3.6 \(\int \frac{A+B \log (e (\frac{a+b x}{c+d x})^n)}{(a g+b g x)^2} \, dx\)

Optimal. Leaf size=67 \[ -\frac{(c+d x) \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{g^2 (a+b x) (b c-a d)}-\frac{B n}{b g^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0904321, antiderivative size = 108, normalized size of antiderivative = 1.61, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2525, 12, 44} \[ -\frac{B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A}{b g^2 (a+b x)}-\frac{B d n \log (a+b x)}{b g^2 (b c-a d)}+\frac{B d n \log (c+d x)}{b g^2 (b c-a d)}-\frac{B n}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 12

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0633145, size = 115, normalized size = 1.72 \[ \frac{B n (b c-a d) \left (-\frac{1}{(a+b x) (b c-a d)}-\frac{d \log (a+b x)}{(b c-a d)^2}+\frac{d \log (c+d x)}{(b c-a d)^2}\right )}{b g^2}-\frac{B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A}{b g (a g+b g x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(b*g*(a*g + b*g*x))) + (B*(b*c - a*d)*n*(-(1/((b*c - a*d)*(a + b*x)))
- (d*Log[a + b*x])/(b*c - a*d)^2 + (d*Log[c + d*x])/(b*c - a*d)^2))/(b*g^2)

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Maple [F]  time = 0.449, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( bgx+ag \right ) ^{2}} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.21445, size = 185, normalized size = 2.76 \begin{align*} -B n{\left (\frac{1}{b^{2} g^{2} x + a b g^{2}} + \frac{d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} - \frac{d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - \frac{B \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac{A}{b^{2} g^{2} x + a b g^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.796695, size = 221, normalized size = 3.3 \begin{align*} -\frac{A b c - A a d +{\left (B b c - B a d\right )} n +{\left (B b c - B a d\right )} \log \left (e\right ) +{\left (B b d n x + B b c n\right )} \log \left (\frac{b x + a}{d x + c}\right )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x +{\left (a b^{2} c - a^{2} b d\right )} g^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.42295, size = 162, normalized size = 2.42 \begin{align*} -\frac{B d n \log \left (b x + a\right )}{b^{2} c g^{2} - a b d g^{2}} + \frac{B d n \log \left (d x + c\right )}{b^{2} c g^{2} - a b d g^{2}} - \frac{B n \log \left (\frac{b x + a}{d x + c}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac{B n + A + B}{b^{2} g^{2} x + a b g^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*d*n*log(b*x + a)/(b^2*c*g^2 - a*b*d*g^2) + B*d*n*log(d*x + c)/(b^2*c*g^2 - a*b*d*g^2) - B*n*log((b*x + a)/(
d*x + c))/(b^2*g^2*x + a*b*g^2) - (B*n + A + B)/(b^2*g^2*x + a*b*g^2)