\(\int \frac {F^{c (a+b x)}}{(d+e x+f x^2)^2} \, dx\) [26]

Optimal result

Integrand size = 22, antiderivative size = 304 \[ \int \frac {F^{c (a+b x)}}{\left (d+e x+f x^2\right )^2} \, dx=-\frac {F^{c \left (a-\frac {b e}{f}\right )+\frac {b c \left (e-\sqrt {e^2-4 d f}\right )}{2 f}+\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 f}} (e+2 f x)}{\left (e^2-4 d f\right ) (d+x (e+f x))}-\frac {F^{c \left (a-\frac {b e}{f}\right )+\frac {b c \left (e+\sqrt {e^2-4 d f}\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right ) \left (2 f-b c \sqrt {e^2-4 d f} \log (F)\right )}{\left (e^2-4 d f\right )^{3/2}}+\frac {F^{c \left (a-\frac {b e}{f}\right )+\frac {b c \left (e-\sqrt {e^2-4 d f}\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right ) \left (2 f+b c \sqrt {e^2-4 d f} \log (F)\right )}{\left (e^2-4 d f\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.82 \[ \int \frac {F^{c (a+b x)}}{\left (d+e x+f x^2\right )^2} \, dx=\frac {F^{a c-\frac {b c \left (e+\sqrt {e^2-4 d f}\right )}{2 f}} \left (-\sqrt {e^2-4 d f} F^{\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 f}} (e+2 f x)-F^{\frac {b c \sqrt {e^2-4 d f}}{f}} (d+x (e+f x)) \operatorname {ExpIntegralEi}\left (\frac {b c \left (e-\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right ) \left (2 f-b c \sqrt {e^2-4 d f} \log (F)\right )+(d+x (e+f x)) \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+\sqrt {e^2-4 d f}+2 f x\right ) \log (F)}{2 f}\right ) \left (2 f+b c \sqrt {e^2-4 d f} \log (F)\right )\right )}{\left (e^2-4 d f\right )^{3/2} (d+x (e+f x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {7292, 7293, 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 f F^{a c-\frac {b c \left (e-\sqrt {e^2-4 d f}\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x-\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{\left (e^2-4 d f\right )^{3/2}}+\frac {b c \log (F) F^{a c-\frac {b c \left (e-\sqrt {e^2-4 d f}\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x-\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{e^2-4 d f}+\frac {2 f F^{a c-\frac {b c \left (\sqrt {e^2-4 d f}+e\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x+\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{\left (e^2-4 d f\right )^{3/2}}+\frac {b c \log (F) F^{a c-\frac {b c \left (\sqrt {e^2-4 d f}+e\right )}{2 f}} \operatorname {ExpIntegralEi}\left (\frac {b c \left (e+2 f x+\sqrt {e^2-4 d f}\right ) \log (F)}{2 f}\right )}{e^2-4 d f}-\frac {2 f F^{a c+b c x}}{\left (e^2-4 d f\right ) \left (-\sqrt {e^2-4 d f}+e+2 f x\right )}-\frac {2 f F^{a c+b c x}}{\left (e^2-4 d f\right ) \left (\sqrt {e^2-4 d f}+e+2 f x\right )}\)

Input:

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

Output:

(-2*f*F^(a*c + b*c*x))/((e^2 - 4*d*f)*(e - Sqrt[e^2 - 4*d*f] + 2*f*x)) - ( 
2*f*F^(a*c + b*c*x))/((e^2 - 4*d*f)*(e + Sqrt[e^2 - 4*d*f] + 2*f*x)) - (2* 
f*F^(a*c - (b*c*(e - Sqrt[e^2 - 4*d*f]))/(2*f))*ExpIntegralEi[(b*c*(e - Sq 
rt[e^2 - 4*d*f] + 2*f*x)*Log[F])/(2*f)])/(e^2 - 4*d*f)^(3/2) + (2*f*F^(a*c 
 - (b*c*(e + Sqrt[e^2 - 4*d*f]))/(2*f))*ExpIntegralEi[(b*c*(e + Sqrt[e^2 - 
 4*d*f] + 2*f*x)*Log[F])/(2*f)])/(e^2 - 4*d*f)^(3/2) + (b*c*F^(a*c - (b*c* 
(e - Sqrt[e^2 - 4*d*f]))/(2*f))*ExpIntegralEi[(b*c*(e - Sqrt[e^2 - 4*d*f] 
+ 2*f*x)*Log[F])/(2*f)]*Log[F])/(e^2 - 4*d*f) + (b*c*F^(a*c - (b*c*(e + Sq 
rt[e^2 - 4*d*f]))/(2*f))*ExpIntegralEi[(b*c*(e + Sqrt[e^2 - 4*d*f] + 2*f*x 
)*Log[F])/(2*f)]*Log[F])/(e^2 - 4*d*f)
 

Defintions of rubi rules used

rule 2009

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

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(272)=544\).

