Integrand size = 18, antiderivative size = 133 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {3 d x}{16 a^2 f}-\frac {d x^2}{8 a^2}+\frac {x (c+d x)}{4 a^2}-\frac {d}{16 f^2 (a+a \tanh (e+f x))^2}-\frac {c+d x}{4 f (a+a \tanh (e+f x))^2}-\frac {3 d}{16 f^2 \left (a^2+a^2 \tanh (e+f x)\right )}-\frac {c+d x}{4 f \left (a^2+a^2 \tanh (e+f x)\right )} \]
3/16*d*x/a^2/f-1/8*d*x^2/a^2+1/4*x*(d*x+c)/a^2-1/16*d/f^2/(a+a*tanh(f*x+e) )^2+1/4*(-d*x-c)/f/(a+a*tanh(f*x+e))^2-3/16*d/f^2/(a^2+a^2*tanh(f*x+e))+1/ 4*(-d*x-c)/f/(a^2+a^2*tanh(f*x+e))
Time = 0.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {\text {sech}^2(e+f x) \left (-8 (d+2 c f+2 d f x)+\left (4 c f (-1+4 f x)+d \left (-1-4 f x+8 f^2 x^2\right )\right ) \cosh (2 (e+f x))+\left (4 c f (1+4 f x)+d \left (1+4 f x+8 f^2 x^2\right )\right ) \sinh (2 (e+f x))\right )}{64 a^2 f^2 (1+\tanh (e+f x))^2} \]
(Sech[e + f*x]^2*(-8*(d + 2*c*f + 2*d*f*x) + (4*c*f*(-1 + 4*f*x) + d*(-1 - 4*f*x + 8*f^2*x^2))*Cosh[2*(e + f*x)] + (4*c*f*(1 + 4*f*x) + d*(1 + 4*f*x + 8*f^2*x^2))*Sinh[2*(e + f*x)]))/(64*a^2*f^2*(1 + Tanh[e + f*x])^2)
Time = 0.39 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4213, 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.
\(\displaystyle \int \frac {c+d x}{(a \tanh (e+f x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {c+d x}{(a-i a \tan (i e+i f x))^2}dx\) |
\(\Big \downarrow \) 4213 |
\(\displaystyle -d \int \left (\frac {x}{4 a^2}-\frac {1}{4 f \left (\tanh (e+f x) a^2+a^2\right )}-\frac {1}{4 f (\tanh (e+f x) a+a)^2}\right )dx-\frac {c+d x}{4 f \left (a^2 \tanh (e+f x)+a^2\right )}+\frac {x (c+d x)}{4 a^2}-\frac {c+d x}{4 f (a \tanh (e+f x)+a)^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c+d x}{4 f \left (a^2 \tanh (e+f x)+a^2\right )}+\frac {x (c+d x)}{4 a^2}-d \left (\frac {3}{16 f^2 \left (a^2 \tanh (e+f x)+a^2\right )}-\frac {3 x}{16 a^2 f}+\frac {x^2}{8 a^2}+\frac {1}{16 f^2 (a \tanh (e+f x)+a)^2}\right )-\frac {c+d x}{4 f (a \tanh (e+f x)+a)^2}\) |
(x*(c + d*x))/(4*a^2) - (c + d*x)/(4*f*(a + a*Tanh[e + f*x])^2) - (c + d*x )/(4*f*(a^2 + a^2*Tanh[e + f*x])) - d*((-3*x)/(16*a^2*f) + x^2/(8*a^2) + 1 /(16*f^2*(a + a*Tanh[e + f*x])^2) + 3/(16*f^2*(a^2 + a^2*Tanh[e + f*x])))
3.1.40.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{u = IntHide[(a + b*Tan[e + f*x])^n, x]}, Simp[(c + d*x) ^m u, x] - Simp[d*m Int[(c + d*x)^(m - 1) u, x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0] && ILtQ[n, -1] && GtQ[m, 0]
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.56
method | result | size |
risch | \(\frac {d \,x^{2}}{8 a^{2}}+\frac {c x}{4 a^{2}}-\frac {\left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{8 a^{2} f^{2}}-\frac {\left (4 d x f +4 c f +d \right ) {\mathrm e}^{-4 f x -4 e}}{64 a^{2} f^{2}}\) | \(74\) |
parallelrisch | \(\frac {-3 d \tanh \left (f x +e \right )-4 d +4 c x \,f^{2}+2 d \,x^{2} f^{2}-4 c \tanh \left (f x +e \right ) f +2 d \tanh \left (f x +e \right ) x f -8 c f -5 d x f +4 d \tanh \left (f x +e \right ) x^{2} f^{2}+8 x \tanh \left (f x +e \right ) c \,f^{2}+2 d \tanh \left (f x +e \right )^{2} x^{2} f^{2}+4 x \tanh \left (f x +e \right )^{2} c \,f^{2}+3 x \tanh \left (f x +e \right )^{2} d f}{16 f^{2} a^{2} \left (1+\tanh \left (f x +e \right )\right )^{2}}\) | \(151\) |
