3.7 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(a g+b g x)^3} \, dx\)

Optimal. Leaf size=151 \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d n}{2 b g^3 (a+b x) (b c-a d)}-\frac {B n}{4 b g^3 (a+b x)^2} \]

[Out]

-1/4*B*n/b/g^3/(b*x+a)^2+1/2*B*d*n/b/(-a*d+b*c)/g^3/(b*x+a)+1/2*B*d^2*n*ln(b*x+a)/b/(-a*d+b*c)^2/g^3+1/2*(-A-B
*ln(e*((b*x+a)/(d*x+c))^n))/b/g^3/(b*x+a)^2-1/2*B*d^2*n*ln(d*x+c)/b/(-a*d+b*c)^2/g^3

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Rubi [A]  time = 0.12, antiderivative size = 151, normalized size of antiderivative = 1.00, 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} \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 b g^3 (a+b x)^2}+\frac {B d^2 n \log (a+b x)}{2 b g^3 (b c-a d)^2}-\frac {B d^2 n \log (c+d x)}{2 b g^3 (b c-a d)^2}+\frac {B d n}{2 b g^3 (a+b x) (b c-a d)}-\frac {B n}{4 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(a*g + b*g*x)^3,x]

[Out]

-(B*n)/(4*b*g^3*(a + b*x)^2) + (B*d*n)/(2*b*(b*c - a*d)*g^3*(a + b*x)) + (B*d^2*n*Log[a + b*x])/(2*b*(b*c - a*
d)^2*g^3) - (A + B*Log[e*((a + b*x)/(c + d*x))^n])/(2*b*g^3*(a + b*x)^2) - (B*d^2*n*Log[c + d*x])/(2*b*(b*c -
a*d)^2*g^3)

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])

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]

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)^3} \, dx &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B n) \int \frac {b c-a d}{g^2 (a+b x)^3 (c+d x)} \, dx}{2 b g}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x)^3 (c+d x)} \, dx}{2 b g^3}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}+\frac {(B (b c-a d) n) \int \left (\frac {b}{(b c-a d) (a+b x)^3}-\frac {b d}{(b c-a d)^2 (a+b x)^2}+\frac {b d^2}{(b c-a d)^3 (a+b x)}-\frac {d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{2 b g^3}\\ &=-\frac {B n}{4 b g^3 (a+b x)^2}+\frac {B d n}{2 b (b c-a d) g^3 (a+b x)}+\frac {B d^2 n \log (a+b x)}{2 b (b c-a d)^2 g^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 b g^3 (a+b x)^2}-\frac {B d^2 n \log (c+d x)}{2 b (b c-a d)^2 g^3}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 114, normalized size = 0.75 \[ -\frac {2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )+\frac {B n \left (2 d^2 (a+b x)^2 \log (c+d x)+(b c-a d) (b (c-2 d x)-3 a d)-2 d^2 (a+b x)^2 \log (a+b x)\right )}{(b c-a d)^2}}{4 b g^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[e*((a + b*x)/(c + d*x))^n])/(a*g + b*g*x)^3,x]

[Out]

-1/4*(2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + (B*n*((b*c - a*d)*(-3*a*d + b*(c - 2*d*x)) - 2*d^2*(a + b*x)^
2*Log[a + b*x] + 2*d^2*(a + b*x)^2*Log[c + d*x]))/(b*c - a*d)^2)/(b*g^3*(a + b*x)^2)

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fricas [A]  time = 0.92, size = 265, normalized size = 1.75 \[ -\frac {2 \, A b^{2} c^{2} - 4 \, A a b c d + 2 \, A a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} n x + {\left (B b^{2} c^{2} - 4 \, B a b c d + 3 \, B a^{2} d^{2}\right )} n + 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} \log \relax (e) - 2 \, {\left (B b^{2} d^{2} n x^{2} + 2 \, B a b d^{2} n x - {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} g^{3} x + {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} g^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

-1/4*(2*A*b^2*c^2 - 4*A*a*b*c*d + 2*A*a^2*d^2 - 2*(B*b^2*c*d - B*a*b*d^2)*n*x + (B*b^2*c^2 - 4*B*a*b*c*d + 3*B
*a^2*d^2)*n + 2*(B*b^2*c^2 - 2*B*a*b*c*d + B*a^2*d^2)*log(e) - 2*(B*b^2*d^2*n*x^2 + 2*B*a*b*d^2*n*x - (B*b^2*c
^2 - 2*B*a*b*c*d)*n)*log((b*x + a)/(d*x + c)))/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*g^3*x^2 + 2*(a*b^4*c^2 -
2*a^2*b^3*c*d + a^3*b^2*d^2)*g^3*x + (a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*g^3)

