Integrand size = 20, antiderivative size = 230 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=-\frac {3 d^3 e^{-4 e-4 f x}}{512 a^2 f^4}-\frac {3 d^3 e^{-2 e-2 f x}}{16 a^2 f^4}-\frac {3 d^2 e^{-4 e-4 f x} (c+d x)}{128 a^2 f^3}-\frac {3 d^2 e^{-2 e-2 f x} (c+d x)}{8 a^2 f^3}-\frac {3 d e^{-4 e-4 f x} (c+d x)^2}{64 a^2 f^2}-\frac {3 d e^{-2 e-2 f x} (c+d x)^2}{8 a^2 f^2}-\frac {e^{-4 e-4 f x} (c+d x)^3}{16 a^2 f}-\frac {e^{-2 e-2 f x} (c+d x)^3}{4 a^2 f}+\frac {(c+d x)^4}{16 a^2 d} \]
-3/512*d^3*exp(-4*f*x-4*e)/a^2/f^4-3/16*d^3*exp(-2*f*x-2*e)/a^2/f^4-3/128* d^2*exp(-4*f*x-4*e)*(d*x+c)/a^2/f^3-3/8*d^2*exp(-2*f*x-2*e)*(d*x+c)/a^2/f^ 3-3/64*d*exp(-4*f*x-4*e)*(d*x+c)^2/a^2/f^2-3/8*d*exp(-2*f*x-2*e)*(d*x+c)^2 /a^2/f^2-1/16*exp(-4*f*x-4*e)*(d*x+c)^3/a^2/f-1/4*exp(-2*f*x-2*e)*(d*x+c)^ 3/a^2/f+1/16*(d*x+c)^4/a^2/d
Time = 2.02 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.83 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=\frac {\text {sech}^2(e+f x) (\cosh (f x)+\sinh (f x))^2 \left (-\left (\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) \cosh (2 f x)\right )+\frac {1}{32} \left (32 c^3 f^3+24 c^2 d f^2 (1+4 f x)+12 c d^2 f \left (1+4 f x+8 f^2 x^2\right )+d^3 \left (3+12 f x+24 f^2 x^2+32 f^3 x^3\right )\right ) \cosh (4 f x) (-\cosh (2 e)+\sinh (2 e))+f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (\cosh (2 e)+\sinh (2 e))+\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) \sinh (2 f x)+\frac {1}{32} \left (32 c^3 f^3+24 c^2 d f^2 (1+4 f x)+12 c d^2 f \left (1+4 f x+8 f^2 x^2\right )+d^3 \left (3+12 f x+24 f^2 x^2+32 f^3 x^3\right )\right ) (\cosh (2 e)-\sinh (2 e)) \sinh (4 f x)\right )}{16 a^2 f^4 (1+\tanh (e+f x))^2} \]
(Sech[e + f*x]^2*(Cosh[f*x] + Sinh[f*x])^2*(-((4*c^3*f^3 + 6*c^2*d*f^2*(1 + 2*f*x) + 6*c*d^2*f*(1 + 2*f*x + 2*f^2*x^2) + d^3*(3 + 6*f*x + 6*f^2*x^2 + 4*f^3*x^3))*Cosh[2*f*x]) + ((32*c^3*f^3 + 24*c^2*d*f^2*(1 + 4*f*x) + 12* c*d^2*f*(1 + 4*f*x + 8*f^2*x^2) + d^3*(3 + 12*f*x + 24*f^2*x^2 + 32*f^3*x^ 3))*Cosh[4*f*x]*(-Cosh[2*e] + Sinh[2*e]))/32 + f^4*x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(Cosh[2*e] + Sinh[2*e]) + (4*c^3*f^3 + 6*c^2*d*f^2* (1 + 2*f*x) + 6*c*d^2*f*(1 + 2*f*x + 2*f^2*x^2) + d^3*(3 + 6*f*x + 6*f^2*x ^2 + 4*f^3*x^3))*Sinh[2*f*x] + ((32*c^3*f^3 + 24*c^2*d*f^2*(1 + 4*f*x) + 1 2*c*d^2*f*(1 + 4*f*x + 8*f^2*x^2) + d^3*(3 + 12*f*x + 24*f^2*x^2 + 32*f^3* x^3))*(Cosh[2*e] - Sinh[2*e])*Sinh[4*f*x])/32))/(16*a^2*f^4*(1 + Tanh[e + f*x])^2)
Time = 0.54 (sec) , antiderivative size = 230, 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.
