac {2 \, {\left (a^{4} + a^{3} b - a^{2} b^{2} - 2 \, a b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{a + b} + \frac {{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{a + b}\right )}}{{\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )} {\left (a + b\right )} a^{3} {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \]
1/2*((3*a*b^2 + 2*b^3)*log(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^( 2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)/(a^5 + 2*a^4*b + a^3*b^2) - 2* (d*x + c)/(a^2 + 2*a*b + b^2) + 2*(a - 2*b)*log(abs(e^(2*d*x + 2*c) - 1))/ a^3 - 4*((a^4 + 3*a^3*b + 3*a^2*b^2 + 2*a*b^3)*e^(6*d*x + 6*c)/(a + b) + 2 *(a^4 + a^3*b - a^2*b^2 - 2*a*b^3)*e^(4*d*x + 4*c)/(a + b) + (a^4 + 3*a^3* b + 3*a^2*b^2 + 2*a*b^3)*e^(2*d*x + 2*c)/(a + b))/((a*e^(4*d*x + 4*c) + b* e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)*(a + b)*a^3*(e^(2*d*x + 2*c) - 1)^2))/d
Timed out. \[ \int \frac {\coth ^3(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {coth}\left (c+d\,x\right )}^3}{{\left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \]