Integrand size = 21, antiderivative size = 68 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^2 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^2 d}-\frac {1}{2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )} \] Output:
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 = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {-2 \log (\cosh (c+d x))-\log \left (a+b \tanh ^2(c+d x)\right )+\frac {a+b}{a+b \tanh ^2(c+d x)}}{2 (a+b)^2 d} \] Input:
Integrate[Tanh[c + d*x]/(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-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 = 0.36 (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 |
\(\displaystyle \frac {-\frac {1}{(a+b) \left (a+b \tanh ^2(c+d x)\right )}-\frac {\log \left (1-\tanh ^2(c+d x)\right )}{(a+b)^2}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{(a+b)^2}}{2 d}\) |
Input:
Int[Tanh[c + d*x]/(a + b*Tanh[c + d*x]^2)^2,x]
Output:
(-(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)
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 = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {\frac {b \left (\frac {\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a +b}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(86\) |
default | \(\frac {\frac {b \left (\frac {\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{b}-\frac {a +b}{b \left (a +b \tanh \left (d x +c \right )^{2}\right )}\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}}{d}\) | \(86\) |
parallelrisch | \(-\frac {-b^{2} \tanh \left (d x +c \right )^{2}+2 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2}+2 a^{2} d x +2 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} a b -\ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a b -\tanh \left (d x +c \right )^{2} a b +2 b \tanh \left (d x +c \right )^{2} x a d -a^{2} \ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \tanh \left (d x +c \right )^{2}\right ) d \left (a +b \right )^{2} a}\) | \(157\) |
risch | \(-\frac {x}{a^{2}+2 a b +b^{2}}-\frac {2 c}{d \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 b \,{\mathrm e}^{2 d x +2 c}}{d \left (a +b \right )^{2} \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 a \,{\mathrm e}^{2 d x +2 c}-2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}\) | \(159\) |
Input:
int(tanh(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/(a+b)^2*b*(1/b*ln(a+b*tanh(d*x+c)^2)-(a+b)/b/(a+b*tanh(d*x+c)^2)) -1/2/(a+b)^2*ln(1+tanh(d*x+c))-1/2/(a+b)^2*ln(-1+tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 623 vs. \(2 (64) = 128\).
Time = 0.09 (sec) , antiderivative size = 623, normalized size of antiderivative = 9.16 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:
integrate(tanh(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-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))
Timed out. \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:
integrate(tanh(d*x+c)/(a+b*tanh(d*x+c)**2)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (64) = 128\).
Time = 0.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.50 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=-\frac {2 \, b e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d} \] Input:
integrate(tanh(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
-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 = 0.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.19 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {\log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2}{{\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )} {\left (a + b\right )}}}{2 \, d} \] Input:
integrate(tanh(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
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 = 2.75 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx=\frac {\frac {a\,x}{a^2+2\,a\,b+b^2}+\frac {b\,x\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{a^2+2\,a\,b+b^2}+\frac {b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}{2\,a\,d\,\left (a+b\right )}}{b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a}+\frac {\ln \left (b\,{\mathrm {tanh}\left (c+d\,x\right )}^2+a\right )}{2\,d\,\left (a^2+2\,a\,b+b^2\right )}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )}{d\,{\left (a+b\right )}^2} \] Input:
int(tanh(c + d*x)/(a + b*tanh(c + d*x)^2)^2,x)
Output:
((a*x)/(2*a*b + a^2 + b^2) + (b*x*tanh(c + d*x)^2)/(2*a*b + a^2 + b^2) + ( b*tanh(c + d*x)^2)/(2*a*d*(a + b)))/(a + b*tanh(c + d*x)^2) + log(a + b*ta nh(c + d*x)^2)/(2*d*(2*a*b + a^2 + b^2)) - log(tanh(c + d*x) + 1)/(d*(a + b)^2)
Time = 0.24 (sec) , antiderivative size = 2288, normalized size of antiderivative = 33.65 \[ \int \frac {\tanh (c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:
int(tanh(d*x+c)/(a+b*tanh(d*x+c)^2)^2,x)
Output:
(e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x)**2*a**3*b - e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x)** 2*a*b**3 + e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**4 - e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sq rt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*a**2*b**2 + e**(4*c + 4* d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt( b))*tanh(c + d*x)**2*a**3*b - e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x)**2*a*b**3 + e* *(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a**4 - e**(4*c + 4*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + s qrt(a + b) + 2*e**(c + d*x)*sqrt(b))*a**2*b**2 + e**(4*c + 4*d*x)*tanh(c + d*x)**2*a**4 - 2*e**(4*c + 4*d*x)*tanh(c + d*x)**2*a**3*b*d*x - 2*e**(4*c + 4*d*x)*tanh(c + d*x)**2*a**2*b**2 + 2*e**(4*c + 4*d*x)*tanh(c + d*x)**2 *a*b**3*d*x - 2*e**(4*c + 4*d*x)*tanh(c + d*x)**2*a*b**3 - e**(4*c + 4*d*x )*tanh(c + d*x)**2*b**4 - 2*e**(4*c + 4*d*x)*a**4*d*x - 2*e**(4*c + 4*d*x) *a**4 - 2*e**(4*c + 4*d*x)*a**3*b + 2*e**(4*c + 4*d*x)*a**2*b**2*d*x + 2*e **(2*c + 2*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x)**2*a**3*b - 4*e**(2*c + 2*d*x)*log(e**(2*c + 2*d*x)*sqrt(a + b) + sqrt(a + b) - 2*e**(c + d*x)*sqrt(b))*tanh(c + d*x...