Integrand size = 20, antiderivative size = 170 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=-\frac {d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {d^2 e^{-2 e-2 f x}}{8 a^2 f^3}-\frac {d e^{-4 e-4 f x} (c+d x)}{32 a^2 f^2}-\frac {d e^{-2 e-2 f x} (c+d x)}{4 a^2 f^2}-\frac {e^{-4 e-4 f x} (c+d x)^2}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^2}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d} \]
-1/128*d^2*exp(-4*f*x-4*e)/a^2/f^3-1/8*d^2*exp(-2*f*x-2*e)/a^2/f^3-1/32*d* exp(-4*f*x-4*e)*(d*x+c)/a^2/f^2-1/4*d*exp(-2*f*x-2*e)*(d*x+c)/a^2/f^2-1/16 *exp(-4*f*x-4*e)*(d*x+c)^2/a^2/f-1/4*exp(-2*f*x-2*e)*(d*x+c)^2/a^2/f+1/12* (d*x+c)^3/a^2/d
Time = 1.53 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=\frac {\text {sech}^2(e+f x) \left (-48 \left (2 c^2 f^2+2 c d f (1+2 f x)+d^2 \left (1+2 f x+2 f^2 x^2\right )\right )+\left (24 c^2 f^2 (-1+4 f x)+12 c d f \left (-1-4 f x+8 f^2 x^2\right )+d^2 \left (-3-12 f x-24 f^2 x^2+32 f^3 x^3\right )\right ) \cosh (2 (e+f x))+\left (24 c^2 f^2 (1+4 f x)+12 c d f \left (1+4 f x+8 f^2 x^2\right )+d^2 \left (3+12 f x+24 f^2 x^2+32 f^3 x^3\right )\right ) \sinh (2 (e+f x))\right )}{384 a^2 f^3 (1+\tanh (e+f x))^2} \]
(Sech[e + f*x]^2*(-48*(2*c^2*f^2 + 2*c*d*f*(1 + 2*f*x) + d^2*(1 + 2*f*x + 2*f^2*x^2)) + (24*c^2*f^2*(-1 + 4*f*x) + 12*c*d*f*(-1 - 4*f*x + 8*f^2*x^2) + d^2*(-3 - 12*f*x - 24*f^2*x^2 + 32*f^3*x^3))*Cosh[2*(e + f*x)] + (24*c^ 2*f^2*(1 + 4*f*x) + 12*c*d*f*(1 + 4*f*x + 8*f^2*x^2) + d^2*(3 + 12*f*x + 2 4*f^2*x^2 + 32*f^3*x^3))*Sinh[2*(e + f*x)]))/(384*a^2*f^3*(1 + Tanh[e + f* x])^2)
Time = 0.43 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4212, 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.
| \(\displaystyle \int \frac {(c+d x)^2}{(a \tanh (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d x)^2}{(a-i a \tan (i e+i f x))^2}dx\) |
| \(\Big \downarrow \) 4212 |
| \(\displaystyle \int \left (\frac {(c+d x)^2 e^{-4 e-4 f x}}{4 a^2}+\frac {(c+d x)^2 e^{-2 e-2 f x}}{2 a^2}+\frac {(c+d x)^2}{4 a^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d (c+d x) e^{-4 e-4 f x}}{32 a^2 f^2}-\frac {d (c+d x) e^{-2 e-2 f x}}{4 a^2 f^2}-\frac {(c+d x)^2 e^{-4 e-4 f x}}{16 a^2 f}-\frac {(c+d x)^2 e^{-2 e-2 f x}}{4 a^2 f}+\frac {(c+d x)^3}{12 a^2 d}-\frac {d^2 e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {d^2 e^{-2 e-2 f x}}{8 a^2 f^3}\) |
-1/128*(d^2*E^(-4*e - 4*f*x))/(a^2*f^3) - (d^2*E^(-2*e - 2*f*x))/(8*a^2*f^ 3) - (d*E^(-4*e - 4*f*x)*(c + d*x))/(32*a^2*f^2) - (d*E^(-2*e - 2*f*x)*(c + d*x))/(4*a^2*f^2) - (E^(-4*e - 4*f*x)*(c + d*x)^2)/(16*a^2*f) - (E^(-2*e - 2*f*x)*(c + d*x)^2)/(4*a^2*f) + (c + d*x)^3/(12*a^2*d)
3.1.39.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(2*a) + E^(2*(a/b)*(e + f* x))/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 + b^2 , 0] && ILtQ[n, 0]
Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96
| method | result | size |
| risch | \(\frac {d^{2} x^{3}}{12 a^{2}}+\frac {d c \,x^{2}}{4 a^{2}}+\frac {c^{2} x}{4 a^{2}}+\frac {c^{3}}{12 a^{2} d}-\frac {\left (2 d^{2} x^{2} f^{2}+4 c d \,f^{2} x +2 c^{2} f^{2}+2 d^{2} f x +2 c d f +d^{2}\right ) {\mathrm e}^{-2 f x -2 e}}{8 a^{2} f^{3}}-\frac {\left (8 d^{2} x^{2} f^{2}+16 c d \,f^{2} x +8 c^{2} f^{2}+4 d^{2} f x +4 c d f +d^{2}\right ) {\mathrm e}^{-4 f x -4 e}}{128 a^{2} f^{3}}\) | \(163\) |
