\(\int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx\) [112]

Optimal result

Integrand size = 22, antiderivative size = 321 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d e^2 F^{a+b c+b d x} \log (F)}{12 x^3}-\frac {b d e f F^{a+b c+b d x} \log (F)}{3 x^2}-\frac {b d f^2 F^{a+b c+b d x} \log (F)}{2 x}-\frac {b^2 d^2 e^2 F^{a+b c+b d x} \log ^2(F)}{24 x^2}-\frac {b^2 d^2 e f F^{a+b c+b d x} \log ^2(F)}{3 x}+\frac {1}{2} b^2 d^2 f^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F)-\frac {b^3 d^3 e^2 F^{a+b c+b d x} \log ^3(F)}{24 x}+\frac {1}{3} b^3 d^3 e f F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^3(F)+\frac {1}{24} b^4 d^4 e^2 F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^4(F) \] Output:

-1/4*e^2*F^(b*d*x+b*c+a)/x^4-2/3*e*f*F^(b*d*x+b*c+a)/x^3-1/2*f^2*F^(b*d*x+ 
b*c+a)/x^2-1/12*b*d*e^2*F^(b*d*x+b*c+a)*ln(F)/x^3-1/3*b*d*e*f*F^(b*d*x+b*c 
+a)*ln(F)/x^2-1/2*b*d*f^2*F^(b*d*x+b*c+a)*ln(F)/x-1/24*b^2*d^2*e^2*F^(b*d* 
x+b*c+a)*ln(F)^2/x^2-1/3*b^2*d^2*e*f*F^(b*d*x+b*c+a)*ln(F)^2/x+1/2*b^2*d^2 
*f^2*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^2-1/24*b^3*d^3*e^2*F^(b*d*x+b*c+a)*ln 
(F)^3/x+1/3*b^3*d^3*e*f*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^3+1/24*b^4*d^4*e^2 
*F^(b*c+a)*Ei(b*d*x*ln(F))*ln(F)^4
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.49 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=\frac {F^{a+b c} \left (b^2 d^2 x^4 \operatorname {ExpIntegralEi}(b d x \log (F)) \log ^2(F) \left (12 f^2+8 b d e f \log (F)+b^2 d^2 e^2 \log ^2(F)\right )-F^{b d x} \left (2 \left (3 e^2+8 e f x+6 f^2 x^2\right )+2 b d x \left (e^2+4 e f x+6 f^2 x^2\right ) \log (F)+b^2 d^2 e x^2 (e+8 f x) \log ^2(F)+b^3 d^3 e^2 x^3 \log ^3(F)\right )\right )}{24 x^4} \] Input:

Integrate[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]
 

Output:

(F^(a + b*c)*(b^2*d^2*x^4*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2*(12*f^2 + 8 
*b*d*e*f*Log[F] + b^2*d^2*e^2*Log[F]^2) - F^(b*d*x)*(2*(3*e^2 + 8*e*f*x + 
6*f^2*x^2) + 2*b*d*x*(e^2 + 4*e*f*x + 6*f^2*x^2)*Log[F] + b^2*d^2*e*x^2*(e 
 + 8*f*x)*Log[F]^2 + b^3*d^3*e^2*x^3*Log[F]^3)))/(24*x^4)
 

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2629, 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.

\(\displaystyle \int \frac {(e+f x)^2 F^{a+b (c+d x)}}{x^5} \, dx\)