1/8*d*x^2/a^2+1/4/a^2*c*x-1/8*(2*d*f*x+2*c*f+d)/a^2/f^2*exp(-2*f*x-2*e)-1/ 64*(4*d*f*x+4*c*f+d)/a^2/f^2*exp(-4*f*x-4*e)
Time = 0.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.44 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=-\frac {16 \, d f x - {\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \, {\left (4 \, c f^{2} - d f\right )} x - d\right )} \cosh \left (f x + e\right )^{2} - 2 \, {\left (8 \, d f^{2} x^{2} + 4 \, c f + 4 \, {\left (4 \, c f^{2} + d f\right )} x + d\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) - {\left (8 \, d f^{2} x^{2} - 4 \, c f + 4 \, {\left (4 \, c f^{2} - d f\right )} x - d\right )} \sinh \left (f x + e\right )^{2} + 16 \, c f + 8 \, d}{64 \, {\left (a^{2} f^{2} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{2} \sinh \left (f x + e\right )^{2}\right )}} \]
-1/64*(16*d*f*x - (8*d*f^2*x^2 - 4*c*f + 4*(4*c*f^2 - d*f)*x - d)*cosh(f*x + e)^2 - 2*(8*d*f^2*x^2 + 4*c*f + 4*(4*c*f^2 + d*f)*x + d)*cosh(f*x + e)* sinh(f*x + e) - (8*d*f^2*x^2 - 4*c*f + 4*(4*c*f^2 - d*f)*x - d)*sinh(f*x + e)^2 + 16*c*f + 8*d)/(a^2*f^2*cosh(f*x + e)^2 + 2*a^2*f^2*cosh(f*x + e)*s inh(f*x + e) + a^2*f^2*sinh(f*x + e)^2)
\[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {\int \frac {c}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
(Integral(c/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(d*x/(t anh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x))/a**2
Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {1}{16} \, c {\left (\frac {4 \, {\left (f x + e\right )}}{a^{2} f} - \frac {4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac {{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \, {\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d e^{\left (-4 \, e\right )}}{64 \, a^{2} f^{2}} \]
1/16*c*(4*(f*x + e)/(a^2*f) - (4*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e))/(a^2 *f)) + 1/64*(8*f^2*x^2*e^(4*e) - 8*(2*f*x*e^(2*e) + e^(2*e))*e^(-2*f*x) - (4*f*x + 1)*e^(-4*f*x))*d*e^(-4*e)/(a^2*f^2)
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.77 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {{\left (8 \, d f^{2} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 16 \, c f^{2} x e^{\left (4 \, f x + 4 \, e\right )} - 16 \, d f x e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d f x - 16 \, c f e^{\left (2 \, f x + 2 \, e\right )} - 4 \, c f - 8 \, d e^{\left (2 \, f x + 2 \, e\right )} - d\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{64 \, a^{2} f^{2}} \]
1/64*(8*d*f^2*x^2*e^(4*f*x + 4*e) + 16*c*f^2*x*e^(4*f*x + 4*e) - 16*d*f*x* e^(2*f*x + 2*e) - 4*d*f*x - 16*c*f*e^(2*f*x + 2*e) - 4*c*f - 8*d*e^(2*f*x + 2*e) - d)*e^(-4*f*x - 4*e)/(a^2*f^2)
Time = 1.79 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67 \[ \int \frac {c+d x}{(a+a \tanh (e+f x))^2} \, dx=\frac {d\,x^2}{8\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {d+4\,c\,f}{64\,a^2\,f^2}+\frac {d\,x}{16\,a^2\,f}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {d+2\,c\,f}{8\,a^2\,f^2}+\frac {d\,x}{4\,a^2\,f}\right )+\frac {c\,x}{4\,a^2} \]