| parallelrisch | \(\frac {-24 d^{2}-27 d^{2} f x -21 \tanh \left (f x +e \right ) d^{2}-48 c^{2} f^{2}+24 c d \,x^{2} f^{3}-60 c d \,f^{2} x -30 d^{2} x^{2} f^{2}-48 c d f -24 \tanh \left (f x +e \right ) c^{2} f^{2}+24 x \,c^{2} f^{3}+8 d^{2} x^{3} f^{3}+16 d^{2} \tanh \left (f x +e \right ) x^{3} f^{3}+48 x \tanh \left (f x +e \right ) c^{2} f^{3}+6 x \tanh \left (f x +e \right ) d^{2} f +12 x^{2} \tanh \left (f x +e \right ) d^{2} f^{2}+36 x \tanh \left (f x +e \right )^{2} c d \,f^{2}+24 x^{2} \tanh \left (f x +e \right )^{2} c d \,f^{3}+24 x \tanh \left (f x +e \right ) c d \,f^{2}+8 d^{2} \tanh \left (f x +e \right )^{2} x^{3} f^{3}-36 \tanh \left (f x +e \right ) c d f +24 x \tanh \left (f x +e \right )^{2} c^{2} f^{3}+21 x \tanh \left (f x +e \right )^{2} d^{2} f +18 x^{2} \tanh \left (f x +e \right )^{2} d^{2} f^{2}+48 x^{2} \tanh \left (f x +e \right ) c d \,f^{3}}{96 f^{3} a^{2} \left (1+\tanh \left (f x +e \right )\right )^{2}}\) | \(326\) |
1/12/a^2*d^2*x^3+1/4/a^2*d*c*x^2+1/4/a^2*c^2*x+1/12/a^2/d*c^3-1/8*(2*d^2*f ^2*x^2+4*c*d*f^2*x+2*c^2*f^2+2*d^2*f*x+2*c*d*f+d^2)/a^2/f^3*exp(-2*f*x-2*e )-1/128*(8*d^2*f^2*x^2+16*c*d*f^2*x+8*c^2*f^2+4*d^2*f*x+4*c*d*f+d^2)/a^2/f ^3*exp(-4*f*x-4*e)
Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (150) = 300\).
Time = 0.26 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.12 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=-\frac {96 \, d^{2} f^{2} x^{2} + 96 \, c^{2} f^{2} + 96 \, c d f - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \cosh \left (f x + e\right )^{2} - 2 \, {\left (32 \, d^{2} f^{3} x^{3} + 24 \, c^{2} f^{2} + 12 \, c d f + 24 \, {\left (4 \, c d f^{3} + d^{2} f^{2}\right )} x^{2} + 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} + 4 \, c d f^{2} + d^{2} f\right )} x\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) - {\left (32 \, d^{2} f^{3} x^{3} - 24 \, c^{2} f^{2} - 12 \, c d f + 24 \, {\left (4 \, c d f^{3} - d^{2} f^{2}\right )} x^{2} - 3 \, d^{2} + 12 \, {\left (8 \, c^{2} f^{3} - 4 \, c d f^{2} - d^{2} f\right )} x\right )} \sinh \left (f x + e\right )^{2} + 48 \, d^{2} + 96 \, {\left (2 \, c d f^{2} + d^{2} f\right )} x}{384 \, {\left (a^{2} f^{3} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{3} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{3} \sinh \left (f x + e\right )^{2}\right )}} \]
-1/384*(96*d^2*f^2*x^2 + 96*c^2*f^2 + 96*c*d*f - (32*d^2*f^3*x^3 - 24*c^2* f^2 - 12*c*d*f + 24*(4*c*d*f^3 - d^2*f^2)*x^2 - 3*d^2 + 12*(8*c^2*f^3 - 4* c*d*f^2 - d^2*f)*x)*cosh(f*x + e)^2 - 2*(32*d^2*f^3*x^3 + 24*c^2*f^2 + 12* c*d*f + 24*(4*c*d*f^3 + d^2*f^2)*x^2 + 3*d^2 + 12*(8*c^2*f^3 + 4*c*d*f^2 + d^2*f)*x)*cosh(f*x + e)*sinh(f*x + e) - (32*d^2*f^3*x^3 - 24*c^2*f^2 - 12 *c*d*f + 24*(4*c*d*f^3 - d^2*f^2)*x^2 - 3*d^2 + 12*(8*c^2*f^3 - 4*c*d*f^2 - d^2*f)*x)*sinh(f*x + e)^2 + 48*d^2 + 96*(2*c*d*f^2 + d^2*f)*x)/(a^2*f^3* cosh(f*x + e)^2 + 2*a^2*f^3*cosh(f*x + e)*sinh(f*x + e) + a^2*f^3*sinh(f*x + e)^2)
\[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=\frac {\int \frac {c^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} x^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