\(\Big \downarrow \) 2629

\(\displaystyle \int \left (\frac {e^2 F^{a+b (c+d x)}}{x^5}+\frac {2 e f F^{a+b (c+d x)}}{x^4}+\frac {f^2 F^{a+b (c+d x)}}{x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{24} b^4 d^4 e^2 \log ^4(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b^3 d^3 e^2 \log ^3(F) F^{a+b c+b d x}}{24 x}+\frac {1}{3} b^3 d^3 e f \log ^3(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {b^2 d^2 e^2 \log ^2(F) F^{a+b c+b d x}}{24 x^2}-\frac {b^2 d^2 e f \log ^2(F) F^{a+b c+b d x}}{3 x}+\frac {1}{2} b^2 d^2 f^2 \log ^2(F) F^{a+b c} \operatorname {ExpIntegralEi}(b d x \log (F))-\frac {e^2 F^{a+b c+b d x}}{4 x^4}-\frac {b d e^2 \log (F) F^{a+b c+b d x}}{12 x^3}-\frac {2 e f F^{a+b c+b d x}}{3 x^3}-\frac {b d e f \log (F) F^{a+b c+b d x}}{3 x^2}-\frac {f^2 F^{a+b c+b d x}}{2 x^2}-\frac {b d f^2 \log (F) F^{a+b c+b d x}}{2 x}\)

Input:

Int[(F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x]
 

Output:

-1/4*(e^2*F^(a + b*c + b*d*x))/x^4 - (2*e*f*F^(a + b*c + b*d*x))/(3*x^3) - 
 (f^2*F^(a + b*c + b*d*x))/(2*x^2) - (b*d*e^2*F^(a + b*c + b*d*x)*Log[F])/ 
(12*x^3) - (b*d*e*f*F^(a + b*c + b*d*x)*Log[F])/(3*x^2) - (b*d*f^2*F^(a + 
b*c + b*d*x)*Log[F])/(2*x) - (b^2*d^2*e^2*F^(a + b*c + b*d*x)*Log[F]^2)/(2 
4*x^2) - (b^2*d^2*e*f*F^(a + b*c + b*d*x)*Log[F]^2)/(3*x) + (b^2*d^2*f^2*F 
^(a + b*c)*ExpIntegralEi[b*d*x*Log[F]]*Log[F]^2)/2 - (b^3*d^3*e^2*F^(a + b 
*c + b*d*x)*Log[F]^3)/(24*x) + (b^3*d^3*e*f*F^(a + b*c)*ExpIntegralEi[b*d* 
x*Log[F]]*Log[F]^3)/3 + (b^4*d^4*e^2*F^(a + b*c)*ExpIntegralEi[b*d*x*Log[F 
]]*Log[F]^4)/24
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2629

Int[(F_)^(v_)*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandInte 
grand[F^v, Px*(d + e*x)^m, x], x] /; FreeQ[{F, d, e, m}, x] && PolynomialQ[ 
Px, x] && LinearQ[v, x] &&  !TrueQ[$UseGamma]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.20

