3.1.90 \(\int \frac {F^{c+d x} x}{(a+b F^{c+d x})^3} \, dx\) [90]

Optimal. Leaf size=106 \[ \frac {1}{2 a b d^2 \left (a+b F^{c+d x}\right ) \log ^2(F)}+\frac {x}{2 a^2 b d \log (F)}-\frac {x}{2 b d \left (a+b F^{c+d x}\right )^2 \log (F)}-\frac {\log \left (a+b F^{c+d x}\right )}{2 a^2 b d^2 \log ^2(F)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2222, 2320, 46} \begin {gather*} -\frac {\log \left (a+b F^{c+d x}\right )}{2 a^2 b d^2 \log ^2(F)}+\frac {x}{2 a^2 b d \log (F)}+\frac {1}{2 a b d^2 \log ^2(F) \left (a+b F^{c+d x}\right )}-\frac {x}{2 b d \log (F) \left (a+b F^{c+d x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(2*a*b*d^2*(a + b*F^(c + d*x))*Log[F]^2) + x/(2*a^2*b*d*Log[F]) - x/(2*b*d*(a + b*F^(c + d*x))^2*Log[F]) - L
og[a + b*F^(c + d*x)]/(2*a^2*b*d^2*Log[F]^2)

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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 2320

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]]]

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 98, normalized size = 0.92 \begin {gather*} \frac {b d F^{c+d x} \left (2 a+b F^{c+d x}\right ) x \log (F)-\left (a+b F^{c+d x}\right ) \left (-a+\left (a+b F^{c+d x}\right ) \log \left (a+b F^{c+d x}\right )\right )}{2 a^2 b d^2 \left (a+b F^{c+d x}\right )^2 \log ^2(F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 111, normalized size = 1.05

method	result	size
risch	\(\frac {x}{2 a^{2} b d \ln \left (F \right )}+\frac {c}{2 \ln \left (F \right ) b \,d^{2} a^{2}}-\frac {\ln \left (F \right ) a d x -b \,F^{d x +c}-a}{2 \ln \left (F \right )^{2} d^{2} b \left (a +b \,F^{d x +c}\right )^{2} a}-\frac {\ln \left (F^{d x +c}+\frac {a}{b}\right )}{2 \ln \left (F \right )^{2} b \,d^{2} a^{2}}\)	\(111\)
norman	\(\frac {\frac {{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}}{2 \ln \left (F \right )^{2} a \,d^{2}}+\frac {x \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}}{\ln \left (F \right ) a d}+\frac {b x \,{\mathrm e}^{\left (2 d x +2 c \right ) \ln \left (F \right )}}{2 \ln \left (F \right ) a^{2} d}+\frac {1}{2 \ln \left (F \right )^{2} b \,d^{2}}}{\left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )^{2}}-\frac {\ln \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}{2 \ln \left (F \right )^{2} b \,d^{2} a^{2}}\)	\(127\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 150, normalized size = 1.42 \begin {gather*} \frac {F^{2 \, d x} F^{2 \, c} b^{2} d x \log \left (F\right ) + {\left (2 \, F^{c} a b d x \log \left (F\right ) + F^{c} a b\right )} F^{d x} + a^{2}}{2 \, {\left (2 \, F^{d x} F^{c} a^{3} b^{2} d^{2} \log \left (F\right )^{2} + F^{2 \, d x} F^{2 \, c} a^{2} b^{3} d^{2} \log \left (F\right )^{2} + a^{4} b d^{2} \log \left (F\right )^{2}\right )}} - \frac {\log \left (\frac {F^{d x} F^{c} b + a}{F^{c} b}\right )}{2 \, a^{2} 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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 148, normalized size = 1.40 \begin {gather*} \frac {F^{2 \, d x + 2 \, c} b^{2} d x \log \left (F\right ) + {\left (2 \, a b d x \log \left (F\right ) + a b\right )} F^{d x + c} + a^{2} - {\left (2 \, F^{d x + c} a b + F^{2 \, d x + 2 \, c} b^{2} + a^{2}\right )} \log \left (F^{d x + c} b + a\right )}{2 \, {\left (2 \, F^{d x + c} a^{3} b^{2} d^{2} \log \left (F\right )^{2} + F^{2 \, d x + 2 \, c} a^{2} b^{3} d^{2} \log \left (F\right )^{2} + a^{4} b d^{2} \log \left (F\right )^{2}\right )}} \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))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.09, size = 122, normalized size = 1.15 \begin {gather*} \frac {F^{c + d x} b - a d x \log {\left (F \right )} + a}{4 F^{c + d x} a^{2} b^{2} d^{2} \log {\left (F \right )}^{2} + 2 F^{2 c + 2 d x} a b^{3} d^{2} \log {\left (F \right )}^{2} + 2 a^{3} b d^{2} \log {\left (F \right )}^{2}} + \frac {x}{2 a^{2} b d \log {\left (F \right )}} - \frac {\log {\left (F^{c + d x} + \frac {a}{b} \right )}}{2 a^{2} 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))**3,x)

[Out]

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

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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))^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.75, size = 155, normalized size = 1.46 \begin {gather*} -\frac {\frac {F^c\,F^{d\,x}}{2\,a\,d^2\,{\ln \left (F\right )}^2}-\frac {F^c\,F^{d\,x}\,x}{a\,d\,\ln \left (F\right )}+\frac {F^{2\,c}\,F^{2\,d\,x}\,b}{2\,a^2\,d^2\,{\ln \left (F\right )}^2}-\frac {F^{2\,c}\,F^{2\,d\,x}\,b\,x}{2\,a^2\,d\,\ln \left (F\right )}}{a^2+F^{2\,c}\,F^{2\,d\,x}\,b^2+2\,F^c\,F^{d\,x}\,a\,b}-\frac {\ln \left (a+F^c\,F^{d\,x}\,b\right )}{2\,a^2\,b\,d^2\,{\ln \left (F\right )}^2} \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)^3,x)

[Out]

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

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