(Integral(c**2/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(d** 2*x**2/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(2*c*d*x/(ta nh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x))/a**2
Time = 0.43 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.12 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=\frac {1}{16} \, c^{2} {\left (\frac {4 \, {\left (f x + e\right )}}{a^{2} f} - \frac {4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac {{\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \, {\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} c d e^{\left (-4 \, e\right )}}{32 \, a^{2} f^{2}} + \frac {{\left (32 \, f^{3} x^{3} e^{\left (4 \, e\right )} - 48 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 3 \, {\left (8 \, f^{2} x^{2} + 4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} d^{2} e^{\left (-4 \, e\right )}}{384 \, a^{2} f^{3}} \]
1/16*c^2*(4*(f*x + e)/(a^2*f) - (4*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e))/(a ^2*f)) + 1/32*(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))*c*d*e^(-4*e)/(a^2*f^2) + 1/384*(32*f^3*x^3*e^(4* e) - 48*(2*f^2*x^2*e^(2*e) + 2*f*x*e^(2*e) + e^(2*e))*e^(-2*f*x) - 3*(8*f^ 2*x^2 + 4*f*x + 1)*e^(-4*f*x))*d^2*e^(-4*e)/(a^2*f^3)
Time = 0.27 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.28 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=\frac {{\left (32 \, d^{2} f^{3} x^{3} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c d f^{3} x^{2} e^{\left (4 \, f x + 4 \, e\right )} + 96 \, c^{2} f^{3} x e^{\left (4 \, f x + 4 \, e\right )} - 96 \, d^{2} f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 24 \, d^{2} f^{2} x^{2} - 192 \, c d f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 48 \, c d f^{2} x - 96 \, c^{2} f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, d^{2} f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c^{2} f^{2} - 12 \, d^{2} f x - 96 \, c d f e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d f - 48 \, d^{2} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{384 \, a^{2} f^{3}} \]
1/384*(32*d^2*f^3*x^3*e^(4*f*x + 4*e) + 96*c*d*f^3*x^2*e^(4*f*x + 4*e) + 9 6*c^2*f^3*x*e^(4*f*x + 4*e) - 96*d^2*f^2*x^2*e^(2*f*x + 2*e) - 24*d^2*f^2* x^2 - 192*c*d*f^2*x*e^(2*f*x + 2*e) - 48*c*d*f^2*x - 96*c^2*f^2*e^(2*f*x + 2*e) - 96*d^2*f*x*e^(2*f*x + 2*e) - 24*c^2*f^2 - 12*d^2*f*x - 96*c*d*f*e^ (2*f*x + 2*e) - 12*c*d*f - 48*d^2*e^(2*f*x + 2*e) - 3*d^2)*e^(-4*f*x - 4*e )/(a^2*f^3)
Time = 1.83 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.97 \[ \int \frac {(c+d x)^2}{(a+a \tanh (e+f x))^2} \, dx=\frac {c^2\,x}{4\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {8\,c^2\,f^2+4\,c\,d\,f+d^2}{128\,a^2\,f^3}+\frac {d^2\,x^2}{16\,a^2\,f}+\frac {d\,x\,\left (d+4\,c\,f\right )}{32\,a^2\,f^2}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {2\,c^2\,f^2+2\,c\,d\,f+d^2}{8\,a^2\,f^3}+\frac {d^2\,x^2}{4\,a^2\,f}+\frac {d\,x\,\left (d+2\,c\,f\right )}{4\,a^2\,f^2}\right )+\frac {d^2\,x^3}{12\,a^2}+\frac {c\,d\,x^2}{4\,a^2} \]
(c^2*x)/(4*a^2) - exp(- 4*e - 4*f*x)*((d^2 + 8*c^2*f^2 + 4*c*d*f)/(128*a^2 *f^3) + (d^2*x^2)/(16*a^2*f) + (d*x*(d + 4*c*f))/(32*a^2*f^2)) - exp(- 2*e - 2*f*x)*((d^2 + 2*c^2*f^2 + 2*c*d*f)/(8*a^2*f^3) + (d^2*x^2)/(4*a^2*f) + (d*x*(d + 2*c*f))/(4*a^2*f^2)) + (d^2*x^3)/(12*a^2) + (c*d*x^2)/(4*a^2)