method	result	size
risch	\(-\frac {\ln \left (F \right )^{4} F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b^{4} d^{4} e^{2} x^{4}+8 \ln \left (F \right )^{3} F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b^{3} d^{3} e f \,x^{4}+F^{b d x} F^{b c +a} b^{3} d^{3} e^{2} x^{3} \ln \left (F \right )^{3}+12 \ln \left (F \right )^{2} F^{b c} F^{a} \operatorname {expIntegral}_{1}\left (\ln \left (F \right ) b c +\ln \left (F \right ) a -b d x \ln \left (F \right )-\left (b c +a \right ) \ln \left (F \right )\right ) b^{2} d^{2} f^{2} x^{4}+8 F^{b d x} F^{b c +a} b^{2} d^{2} e f \,x^{3} \ln \left (F \right )^{2}+F^{b d x} F^{b c +a} b^{2} d^{2} e^{2} x^{2} \ln \left (F \right )^{2}+12 F^{b d x} F^{b c +a} b d \,f^{2} x^{3} \ln \left (F \right )+8 F^{b d x} F^{b c +a} b d e f \,x^{2} \ln \left (F \right )+2 F^{b d x} F^{b c +a} b d \,e^{2} x \ln \left (F \right )+12 F^{b d x} F^{b c +a} f^{2} x^{2}+16 F^{b d x} F^{b c +a} e f x +6 F^{b d x} F^{b c +a} e^{2}}{24 x^{4}}\)	\(386\)
meijerg	\(\ln \left (F \right )^{2} b^{2} d^{2} F^{b c +a} f^{2} \left (-\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{b d x \ln \left (F \right )}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (-b d \right )}{2}+\frac {\ln \left (\ln \left (F \right )\right )}{2}+\frac {9 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+12 b d x \ln \left (F \right )+6}{12 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\left (3 b d x \ln \left (F \right )+3\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{6 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{2}-\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{2}\right )-2 \ln \left (F \right )^{3} b^{3} d^{3} F^{b c +a} f e \left (\frac {1}{3 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {1}{2 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}+\frac {1}{2 b d x \ln \left (F \right )}+\frac {11}{36}-\frac {\ln \left (x \right )}{6}-\frac {\ln \left (-b d \right )}{6}-\frac {\ln \left (\ln \left (F \right )\right )}{6}-\frac {22 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}+36 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+36 b d x \ln \left (F \right )+24}{72 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {\left (4 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+4 b d x \ln \left (F \right )+8\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{24 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}+\frac {\ln \left (-b d x \ln \left (F \right )\right )}{6}+\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{6}\right )+F^{b c +a} e^{2} \ln \left (F \right )^{4} b^{4} d^{4} \left (-\frac {1}{4 b^{4} d^{4} x^{4} \ln \left (F \right )^{4}}-\frac {1}{3 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}}-\frac {1}{4 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}}-\frac {1}{6 b d x \ln \left (F \right )}-\frac {25}{288}+\frac {\ln \left (x \right )}{24}+\frac {\ln \left (-b d \right )}{24}+\frac {\ln \left (\ln \left (F \right )\right )}{24}+\frac {125 b^{4} d^{4} x^{4} \ln \left (F \right )^{4}+240 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}+360 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+480 b d x \ln \left (F \right )+360}{1440 b^{4} d^{4} x^{4} \ln \left (F \right )^{4}}-\frac {\left (5 b^{3} d^{3} x^{3} \ln \left (F \right )^{3}+5 b^{2} d^{2} x^{2} \ln \left (F \right )^{2}+10 b d x \ln \left (F \right )+30\right ) {\mathrm e}^{b d x \ln \left (F \right )}}{120 b^{4} d^{4} x^{4} \ln \left (F \right )^{4}}-\frac {\ln \left (-b d x \ln \left (F \right )\right )}{24}-\frac {\operatorname {expIntegral}_{1}\left (-b d x \ln \left (F \right )\right )}{24}\right )\)	\(617\)

Input:

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/24*(ln(F)^4*F^(b*c)*F^a*Ei(1,ln(F)*b*c+ln(F)*a-b*d*x*ln(F)-(b*c+a)*ln(F 
))*b^4*d^4*e^2*x^4+8*ln(F)^3*F^(b*c)*F^a*Ei(1,ln(F)*b*c+ln(F)*a-b*d*x*ln(F 
)-(b*c+a)*ln(F))*b^3*d^3*e*f*x^4+F^(b*d*x)*F^(b*c+a)*b^3*d^3*e^2*x^3*ln(F) 
^3+12*ln(F)^2*F^(b*c)*F^a*Ei(1,ln(F)*b*c+ln(F)*a-b*d*x*ln(F)-(b*c+a)*ln(F) 
)*b^2*d^2*f^2*x^4+8*F^(b*d*x)*F^(b*c+a)*b^2*d^2*e*f*x^3*ln(F)^2+F^(b*d*x)* 
F^(b*c+a)*b^2*d^2*e^2*x^2*ln(F)^2+12*F^(b*d*x)*F^(b*c+a)*b*d*f^2*x^3*ln(F) 
+8*F^(b*d*x)*F^(b*c+a)*b*d*e*f*x^2*ln(F)+2*F^(b*d*x)*F^(b*c+a)*b*d*e^2*x*l 
n(F)+12*F^(b*d*x)*F^(b*c+a)*f^2*x^2+16*F^(b*d*x)*F^(b*c+a)*e*f*x+6*F^(b*d* 
x)*F^(b*c+a)*e^2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.58 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=\frac {{\left (b^{4} d^{4} e^{2} x^{4} \log \left (F\right )^{4} + 8 \, b^{3} d^{3} e f x^{4} \log \left (F\right )^{3} + 12 \, b^{2} d^{2} f^{2} x^{4} \log \left (F\right )^{2}\right )} F^{b c + a} {\rm Ei}\left (b d x \log \left (F\right )\right ) - {\left (b^{3} d^{3} e^{2} x^{3} \log \left (F\right )^{3} + 12 \, f^{2} x^{2} + 16 \, e f x + {\left (8 \, b^{2} d^{2} e f x^{3} + b^{2} d^{2} e^{2} x^{2}\right )} \log \left (F\right )^{2} + 6 \, e^{2} + 2 \, {\left (6 \, b d f^{2} x^{3} + 4 \, b d e f x^{2} + b d e^{2} x\right )} \log \left (F\right )\right )} F^{b d x + b c + a}}{24 \, x^{4}} \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="fricas")
 

