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

Optimal. Leaf size=107 \[ \frac {x^2}{a b d \log (F)}-\frac {x^2}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {2 x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {2 \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 107, 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 = {2222, 2215, 2221, 2317, 2438} \begin {gather*} -\frac {2 \text {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}-\frac {2 x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{a b d^2 \log ^2(F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}+\frac {x^2}{a b d \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2215

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 2221

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]
- 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 2222

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 2317

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 2438

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

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Mathematica [A]
time = 0.11, size = 103, normalized size = 0.96 \begin {gather*} \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 ) \text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \left (a+b F^{c+d x}\right ) \log ^3(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(230\) vs. \(2(107)=214\).
time = 0.02, size = 231, normalized size = 2.16

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 99, normalized size = 0.93 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Fricas [A]
time = 0.39, size = 186, normalized size = 1.74 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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