3.89 \(\int \frac {F^{c+d x} x^2}{(a+b F^{c+d x})^3} \, dx\)

Optimal. Leaf size=182 \[ -\frac {\text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a^2 b d^3 \log ^3(F)}+\frac {\log \left (a+b F^{c+d x}\right )}{a^2 b d^3 \log ^3(F)}-\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a^2 b d^2 \log ^2(F)}-\frac {x}{a^2 b d^2 \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}+\frac {x}{a b d^2 \log ^2(F) \left (a+b F^{c+d x}\right )}-\frac {x^2}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2191, 2185, 2184, 2190, 2279, 2391, 2282, 36, 29, 31} \[ -\frac {\text {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a^2 b d^3 \log ^3(F)}-\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a^2 b d^2 \log ^2(F)}+\frac {\log \left (a+b F^{c+d x}\right )}{a^2 b d^3 \log ^3(F)}-\frac {x}{a^2 b d^2 \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}+\frac {x}{a b d^2 \log ^2(F) \left (a+b F^{c+d x}\right )}-\frac {x^2}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(F^(c + d*x)*x^2)/(a + b*F^(c + d*x))^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

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] - Dist[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 2185

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2191

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] - Dist[(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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

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]

Rubi steps

\begin {align*} \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^3} \, dx &=-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}+\frac {\int \frac {x}{\left (a+b F^{c+d x}\right )^2} \, dx}{b d \log (F)}\\ &=-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {\int \frac {F^{c+d x} x}{\left (a+b F^{c+d x}\right )^2} \, dx}{a d \log (F)}+\frac {\int \frac {x}{a+b F^{c+d x}} \, dx}{a b d \log (F)}\\ &=\frac {x}{a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {\int \frac {1}{a+b F^{c+d x}} \, dx}{a b d^2 \log ^2(F)}-\frac {\int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx}{a^2 d \log (F)}\\ &=\frac {x}{a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a^2 b d^2 \log ^2(F)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x (a+b x)} \, dx,x,F^{c+d x}\right )}{a b d^3 \log ^3(F)}+\frac {\int \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{a^2 b d^2 \log ^2(F)}\\ &=\frac {x}{a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a^2 b d^2 \log ^2(F)}+\frac {\operatorname {Subst}\left (\int \frac {1}{a+b x} \, dx,x,F^{c+d x}\right )}{a^2 d^3 \log ^3(F)}-\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,F^{c+d x}\right )}{a^2 b d^3 \log ^3(F)}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{a^2 b d^3 \log ^3(F)}\\ &=-\frac {x}{a^2 b d^2 \log ^2(F)}+\frac {x}{a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x^2}{2 a^2 b d \log (F)}-\frac {x^2}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}+\frac {\log \left (a+b F^{c+d x}\right )}{a^2 b d^3 \log ^3(F)}-\frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a^2 b d^2 \log ^2(F)}-\frac {\text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a^2 b d^3 \log ^3(F)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 177, normalized size = 0.97 \[ \frac {b d^2 x^2 \log ^2(F) F^{c+d x} \left (2 a+b F^{c+d x}\right )-2 \left (a+b F^{c+d x}\right )^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )+2 \left (a+b F^{c+d x}\right )^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )-2 d x \log (F) \left (a+b F^{c+d x}\right ) \left (\left (a+b F^{c+d x}\right ) \log \left (\frac {b F^{c+d x}}{a}+1\right )+b F^{c+d x}\right )}{2 a^2 b d^3 \log ^3(F) \left (a+b F^{c+d x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(F^(c + d*x)*x^2)/(a + b*F^(c + d*x))^3,x]

[Out]

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

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fricas [B]  time = 0.42, size = 379, normalized size = 2.08 \[ -\frac {a^{2} c^{2} \log \relax (F)^{2} + 2 \, a^{2} c \log \relax (F) - {\left ({\left (b^{2} d^{2} x^{2} - b^{2} c^{2}\right )} \log \relax (F)^{2} - 2 \, {\left (b^{2} d x + b^{2} c\right )} \log \relax (F)\right )} F^{2 \, d x + 2 \, c} - 2 \, {\left ({\left (a b d^{2} x^{2} - a b c^{2}\right )} \log \relax (F)^{2} - {\left (a b d x + 2 \, a b c\right )} \log \relax (F)\right )} F^{d x + c} + 2 \, {\left (2 \, F^{d x + c} a b + F^{2 \, d x + 2 \, c} b^{2} + a^{2}\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 2 \, {\left (a^{2} c \log \relax (F) + {\left (b^{2} c \log \relax (F) + b^{2}\right )} F^{2 \, d x + 2 \, c} + 2 \, {\left (a b c \log \relax (F) + a b\right )} F^{d x + c} + a^{2}\right )} \log \left (F^{d x + c} b + a\right ) + 2 \, {\left ({\left (b^{2} d x + b^{2} c\right )} F^{2 \, d x + 2 \, c} \log \relax (F) + 2 \, {\left (a b d x + a b c\right )} F^{d x + c} \log \relax (F) + {\left (a^{2} d x + a^{2} c\right )} \log \relax (F)\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right )}{2 \, {\left (2 \, F^{d x + c} a^{3} b^{2} d^{3} \log \relax (F)^{3} + F^{2 \, d x + 2 \, c} a^{2} b^{3} d^{3} \log \relax (F)^{3} + a^{4} b d^{3} \log \relax (F)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x^2/(a+b*F^(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(a^2*c^2*log(F)^2 + 2*a^2*c*log(F) - ((b^2*d^2*x^2 - b^2*c^2)*log(F)^2 - 2*(b^2*d*x + b^2*c)*log(F))*F^(2
*d*x + 2*c) - 2*((a*b*d^2*x^2 - a*b*c^2)*log(F)^2 - (a*b*d*x + 2*a*b*c)*log(F))*F^(d*x + c) + 2*(2*F^(d*x + c)
*a*b + F^(2*d*x + 2*c)*b^2 + a^2)*dilog(-(F^(d*x + c)*b + a)/a + 1) - 2*(a^2*c*log(F) + (b^2*c*log(F) + b^2)*F
^(2*d*x + 2*c) + 2*(a*b*c*log(F) + a*b)*F^(d*x + c) + a^2)*log(F^(d*x + c)*b + a) + 2*((b^2*d*x + b^2*c)*F^(2*
d*x + 2*c)*log(F) + 2*(a*b*d*x + a*b*c)*F^(d*x + c)*log(F) + (a^2*d*x + a^2*c)*log(F))*log((F^(d*x + c)*b + a)
/a))/(2*F^(d*x + c)*a^3*b^2*d^3*log(F)^3 + F^(2*d*x + 2*c)*a^2*b^3*d^3*log(F)^3 + a^4*b*d^3*log(F)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{d x + c} x^{2}}{{\left (F^{d x + c} b + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x^2/(a+b*F^(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(F^(d*x + c)*x^2/(F^(d*x + c)*b + a)^3, x)

