3.36 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(c g+d g x)^4} \, dx\)

Optimal. Leaf size=183 \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d g^4 (c+d x)^3}+\frac {b^3 B n \log (a+b x)}{3 d g^4 (b c-a d)^3}-\frac {b^3 B n \log (c+d x)}{3 d g^4 (b c-a d)^3}+\frac {b^2 B n}{3 d g^4 (c+d x) (b c-a d)^2}+\frac {b B n}{6 d g^4 (c+d x)^2 (b c-a d)}+\frac {B n}{9 d g^4 (c+d x)^3} \]

[Out]

1/9*B*n/d/g^4/(d*x+c)^3+1/6*b*B*n/d/(-a*d+b*c)/g^4/(d*x+c)^2+1/3*b^2*B*n/d/(-a*d+b*c)^2/g^4/(d*x+c)+1/3*b^3*B*
n*ln(b*x+a)/d/(-a*d+b*c)^3/g^4+1/3*(-A-B*ln(e*((b*x+a)/(d*x+c))^n))/d/g^4/(d*x+c)^3-1/3*b^3*B*n*ln(d*x+c)/d/(-
a*d+b*c)^3/g^4

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Rubi [A]  time = 0.14, antiderivative size = 183, 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} \[ -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{3 d g^4 (c+d x)^3}+\frac {b^2 B n}{3 d g^4 (c+d x) (b c-a d)^2}+\frac {b^3 B n \log (a+b x)}{3 d g^4 (b c-a d)^3}-\frac {b^3 B n \log (c+d x)}{3 d g^4 (b c-a d)^3}+\frac {b B n}{6 d g^4 (c+d x)^2 (b c-a d)}+\frac {B n}{9 d g^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*n)/(9*d*g^4*(c + d*x)^3) + (b*B*n)/(6*d*(b*c - a*d)*g^4*(c + d*x)^2) + (b^2*B*n)/(3*d*(b*c - a*d)^2*g^4*(c
+ d*x)) + (b^3*B*n*Log[a + b*x])/(3*d*(b*c - a*d)^3*g^4) - (A + B*Log[e*((a + b*x)/(c + d*x))^n])/(3*d*g^4*(c
+ d*x)^3) - (b^3*B*n*Log[c + d*x])/(3*d*(b*c - a*d)^3*g^4)

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 )}{(c g+d g x)^4} \, dx &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{3 d g^4 (c+d x)^3}+\frac {(B n) \int \frac {b c-a d}{g^3 (a+b x) (c+d x)^4} \, dx}{3 d g}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{3 d g^4 (c+d x)^3}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x)^4} \, dx}{3 d g^4}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{3 d g^4 (c+d x)^3}+\frac {(B (b c-a d) n) \int \left (\frac {b^4}{(b c-a d)^4 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^4}-\frac {b d}{(b c-a d)^2 (c+d x)^3}-\frac {b^2 d}{(b c-a d)^3 (c+d x)^2}-\frac {b^3 d}{(b c-a d)^4 (c+d x)}\right ) \, dx}{3 d g^4}\\ &=\frac {B n}{9 d g^4 (c+d x)^3}+\frac {b B n}{6 d (b c-a d) g^4 (c+d x)^2}+\frac {b^2 B n}{3 d (b c-a d)^2 g^4 (c+d x)}+\frac {b^3 B n \log (a+b x)}{3 d (b c-a d)^3 g^4}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{3 d g^4 (c+d x)^3}-\frac {b^3 B n \log (c+d x)}{3 d (b c-a d)^3 g^4}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 146, normalized size = 0.80 \[ \frac {\frac {B n \left ((b c-a d) \left (2 a^2 d^2-a b d (7 c+3 d x)+b^2 \left (11 c^2+15 c d x+6 d^2 x^2\right )\right )+6 b^3 (c+d x)^3 \log (a+b x)-6 b^3 (c+d x)^3 \log (c+d x)\right )}{(b c-a d)^3}-6 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{18 d g^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + (B*n*((b*c - a*d)*(2*a^2*d^2 - a*b*d*(7*c + 3*d*x) + b^2*(11*c^2
+ 15*c*d*x + 6*d^2*x^2)) + 6*b^3*(c + d*x)^3*Log[a + b*x] - 6*b^3*(c + d*x)^3*Log[c + d*x]))/(b*c - a*d)^3)/(1
8*d*g^4*(c + d*x)^3)

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fricas [B]  time = 0.93, size = 483, normalized size = 2.64 \[ -\frac {6 \, A b^{3} c^{3} - 18 \, A a b^{2} c^{2} d + 18 \, A a^{2} b c d^{2} - 6 \, A a^{3} d^{3} - 6 \, {\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} n x^{2} - 3 \, {\left (5 \, B b^{3} c^{2} d - 6 \, B a b^{2} c d^{2} + B a^{2} b d^{3}\right )} n x - {\left (11 \, B b^{3} c^{3} - 18 \, B a b^{2} c^{2} d + 9 \, B a^{2} b c d^{2} - 2 \, B a^{3} d^{3}\right )} n + 6 \, {\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2} - B a^{3} d^{3}\right )} \log \relax (e) - 6 \, {\left (B b^{3} d^{3} n x^{3} + 3 \, B b^{3} c d^{2} n x^{2} + 3 \, B b^{3} c^{2} d n x + {\left (3 \, B a b^{2} c^{2} d - 3 \, B a^{2} b c d^{2} + B a^{3} d^{3}\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{18 \, {\left ({\left (b^{3} c^{3} d^{4} - 3 \, a b^{2} c^{2} d^{5} + 3 \, a^{2} b c d^{6} - a^{3} d^{7}\right )} g^{4} x^{3} + 3 \, {\left (b^{3} c^{4} d^{3} - 3 \, a b^{2} c^{3} d^{4} + 3 \, a^{2} b c^{2} d^{5} - a^{3} c d^{6}\right )} g^{4} x^{2} + 3 \, {\left (b^{3} c^{5} d^{2} - 3 \, a b^{2} c^{4} d^{3} + 3 \, a^{2} b c^{3} d^{4} - a^{3} c^{2} d^{5}\right )} g^{4} x + {\left (b^{3} c^{6} d - 3 \, a b^{2} c^{5} d^{2} + 3 \, a^{2} b c^{4} d^{3} - a^{3} c^{3} d^{4}\right )} g^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^4,x, algorithm="fricas")

