3.1.6 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x)^2} \, dx\) [6]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2549, 2341} \begin {gather*} -\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 (c+d x)}{g^2 (a+b x) (b c-a d)} \end {gather*}

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*(c + d*x))/((b*c - a*d)*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))

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 2549

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/b)^m, Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + 2)), x]
, x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m,
p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

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*}

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Mathematica [A]
time = 0.05, size = 115, normalized size = 1.72 \begin {gather*} -\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{b g (a g+b g x)}+\frac {B (b c-a d) n \left (-\frac {1}{(b c-a d) (a+b x)}-\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} \end {gather*}

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 = 0.05, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{\left (b g x +a g \right )^{2}}\, dx\]

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] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (68) = 136\).
time = 0.30, size = 138, normalized size = 2.06 \begin {gather*} -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 ({\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n} e\right )}{b^{2} g^{2} x + a b g^{2}} - \frac {A}{b^{2} g^{2} x + a b g^{2}} \end {gather*}

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((b*x/(d*x + c) + a/(d*x + c))^n*e)/(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 = 0.43, size = 94, normalized size = 1.40 \begin {gather*} -\frac {{\left (A + B\right )} b c - {\left (A + B\right )} a d + {\left (B b c - B a d\right )} n + {\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 {gather*}

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)*b*c - (A + B)*a*d + (B*b*c - B*a*d)*n + (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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

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]

Exception raised: NotImplementedError >> no valid subset found

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Giac [A]
time = 3.82, size = 85, normalized size = 1.27 \begin {gather*} -{\left (\frac {{\left (d x + c\right )} B n \log \left (\frac {b x + a}{d x + c}\right )}{{\left (b x + a\right )} g^{2}} + \frac {{\left (B n + A + B\right )} {\left (d x + c\right )}}{{\left (b x + a\right )} g^{2}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \end {gather*}

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]

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

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Mupad [B]
time = 5.65, size = 112, normalized size = 1.67 \begin {gather*} -\frac {A+B\,n}{x\,b^2\,g^2+a\,b\,g^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{b\,\left (a\,g^2+b\,g^2\,x\right )}-\frac {B\,d\,n\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{b\,g^2\,\left (a\,d-b\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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