Integrand size = 21, antiderivative size = 68 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^2 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^2 d}-\frac {1}{2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \]
ln(cosh(d*x+c))/(a+b)^2/d+1/2*ln(a+b*tanh(d*x+c)^2)/(a+b)^2/d-1/2/(a+b)/d/ (a+b*tanh(d*x+c)^2)
Time = 0.45 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {-2 \log (\cosh (c+d x))-\log \left (a+b \tanh ^2(c+d x)\right )+\frac {a+b}{a+b \tanh ^2(c+d x)}}{2 (a+b)^2 d} \]
-1/2*(-2*Log[Cosh[c + d*x]] - Log[a + b*Tanh[c + d*x]^2] + (a + b)/(a + b* Tanh[c + d*x]^2))/((a + b)^2*d)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 4153, 26, 353, 54, 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.
| \(\displaystyle \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {i \tan (i c+i d x)}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \frac {\tan (i c+i d x)}{\left (a-b \tan (i c+i d x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {i \int \frac {i \tanh (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\tanh (c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\int \left (\frac {b}{(a+b)^2 \left (b \tanh ^2(c+d x)+a\right )}+\frac {b}{(a+b) \left (b \tanh ^2(c+d x)+a\right )^2}-\frac {1}{(a+b)^2 \left (\tanh ^2(c+d x)-1\right )}\right )d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{(a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^2}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\) |
(-(Log[1 - Tanh[c + d*x]^2]/(a + b)^2) + Log[a + b*Tanh[c + d*x]^2]/(a + b )^2 - 1/((a + b)*(a + b*Tanh[c + d*x]^2)))/(2*d)
3.2.84.3.1 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_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(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]
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 = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26
| method | result | size |
| derivativedivides | \(\frac {\frac {b \left (-\frac {a +b}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(86\) |
| default | \(\frac {\frac {b \left (-\frac {a +b}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(86\) |
| parallelrisch | \(-\frac {-\tanh \left (d x +c \right )^{2} a b -a^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )-b^{2} \tanh \left (d x +c \right )^{2}+2 a^{2} d x -\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a b +2 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2}+2 x \tanh \left (d x +c \right )^{2} a b d +2 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} a b}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} a}\) | \(157\) |
| risch | \(-\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b \,{\mathrm e}^{2 d x +2 c}}{d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}\) | \(159\) |
1/d*(1/2*b/(a+b)^2*(-(a+b)/b/(a+b*tanh(d*x+c)^2)+1/b*ln(a+b*tanh(d*x+c)^2) )-1/2/(a+b)^2*ln(tanh(d*x+c)+1)-1/2/(a+b)^2*ln(tanh(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 623 vs. \(2 (64) = 128\).
Time = 0.28 (sec) , antiderivative size = 623, normalized size of antiderivative = 9.16 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, {\left (a + b\right )} d x \cosh \left (d x + c\right )^{4} + 8 \, {\left (a + b\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 2 \, {\left (a + b\right )} d x \sinh \left (d x + c\right )^{4} + 2 \, {\left (a + b\right )} d x + 4 \, {\left ({\left (a - b\right )} d x + b\right )} \cosh \left (d x + c\right )^{2} + 4 \, {\left (3 \, {\left (a + b\right )} d x \cosh \left (d x + c\right )^{2} + {\left (a - b\right )} d x + b\right )} \sinh \left (d x + c\right )^{2} - {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a + b\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a + b\right )} \cosh \left (d x + c\right )^{2} + a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{3} + {\left (a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + a + b\right )} \log \left (\frac {2 \, {\left ({\left (a + b\right )} \cosh \left (d x + c\right )^{2} + {\left (a + b\right )} \sinh \left (d x + c\right )^{2} + a - b\right )}}{\cosh \left (d x + c\right )^{2} - 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) + 8 \, {\left ({\left (a + b\right )} d x \cosh \left (d x + c\right )^{3} + {\left ({\left (a - b\right )} d x + b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{2} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d\right )} \sinh \left (d x + c\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d + 4 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d \cosh \left (d x + c\right )^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}} \]