Time = 0.12 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.05

method	result	size
risch	\(\frac {2 \ln \left (F \right )^{2} b^{2} c^{2} F^{b c x} F^{a c} x f}{\left (4 d f -e^{2}\right ) \left (f \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +\ln \left (F \right )^{2} b^{2} c^{2} d \right )}+\frac {\ln \left (F \right )^{2} b^{2} c^{2} F^{b c x} F^{a c} e}{\left (4 d f -e^{2}\right ) \left (f \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} e x +\ln \left (F \right )^{2} b^{2} c^{2} d \right )}+\frac {\ln \left (F \right ) b c \,F^{\frac {\left (2 f a -b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (\frac {2 f a \ln \left (F \right ) c -\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c -2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right )}{4 d f -e^{2}}+\frac {\ln \left (F \right ) b c \,F^{-\frac {\left (-2 f a +b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (-\frac {-2 f a \ln \left (F \right ) c +\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c +2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right )}{4 d f -e^{2}}-\frac {2 b \,F^{\frac {\left (2 f a -b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (\frac {2 f a \ln \left (F \right ) c -\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c -2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right ) f}{\left (4 d f -e^{2}\right ) \sqrt {-4 b^{2} d f +b^{2} e^{2}}}+\frac {2 b \,F^{-\frac {\left (-2 f a +b e +\sqrt {-b^{2} \left (4 d f -e^{2}\right )}\right ) c}{2 f}} \operatorname {expIntegral}_{1}\left (-\frac {-2 f a \ln \left (F \right ) c +\ln \left (F \right ) b c e +\sqrt {-4 b^{2} d f +b^{2} e^{2}}\, \ln \left (F \right ) c +2 f \left (b c x \ln \left (F \right )+a c \ln \left (F \right )\right )}{2 f}\right ) f}{\left (4 d f -e^{2}\right ) \sqrt {-4 b^{2} d f +b^{2} e^{2}}}\)	\(624\)

Input:

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

Output:

2*ln(F)^2*b^2*c^2*F^(b*c*x)*F^(a*c)/(4*d*f-e^2)/(f*x^2*ln(F)^2*b^2*c^2+ln( 
F)^2*b^2*c^2*e*x+ln(F)^2*b^2*c^2*d)*x*f+ln(F)^2*b^2*c^2*F^(b*c*x)*F^(a*c)/ 
(4*d*f-e^2)/(f*x^2*ln(F)^2*b^2*c^2+ln(F)^2*b^2*c^2*e*x+ln(F)^2*b^2*c^2*d)* 
e+ln(F)*b*c/(4*d*f-e^2)*F^(1/2/f*(2*f*a-b*e+(-b^2*(4*d*f-e^2))^(1/2))*c)*E 
i(1,1/2*(2*f*a*ln(F)*c-ln(F)*b*c*e+(-4*b^2*d*f+b^2*e^2)^(1/2)*ln(F)*c-2*f* 
(b*c*x*ln(F)+a*c*ln(F)))/f)+ln(F)*b*c/(4*d*f-e^2)*F^(-1/2*(-2*f*a+b*e+(-b^ 
2*(4*d*f-e^2))^(1/2))*c/f)*Ei(1,-1/2*(-2*f*a*ln(F)*c+ln(F)*b*c*e+(-4*b^2*d 
*f+b^2*e^2)^(1/2)*ln(F)*c+2*f*(b*c*x*ln(F)+a*c*ln(F)))/f)-2*b/(4*d*f-e^2)/ 
(-4*b^2*d*f+b^2*e^2)^(1/2)*F^(1/2/f*(2*f*a-b*e+(-b^2*(4*d*f-e^2))^(1/2))*c 
)*Ei(1,1/2*(2*f*a*ln(F)*c-ln(F)*b*c*e+(-4*b^2*d*f+b^2*e^2)^(1/2)*ln(F)*c-2 
*f*(b*c*x*ln(F)+a*c*ln(F)))/f)*f+2*b/(4*d*f-e^2)/(-4*b^2*d*f+b^2*e^2)^(1/2 
)*F^(-1/2*(-2*f*a+b*e+(-b^2*(4*d*f-e^2))^(1/2))*c/f)*Ei(1,-1/2*(-2*f*a*ln( 
F)*c+ln(F)*b*c*e+(-4*b^2*d*f+b^2*e^2)^(1/2)*ln(F)*c+2*f*(b*c*x*ln(F)+a*c*l 
n(F)))/f)*f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (270) = 540\).

