\(\int \frac {F^{c+d x} x^2}{(a+b F^{c+d x})^2} \, dx\) [83]

Optimal result

Integrand size = 24, antiderivative size = 107 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {x^2}{a b d \log (F)}-\frac {x^2}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {2 x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)} \] Output:

x^2/a/b/d/ln(F)-x^2/b/d/(a+b*F^(d*x+c))/ln(F)-2*x*ln(1+b*F^(d*x+c)/a)/a/b/ 
d^2/ln(F)^2-2*polylog(2,-b*F^(d*x+c)/a)/a/b/d^3/ln(F)^3
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {d x \log (F) \left (b d F^{c+d x} x \log (F)-2 \left (a+b F^{c+d x}\right ) \log \left (1+\frac {b F^{c+d x}}{a}\right )\right )-2 \left (a+b F^{c+d x}\right ) \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \left (a+b F^{c+d x}\right ) \log ^3(F)} \] Input:

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

Output:

(d*x*Log[F]*(b*d*F^(c + d*x)*x*Log[F] - 2*(a + b*F^(c + d*x))*Log[1 + (b*F 
^(c + d*x))/a]) - 2*(a + b*F^(c + d*x))*PolyLog[2, -((b*F^(c + d*x))/a)])/ 
(a*b*d^3*(a + b*F^(c + d*x))*Log[F]^3)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2621, 2615, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 2621

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

\(\Big \downarrow \) 2615

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

\(\Big \downarrow \) 2620

\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {\int \log \left (\frac {b F^{c+d x}}{a}+1\right )dx}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {\int F^{-c-d x} \log \left (\frac {b F^{c+d x}}{a}+1\right )dF^{c+d x}}{b d^2 \log ^2(F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {\operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\)

Input:

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

Output:

-(x^2/(b*d*(a + b*F^(c + d*x))*Log[F])) + (2*(x^2/(2*a) - (b*((x*Log[1 + ( 
b*F^(c + d*x))/a])/(b*d*Log[F]) + PolyLog[2, -((b*F^(c + d*x))/a)]/(b*d^2* 
Log[F]^2)))/a))/(b*d*Log[F])
 

Defintions of rubi rules used

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2621

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*( 
(e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> 
 Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1)*Log 
[F])), x] - Simp[d*(m/(b*f*g*n*(p + 1)*Log[F]))   Int[(c + d*x)^(m - 1)*(a 
+ b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, 
m, n, p}, x] && NeQ[p, -1]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(230\) vs. \(2(107)=214\).

Time = 0.06 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.16

method	result	size
risch	\(-\frac {x^{2}}{b d \left (a +b \,F^{d x +c}\right ) \ln \left (F \right )}+\frac {x^{2}}{a b d \ln \left (F \right )}+\frac {2 c x}{b \,d^{2} \ln \left (F \right ) a}+\frac {c^{2}}{b \,d^{3} \ln \left (F \right ) a}-\frac {2 \ln \left (1+\frac {b \,F^{c} F^{d x}}{a}\right ) x}{b \,d^{2} \ln \left (F \right )^{2} a}-\frac {2 \ln \left (1+\frac {b \,F^{c} F^{d x}}{a}\right ) c}{b \,d^{3} \ln \left (F \right )^{2} a}-\frac {2 \operatorname {polylog}\left (2, -\frac {b \,F^{c} F^{d x}}{a}\right )}{b \,d^{3} \ln \left (F \right )^{3} a}-\frac {2 c \ln \left (F^{c} F^{d x}\right )}{b \,d^{3} \ln \left (F \right )^{2} a}+\frac {2 c \ln \left (F^{d x} F^{c} b +a \right )}{b \,d^{3} \ln \left (F \right )^{2} a}\)	\(231\)

Input:

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

Output:

-x^2/b/d/(a+b*F^(d*x+c))/ln(F)+x^2/a/b/d/ln(F)+2/b/d^2/ln(F)/a*c*x+1/b/d^3 
/ln(F)/a*c^2-2/b/d^2/ln(F)^2/a*ln(1+b*F^c*F^(d*x)/a)*x-2/b/d^3/ln(F)^2/a*l 
n(1+b*F^c*F^(d*x)/a)*c-2/b/d^3/ln(F)^3/a*polylog(2,-b*F^c*F^(d*x)/a)-2/b/d 
^3/ln(F)^2*c/a*ln(F^c*F^(d*x))+2/b/d^3/ln(F)^2*c/a*ln(F^(d*x)*F^c*b+a)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.74 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {a c^{2} \log \left (F\right )^{2} - {\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \left (F\right )^{2} + 2 \, {\left (F^{d x + c} b + a\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 2 \, {\left (F^{d x + c} b c \log \left (F\right ) + a c \log \left (F\right )\right )} \log \left (F^{d x + c} b + a\right ) + 2 \, {\left ({\left (b d x + b c\right )} F^{d x + c} \log \left (F\right ) + {\left (a d x + a c\right )} \log \left (F\right )\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right )}{F^{d x + c} a b^{2} d^{3} \log \left (F\right )^{3} + a^{2} b d^{3} \log \left (F\right )^{3}} \] Input:

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

Output:

-(a*c^2*log(F)^2 - (b*d^2*x^2 - b*c^2)*F^(d*x + c)*log(F)^2 + 2*(F^(d*x + 
c)*b + a)*dilog(-(F^(d*x + c)*b + a)/a + 1) - 2*(F^(d*x + c)*b*c*log(F) + 
a*c*log(F))*log(F^(d*x + c)*b + a) + 2*((b*d*x + b*c)*F^(d*x + c)*log(F) + 
 (a*d*x + a*c)*log(F))*log((F^(d*x + c)*b + a)/a))/(F^(d*x + c)*a*b^2*d^3* 
log(F)^3 + a^2*b*d^3*log(F)^3)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

-x**2/(F**(c + d*x)*b**2*d*log(F) + a*b*d*log(F)) + 2*Integral(x/(a + b*ex 
p(c*log(F))*exp(d*x*log(F))), x)/(b*d*log(F))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^{2}}{F^{d x} F^{c} b^{2} d \log \left (F\right ) + a b d \log \left (F\right )} + \frac {x^{2}}{a b d \log \left (F\right )} - \frac {2 \, {\left (d x \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right )\right )}}{a b d^{3} \log \left (F\right )^{3}} \] Input:

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

Output:

-x^2/(F^(d*x)*F^c*b^2*d*log(F) + a*b*d*log(F)) + x^2/(a*b*d*log(F)) - 2*(d 
*x*log(F^(d*x)*F^c*b/a + 1)*log(F) + dilog(-F^(d*x)*F^c*b/a))/(a*b*d^3*log 
(F)^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {2 f^{d x +c} \left (\int \frac {x}{f^{2 d x +2 c} b^{2}+2 f^{d x +c} a b +a^{2}}d x \right ) \mathrm {log}\left (f \right )^{2} a^{2} b \,d^{2}-2 f^{d x +c} \mathrm {log}\left (f^{d x +c} b +a \right ) b +2 f^{d x +c} \mathrm {log}\left (f \right ) b d x +2 \left (\int \frac {x}{f^{2 d x +2 c} b^{2}+2 f^{d x +c} a b +a^{2}}d x \right ) \mathrm {log}\left (f \right )^{2} a^{3} d^{2}-2 \,\mathrm {log}\left (f^{d x +c} b +a \right ) a -\mathrm {log}\left (f \right )^{2} a \,d^{2} x^{2}}{\mathrm {log}\left (f \right )^{3} a b \,d^{3} \left (f^{d x +c} b +a \right )} \] Input:

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

Output:

(2*f**(c + d*x)*int(x/(f**(2*c + 2*d*x)*b**2 + 2*f**(c + d*x)*a*b + a**2), 
x)*log(f)**2*a**2*b*d**2 - 2*f**(c + d*x)*log(f**(c + d*x)*b + a)*b + 2*f* 
*(c + d*x)*log(f)*b*d*x + 2*int(x/(f**(2*c + 2*d*x)*b**2 + 2*f**(c + d*x)* 
a*b + a**2),x)*log(f)**2*a**3*d**2 - 2*log(f**(c + d*x)*b + a)*a - log(f)* 
*2*a*d**2*x**2)/(log(f)**3*a*b*d**3*(f**(c + d*x)*b + a))