\(\displaystyle \int \frac {(c+d x)^3}{(a \tanh (e+f x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d x)^3}{(a-i a \tan (i e+i f x))^2}dx\) |
\(\Big \downarrow \) 4212 |
\(\displaystyle \int \left (\frac {(c+d x)^3 e^{-4 e-4 f x}}{4 a^2}+\frac {(c+d x)^3 e^{-2 e-2 f x}}{2 a^2}+\frac {(c+d x)^3}{4 a^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 d^2 (c+d x) e^{-4 e-4 f x}}{128 a^2 f^3}-\frac {3 d^2 (c+d x) e^{-2 e-2 f x}}{8 a^2 f^3}-\frac {3 d (c+d x)^2 e^{-4 e-4 f x}}{64 a^2 f^2}-\frac {3 d (c+d x)^2 e^{-2 e-2 f x}}{8 a^2 f^2}-\frac {(c+d x)^3 e^{-4 e-4 f x}}{16 a^2 f}-\frac {(c+d x)^3 e^{-2 e-2 f x}}{4 a^2 f}+\frac {(c+d x)^4}{16 a^2 d}-\frac {3 d^3 e^{-4 e-4 f x}}{512 a^2 f^4}-\frac {3 d^3 e^{-2 e-2 f x}}{16 a^2 f^4}\) |
(-3*d^3*E^(-4*e - 4*f*x))/(512*a^2*f^4) - (3*d^3*E^(-2*e - 2*f*x))/(16*a^2 *f^4) - (3*d^2*E^(-4*e - 4*f*x)*(c + d*x))/(128*a^2*f^3) - (3*d^2*E^(-2*e - 2*f*x)*(c + d*x))/(8*a^2*f^3) - (3*d*E^(-4*e - 4*f*x)*(c + d*x)^2)/(64*a ^2*f^2) - (3*d*E^(-2*e - 2*f*x)*(c + d*x)^2)/(8*a^2*f^2) - (E^(-4*e - 4*f* x)*(c + d*x)^3)/(16*a^2*f) - (E^(-2*e - 2*f*x)*(c + d*x)^3)/(4*a^2*f) + (c + d*x)^4/(16*a^2*d)
3.1.38.3.1 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 = 0.30 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.19
method | result | size |
risch | \(\frac {d^{3} x^{4}}{16 a^{2}}+\frac {d^{2} c \,x^{3}}{4 a^{2}}+\frac {3 d \,c^{2} x^{2}}{8 a^{2}}+\frac {c^{3} x}{4 a^{2}}+\frac {c^{4}}{16 a^{2} d}-\frac {\left (4 d^{3} x^{3} f^{3}+12 c \,d^{2} f^{3} x^{2}+12 c^{2} d \,f^{3} x +6 d^{3} f^{2} x^{2}+4 c^{3} f^{3}+12 c \,d^{2} f^{2} x +6 c^{2} d \,f^{2}+6 d^{3} f x +6 c \,d^{2} f +3 d^{3}\right ) {\mathrm e}^{-2 f x -2 e}}{16 a^{2} f^{4}}-\frac {\left (32 d^{3} x^{3} f^{3}+96 c \,d^{2} f^{3} x^{2}+96 c^{2} d \,f^{3} x +24 d^{3} f^{2} x^{2}+32 c^{3} f^{3}+48 c \,d^{2} f^{2} x +24 c^{2} d \,f^{2}+12 d^{3} f x +12 c \,d^{2} f +3 d^{3}\right ) {\mathrm e}^{-4 f x -4 e}}{512 a^{2} f^{4}}\) | \(273\) |
