3.5.3 \(\int \frac {(a+b \log (c (d+e x))) (f+g \log (c (d+e x)))}{(d+e x)^2} \, dx\) [403]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 = {2416, 12, 2341, 2413} \begin {gather*} -\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)} \end {gather*}

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

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Mathematica [A]
time = 0.05, size = 58, normalized size = 0.57 \begin {gather*} -\frac {a (f+g)+b (f+2 g)+(a g+b (f+2 g)) \log (c (d+e x))+b g \log ^2(c (d+e x))}{e (d+e x)} \end {gather*}

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)))

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Maple [A]
time = 0.37, size = 169, normalized size = 1.66

method	result	size
norman	\(\frac {-\frac {a f +a g +b f +2 b g}{e}-\frac {\left (a g +b f +2 b g \right ) \ln \left (c \left (e x +d \right )\right )}{e}-\frac {b g \ln \left (c \left (e x +d \right )\right )^{2}}{e}}{e x +d}\)	\(70\)
risch	\(-\frac {b g \ln \left (c \left (e x +d \right )\right )^{2}}{e \left (e x +d \right )}-\frac {\left (a g +b f +2 b g \right ) \ln \left (c \left (e x +d \right )\right )}{e \left (e x +d \right )}-\frac {a f}{e \left (e x +d \right )}-\frac {a g}{e \left (e x +d \right )}-\frac {b f}{e \left (e x +d \right )}-\frac {2 b g}{e \left (e x +d \right )}\)	\(113\)
derivativedivides	\(\frac {-\frac {c^{2} a f}{c e x +c d}+c^{2} a g \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b f \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b g \left (-\frac {\ln \left (c e x +c d \right )^{2}}{c e x +c d}-\frac {2 \ln \left (c e x +c d \right )}{c e x +c d}-\frac {2}{c e x +c d}\right )}{c e}\)	\(169\)
default	\(\frac {-\frac {c^{2} a f}{c e x +c d}+c^{2} a g \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b f \left (-\frac {\ln \left (c e x +c d \right )}{c e x +c d}-\frac {1}{c e x +c d}\right )+c^{2} b g \left (-\frac {\ln \left (c e x +c d \right )^{2}}{c e x +c d}-\frac {2 \ln \left (c e x +c d \right )}{c e x +c d}-\frac {2}{c e x +c d}\right )}{c e}\)	\(169\)

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,method=_RETURNVERBOSE)

[Out]

1/c/e*(-c^2*a*f/(c*e*x+c*d)+c^2*a*g*(-1/(c*e*x+c*d)*ln(c*e*x+c*d)-1/(c*e*x+c*d))+c^2*b*f*(-1/(c*e*x+c*d)*ln(c*
e*x+c*d)-1/(c*e*x+c*d))+c^2*b*g*(-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 = 0.28, size = 161, normalized size = 1.58 \begin {gather*} -b f {\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 )} - a g {\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 {{\left (c^{2} \log \left (c x e + c d\right )^{2} + 2 \, c^{2} \log \left (c x e + c d\right ) + 2 \, c^{2}\right )} b g e^{\left (-1\right )}}{{\left (c x e + c d\right )} c} - \frac {a f}{x e^{2} + d e} \end {gather*}

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*f*(c*e/(c*x*e^3 + c*d*e^2) + log(c*x*e + c*d)/(x*e^2 + d*e)) - a*g*(c*e/(c*x*e^3 + c*d*e^2) + log(c*x*e + c
*d)/(x*e^2 + d*e)) - (c^2*log(c*x*e + c*d)^2 + 2*c^2*log(c*x*e + c*d) + 2*c^2)*b*g*e^(-1)/((c*x*e + c*d)*c) -
a*f/(x*e^2 + d*e)

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Fricas [A]
time = 0.35, size = 63, normalized size = 0.62 \begin {gather*} -\frac {b g \log \left (c x e + 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 x e + c d\right )}{x e^{2} + d e} \end {gather*}

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

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Sympy [A]
time = 0.16, size = 75, normalized size = 0.74 \begin {gather*} - \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 {gather*}

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)

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Giac [A]
time = 4.99, size = 104, normalized size = 1.02 \begin {gather*} -\frac {{\left (b c^{2} g \log \left ({\left (x e + d\right )} c\right )^{2} + b c^{2} f \log \left ({\left (x e + d\right )} c\right ) + a c^{2} g \log \left ({\left (x e + d\right )} c\right ) + 2 \, b c^{2} g \log \left ({\left (x e + d\right )} c\right ) + a c^{2} f + b c^{2} f + a c^{2} g + 2 \, b c^{2} g\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((a+b*log(c*(e*x+d)))*(f+g*log(c*(e*x+d)))/(e*x+d)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.50, size = 92, normalized size = 0.90 \begin {gather*} -\frac {d\,\left (b\,g\,{\ln \left (c\,d+c\,e\,x\right )}^2+a\,g\,\ln \left (c\,d+c\,e\,x\right )+b\,f\,\ln \left (c\,d+c\,e\,x\right )+2\,b\,g\,\ln \left (c\,d+c\,e\,x\right )\right )-e\,\left (a\,f\,x+a\,g\,x+b\,f\,x+2\,b\,g\,x\right )}{d^2\,e+x\,d\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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