Output:

1/24*((b^4*d^4*e^2*x^4*log(F)^4 + 8*b^3*d^3*e*f*x^4*log(F)^3 + 12*b^2*d^2* 
f^2*x^4*log(F)^2)*F^(b*c + a)*Ei(b*d*x*log(F)) - (b^3*d^3*e^2*x^3*log(F)^3 
 + 12*f^2*x^2 + 16*e*f*x + (8*b^2*d^2*e*f*x^3 + b^2*d^2*e^2*x^2)*log(F)^2 
+ 6*e^2 + 2*(6*b*d*f^2*x^3 + 4*b*d*e*f*x^2 + b*d*e^2*x)*log(F))*F^(b*d*x + 
 b*c + a))/x^4
 

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=\int \frac {F^{a + b \left (c + d x\right )} \left (e + f x\right )^{2}}{x^{5}}\, dx \] Input:

integrate(F**(a+b*(d*x+c))*(f*x+e)**2/x**5,x)
 

Output:

Integral(F**(a + b*(c + d*x))*(e + f*x)**2/x**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.29 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=-F^{b c + a} b^{4} d^{4} e^{2} \Gamma \left (-4, -b d x \log \left (F\right )\right ) \log \left (F\right )^{4} + 2 \, F^{b c + a} b^{3} d^{3} e f \Gamma \left (-3, -b d x \log \left (F\right )\right ) \log \left (F\right )^{3} - F^{b c + a} b^{2} d^{2} f^{2} \Gamma \left (-2, -b d x \log \left (F\right )\right ) \log \left (F\right )^{2} \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="maxima")
 

Output:

-F^(b*c + a)*b^4*d^4*e^2*gamma(-4, -b*d*x*log(F))*log(F)^4 + 2*F^(b*c + a) 
*b^3*d^3*e*f*gamma(-3, -b*d*x*log(F))*log(F)^3 - F^(b*c + a)*b^2*d^2*f^2*g 
amma(-2, -b*d*x*log(F))*log(F)^2
 

Giac [F]

\[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=\int { \frac {{\left (f x + e\right )}^{2} F^{{\left (d x + c\right )} b + a}}{x^{5}} \,d x } \] Input:

integrate(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x, algorithm="giac")
 

Output:

integrate((f*x + e)^2*F^((d*x + c)*b + a)/x^5, x)
 

Mupad [B] (verification not implemented)

Time = 22.96 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=-F^{a+b\,c}\,b^2\,d^2\,f^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{2}+F^{b\,d\,x}\,\left (\frac {1}{2\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{2\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}\right )\right )-F^{a+b\,c}\,b^4\,d^4\,e^2\,{\ln \left (F\right )}^4\,\left (F^{b\,d\,x}\,\left (\frac {1}{24\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{24\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}+\frac {1}{12\,b^3\,d^3\,x^3\,{\ln \left (F\right )}^3}+\frac {1}{4\,b^4\,d^4\,x^4\,{\ln \left (F\right )}^4}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{24}\right )-2\,F^{a+b\,c}\,b^3\,d^3\,e\,f\,{\ln \left (F\right )}^3\,\left (F^{b\,d\,x}\,\left (\frac {1}{6\,b\,d\,x\,\ln \left (F\right )}+\frac {1}{6\,b^2\,d^2\,x^2\,{\ln \left (F\right )}^2}+\frac {1}{3\,b^3\,d^3\,x^3\,{\ln \left (F\right )}^3}\right )+\frac {\mathrm {expint}\left (-b\,d\,x\,\ln \left (F\right )\right )}{6}\right ) \] Input:

