3.403 \(\int \frac{(a+b \log (c (d+e x))) (f+g \log (c (d+e x)))}{(d+e x)^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac{(a+b \log (c (d+e x))) (g \log (c (d+e x))+f)}{e (d+e x)}-\frac{g (a+b \log (c (d+e x))+b)}{e (d+e x)}-\frac{b (g \log (c (d+e x))+f)}{e (d+e x)}-\frac{b g}{e (d+e x)} \]

[Out]

-((b*g)/(e*(d + e*x))) - (g*(a + b + b*Log[c*(d + e*x)]))/(e*(d + e*x)) - (b*(f + g*Log[c*(d + e*x)]))/(e*(d +
e*x)) - ((a + b*Log[c*(d + e*x)])*(f + g*Log[c*(d + e*x)]))/(e*(d + e*x))

________________________________________________________________________________________

Rubi [A]  time = 0.110858, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2369, 12, 2304, 2366} \[ -\frac{(a+b \log (c (d+e x))) (g \log (c (d+e x))+f)}{e (d+e x)}-\frac{g (a+b \log (c (d+e x))+b)}{e (d+e x)}-\frac{b (g \log (c (d+e x))+f)}{e (d+e x)}-\frac{b g}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*(d + e*x)])*(f + g*Log[c*(d + e*x)]))/(d + e*x)^2,x]

[Out]

-((b*g)/(e*(d + e*x))) - (g*(a + b + b*Log[c*(d + e*x)]))/(e*(d + e*x)) - (b*(f + g*Log[c*(d + e*x)]))/(e*(d +
e*x)) - ((a + b*Log[c*(d + e*x)])*(f + g*Log[c*(d + e*x)]))/(e*(d + e*x))

Rule 2369

Int[((a_.) + Log[v_]*(b_.))^(p_.)*((c_.) + Log[v_]*(d_.))^(q_.)*(u_)^(m_.), x_Symbol] :> With[{e = Coeff[u, x,
0], f = Coeff[u, x, 1], g = Coeff[v, x, 0], h = Coeff[v, x, 1]}, Dist[1/h, Subst[Int[((f*x)/h)^m*(a + b*Log[x
])^p*(c + d*Log[x])^q, x], x, v], x] /; EqQ[f*g - e*h, 0] && NeQ[g, 0]] /; FreeQ[{a, b, c, d, m, p, q}, x] &&
LinearQ[{u, v}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps

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

Mathematica [A]  time = 0.103999, size = 58, normalized size = 0.57 \[ -\frac{(a g+b (f+2 g)) \log (c (d+e x))+a (f+g)+b g \log ^2(c (d+e x))+b (f+2 g)}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*(d + e*x)])*(f + g*Log[c*(d + e*x)]))/(d + e*x)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.065, size = 184, normalized size = 1.8 \begin{align*} -{\frac{acf}{e \left ( cex+cd \right ) }}-{\frac{acg\ln \left ( cex+cd \right ) }{e \left ( cex+cd \right ) }}-{\frac{acg}{e \left ( cex+cd \right ) }}-{\frac{bcf\ln \left ( cex+cd \right ) }{e \left ( cex+cd \right ) }}-{\frac{bcf}{e \left ( cex+cd \right ) }}-{\frac{bcg \left ( \ln \left ( cex+cd \right ) \right ) ^{2}}{e \left ( cex+cd \right ) }}-2\,{\frac{bcg\ln \left ( cex+cd \right ) }{e \left ( cex+cd \right ) }}-2\,{\frac{bcg}{e \left ( cex+cd \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)))*(f+g*ln(c*(e*x+d)))/(e*x+d)^2,x)

[Out]

