Integrand size = 21, antiderivative size = 119 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a \left (a^2+3 b^2\right ) \arctan (\sinh (c+d x))}{2 \left (a^2+b^2\right )^2 d}-\frac {b^3 \log (\cosh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {b^3 \log (a+b \sinh (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {\text {sech}^2(c+d x) (b+a \sinh (c+d x))}{2 \left (a^2+b^2\right ) d} \]
1/2*a*(a^2+3*b^2)*arctan(sinh(d*x+c))/(a^2+b^2)^2/d-b^3*ln(cosh(d*x+c))/(a ^2+b^2)^2/d+b^3*ln(a+b*sinh(d*x+c))/(a^2+b^2)^2/d+1/2*sech(d*x+c)^2*(b+a*s inh(d*x+c))/(a^2+b^2)/d
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a \left (a^2+b^2\right ) \arctan (\sinh (c+d x))-b^2 ((i a+b) \log (i-\sinh (c+d x))+(-i a+b) \log (i+\sinh (c+d x))-2 b \log (a+b \sinh (c+d x)))+b \left (a^2+b^2\right ) \text {sech}^2(c+d x)+a \left (a^2+b^2\right ) \text {sech}(c+d x) \tanh (c+d x)}{2 \left (a^2+b^2\right )^2 d} \]
(a*(a^2 + b^2)*ArcTan[Sinh[c + d*x]] - b^2*((I*a + b)*Log[I - Sinh[c + d*x ]] + ((-I)*a + b)*Log[I + Sinh[c + d*x]] - 2*b*Log[a + b*Sinh[c + d*x]]) + b*(a^2 + b^2)*Sech[c + d*x]^2 + a*(a^2 + b^2)*Sech[c + d*x]*Tanh[c + d*x] )/(2*(a^2 + b^2)^2*d)
Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.33, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3147, 496, 25, 657, 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.
\(\displaystyle \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (i c+i d x)^3 (a-i b \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b^3 \int \frac {1}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )^2}d(b \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 496 |
\(\displaystyle \frac {b^3 \left (\frac {a b \sinh (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(c+d x)+b^2\right )}-\frac {\int -\frac {a^2+b \sinh (c+d x) a+2 b^2}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{2 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b^3 \left (\frac {\int \frac {a^2+b \sinh (c+d x) a+2 b^2}{(a+b \sinh (c+d x)) \left (\sinh ^2(c+d x) b^2+b^2\right )}d(b \sinh (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {b^3 \left (\frac {\int \left (\frac {2 b^2}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {a^3+3 b^2 a-2 b^3 \sinh (c+d x)}{\left (a^2+b^2\right ) \left (\sinh ^2(c+d x) b^2+b^2\right )}\right )d(b \sinh (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b^3 \left (\frac {\frac {a \left (a^2+3 b^2\right ) \arctan (\sinh (c+d x))}{b \left (a^2+b^2\right )}-\frac {b^2 \log \left (b^2 \sinh ^2(c+d x)+b^2\right )}{a^2+b^2}+\frac {2 b^2 \log (a+b \sinh (c+d x))}{a^2+b^2}}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \sinh (c+d x)+b^2}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \sinh ^2(c+d x)+b^2\right )}\right )}{d}\) |
(b^3*(((a*(a^2 + 3*b^2)*ArcTan[Sinh[c + d*x]])/(b*(a^2 + b^2)) + (2*b^2*Lo g[a + b*Sinh[c + d*x]])/(a^2 + b^2) - (b^2*Log[b^2 + b^2*Sinh[c + d*x]^2]) /(a^2 + b^2))/(2*b^2*(a^2 + b^2)) + (b^2 + a*b*Sinh[c + d*x])/(2*b^2*(a^2 + b^2)*(b^2 + b^2*Sinh[c + d*x]^2))))/d
3.4.16.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(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] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 19.28 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.72
method | result | size |
derivativedivides | \(\frac {\frac {b^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{3}-\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{2} b -b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {1}{2} a^{3}+\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-b^{3} \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (a^{3}+3 a \,b^{2}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}}{d}\) | \(205\) |
default | \(\frac {\frac {b^{3} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}{a^{4}+2 a^{2} b^{2}+b^{4}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{3}-\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a^{2} b -b^{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {1}{2} a^{3}+\frac {1}{2} a \,b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-b^{3} \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (a^{3}+3 a \,b^{2}\right ) \arctan \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}+2 a^{2} b^{2}+b^{4}}}{d}\) | \(205\) |