-1/2*(2*(a + b)*d*x*cosh(d*x + c)^4 + 8*(a + b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a + b)*d*x*sinh(d*x + c)^4 + 2*(a + b)*d*x + 4*((a - b)*d*x + b)*cosh(d*x + c)^2 + 4*(3*(a + b)*d*x*cosh(d*x + c)^2 + (a - b)*d*x + b)* sinh(d*x + c)^2 - ((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh( d*x + c)^3 + (a + b)*sinh(d*x + c)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x + c)^ 3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)*log(2*((a + b)*cosh(d*x + c)^2 + (a + b)*sinh(d*x + c)^2 + a - b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)) + 8*((a + b)*d*x*cosh(d*x + c)^3 + (( a - b)*d*x + b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c) *sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)^4 + 2*( a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + 3*a^2*b + 3*a*b ^2 + b^3)*d*cosh(d*x + c)^2 + (a^3 + a^2*b - a*b^2 - b^3)*d)*sinh(d*x + c) ^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d + 4*((a^3 + 3*a^2*b + 3*a*b^2 + b^3 )*d*cosh(d*x + c)^3 + (a^3 + a^2*b - a*b^2 - b^3)*d*cosh(d*x + c))*sinh(d* x + c))
Time = 58.58 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {b \left (\begin {cases} \frac {\tanh ^{2}{\left (c + d x \right )}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )} & \text {otherwise} \end {cases}\right )}{2 d \left (a + b\right )} + \frac {b \left (\begin {cases} \frac {\tanh ^{2}{\left (c + d x \right )}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \tanh ^{2}{\left (c + d x \right )} \right )}}{b} & \text {otherwise} \end {cases}\right )}{2 d \left (a + b\right )^{2}} - \frac {\log {\left (\tanh ^{2}{\left (c + d x \right )} - 1 \right )}}{2 d \left (a + b\right )^{2}} \]
b*Piecewise((tanh(c + d*x)**2/a**2, Eq(b, 0)), (-1/(b*(a + b*tanh(c + d*x) **2)), True))/(2*d*(a + b)) + b*Piecewise((tanh(c + d*x)**2/a, Eq(b, 0)), (log(a + b*tanh(c + d*x)**2)/b, True))/(2*d*(a + b)**2) - log(tanh(c + d*x )**2 - 1)/(2*d*(a + b)**2)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (64) = 128\).
Time = 0.22 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.50 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, b e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} \]
-2*b*e^(-2*d*x - 2*c)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3 + 2*(a^3 + a^2*b - a *b^2 - b^3)*e^(-2*d*x - 2*c) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-4*d*x - 4*c))*d) + (d*x + c)/((a^2 + 2*a*b + b^2)*d) + 1/2*log(2*(a - b)*e^(-2*d* x - 2*c) + (a + b)*e^(-4*d*x - 4*c) + a + b)/((a^2 + 2*a*b + b^2)*d)
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (64) = 128\).
Time = 0.34 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.19 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {\log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2}{{\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )} {\left (a + b\right )}}}{2 \, d} \]
1/2*(log(abs(a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2*b))/(a^2 + 2*a*b + b^2) - (e^(2*d*x + 2*c) + e^(-2*d*x - 2*c) + 2)/((a*(e^(2*d*x + 2*c) + e^(-2*d*x - 2*c)) + b*(e^(2*d *x + 2*c) + e^(-2*d*x - 2*c)) + 2*a - 2*b)*(a + b)))/d
Time = 2.05 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {a\,x}{a^2+2\,a\,b+b^2}+\frac {b\,x\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{a^2+2\,a\,b+b^2}+\frac {b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{2\,a\,d\,\left (a+b\right )}}{b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a}+\frac {\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2\,d\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )}{d\,{\left (a+b\right )}^2} \]