-c/e*a*f/(c*e*x+c*d)-c/e*a*g*ln(c*e*x+c*d)/(c*e*x+c*d)-c/e*a*g/(c*e*x+c*d)-c/e*b*f*ln(c*e*x+c*d)/(c*e*x+c*d)-c
/e*b*f/(c*e*x+c*d)-c/e*b*g/(c*e*x+c*d)*ln(c*e*x+c*d)^2-2*c/e*b*g*ln(c*e*x+c*d)/(c*e*x+c*d)-2*c/e*b*g/(c*e*x+c*
d)

________________________________________________________________________________________

Maxima [A]  time = 1.08332, size = 215, normalized size = 2.11 \begin{align*} -b{\left (\frac{c e}{c e^{3} x + c d e^{2}} + \frac{\log \left (c e x + c d\right )}{e^{2} x + d e}\right )} f - a{\left (\frac{c e}{c e^{3} x + c d e^{2}} + \frac{\log \left (c e x + c d\right )}{e^{2} x + d e}\right )} g - \frac{a f}{e^{2} x + d e} - \frac{{\left (c^{2} \log \left (c e x + c d\right )^{2} + 2 \, c^{2} \log \left (c e x + c d\right ) + 2 \, c^{2}\right )} b g}{{\left (c e x + c d\right )} c e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)))*(f+g*log(c*(e*x+d)))/(e*x+d)^2,x, algorithm="maxima")

[Out]

-b*(c*e/(c*e^3*x + c*d*e^2) + log(c*e*x + c*d)/(e^2*x + d*e))*f - a*(c*e/(c*e^3*x + c*d*e^2) + log(c*e*x + c*d
)/(e^2*x + d*e))*g - a*f/(e^2*x + d*e) - (c^2*log(c*e*x + c*d)^2 + 2*c^2*log(c*e*x + c*d) + 2*c^2)*b*g/((c*e*x
+ c*d)*c*e)

________________________________________________________________________________________

Fricas [A]  time = 2.29366, size = 143, normalized size = 1.4 \begin{align*} -\frac{b g \log \left (c e x + c d\right )^{2} +{\left (a + b\right )} f +{\left (a + 2 \, b\right )} g +{\left (b f +{\left (a + 2 \, b\right )} g\right )} \log \left (c e x + c d\right )}{e^{2} x + d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)))*(f+g*log(c*(e*x+d)))/(e*x+d)^2,x, algorithm="fricas")

[Out]

-(b*g*log(c*e*x + c*d)^2 + (a + b)*f + (a + 2*b)*g + (b*f + (a + 2*b)*g)*log(c*e*x + c*d))/(e^2*x + d*e)

________________________________________________________________________________________

Sympy [A]  time = 0.448629, size = 75, normalized size = 0.74 \begin{align*} - \frac{b g \log{\left (c \left (d + e x\right ) \right )}^{2}}{d e + e^{2} x} + \frac{\left (- a g - b f - 2 b g\right ) \log{\left (c \left (d + e x\right ) \right )}}{d e + e^{2} x} - \frac{a f + a g + b f + 2 b g}{d e + e^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)))*(f+g*ln(c*(e*x+d)))/(e*x+d)**2,x)

[Out]

-b*g*log(c*(d + e*x))**2/(d*e + e**2*x) + (-a*g - b*f - 2*b*g)*log(c*(d + e*x))/(d*e + e**2*x) - (a*f + a*g +
b*f + 2*b*g)/(d*e + e**2*x)

________________________________________________________________________________________

Giac [A]  time = 1.2507, size = 104, normalized size = 1.02 \begin{align*} -\frac{b g e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c e\right )^{2}}{x e + d} - \frac{{\left (b f + a g\right )} e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c e\right )}{x e + d} - \frac{{\left (a f + b g\right )} e^{\left (-1\right )}}{x e + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)))*(f+g*log(c*(e*x+d)))/(e*x+d)^2,x, algorithm="giac")

[Out]

-b*g*e^(-1)*log((x*e + d)*c*e)^2/(x*e + d) - (b*f + a*g)*e^(-1)*log((x*e + d)*c*e)/(x*e + d) - (a*f + b*g)*e^(
-1)/(x*e + d)