Time = 0.09 (sec) , antiderivative size = 632, normalized size of antiderivative = 2.08 \[ \int \frac {F^{c (a+b x)}}{\left (d+e x+f x^2\right )^2} \, dx=\frac {{\left ({\left (b^{2} c^{2} d e^{2} - 4 \, b^{2} c^{2} d^{2} f + {\left (b^{2} c^{2} e^{2} f - 4 \, b^{2} c^{2} d f^{2}\right )} x^{2} + {\left (b^{2} c^{2} e^{3} - 4 \, b^{2} c^{2} d e f\right )} x\right )} \log \left (F\right )^{2} + 2 \, {\left (f^{3} x^{2} + e f^{2} x + d f^{2}\right )} \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}\right )} {\rm Ei}\left (\frac {{\left (2 \, b c f x + b c e\right )} \log \left (F\right ) + f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right ) e^{\left (-\frac {{\left (b c e - 2 \, a c f\right )} \log \left (F\right ) + f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right )} + {\left ({\left (b^{2} c^{2} d e^{2} - 4 \, b^{2} c^{2} d^{2} f + {\left (b^{2} c^{2} e^{2} f - 4 \, b^{2} c^{2} d f^{2}\right )} x^{2} + {\left (b^{2} c^{2} e^{3} - 4 \, b^{2} c^{2} d e f\right )} x\right )} \log \left (F\right )^{2} - 2 \, {\left (f^{3} x^{2} + e f^{2} x + d f^{2}\right )} \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}\right )} {\rm Ei}\left (\frac {{\left (2 \, b c f x + b c e\right )} \log \left (F\right ) - f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right ) e^{\left (-\frac {{\left (b c e - 2 \, a c f\right )} \log \left (F\right ) - f \sqrt {\frac {{\left (b^{2} c^{2} e^{2} - 4 \, b^{2} c^{2} d f\right )} \log \left (F\right )^{2}}{f^{2}}}}{2 \, f}\right )} - {\left (b c e^{3} - 4 \, b c d e f + 2 \, {\left (b c e^{2} f - 4 \, b c d f^{2}\right )} x\right )} F^{b c x + a c} \log \left (F\right )}{{\left (b c d e^{4} - 8 \, b c d^{2} e^{2} f + 16 \, b c d^{3} f^{2} + {\left (b c e^{4} f - 8 \, b c d e^{2} f^{2} + 16 \, b c d^{2} f^{3}\right )} x^{2} + {\left (b c e^{5} - 8 \, b c d e^{3} f + 16 \, b c d^{2} e f^{2}\right )} x\right )} \log \left (F\right )} \] Input:

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

Output:

