3.5.2 \(\int \frac {\log (c (d+e x)) (a+b \log (c (d+e x)))}{(d+e x)^2} \, dx\) [402]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {2416, 12, 2341, 2413} \begin {gather*} -\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)} \end {gather*}

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 12

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

Rule 2341

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 2413

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])

Rule 2416

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]

Rubi steps

\begin {align*} \int \frac {\log (c (d+e x)) (a+b \log (c (d+e x)))}{(d+e x)^2} \, dx &=\frac {\text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.03, size = 43, normalized size = 0.47 \begin {gather*} -\frac {a+2 b+(a+2 b) \log (c (d+e x))+b \log ^2(c (d+e x))}{e (d+e x)} \end {gather*}

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 = 0.34, size = 110, normalized size = 1.20

method	result	size
norman	\(\frac {-\frac {a +2 b}{e}-\frac {b \ln \left (c \left (e x +d \right )\right )^{2}}{e}-\frac {\left (a +2 b \right ) \ln \left (c \left (e x +d \right )\right )}{e}}{e x +d}\)	\(54\)
risch	\(-\frac {b \ln \left (c \left (e x +d \right )\right )^{2}}{e \left (e x +d \right )}-\frac {\left (a +2 b \right ) \ln \left (c \left (e x +d \right )\right )}{e \left (e x +d \right )}-\frac {a}{e \left (e x +d \right )}-\frac {2 b}{e \left (e x +d \right )}\)	\(76\)
derivativedivides	\(\frac {c^{2} a \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b \left (-\frac {\ln \left (c e x +c d \right )^{2}}{c e x +c d}-\frac {2 \ln \left (c e x +c d \right )}{c e x +c d}-\frac {2}{c e x +c d}\right )}{c e}\)	\(110\)
default	\(\frac {c^{2} a \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b \left (-\frac {\ln \left (c e x +c d \right )^{2}}{c e x +c d}-\frac {2 \ln \left (c e x +c d \right )}{c e x +c d}-\frac {2}{c e x +c d}\right )}{c e}\)	\(110\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 100, normalized size = 1.09 \begin {gather*} -{\left (b {\left (\frac {c e}{c x e^{3} + c d e^{2}} + \frac {\log \left (c x e + c d\right )}{x e^{2} + d e}\right )} + \frac {a}{x e^{2} + d e}\right )} \log \left ({\left (x e + d\right )} c\right ) - \frac {{\left (b {\left (\log \left (c\right ) + 2\right )} + b \log \left (x e + d\right ) + a\right )} e}{x e^{3} + d e^{2}} \end {gather*}

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

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Fricas [A]
time = 0.35, size = 48, normalized size = 0.52 \begin {gather*} -\frac {b \log \left (c x e + c d\right )^{2} + {\left (a + 2 \, b\right )} \log \left (c x e + c d\right ) + a + 2 \, b}{x e^{2} + d e} \end {gather*}

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

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Sympy [A]
time = 0.14, size = 56, normalized size = 0.61 \begin {gather*} - \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 {gather*}

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 = 4.45, size = 72, normalized size = 0.78 \begin {gather*} -\frac {{\left (b c^{2} \log \left ({\left (x e + d\right )} c\right )^{2} + a c^{2} \log \left ({\left (x e + d\right )} c\right ) + 2 \, b c^{2} \log \left ({\left (x e + d\right )} c\right ) + a c^{2} + 2 \, b c^{2}\right )} e^{\left (-1\right )}}{{\left (x e + d\right )} c^{2}} \end {gather*}

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

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Mupad [B]
time = 0.47, size = 63, normalized size = 0.68 \begin {gather*} -\frac {d\,\left (b\,{\ln \left (c\,\left (d+e\,x\right )\right )}^2+a\,\ln \left (c\,\left (d+e\,x\right )\right )+2\,b\,\ln \left (c\,\left (d+e\,x\right )\right )\right )-e\,\left (a\,x+2\,b\,x\right )}{d\,e\,\left (d+e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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