parallelrisch | \(\frac {-48 d^{3}-120 c^{2} d \,f^{3} x -45 \tanh \left (f x +e \right ) d^{3}+24 x \tanh \left (f x +e \right ) c \,d^{2} f^{2}+64 x^{3} \tanh \left (f x +e \right ) c \,d^{2} f^{4}+96 x^{2} \tanh \left (f x +e \right ) c^{2} d \,f^{4}+48 x^{2} \tanh \left (f x +e \right ) c \,d^{2} f^{3}+48 x \tanh \left (f x +e \right ) c^{2} d \,f^{3}-64 c^{3} f^{3}+8 d^{3} \tanh \left (f x +e \right )^{2} x^{4} f^{4}-72 \tanh \left (f x +e \right ) c^{2} d \,f^{2}-84 \tanh \left (f x +e \right ) c \,d^{2} f +32 x \,c^{3} f^{4}-108 c \,d^{2} f^{2} x -54 d^{3} f^{2} x^{2}-96 c^{2} d \,f^{2}+72 x \tanh \left (f x +e \right )^{2} c^{2} d \,f^{3}+84 x \tanh \left (f x +e \right )^{2} c \,d^{2} f^{2}+48 x^{2} \tanh \left (f x +e \right )^{2} c^{2} d \,f^{4}+72 x^{2} \tanh \left (f x +e \right )^{2} c \,d^{2} f^{3}+32 x^{3} \tanh \left (f x +e \right )^{2} c \,d^{2} f^{4}+8 d^{3} x^{4} f^{4}+12 x^{2} \tanh \left (f x +e \right ) d^{3} f^{2}+16 x^{3} \tanh \left (f x +e \right ) d^{3} f^{3}+64 x \tanh \left (f x +e \right ) c^{3} f^{4}+6 x \tanh \left (f x +e \right ) d^{3} f +16 d^{3} \tanh \left (f x +e \right ) x^{4} f^{4}+48 x^{2} c^{2} d \,f^{4}+32 x^{3} c \,d^{2} f^{4}-51 d^{3} f x -96 c \,d^{2} f +32 x \tanh \left (f x +e \right )^{2} c^{3} f^{4}+45 x \tanh \left (f x +e \right )^{2} d^{3} f +42 x^{2} \tanh \left (f x +e \right )^{2} d^{3} f^{2}+24 x^{3} \tanh \left (f x +e \right )^{2} d^{3} f^{3}-40 d^{3} x^{3} f^{3}-120 c \,d^{2} f^{3} x^{2}-32 \tanh \left (f x +e \right ) c^{3} f^{3}}{128 f^{4} a^{2} \left (1+\tanh \left (f x +e \right )\right )^{2}}\) | \(557\) |
1/16/a^2*d^3*x^4+1/4/a^2*d^2*c*x^3+3/8/a^2*d*c^2*x^2+1/4/a^2*c^3*x+1/16/a^ 2/d*c^4-1/16*(4*d^3*f^3*x^3+12*c*d^2*f^3*x^2+12*c^2*d*f^3*x+6*d^3*f^2*x^2+ 4*c^3*f^3+12*c*d^2*f^2*x+6*c^2*d*f^2+6*d^3*f*x+6*c*d^2*f+3*d^3)/a^2/f^4*ex p(-2*f*x-2*e)-1/512*(32*d^3*f^3*x^3+96*c*d^2*f^3*x^2+96*c^2*d*f^3*x+24*d^3 *f^2*x^2+32*c^3*f^3+48*c*d^2*f^2*x+24*c^2*d*f^2+12*d^3*f*x+12*c*d^2*f+3*d^ 3)/a^2/f^4*exp(-4*f*x-4*e)
Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (204) = 408\).