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maple [A]  time = 0.08, size = 304, normalized size = 1.67 \[ \frac {x^{2}}{2 a^{2} b d \ln \relax (F )}+\frac {c x}{a^{2} b \,d^{2} \ln \relax (F )}-\frac {\left (a d x \ln \relax (F )-2 b \,F^{d x +c}-2 a \right ) x}{2 \left (b \,F^{d x +c}+a \right )^{2} a b \,d^{2} \ln \relax (F )^{2}}+\frac {c^{2}}{2 a^{2} b \,d^{3} \ln \relax (F )}-\frac {x \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{a^{2} b \,d^{2} \ln \relax (F )^{2}}-\frac {c \ln \left (F^{c} F^{d x}\right )}{a^{2} b \,d^{3} \ln \relax (F )^{2}}-\frac {c \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{a^{2} b \,d^{3} \ln \relax (F )^{2}}+\frac {c \ln \left (b \,F^{c} F^{d x}+a \right )}{a^{2} b \,d^{3} \ln \relax (F )^{2}}-\frac {\polylog \left (2, -\frac {b \,F^{c} F^{d x}}{a}\right )}{a^{2} b \,d^{3} \ln \relax (F )^{3}}-\frac {\ln \left (F^{c} F^{d x}\right )}{a^{2} b \,d^{3} \ln \relax (F )^{3}}+\frac {\ln \left (b \,F^{c} F^{d x}+a \right )}{a^{2} b \,d^{3} \ln \relax (F )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x+c)*x^2/(b*F^(d*x+c)+a)^3,x)

[Out]

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

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maxima [A]  time = 0.51, size = 214, normalized size = 1.18 \[ -\frac {a d x^{2} \log \relax (F) - 2 \, F^{d x} F^{c} b x - 2 \, a x}{2 \, {\left (2 \, F^{d x} F^{c} a^{2} b^{2} d^{2} \log \relax (F)^{2} + F^{2 \, d x} F^{2 \, c} a b^{3} d^{2} \log \relax (F)^{2} + a^{3} b d^{2} \log \relax (F)^{2}\right )}} + \frac {\log \left (F^{d x}\right )^{2}}{2 \, a^{2} b d^{3} \log \relax (F)^{3}} - \frac {\log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F^{d x}\right ) + {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right )}{a^{2} b d^{3} \log \relax (F)^{3}} + \frac {\log \left (F^{d x} F^{c} b + a\right )}{a^{2} b d^{3} \log \relax (F)^{3}} - \frac {\log \left (F^{d x}\right )}{a^{2} b d^{3} \log \relax (F)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(d*x+c)*x^2/(a+b*F^(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c+d\,x}\,x^2}{{\left (a+F^{c+d\,x}\,b\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((F^(c + d*x)*x^2)/(a + F^(c + d*x)*b)^3,x)

[Out]

int((F^(c + d*x)*x^2)/(a + F^(c + d*x)*b)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 F^{c + d x} b x - a d x^{2} \log {\relax (F )} + 2 a x}{4 F^{c + d x} a^{2} b^{2} d^{2} \log {\relax (F )}^{2} + 2 F^{2 c + 2 d x} a b^{3} d^{2} \log {\relax (F )}^{2} + 2 a^{3} b d^{2} \log {\relax (F )}^{2}} + \frac {\int \frac {d x \log {\relax (F )}}{a + b e^{c \log {\relax (F )}} e^{d x \log {\relax (F )}}}\, dx + \int \left (- \frac {1}{a + b e^{c \log {\relax (F )}} e^{d x \log {\relax (F )}}}\right )\, dx}{a b d^{2} \log {\relax (F )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(d*x+c)*x**2/(a+b*F**(d*x+c))**3,x)

[Out]

(2*F**(c + d*x)*b*x - a*d*x**2*log(F) + 2*a*x)/(4*F**(c + d*x)*a**2*b**2*d**2*log(F)**2 + 2*F**(2*c + 2*d*x)*a
*b**3*d**2*log(F)**2 + 2*a**3*b*d**2*log(F)**2) + (Integral(d*x*log(F)/(a + b*exp(c*log(F))*exp(d*x*log(F))),
x) + Integral(-1/(a + b*exp(c*log(F))*exp(d*x*log(F))), x))/(a*b*d**2*log(F)**2)

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