3.1.32 \(\int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx\) [32]

3.1.32.1 Optimal result

Integrand size = 20, antiderivative size = 169 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {3 d^3 x}{8 a f^3}+\frac {3 d (c+d x)^2}{8 a f^2}+\frac {(c+d x)^3}{4 a f}+\frac {(c+d x)^4}{8 a d}-\frac {3 d^3}{8 f^4 (a+a \tanh (e+f x))}-\frac {3 d^2 (c+d x)}{4 f^3 (a+a \tanh (e+f x))}-\frac {3 d (c+d x)^2}{4 f^2 (a+a \tanh (e+f x))}-\frac {(c+d x)^3}{2 f (a+a \tanh (e+f x))} \]

output

3/8*d^3*x/a/f^3+3/8*d*(d*x+c)^2/a/f^2+1/4*(d*x+c)^3/a/f+1/8*(d*x+c)^4/a/d- 
3/8*d^3/f^4/(a+a*tanh(f*x+e))-3/4*d^2*(d*x+c)/f^3/(a+a*tanh(f*x+e))-3/4*d* 
(d*x+c)^2/f^2/(a+a*tanh(f*x+e))-1/2*(d*x+c)^3/f/(a+a*tanh(f*x+e))
 

3.1.32.2 Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.44 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {\text {sech}(e+f x) (\cosh (f x)+\sinh (f x)) \left (\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) \cosh (2 f x) (-\cosh (e)+\sinh (e))+2 f^4 x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (\cosh (e)+\sinh (e))+\left (4 c^3 f^3+6 c^2 d f^2 (1+2 f x)+6 c d^2 f \left (1+2 f x+2 f^2 x^2\right )+d^3 \left (3+6 f x+6 f^2 x^2+4 f^3 x^3\right )\right ) (\cosh (e)-\sinh (e)) \sinh (2 f x)\right )}{16 a f^4 (1+\tanh (e+f x))} \]

input

Integrate[(c + d*x)^3/(a + a*Tanh[e + f*x]),x]
 

output

(Sech[e + f*x]*(Cosh[f*x] + Sinh[f*x])*((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]*(-Cosh[e] + Sinh[e]) + 2*f^4*x*(4*c^3 + 6*c^2*d*x + 4* 
c*d^2*x^2 + d^3*x^3)*(Cosh[e] + Sinh[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))*(Cosh[e] - Sinh[e])*Sinh[2*f*x]))/(16*a*f^4*(1 + Tanh[e + f*x]) 
)
 

3.1.32.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3042, 4206, 3042, 4206, 3042, 4206, 3042, 3960, 24}

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.

\(\displaystyle \int \frac {(c+d x)^3}{a \tanh (e+f x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d x)^3}{a-i a \tan (i e+i f x)}dx\)

\(\Big \downarrow \) 4206

\(\displaystyle \frac {3 d \int \frac {(c+d x)^2}{\tanh (e+f x) a+a}dx}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d \int \frac {(c+d x)^2}{a-i a \tan (i e+i f x)}dx}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 4206

\(\displaystyle \frac {3 d \left (\frac {d \int \frac {c+d x}{\tanh (e+f x) a+a}dx}{f}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d \left (\frac {d \int \frac {c+d x}{a-i a \tan (i e+i f x)}dx}{f}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 4206

\(\displaystyle \frac {3 d \left (\frac {d \left (\frac {d \int \frac {1}{\tanh (e+f x) a+a}dx}{2 f}-\frac {c+d x}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a d}\right )}{f}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d \left (\frac {d \left (\frac {d \int \frac {1}{a-i a \tan (i e+i f x)}dx}{2 f}-\frac {c+d x}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a d}\right )}{f}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 3960

\(\displaystyle \frac {3 d \left (\frac {d \left (\frac {d \left (\frac {\int 1dx}{2 a}-\frac {1}{2 f (a \tanh (e+f x)+a)}\right )}{2 f}-\frac {c+d x}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a d}\right )}{f}-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}-\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^4}{8 a d}\)