Time = 0.26 (sec) , antiderivative size = 573, normalized size of antiderivative = 2.49 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=-\frac {128 \, d^{3} f^{3} x^{3} + 128 \, c^{3} f^{3} + 192 \, c^{2} d f^{2} + 192 \, c d^{2} f + 96 \, d^{3} + 192 \, {\left (2 \, c d^{2} f^{3} + d^{3} f^{2}\right )} x^{2} - {\left (32 \, d^{3} f^{4} x^{4} - 32 \, c^{3} f^{3} - 24 \, c^{2} d f^{2} - 12 \, c d^{2} f + 32 \, {\left (4 \, c d^{2} f^{4} - d^{3} f^{3}\right )} x^{3} - 3 \, d^{3} + 24 \, {\left (8 \, c^{2} d f^{4} - 4 \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} + 4 \, {\left (32 \, c^{3} f^{4} - 24 \, c^{2} d f^{3} - 12 \, c d^{2} f^{2} - 3 \, d^{3} f\right )} x\right )} \cosh \left (f x + e\right )^{2} - 2 \, {\left (32 \, d^{3} f^{4} x^{4} + 32 \, c^{3} f^{3} + 24 \, c^{2} d f^{2} + 12 \, c d^{2} f + 32 \, {\left (4 \, c d^{2} f^{4} + d^{3} f^{3}\right )} x^{3} + 3 \, d^{3} + 24 \, {\left (8 \, c^{2} d f^{4} + 4 \, c d^{2} f^{3} + d^{3} f^{2}\right )} x^{2} + 4 \, {\left (32 \, c^{3} f^{4} + 24 \, c^{2} d f^{3} + 12 \, c d^{2} f^{2} + 3 \, d^{3} f\right )} x\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) - {\left (32 \, d^{3} f^{4} x^{4} - 32 \, c^{3} f^{3} - 24 \, c^{2} d f^{2} - 12 \, c d^{2} f + 32 \, {\left (4 \, c d^{2} f^{4} - d^{3} f^{3}\right )} x^{3} - 3 \, d^{3} + 24 \, {\left (8 \, c^{2} d f^{4} - 4 \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} + 4 \, {\left (32 \, c^{3} f^{4} - 24 \, c^{2} d f^{3} - 12 \, c d^{2} f^{2} - 3 \, d^{3} f\right )} x\right )} \sinh \left (f x + e\right )^{2} + 192 \, {\left (2 \, c^{2} d f^{3} + 2 \, c d^{2} f^{2} + d^{3} f\right )} x}{512 \, {\left (a^{2} f^{4} \cosh \left (f x + e\right )^{2} + 2 \, a^{2} f^{4} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + a^{2} f^{4} \sinh \left (f x + e\right )^{2}\right )}} \]
-1/512*(128*d^3*f^3*x^3 + 128*c^3*f^3 + 192*c^2*d*f^2 + 192*c*d^2*f + 96*d ^3 + 192*(2*c*d^2*f^3 + d^3*f^2)*x^2 - (32*d^3*f^4*x^4 - 32*c^3*f^3 - 24*c ^2*d*f^2 - 12*c*d^2*f + 32*(4*c*d^2*f^4 - d^3*f^3)*x^3 - 3*d^3 + 24*(8*c^2 *d*f^4 - 4*c*d^2*f^3 - d^3*f^2)*x^2 + 4*(32*c^3*f^4 - 24*c^2*d*f^3 - 12*c* d^2*f^2 - 3*d^3*f)*x)*cosh(f*x + e)^2 - 2*(32*d^3*f^4*x^4 + 32*c^3*f^3 + 2 4*c^2*d*f^2 + 12*c*d^2*f + 32*(4*c*d^2*f^4 + d^3*f^3)*x^3 + 3*d^3 + 24*(8* c^2*d*f^4 + 4*c*d^2*f^3 + d^3*f^2)*x^2 + 4*(32*c^3*f^4 + 24*c^2*d*f^3 + 12 *c*d^2*f^2 + 3*d^3*f)*x)*cosh(f*x + e)*sinh(f*x + e) - (32*d^3*f^4*x^4 - 3 2*c^3*f^3 - 24*c^2*d*f^2 - 12*c*d^2*f + 32*(4*c*d^2*f^4 - d^3*f^3)*x^3 - 3 *d^3 + 24*(8*c^2*d*f^4 - 4*c*d^2*f^3 - d^3*f^2)*x^2 + 4*(32*c^3*f^4 - 24*c ^2*d*f^3 - 12*c*d^2*f^2 - 3*d^3*f)*x)*sinh(f*x + e)^2 + 192*(2*c^2*d*f^3 + 2*c*d^2*f^2 + d^3*f)*x)/(a^2*f^4*cosh(f*x + e)^2 + 2*a^2*f^4*cosh(f*x + e )*sinh(f*x + e) + a^2*f^4*sinh(f*x + e)^2)