risch | \(\frac {2 b^{3} d^{2} x}{a^{4} d^{2}+2 a^{2} b^{2} d^{2}+b^{4} d^{2}}+\frac {2 b^{3} d c}{a^{4} d^{2}+2 a^{2} b^{2} d^{2}+b^{4} d^{2}}-\frac {2 b^{3} x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 b^{3} c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {{\mathrm e}^{d x +c} \left (a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{d x +c}-a \right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )^{2}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}+i\right ) a \,b^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}+i\right ) b^{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{3}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}-i\right ) a \,b^{2}}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}-\frac {\ln \left ({\mathrm e}^{d x +c}-i\right ) b^{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \,{\mathrm e}^{d x +c}}{b}-1\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\) | \(443\) |
1/d*(b^3/(a^4+2*a^2*b^2+b^4)*ln(tanh(1/2*d*x+1/2*c)^2*a-2*b*tanh(1/2*d*x+1 /2*c)-a)+2/(a^4+2*a^2*b^2+b^4)*(((-1/2*a^3-1/2*a*b^2)*tanh(1/2*d*x+1/2*c)^ 3+(-a^2*b-b^3)*tanh(1/2*d*x+1/2*c)^2+(1/2*a^3+1/2*a*b^2)*tanh(1/2*d*x+1/2* c))/(1+tanh(1/2*d*x+1/2*c)^2)^2-1/2*b^3*ln(1+tanh(1/2*d*x+1/2*c)^2)+1/2*(a ^3+3*a*b^2)*arctan(tanh(1/2*d*x+1/2*c))))
Leaf count of result is larger than twice the leaf count of optimal. 893 vs. \(2 (115) = 230\).
Time = 0.26 (sec) , antiderivative size = 893, normalized size of antiderivative = 7.50 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\text {Too large to display} \]
((a^3 + a*b^2)*cosh(d*x + c)^3 + (a^3 + a*b^2)*sinh(d*x + c)^3 + 2*(a^2*b + b^3)*cosh(d*x + c)^2 + (2*a^2*b + 2*b^3 + 3*(a^3 + a*b^2)*cosh(d*x + c)) *sinh(d*x + c)^2 + ((a^3 + 3*a*b^2)*cosh(d*x + c)^4 + 4*(a^3 + 3*a*b^2)*co sh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a*b^2)*sinh(d*x + c)^4 + a^3 + 3*a* b^2 + 2*(a^3 + 3*a*b^2)*cosh(d*x + c)^2 + 2*(a^3 + 3*a*b^2 + 3*(a^3 + 3*a* b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*((a^3 + 3*a*b^2)*cosh(d*x + c)^3 + (a^3 + 3*a*b^2)*cosh(d*x + c))*sinh(d*x + c))*arctan(cosh(d*x + c) + si nh(d*x + c)) - (a^3 + a*b^2)*cosh(d*x + c) + (b^3*cosh(d*x + c)^4 + 4*b^3* cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^ 2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d* x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x + c))*log(2*(b*sinh(d*x + c) + a)/( cosh(d*x + c) - sinh(d*x + c))) - (b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 + 2*b^3*cosh(d*x + c)^2 + b^3 + 2 *(3*b^3*cosh(d*x + c)^2 + b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 + b^3*cosh(d*x + c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sin h(d*x + c))) - (a^3 + a*b^2 - 3*(a^3 + a*b^2)*cosh(d*x + c)^2 - 4*(a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^4 + 4*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + 2*a^2*b^2 + b^4)*d*sinh(d*x + c)^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^2 + 2*(3*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)^2 + (a^4 + 2*a^2*...
\[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\int \frac {\operatorname {sech}^{3}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.82 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {b^{3} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {b^{3} \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} + \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} \]
b^3*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^4 + 2*a^2*b^2 + b^ 4)*d) - b^3*log(e^(-2*d*x - 2*c) + 1)/((a^4 + 2*a^2*b^2 + b^4)*d) - (a^3 + 3*a*b^2)*arctan(e^(-d*x - c))/((a^4 + 2*a^2*b^2 + b^4)*d) + (a*e^(-d*x - c) + 2*b*e^(-2*d*x - 2*c) - a*e^(-3*d*x - 3*c))/((a^2 + b^2 + 2*(a^2 + b^2 )*e^(-2*d*x - 2*c) + (a^2 + b^2)*e^(-4*d*x - 4*c))*d)
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (115) = 230\).