\(\Big \downarrow \) 24

\(\displaystyle -\frac {(c+d x)^3}{2 f (a \tanh (e+f x)+a)}+\frac {3 d \left (-\frac {(c+d x)^2}{2 f (a \tanh (e+f x)+a)}+\frac {d \left (-\frac {c+d x}{2 f (a \tanh (e+f x)+a)}+\frac {(c+d x)^2}{4 a d}+\frac {d \left (\frac {x}{2 a}-\frac {1}{2 f (a \tanh (e+f x)+a)}\right )}{2 f}\right )}{f}+\frac {(c+d x)^3}{6 a d}\right )}{2 f}+\frac {(c+d x)^4}{8 a d}\)

input

Int[(c + d*x)^3/(a + a*Tanh[e + f*x]),x]
 

output

(c + d*x)^4/(8*a*d) - (c + d*x)^3/(2*f*(a + a*Tanh[e + f*x])) + (3*d*((c + 
 d*x)^3/(6*a*d) - (c + d*x)^2/(2*f*(a + a*Tanh[e + f*x])) + (d*((c + d*x)^ 
2/(4*a*d) - (c + d*x)/(2*f*(a + a*Tanh[e + f*x])) + (d*(x/(2*a) - 1/(2*f*( 
a + a*Tanh[e + f*x]))))/(2*f)))/f))/(2*f)
 

3.1.32.3.1 Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a)   Int[(a + b*Tan[c + d*x])^ 
(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
 

rule 4206

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(2*a*d*(m + 1)), x] + (Simp[a*d*(m/(2*b*f)) 
   Int[(c + d*x)^(m - 1)/(a + b*Tan[e + f*x]), x], x] - Simp[a*((c + d*x)^m 
/(2*b*f*(a + b*Tan[e + f*x]))), x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[ 
a^2 + b^2, 0] && GtQ[m, 0]
 

3.1.32.4 Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98

method	result	size
risch	\(\frac {d^{3} x^{4}}{8 a}+\frac {d^{2} c \,x^{3}}{2 a}+\frac {3 d \,c^{2} x^{2}}{4 a}+\frac {c^{3} x}{2 a}+\frac {c^{4}}{8 a d}-\frac {\left (4 d^{3} x^{3} f^{3}+12 c \,d^{2} f^{3} x^{2}+12 c^{2} d \,f^{3} x +6 d^{3} f^{2} x^{2}+4 c^{3} f^{3}+12 c \,d^{2} f^{2} x +6 c^{2} d \,f^{2}+6 d^{3} f x +6 c \,d^{2} f +3 d^{3}\right ) {\mathrm e}^{-2 f x -2 e}}{16 a \,f^{4}}\)	\(165\)
parallelrisch	\(\frac {-3 d^{3}-6 c^{2} d \,f^{3} x +6 x \tanh \left (f x +e \right ) c \,d^{2} f^{2}+4 x^{3} \tanh \left (f x +e \right ) c \,d^{2} f^{4}+6 x^{2} \tanh \left (f x +e \right ) c^{2} d \,f^{4}+6 x^{2} \tanh \left (f x +e \right ) c \,d^{2} f^{3}+6 x \tanh \left (f x +e \right ) c^{2} d \,f^{3}-4 c^{3} f^{3}+4 x \,c^{3} f^{4}-6 c \,d^{2} f^{2} x -3 d^{3} f^{2} x^{2}-6 c^{2} d \,f^{2}+d^{3} x^{4} f^{4}+3 x^{2} \tanh \left (f x +e \right ) d^{3} f^{2}+2 x^{3} \tanh \left (f x +e \right ) d^{3} f^{3}+4 x \tanh \left (f x +e \right ) c^{3} f^{4}+3 x \tanh \left (f x +e \right ) d^{3} f +d^{3} \tanh \left (f x +e \right ) x^{4} f^{4}+6 x^{2} c^{2} d \,f^{4}+4 x^{3} c \,d^{2} f^{4}-3 d^{3} f x -6 c \,d^{2} f -2 d^{3} x^{3} f^{3}-6 c \,d^{2} f^{3} x^{2}}{8 f^{4} a \left (1+\tanh \left (f x +e \right )\right )}\)	\(317\)
default	\(\text {Expression too large to display}\)	\(959\)

input

int((d*x+c)^3/(a+a*tanh(f*x+e)),x,method=_RETURNVERBOSE)
 

output

1/8/a*d^3*x^4+1/2/a*d^2*c*x^3+3/4/a*d*c^2*x^2+1/2/a*c^3*x+1/8/a/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/f^4*exp(-2*f*x-2*e)
 

