3.1.78 \(\int \frac {F^{c+d x} x}{a+b F^{c+d x}} \, dx\) [78]

Optimal. Leaf size=54 \[ \frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {\text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2221, 2317, 2438} \begin {gather*} \frac {\text {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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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

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Mathematica [A]
time = 0.09, size = 54, normalized size = 1.00 \begin {gather*} \frac {x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {\text {Li}_2\left (-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(54)=108\).
time = 0.03, size = 154, normalized size = 2.85

method	result	size
risch	\(-\frac {c x}{d b}-\frac {c^{2}}{2 d^{2} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x}{d \ln \left (F \right ) b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c}{d^{2} \ln \left (F \right ) b}+\frac {\polylog \left (2, -\frac {b \,F^{d x} F^{c}}{a}\right )}{d^{2} \ln \left (F \right )^{2} b}-\frac {c \ln \left (a +F^{c} F^{d x} b \right )}{d^{2} \ln \left (F \right ) b}+\frac {c \ln \left (F^{d x} F^{c}\right )}{d^{2} \ln \left (F \right ) b}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 48, normalized size = 0.89 \begin {gather*} \frac {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 )}{b d^{2} \log \left (F\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 75, normalized size = 1.39 \begin {gather*} -\frac {c \log \left (F^{d x + c} b + a\right ) \log \left (F\right ) - {\left (d x + c\right )} \log \left (F\right ) \log \left (\frac {F^{d x + c} b + a}{a}\right ) - {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right )}{b d^{2} \log \left (F\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {F^{c + d x} x}{F^{c} F^{d x} b + a}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(a+b*F^(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {F^{c+d\,x}\,x}{a+F^{c+d\,x}\,b} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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