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giac [A]  time = 4.80, size = 220, normalized size = 1.46 \[ -\frac {1}{4} \, {\left (\frac {2 \, {\left (B b n - \frac {2 \, {\left (b x + a\right )} B d n}{d x + c}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{\frac {{\left (b x + a\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}} + \frac {B b n - \frac {4 \, {\left (b x + a\right )} B d n}{d x + c} + 2 \, A b + 2 \, B b - \frac {4 \, {\left (b x + a\right )} A d}{d x + c} - \frac {4 \, {\left (b x + a\right )} B d}{d x + c}}{\frac {{\left (b x + a\right )}^{2} b c g^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{2} a d g^{3}}{{\left (d x + c\right )}^{2}}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

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)^3,x, algorithm="giac")

[Out]

-1/4*(2*(B*b*n - 2*(b*x + a)*B*d*n/(d*x + c))*log((b*x + a)/(d*x + c))/((b*x + a)^2*b*c*g^3/(d*x + c)^2 - (b*x
+ a)^2*a*d*g^3/(d*x + c)^2) + (B*b*n - 4*(b*x + a)*B*d*n/(d*x + c) + 2*A*b + 2*B*b - 4*(b*x + a)*A*d/(d*x + c
) - 4*(b*x + a)*B*d/(d*x + c))/((b*x + a)^2*b*c*g^3/(d*x + c)^2 - (b*x + a)^2*a*d*g^3/(d*x + c)^2))*(b*c/(b*c
- a*d)^2 - a*d/(b*c - a*d)^2)

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{\left (b g x +a g \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g)^3,x)

[Out]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(b*g*x+a*g)^3,x)

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maxima [A]  time = 1.36, size = 259, normalized size = 1.72 \[ \frac {1}{4} \, B n {\left (\frac {2 \, b d x - b c + 3 \, a d}{{\left (b^{4} c - a b^{3} d\right )} g^{3} x^{2} + 2 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} g^{3} x + {\left (a^{2} b^{2} c - a^{3} b d\right )} g^{3}} + \frac {2 \, d^{2} \log \left (b x + a\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} g^{3}}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} - \frac {A}{2 \, {\left (b^{3} g^{3} x^{2} + 2 \, a b^{2} g^{3} x + a^{2} b g^{3}\right )}} \]

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)^3,x, algorithm="maxima")

[Out]

1/4*B*n*((2*b*d*x - b*c + 3*a*d)/((b^4*c - a*b^3*d)*g^3*x^2 + 2*(a*b^3*c - a^2*b^2*d)*g^3*x + (a^2*b^2*c - a^3
*b*d)*g^3) + 2*d^2*log(b*x + a)/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*g^3) - 2*d^2*log(d*x + c)/((b^3*c^2 - 2*a
*b^2*c*d + a^2*b*d^2)*g^3)) - 1/2*B*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*
b*g^3) - 1/2*A/(b^3*g^3*x^2 + 2*a*b^2*g^3*x + a^2*b*g^3)

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mupad [B]  time = 4.52, size = 222, normalized size = 1.47 \[ -\frac {\frac {2\,A\,a\,d-2\,A\,b\,c+3\,B\,a\,d\,n-B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,x}{a\,d-b\,c}}{2\,a^2\,b\,g^3+4\,a\,b^2\,g^3\,x+2\,b^3\,g^3\,x^2}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,b\,\left (a^2\,g^3+2\,a\,b\,g^3\,x+b^2\,g^3\,x^2\right )}-\frac {B\,d^2\,n\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2\,g^3-2\,a^2\,b\,d^2\,g^3}{2\,b\,g^3\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,g^3\,{\left (a\,d-b\,c\right )}^2} \]

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)^3,x)

[Out]

- ((2*A*a*d - 2*A*b*c + 3*B*a*d*n - B*b*c*n)/(2*(a*d - b*c)) + (B*b*d*n*x)/(a*d - b*c))/(2*a^2*b*g^3 + 2*b^3*g
^3*x^2 + 4*a*b^2*g^3*x) - (B*log(e*((a + b*x)/(c + d*x))^n))/(2*b*(a^2*g^3 + b^2*g^3*x^2 + 2*a*b*g^3*x)) - (B*
d^2*n*atanh((2*b^3*c^2*g^3 - 2*a^2*b*d^2*g^3)/(2*b*g^3*(a*d - b*c)^2) - (2*b*d*x)/(a*d - b*c)))/(b*g^3*(a*d -
b*c)^2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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)**3,x)

[Out]

Exception raised: NotImplementedError

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