3.1.32.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.80 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {{\left (2 \, d^{3} f^{4} x^{4} - 4 \, c^{3} f^{3} - 6 \, c^{2} d f^{2} - 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} - d^{3} f^{3}\right )} x^{3} - 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} - 2 \, c d^{2} f^{3} - d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} - 6 \, c^{2} d f^{3} - 6 \, c d^{2} f^{2} - 3 \, d^{3} f\right )} x\right )} \cosh \left (f x + e\right ) + {\left (2 \, d^{3} f^{4} x^{4} + 4 \, c^{3} f^{3} + 6 \, c^{2} d f^{2} + 6 \, c d^{2} f + 4 \, {\left (2 \, c d^{2} f^{4} + d^{3} f^{3}\right )} x^{3} + 3 \, d^{3} + 6 \, {\left (2 \, c^{2} d f^{4} + 2 \, c d^{2} f^{3} + d^{3} f^{2}\right )} x^{2} + 2 \, {\left (4 \, c^{3} f^{4} + 6 \, c^{2} d f^{3} + 6 \, c d^{2} f^{2} + 3 \, d^{3} f\right )} x\right )} \sinh \left (f x + e\right )}{16 \, {\left (a f^{4} \cosh \left (f x + e\right ) + a f^{4} \sinh \left (f x + e\right )\right )}} \]

input

integrate((d*x+c)^3/(a+a*tanh(f*x+e)),x, algorithm="fricas")
 

output

1/16*((2*d^3*f^4*x^4 - 4*c^3*f^3 - 6*c^2*d*f^2 - 6*c*d^2*f + 4*(2*c*d^2*f^ 
4 - d^3*f^3)*x^3 - 3*d^3 + 6*(2*c^2*d*f^4 - 2*c*d^2*f^3 - d^3*f^2)*x^2 + 2 
*(4*c^3*f^4 - 6*c^2*d*f^3 - 6*c*d^2*f^2 - 3*d^3*f)*x)*cosh(f*x + e) + (2*d 
^3*f^4*x^4 + 4*c^3*f^3 + 6*c^2*d*f^2 + 6*c*d^2*f + 4*(2*c*d^2*f^4 + d^3*f^ 
3)*x^3 + 3*d^3 + 6*(2*c^2*d*f^4 + 2*c*d^2*f^3 + d^3*f^2)*x^2 + 2*(4*c^3*f^ 
4 + 6*c^2*d*f^3 + 6*c*d^2*f^2 + 3*d^3*f)*x)*sinh(f*x + e))/(a*f^4*cosh(f*x 
 + e) + a*f^4*sinh(f*x + e))
 

3.1.32.6 Sympy [F]

\[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {\int \frac {c^{3}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} x^{3}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} x^{2}}{\tanh {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d x}{\tanh {\left (e + f x \right )} + 1}\, dx}{a} \]

input

integrate((d*x+c)**3/(a+a*tanh(f*x+e)),x)
 

output

(Integral(c**3/(tanh(e + f*x) + 1), x) + Integral(d**3*x**3/(tanh(e + f*x) 
 + 1), x) + Integral(3*c*d**2*x**2/(tanh(e + f*x) + 1), x) + Integral(3*c* 
*2*d*x/(tanh(e + f*x) + 1), x))/a
 

3.1.32.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.10 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {1}{4} \, c^{3} {\left (\frac {2 \, {\left (f x + e\right )}}{a f} - \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{a f}\right )} + \frac {3 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} - {\left (2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} c^{2} d e^{\left (-2 \, e\right )}}{8 \, a f^{2}} + \frac {{\left (4 \, f^{3} x^{3} e^{\left (2 \, e\right )} - 3 \, {\left (2 \, f^{2} x^{2} + 2 \, f x + 1\right )} e^{\left (-2 \, f x\right )}\right )} c d^{2} e^{\left (-2 \, e\right )}}{8 \, a f^{3}} + \frac {{\left (2 \, f^{4} x^{4} e^{\left (2 \, e\right )} - {\left (4 \, f^{3} x^{3} + 6 \, f^{2} x^{2} + 6 \, f x + 3\right )} e^{\left (-2 \, f x\right )}\right )} d^{3} e^{\left (-2 \, e\right )}}{16 \, a f^{4}} \]

input

integrate((d*x+c)^3/(a+a*tanh(f*x+e)),x, algorithm="maxima")
 

output

1/4*c^3*(2*(f*x + e)/(a*f) - e^(-2*f*x - 2*e)/(a*f)) + 3/8*(2*f^2*x^2*e^(2 
*e) - (2*f*x + 1)*e^(-2*f*x))*c^2*d*e^(-2*e)/(a*f^2) + 1/8*(4*f^3*x^3*e^(2 
*e) - 3*(2*f^2*x^2 + 2*f*x + 1)*e^(-2*f*x))*c*d^2*e^(-2*e)/(a*f^3) + 1/16* 
(2*f^4*x^4*e^(2*e) - (4*f^3*x^3 + 6*f^2*x^2 + 6*f*x + 3)*e^(-2*f*x))*d^3*e 
^(-2*e)/(a*f^4)
 