(((b^2*c^2*d*e^2 - 4*b^2*c^2*d^2*f + (b^2*c^2*e^2*f - 4*b^2*c^2*d*f^2)*x^2 
 + (b^2*c^2*e^3 - 4*b^2*c^2*d*e*f)*x)*log(F)^2 + 2*(f^3*x^2 + e*f^2*x + d* 
f^2)*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2))*Ei(1/2*((2*b*c*f*x 
+ b*c*e)*log(F) + f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2))/f)*e 
^(-1/2*((b*c*e - 2*a*c*f)*log(F) + f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*lo 
g(F)^2/f^2))/f) + ((b^2*c^2*d*e^2 - 4*b^2*c^2*d^2*f + (b^2*c^2*e^2*f - 4*b 
^2*c^2*d*f^2)*x^2 + (b^2*c^2*e^3 - 4*b^2*c^2*d*e*f)*x)*log(F)^2 - 2*(f^3*x 
^2 + e*f^2*x + d*f^2)*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*log(F)^2/f^2))*Ei 
(1/2*((2*b*c*f*x + b*c*e)*log(F) - f*sqrt((b^2*c^2*e^2 - 4*b^2*c^2*d*f)*lo 
g(F)^2/f^2))/f)*e^(-1/2*((b*c*e - 2*a*c*f)*log(F) - f*sqrt((b^2*c^2*e^2 - 
4*b^2*c^2*d*f)*log(F)^2/f^2))/f) - (b*c*e^3 - 4*b*c*d*e*f + 2*(b*c*e^2*f - 
 4*b*c*d*f^2)*x)*F^(b*c*x + a*c)*log(F))/((b*c*d*e^4 - 8*b*c*d^2*e^2*f + 1 
6*b*c*d^3*f^2 + (b*c*e^4*f - 8*b*c*d*e^2*f^2 + 16*b*c*d^2*f^3)*x^2 + (b*c* 
e^5 - 8*b*c*d*e^3*f + 16*b*c*d^2*e*f^2)*x)*log(F))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(f**(a*c)*( - f**(b*c*x) + int(f**(b*c*x)/(log(f)**2*b**2*c**2*d**3*e + 2* 
log(f)**2*b**2*c**2*d**2*e**2*x + 2*log(f)**2*b**2*c**2*d**2*e*f*x**2 + lo 
g(f)**2*b**2*c**2*d*e**3*x**2 + 2*log(f)**2*b**2*c**2*d*e**2*f*x**3 + log( 
f)**2*b**2*c**2*d*e*f**2*x**4 - 2*log(f)*b*c*d**3*f - log(f)*b*c*d**2*e**2 
 - 4*log(f)*b*c*d**2*e*f*x - 4*log(f)*b*c*d**2*f**2*x**2 - 2*log(f)*b*c*d* 
e**3*x - 4*log(f)*b*c*d*e**2*f*x**2 - 4*log(f)*b*c*d*e*f**2*x**3 - 2*log(f 
)*b*c*d*f**3*x**4 - log(f)*b*c*e**4*x**2 - 2*log(f)*b*c*e**3*f*x**3 - log( 
f)*b*c*e**2*f**2*x**4 + 2*d**2*e*f + 4*d*e**2*f*x + 4*d*e*f**2*x**2 + 2*e* 
*3*f*x**2 + 4*e**2*f**2*x**3 + 2*e*f**3*x**4),x)*log(f)**3*b**3*c**3*d**3* 
e + int(f**(b*c*x)/(log(f)**2*b**2*c**2*d**3*e + 2*log(f)**2*b**2*c**2*d** 
2*e**2*x + 2*log(f)**2*b**2*c**2*d**2*e*f*x**2 + log(f)**2*b**2*c**2*d*e** 
3*x**2 + 2*log(f)**2*b**2*c**2*d*e**2*f*x**3 + log(f)**2*b**2*c**2*d*e*f** 
2*x**4 - 2*log(f)*b*c*d**3*f - log(f)*b*c*d**2*e**2 - 4*log(f)*b*c*d**2*e* 
f*x - 4*log(f)*b*c*d**2*f**2*x**2 - 2*log(f)*b*c*d*e**3*x - 4*log(f)*b*c*d 
*e**2*f*x**2 - 4*log(f)*b*c*d*e*f**2*x**3 - 2*log(f)*b*c*d*f**3*x**4 - log 
(f)*b*c*e**4*x**2 - 2*log(f)*b*c*e**3*f*x**3 - log(f)*b*c*e**2*f**2*x**4 + 
 2*d**2*e*f + 4*d*e**2*f*x + 4*d*e*f**2*x**2 + 2*e**3*f*x**2 + 4*e**2*f**2 
*x**3 + 2*e*f**3*x**4),x)*log(f)**3*b**3*c**3*d**2*e**2*x + int(f**(b*c*x) 
/(log(f)**2*b**2*c**2*d**3*e + 2*log(f)**2*b**2*c**2*d**2*e**2*x + 2*log(f 
)**2*b**2*c**2*d**2*e*f*x**2 + log(f)**2*b**2*c**2*d*e**3*x**2 + 2*log(...