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

Optimal. Leaf size=92 \[ -\frac{\log (c (d+e x)) (a+b \log (c (d+e x)))}{e (d+e x)}-\frac{a+b \log (c (d+e x))+b}{e (d+e x)}-\frac{b \log (c (d+e x))}{e (d+e x)}-\frac{b}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.0945195, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2369, 12, 2304, 2366} \[ -\frac{\log (c (d+e x)) (a+b \log (c (d+e x)))}{e (d+e x)}-\frac{a+b \log (c (d+e x))+b}{e (d+e x)}-\frac{b \log (c (d+e x))}{e (d+e x)}-\frac{b}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b/(e*(d + e*x))) - (b*Log[c*(d + e*x)])/(e*(d + e*x)) - (Log[c*(d + e*x)]*(a + b*Log[c*(d + e*x)]))/(e*(d +
e*x)) - (a + b + b*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{\log (c (d+e x)) (a+b \log (c (d+e x)))}{(d+e x)^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{c^2 \log (x) (a+b \log (x))}{x^2} \, dx,x,c (d+e x)\right )}{c e}\\ &=\frac{c \operatorname{Subst}\left (\int \frac{\log (x) (a+b \log (x))}{x^2} \, dx,x,c (d+e x)\right )}{e}\\ &=-\frac{b \log (c (d+e x))}{e (d+e x)}-\frac{\log (c (d+e x)) (a+b \log (c (d+e x)))}{e (d+e x)}-\frac{c \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}{e (d+e x)}-\frac{b \log (c (d+e x))}{e (d+e x)}-\frac{\log (c (d+e x)) (a+b \log (c (d+e x)))}{e (d+e x)}-\frac{a+b+b \log (c (d+e x))}{e (d+e x)}\\ \end{align*}

Mathematica [A]  time = 0.0650981, size = 43, normalized size = 0.47 \[ -\frac{(a+2 b) \log (c (d+e x))+a+b \log ^2(c (d+e x))+2 b}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.06, size = 116, normalized size = 1.3 \begin{align*} -{\frac{ac\ln \left ( cex+cd \right ) }{e \left ( cex+cd \right ) }}-{\frac{ac}{e \left ( cex+cd \right ) }}-{\frac{bc \left ( \ln \left ( cex+cd \right ) \right ) ^{2}}{e \left ( cex+cd \right ) }}-2\,{\frac{bc\ln \left ( cex+cd \right ) }{e \left ( cex+cd \right ) }}-2\,{\frac{bc}{e \left ( cex+cd \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03457, size = 134, normalized size = 1.46 \begin{align*} -{\left (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 )} + \frac{a}{e^{2} x + d e}\right )} \log \left ({\left (e x + d\right )} c\right ) - \frac{{\left (b{\left (\log \left (c\right ) + 2\right )} + b \log \left (e x + d\right ) + a\right )} e}{e^{3} x + d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 2.32849, size = 105, normalized size = 1.14 \begin{align*} -\frac{b \log \left (c e x + c d\right )^{2} +{\left (a + 2 \, b\right )} \log \left (c e x + c d\right ) + a + 2 \, b}{e^{2} x + d e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.421965, size = 56, normalized size = 0.61 \begin{align*} - \frac{b \log{\left (c \left (d + e x\right ) \right )}^{2}}{d e + e^{2} x} + \frac{\left (- a - 2 b\right ) \log{\left (c \left (d + e x\right ) \right )}}{d e + e^{2} x} - \frac{a + 2 b}{d e + e^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23726, size = 86, normalized size = 0.93 \begin{align*} -\frac{b e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c e\right )^{2}}{x e + d} - \frac{a e^{\left (-1\right )} \log \left ({\left (x e + d\right )} c e\right )}{x e + d} - \frac{b e^{\left (-1\right )}}{x e + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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