3.1.32.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {{\left (2 \, d^{3} f^{4} x^{4} e^{\left (2 \, f x + 2 \, e\right )} + 8 \, c d^{2} f^{4} x^{3} e^{\left (2 \, f x + 2 \, e\right )} + 12 \, c^{2} d f^{4} x^{2} e^{\left (2 \, f x + 2 \, e\right )} - 4 \, d^{3} f^{3} x^{3} + 8 \, c^{3} f^{4} x e^{\left (2 \, f x + 2 \, e\right )} - 12 \, c d^{2} f^{3} x^{2} - 12 \, c^{2} d f^{3} x - 6 \, d^{3} f^{2} x^{2} - 4 \, c^{3} f^{3} - 12 \, c d^{2} f^{2} x - 6 \, c^{2} d f^{2} - 6 \, d^{3} f x - 6 \, c d^{2} f - 3 \, d^{3}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, a f^{4}} \]

input

integrate((d*x+c)^3/(a+a*tanh(f*x+e)),x, algorithm="giac")
 

output

1/16*(2*d^3*f^4*x^4*e^(2*f*x + 2*e) + 8*c*d^2*f^4*x^3*e^(2*f*x + 2*e) + 12 
*c^2*d*f^4*x^2*e^(2*f*x + 2*e) - 4*d^3*f^3*x^3 + 8*c^3*f^4*x*e^(2*f*x + 2* 
e) - 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)*e^(-2*f*x - 2*e)/ 
(a*f^4)
 

3.1.32.9 Mupad [B] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.62 \[ \int \frac {(c+d x)^3}{a+a \tanh (e+f x)} \, dx=\frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (8\,c^3\,x\,{\mathrm {e}}^{2\,e+2\,f\,x}+2\,d^3\,x^4\,{\mathrm {e}}^{2\,e+2\,f\,x}+12\,c^2\,d\,x^2\,{\mathrm {e}}^{2\,e+2\,f\,x}+8\,c\,d^2\,x^3\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16\,a}-\frac {\frac {{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (3\,d^3-3\,d^3\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16}+\frac {f^2\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c^2\,d+6\,d^3\,x^2-6\,c^2\,d\,{\mathrm {e}}^{2\,e+2\,f\,x}+12\,c\,d^2\,x\right )}{16}+\frac {f\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (6\,c\,d^2+6\,d^3\,x-6\,c\,d^2\,{\mathrm {e}}^{2\,e+2\,f\,x}\right )}{16}+\frac {f^3\,{\mathrm {e}}^{-2\,e-2\,f\,x}\,\left (4\,c^3-4\,c^3\,{\mathrm {e}}^{2\,e+2\,f\,x}+4\,d^3\,x^3+12\,c\,d^2\,x^2+12\,c^2\,d\,x\right )}{16}}{a\,f^4} \]

input

int((c + d*x)^3/(a + a*tanh(e + f*x)),x)
 

output

(exp(- 2*e - 2*f*x)*(8*c^3*x*exp(2*e + 2*f*x) + 2*d^3*x^4*exp(2*e + 2*f*x) 
 + 12*c^2*d*x^2*exp(2*e + 2*f*x) + 8*c*d^2*x^3*exp(2*e + 2*f*x)))/(16*a) - 
 ((exp(- 2*e - 2*f*x)*(3*d^3 - 3*d^3*exp(2*e + 2*f*x)))/16 + (f^2*exp(- 2* 
e - 2*f*x)*(6*c^2*d + 6*d^3*x^2 - 6*c^2*d*exp(2*e + 2*f*x) + 12*c*d^2*x))/ 
16 + (f*exp(- 2*e - 2*f*x)*(6*c*d^2 + 6*d^3*x - 6*c*d^2*exp(2*e + 2*f*x))) 
/16 + (f^3*exp(- 2*e - 2*f*x)*(4*c^3 - 4*c^3*exp(2*e + 2*f*x) + 4*d^3*x^3 
+ 12*c*d^2*x^2 + 12*c^2*d*x))/16)/(a*f^4)