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

Optimal. Leaf size=140 \[ \frac {6 \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}-\frac {3 x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac {x^3}{a b d \log (F)} \]

[Out]

x^3/a/b/d/ln(F)-x^3/b/d/(a+b*F^(d*x+c))/ln(F)-3*x^2*ln(1+b*F^(d*x+c)/a)/a/b/d^2/ln(F)^2-6*x*polylog(2,-b*F^(d*
x+c)/a)/a/b/d^3/ln(F)^3+6*polylog(3,-b*F^(d*x+c)/a)/a/b/d^4/ln(F)^4

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Rubi [A]  time = 0.25, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2191, 2184, 2190, 2531, 2282, 6589} \[ -\frac {6 x \text {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \text {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)}-\frac {3 x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac {x^3}{a b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^3/(a*b*d*Log[F]) - x^3/(b*d*(a + b*F^(c + d*x))*Log[F]) - (3*x^2*Log[1 + (b*F^(c + d*x))/a])/(a*b*d^2*Log[F]
^2) - (6*x*PolyLog[2, -((b*F^(c + d*x))/a)])/(a*b*d^3*Log[F]^3) + (6*PolyLog[3, -((b*F^(c + d*x))/a)])/(a*b*d^
4*Log[F]^4)

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 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 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx &=-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac {3 \int \frac {x^2}{a+b F^{c+d x}} \, dx}{b d \log (F)}\\ &=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 \int \frac {F^{c+d x} x^2}{a+b F^{c+d x}} \, dx}{a d \log (F)}\\ &=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}+\frac {6 \int x \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{a b d^2 \log ^2(F)}\\ &=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \int \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{a b d^3 \log ^3(F)}\\ &=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{a b d^4 \log ^4(F)}\\ &=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 137, normalized size = 0.98 \[ \frac {3 \left (\frac {2 \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{a d^3 \log ^3(F)}-\frac {2 x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a d^2 \log ^2(F)}-\frac {x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a d \log (F)}+\frac {x^3}{3 a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.42, size = 246, normalized size = 1.76 \[ \frac {a c^{3} \log \relax (F)^{3} + {\left (b d^{3} x^{3} + b c^{3}\right )} F^{d x + c} \log \relax (F)^{3} - 6 \, {\left (F^{d x + c} b d x \log \relax (F) + a d x \log \relax (F)\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 3 \, {\left (F^{d x + c} b c^{2} \log \relax (F)^{2} + a c^{2} \log \relax (F)^{2}\right )} \log \left (F^{d x + c} b + a\right ) - 3 \, {\left ({\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \relax (F)^{2} + {\left (a d^{2} x^{2} - a c^{2}\right )} \log \relax (F)^{2}\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right ) + 6 \, {\left (F^{d x + c} b + a\right )} {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right )}{F^{d x + c} a b^{2} d^{4} \log \relax (F)^{4} + a^{2} b d^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*c^3*log(F)^3 + (b*d^3*x^3 + b*c^3)*F^(d*x + c)*log(F)^3 - 6*(F^(d*x + c)*b*d*x*log(F) + a*d*x*log(F))*dilog
(-(F^(d*x + c)*b + a)/a + 1) - 3*(F^(d*x + c)*b*c^2*log(F)^2 + a*c^2*log(F)^2)*log(F^(d*x + c)*b + a) - 3*((b*
d^2*x^2 - b*c^2)*F^(d*x + c)*log(F)^2 + (a*d^2*x^2 - a*c^2)*log(F)^2)*log((F^(d*x + c)*b + a)/a) + 6*(F^(d*x +
c)*b + a)*polylog(3, -F^(d*x + c)*b/a))/(F^(d*x + c)*a*b^2*d^4*log(F)^4 + a^2*b*d^4*log(F)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 274, normalized size = 1.96 \[ -\frac {x^{3}}{\left (b \,F^{d x +c}+a \right ) b d \ln \relax (F )}+\frac {x^{3}}{a b d \ln \relax (F )}-\frac {3 c^{2} x}{a b \,d^{3} \ln \relax (F )}-\frac {3 x^{2} \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{a b \,d^{2} \ln \relax (F )^{2}}-\frac {2 c^{3}}{a b \,d^{4} \ln \relax (F )}+\frac {3 c^{2} \ln \left (F^{c} F^{d x}\right )}{a b \,d^{4} \ln \relax (F )^{2}}+\frac {3 c^{2} \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{a b \,d^{4} \ln \relax (F )^{2}}-\frac {3 c^{2} \ln \left (b \,F^{c} F^{d x}+a \right )}{a b \,d^{4} \ln \relax (F )^{2}}-\frac {6 x \polylog \left (2, -\frac {b \,F^{c} F^{d x}}{a}\right )}{a b \,d^{3} \ln \relax (F )^{3}}+\frac {6 \polylog \left (3, -\frac {b \,F^{c} F^{d x}}{a}\right )}{a b \,d^{4} \ln \relax (F )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3/b/d/(b*F^(d*x+c)+a)/ln(F)+x^3/a/b/d/ln(F)-3/d^3/ln(F)/b/a*c^2*x-2/d^4/ln(F)/b/a*c^3-3/d^2/ln(F)^2/b/a*ln(
1/a*b*F^c*F^(d*x)+1)*x^2+3/d^4/ln(F)^2/b/a*ln(1/a*b*F^c*F^(d*x)+1)*c^2-6/d^3/ln(F)^3/b/a*polylog(2,-1/a*b*F^c*
F^(d*x))*x+6/d^4/ln(F)^4/b/a*polylog(3,-1/a*b*F^c*F^(d*x))+3/d^4/ln(F)^2/b*c^2/a*ln(F^c*F^(d*x))-3/d^4/ln(F)^2
/b*c^2/a*ln(b*F^c*F^(d*x)+a)

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maxima [A]  time = 0.52, size = 134, normalized size = 0.96 \[ -\frac {x^{3}}{F^{d x} F^{c} b^{2} d \log \relax (F) + a b d \log \relax (F)} + \frac {\log \left (F^{d x}\right )^{3}}{a b d^{4} \log \relax (F)^{4}} - \frac {3 \, {\left (\log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F^{d x}\right )^{2} + 2 \, {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F^{d x}\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a})\right )}}{a b d^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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