\[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=\frac {\int \frac {c^{3}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} x^{3}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} x^{2}}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d x}{\tanh ^{2}{\left (e + f x \right )} + 2 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]
(Integral(c**3/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(d** 3*x**3/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(3*c*d**2*x* *2/(tanh(e + f*x)**2 + 2*tanh(e + f*x) + 1), x) + Integral(3*c**2*d*x/(tan h(e + f*x)**2 + 2*tanh(e + f*x) + 1), x))/a**2
Time = 0.51 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.29 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=\frac {1}{16} \, c^{3} {\left (\frac {4 \, {\left (f x + e\right )}}{a^{2} f} - \frac {4 \, e^{\left (-2 \, f x - 2 \, e\right )} + e^{\left (-4 \, f x - 4 \, e\right )}}{a^{2} f}\right )} + \frac {3 \, {\left (8 \, f^{2} x^{2} e^{\left (4 \, e\right )} - 8 \, {\left (2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} c^{2} d e^{\left (-4 \, e\right )}}{64 \, a^{2} f^{2}} + \frac {{\left (32 \, f^{3} x^{3} e^{\left (4 \, e\right )} - 48 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - 3 \, {\left (8 \, f^{2} x^{2} + 4 \, f x + 1\right )} e^{\left (-4 \, f x\right )}\right )} c d^{2} e^{\left (-4 \, e\right )}}{128 \, a^{2} f^{3}} + \frac {{\left (32 \, f^{4} x^{4} e^{\left (4 \, e\right )} - 32 \, {\left (4 \, f^{3} x^{3} e^{\left (2 \, e\right )} + 6 \, f^{2} x^{2} e^{\left (2 \, e\right )} + 6 \, f x e^{\left (2 \, e\right )} + 3 \, e^{\left (2 \, e\right )}\right )} e^{\left (-2 \, f x\right )} - {\left (32 \, f^{3} x^{3} + 24 \, f^{2} x^{2} + 12 \, f x + 3\right )} e^{\left (-4 \, f x\right )}\right )} d^{3} e^{\left (-4 \, e\right )}}{512 \, a^{2} f^{4}} \]
1/16*c^3*(4*(f*x + e)/(a^2*f) - (4*e^(-2*f*x - 2*e) + e^(-4*f*x - 4*e))/(a ^2*f)) + 3/64*(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))*c^2*d*e^(-4*e)/(a^2*f^2) + 1/128*(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))*c*d^2*e^(-4*e)/(a^2*f^3) + 1/512*(32*f^4* x^4*e^(4*e) - 32*(4*f^3*x^3*e^(2*e) + 6*f^2*x^2*e^(2*e) + 6*f*x*e^(2*e) + 3*e^(2*e))*e^(-2*f*x) - (32*f^3*x^3 + 24*f^2*x^2 + 12*f*x + 3)*e^(-4*f*x)) *d^3*e^(-4*e)/(a^2*f^4)
Time = 0.27 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.60 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=\frac {{\left (32 \, d^{3} f^{4} x^{4} e^{\left (4 \, f x + 4 \, e\right )} + 128 \, c d^{2} f^{4} x^{3} e^{\left (4 \, f x + 4 \, e\right )} + 192 \, c^{2} d f^{4} x^{2} e^{\left (4 \, f x + 4 \, e\right )} - 128 \, d^{3} f^{3} x^{3} e^{\left (2 \, f x + 2 \, e\right )} - 32 \, d^{3} f^{3} x^{3} + 128 \, c^{3} f^{4} x e^{\left (4 \, f x + 4 \, e\right )} - 384 \, c d^{2} f^{3} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, c d^{2} f^{3} x^{2} - 384 \, c^{2} d f^{3} x e^{\left (2 \, f x + 2 \, e\right )} - 192 \, d^{3} f^{2} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 96 \, c^{2} d f^{3} x - 24 \, d^{3} f^{2} x^{2} - 128 \, c^{3} f^{3} e^{\left (2 \, f x + 2 \, e\right )} - 384 \, c d^{2} f^{2} x e^{\left (2 \, f x + 2 \, e\right )} - 32 \, c^{3} f^{3} - 48 \, c d^{2} f^{2} x - 192 \, c^{2} d f^{2} e^{\left (2 \, f x + 2 \, e\right )} - 192 \, d^{3} f x e^{\left (2 \, f x + 2 \, e\right )} - 24 \, c^{2} d f^{2} - 12 \, d^{3} f x - 192 \, c d^{2} f e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d^{2} f - 96 \, d^{3} e^{\left (2 \, f x + 2 \, e\right )} - 3 \, d^{3}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{512 \, a^{2} f^{4}} \]
1/512*(32*d^3*f^4*x^4*e^(4*f*x + 4*e) + 128*c*d^2*f^4*x^3*e^(4*f*x + 4*e) + 192*c^2*d*f^4*x^2*e^(4*f*x + 4*e) - 128*d^3*f^3*x^3*e^(2*f*x + 2*e) - 32 *d^3*f^3*x^3 + 128*c^3*f^4*x*e^(4*f*x + 4*e) - 384*c*d^2*f^3*x^2*e^(2*f*x + 2*e) - 96*c*d^2*f^3*x^2 - 384*c^2*d*f^3*x*e^(2*f*x + 2*e) - 192*d^3*f^2* x^2*e^(2*f*x + 2*e) - 96*c^2*d*f^3*x - 24*d^3*f^2*x^2 - 128*c^3*f^3*e^(2*f *x + 2*e) - 384*c*d^2*f^2*x*e^(2*f*x + 2*e) - 32*c^3*f^3 - 48*c*d^2*f^2*x - 192*c^2*d*f^2*e^(2*f*x + 2*e) - 192*d^3*f*x*e^(2*f*x + 2*e) - 24*c^2*d*f ^2 - 12*d^3*f*x - 192*c*d^2*f*e^(2*f*x + 2*e) - 12*c*d^2*f - 96*d^3*e^(2*f *x + 2*e) - 3*d^3)*e^(-4*f*x - 4*e)/(a^2*f^4)
Time = 1.85 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x)^3}{(a+a \tanh (e+f x))^2} \, dx=\frac {c^3\,x}{4\,a^2}-{\mathrm {e}}^{-4\,e-4\,f\,x}\,\left (\frac {32\,c^3\,f^3+24\,c^2\,d\,f^2+12\,c\,d^2\,f+3\,d^3}{512\,a^2\,f^4}+\frac {d^3\,x^3}{16\,a^2\,f}+\frac {3\,d\,x\,\left (8\,c^2\,f^2+4\,c\,d\,f+d^2\right )}{128\,a^2\,f^3}+\frac {3\,d^2\,x^2\,\left (d+4\,c\,f\right )}{64\,a^2\,f^2}\right )-{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (\frac {4\,c^3\,f^3+6\,c^2\,d\,f^2+6\,c\,d^2\,f+3\,d^3}{16\,a^2\,f^4}+\frac {d^3\,x^3}{4\,a^2\,f}+\frac {3\,d\,x\,\left (2\,c^2\,f^2+2\,c\,d\,f+d^2\right )}{8\,a^2\,f^3}+\frac {3\,d^2\,x^2\,\left (d+2\,c\,f\right )}{8\,a^2\,f^2}\right )+\frac {d^3\,x^4}{16\,a^2}+\frac {3\,c^2\,d\,x^2}{8\,a^2}+\frac {c\,d^2\,x^3}{4\,a^2} \]
(c^3*x)/(4*a^2) - exp(- 4*e - 4*f*x)*((3*d^3 + 32*c^3*f^3 + 24*c^2*d*f^2 + 12*c*d^2*f)/(512*a^2*f^4) + (d^3*x^3)/(16*a^2*f) + (3*d*x*(d^2 + 8*c^2*f^ 2 + 4*c*d*f))/(128*a^2*f^3) + (3*d^2*x^2*(d + 4*c*f))/(64*a^2*f^2)) - exp( - 2*e - 2*f*x)*((3*d^3 + 4*c^3*f^3 + 6*c^2*d*f^2 + 6*c*d^2*f)/(16*a^2*f^4) + (d^3*x^3)/(4*a^2*f) + (3*d*x*(d^2 + 2*c^2*f^2 + 2*c*d*f))/(8*a^2*f^3) + (3*d^2*x^2*(d + 2*c*f))/(8*a^2*f^2)) + (d^3*x^4)/(16*a^2) + (3*c^2*d*x^2) /(8*a^2) + (c*d^2*x^3)/(4*a^2)