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

Optimal. Leaf size=115 \[ \frac {6 \text {Li}_4\left (-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2190, 2531, 6609, 2282, 6589} \[ \frac {3 x^2 \text {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \text {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)}+\frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx &=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}-\frac {3 \int x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right ) \, dx}{b d \log (F)}\\ &=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 \int x \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{b d^2 \log ^2(F)}\\ &=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \int \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right ) \, dx}{b d^3 \log ^3(F)}\\ &=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a}\right )}{x} \, dx,x,F^{c+d x}\right )}{b d^4 \log ^4(F)}\\ &=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \text {Li}_4\left (-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 115, normalized size = 1.00 \[ \frac {6 \text {Li}_4\left (-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)}-\frac {6 x \text {Li}_3\left (-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {3 x^2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.44, size = 134, normalized size = 1.17 \[ \frac {3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) \log \relax (F)^{2} - c^{3} \log \left (F^{d x + c} b + a\right ) \log \relax (F)^{3} + {\left (d^{3} x^{3} + c^{3}\right )} \log \relax (F)^{3} \log \left (\frac {F^{d x + c} b + a}{a}\right ) - 6 \, d x \log \relax (F) {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right ) + 6 \, {\rm polylog}\left (4, -\frac {F^{d x + c} b}{a}\right )}{b d^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 225, normalized size = 1.96 \[ \frac {x^{3} \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{b d \ln \relax (F )}-\frac {c^{3} x}{b \,d^{3}}-\frac {3 c^{4}}{4 b \,d^{4}}+\frac {c^{3} \ln \left (F^{c} F^{d x}\right )}{b \,d^{4} \ln \relax (F )}+\frac {c^{3} \ln \left (\frac {b \,F^{c} F^{d x}}{a}+1\right )}{b \,d^{4} \ln \relax (F )}-\frac {c^{3} \ln \left (b \,F^{c} F^{d x}+a \right )}{b \,d^{4} \ln \relax (F )}+\frac {3 x^{2} \polylog \left (2, -\frac {b \,F^{c} F^{d x}}{a}\right )}{b \,d^{2} \ln \relax (F )^{2}}-\frac {6 x \polylog \left (3, -\frac {b \,F^{c} F^{d x}}{a}\right )}{b \,d^{3} \ln \relax (F )^{3}}+\frac {6 \polylog \left (4, -\frac {b \,F^{c} F^{d x}}{a}\right )}{b \,d^{4} \ln \relax (F )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 133, normalized size = 1.16 \[ \frac {x^{4}}{4 \, b} - \frac {\log \left (F^{d x}\right )^{4}}{4 \, b d^{4} \log \relax (F)^{4}} + \frac {\log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F^{d x}\right )^{3} + 3 \, {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F^{d x}\right )^{2} - 6 \, \log \left (F^{d x}\right ) {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a}) + 6 \, {\rm Li}_{4}(-\frac {F^{d x} F^{c} b}{a})}{b d^{4} \log \relax (F)^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{c + d x} x^{3}}{F^{c} F^{d x} b + a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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