int((F^(a + b*(c + d*x))*(e + f*x)^2)/x^5,x)
 

Output:

- F^(a + b*c)*b^2*d^2*f^2*log(F)^2*(expint(-b*d*x*log(F))/2 + F^(b*d*x)*(1 
/(2*b*d*x*log(F)) + 1/(2*b^2*d^2*x^2*log(F)^2))) - F^(a + b*c)*b^4*d^4*e^2 
*log(F)^4*(F^(b*d*x)*(1/(24*b*d*x*log(F)) + 1/(24*b^2*d^2*x^2*log(F)^2) + 
1/(12*b^3*d^3*x^3*log(F)^3) + 1/(4*b^4*d^4*x^4*log(F)^4)) + expint(-b*d*x* 
log(F))/24) - 2*F^(a + b*c)*b^3*d^3*e*f*log(F)^3*(F^(b*d*x)*(1/(6*b*d*x*lo 
g(F)) + 1/(6*b^2*d^2*x^2*log(F)^2) + 1/(3*b^3*d^3*x^3*log(F)^3)) + expint( 
-b*d*x*log(F))/6)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.76 \[ \int \frac {F^{a+b (c+d x)} (e+f x)^2}{x^5} \, dx=\frac {f^{b c +a} \left (\mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{4} b^{4} d^{4} e^{2} x^{4}+8 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{3} b^{3} d^{3} e f \,x^{4}+12 \mathit {ei} \left (\mathrm {log}\left (f \right ) b d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} d^{2} f^{2} x^{4}-f^{b d x} \mathrm {log}\left (f \right )^{3} b^{3} d^{3} e^{2} x^{3}-f^{b d x} \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e^{2} x^{2}-8 f^{b d x} \mathrm {log}\left (f \right )^{2} b^{2} d^{2} e f \,x^{3}-2 f^{b d x} \mathrm {log}\left (f \right ) b d \,e^{2} x -8 f^{b d x} \mathrm {log}\left (f \right ) b d e f \,x^{2}-12 f^{b d x} \mathrm {log}\left (f \right ) b d \,f^{2} x^{3}-6 f^{b d x} e^{2}-16 f^{b d x} e f x -12 f^{b d x} f^{2} x^{2}\right )}{24 x^{4}} \] Input:

int(F^(a+b*(d*x+c))*(f*x+e)^2/x^5,x)
 

Output:

(f**(a + b*c)*(ei(log(f)*b*d*x)*log(f)**4*b**4*d**4*e**2*x**4 + 8*ei(log(f 
)*b*d*x)*log(f)**3*b**3*d**3*e*f*x**4 + 12*ei(log(f)*b*d*x)*log(f)**2*b**2 
*d**2*f**2*x**4 - f**(b*d*x)*log(f)**3*b**3*d**3*e**2*x**3 - f**(b*d*x)*lo 
g(f)**2*b**2*d**2*e**2*x**2 - 8*f**(b*d*x)*log(f)**2*b**2*d**2*e*f*x**3 - 
2*f**(b*d*x)*log(f)*b*d*e**2*x - 8*f**(b*d*x)*log(f)*b*d*e*f*x**2 - 12*f** 
(b*d*x)*log(f)*b*d*f**2*x**3 - 6*f**(b*d*x)*e**2 - 16*f**(b*d*x)*e*f*x - 1 
2*f**(b*d*x)*f**2*x**2))/(24*x**4)