Time = 0.33 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.37 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {4 \, b^{4} \log \left ({\left | b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, b^{3} \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (a^{3} + 3 \, a b^{2}\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 2 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 4 \, a^{2} b + 8 \, b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}}}{4 \, d} \]
1/4*(4*b^4*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^4*b + 2*a^2*b ^3 + b^5) - 2*b^3*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^4 + 2*a^2*b^2 + b^4) + (pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(a^3 + 3 *a*b^2)/(a^4 + 2*a^2*b^2 + b^4) + 2*(b^3*(e^(d*x + c) - e^(-d*x - c))^2 + 2*a^3*(e^(d*x + c) - e^(-d*x - c)) + 2*a*b^2*(e^(d*x + c) - e^(-d*x - c)) + 4*a^2*b + 8*b^3)/((a^4 + 2*a^2*b^2 + b^4)*((e^(d*x + c) - e^(-d*x - c))^ 2 + 4)))/d
Time = 3.11 (sec) , antiderivative size = 381, normalized size of antiderivative = 3.20 \[ \int \frac {\text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {\frac {2\,\left (a^2\,b+b^3\right )}{d\,{\left (a^2+b^2\right )}^2}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (a^3+a\,b^2\right )}{d\,{\left (a^2+b^2\right )}^2}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {2\,b}{d\,\left (a^2+b^2\right )}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\ln \left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )\,\left (2\,b+a\,1{}\mathrm {i}\right )}{2\,\left (-d\,a^2+2{}\mathrm {i}\,d\,a\,b+d\,b^2\right )}-\frac {\ln \left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )\,\left (a+b\,2{}\mathrm {i}\right )}{2\,\left (-1{}\mathrm {i}\,d\,a^2+2\,d\,a\,b+1{}\mathrm {i}\,d\,b^2\right )}+\frac {b^3\,\ln \left (2\,a^7\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-16\,b^7-9\,a^2\,b^5-6\,a^4\,b^3-a^6\,b+16\,b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a^6\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+18\,a^3\,b^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+12\,a^5\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+9\,a^2\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+6\,a^4\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+32\,a\,b^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d\,a^4+2\,d\,a^2\,b^2+d\,b^4} \]
((2*(a^2*b + b^3))/(d*(a^2 + b^2)^2) + (exp(c + d*x)*(a*b^2 + a^3))/(d*(a^ 2 + b^2)^2))/(exp(2*c + 2*d*x) + 1) - ((2*b)/(d*(a^2 + b^2)) + (2*a*exp(c + d*x))/(d*(a^2 + b^2)))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - (lo g(exp(c + d*x) + 1i)*(a*1i + 2*b))/(2*(b^2*d - a^2*d + a*b*d*2i)) - (log(e xp(c + d*x)*1i + 1)*(a + b*2i))/(2*(b^2*d*1i - a^2*d*1i + 2*a*b*d)) + (b^3 *log(2*a^7*exp(d*x)*exp(c) - 16*b^7 - 9*a^2*b^5 - 6*a^4*b^3 - a^6*b + 16*b ^7*exp(2*c)*exp(2*d*x) + a^6*b*exp(2*c)*exp(2*d*x) + 18*a^3*b^4*exp(d*x)*e xp(c) + 12*a^5*b^2*exp(d*x)*exp(c) + 9*a^2*b^5*exp(2*c)*exp(2*d*x) + 6*a^4 *b^3*exp(2*c)*exp(2*d*x) + 32*a*b^6*exp(d*x)*exp(c)))/(a^4*d + b^4*d + 2*a ^2*b^2*d)