[Out]

-1/18*(6*A*b^3*c^3 - 18*A*a*b^2*c^2*d + 18*A*a^2*b*c*d^2 - 6*A*a^3*d^3 - 6*(B*b^3*c*d^2 - B*a*b^2*d^3)*n*x^2 -
3*(5*B*b^3*c^2*d - 6*B*a*b^2*c*d^2 + B*a^2*b*d^3)*n*x - (11*B*b^3*c^3 - 18*B*a*b^2*c^2*d + 9*B*a^2*b*c*d^2 -
2*B*a^3*d^3)*n + 6*(B*b^3*c^3 - 3*B*a*b^2*c^2*d + 3*B*a^2*b*c*d^2 - B*a^3*d^3)*log(e) - 6*(B*b^3*d^3*n*x^3 + 3
*B*b^3*c*d^2*n*x^2 + 3*B*b^3*c^2*d*n*x + (3*B*a*b^2*c^2*d - 3*B*a^2*b*c*d^2 + B*a^3*d^3)*n)*log((b*x + a)/(d*x
+ c)))/((b^3*c^3*d^4 - 3*a*b^2*c^2*d^5 + 3*a^2*b*c*d^6 - a^3*d^7)*g^4*x^3 + 3*(b^3*c^4*d^3 - 3*a*b^2*c^3*d^4
+ 3*a^2*b*c^2*d^5 - a^3*c*d^6)*g^4*x^2 + 3*(b^3*c^5*d^2 - 3*a*b^2*c^4*d^3 + 3*a^2*b*c^3*d^4 - a^3*c^2*d^5)*g^4
*x + (b^3*c^6*d - 3*a*b^2*c^5*d^2 + 3*a^2*b*c^4*d^3 - a^3*c^3*d^4)*g^4)

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giac [B]  time = 7.70, size = 399, normalized size = 2.18 \[ \frac {1}{18} \, {\left (6 \, {\left (\frac {3 \, {\left (b x + a\right )} B b^{2} n}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}} - \frac {3 \, {\left (b x + a\right )}^{2} B b d n}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}^{2}} + \frac {{\left (b x + a\right )}^{3} B d^{2} n}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}^{3}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) - \frac {2 \, {\left (B d^{2} n - 3 \, A d^{2} - 3 \, B d^{2}\right )} {\left (b x + a\right )}^{3}}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}^{3}} + \frac {9 \, {\left (B b d n - 2 \, A b d - 2 \, B b d\right )} {\left (b x + a\right )}^{2}}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}^{2}} - \frac {18 \, {\left (B b^{2} n - A b^{2} - B b^{2}\right )} {\left (b x + a\right )}}{{\left (b^{2} c^{2} g^{4} - 2 \, a b c d g^{4} + a^{2} d^{2} g^{4}\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^4,x, algorithm="giac")

[Out]

1/18*(6*(3*(b*x + a)*B*b^2*n/((b^2*c^2*g^4 - 2*a*b*c*d*g^4 + a^2*d^2*g^4)*(d*x + c)) - 3*(b*x + a)^2*B*b*d*n/(
(b^2*c^2*g^4 - 2*a*b*c*d*g^4 + a^2*d^2*g^4)*(d*x + c)^2) + (b*x + a)^3*B*d^2*n/((b^2*c^2*g^4 - 2*a*b*c*d*g^4 +
a^2*d^2*g^4)*(d*x + c)^3))*log((b*x + a)/(d*x + c)) - 2*(B*d^2*n - 3*A*d^2 - 3*B*d^2)*(b*x + a)^3/((b^2*c^2*g
^4 - 2*a*b*c*d*g^4 + a^2*d^2*g^4)*(d*x + c)^3) + 9*(B*b*d*n - 2*A*b*d - 2*B*b*d)*(b*x + a)^2/((b^2*c^2*g^4 - 2
*a*b*c*d*g^4 + a^2*d^2*g^4)*(d*x + c)^2) - 18*(B*b^2*n - A*b^2 - B*b^2)*(b*x + a)/((b^2*c^2*g^4 - 2*a*b*c*d*g^
4 + a^2*d^2*g^4)*(d*x + c)))*(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 \[ \int \frac {B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A}{\left (d g x +c g \right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(d*g*x+c*g)^4,x)

[Out]

int((B*ln(e*((b*x+a)/(d*x+c))^n)+A)/(d*g*x+c*g)^4,x)

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maxima [B]  time = 1.37, size = 433, normalized size = 2.37 \[ \frac {1}{18} \, B n {\left (\frac {6 \, b^{2} d^{2} x^{2} + 11 \, b^{2} c^{2} - 7 \, a b c d + 2 \, a^{2} d^{2} + 3 \, {\left (5 \, b^{2} c d - a b d^{2}\right )} x}{{\left (b^{2} c^{2} d^{4} - 2 \, a b c d^{5} + a^{2} d^{6}\right )} g^{4} x^{3} + 3 \, {\left (b^{2} c^{3} d^{3} - 2 \, a b c^{2} d^{4} + a^{2} c d^{5}\right )} g^{4} x^{2} + 3 \, {\left (b^{2} c^{4} d^{2} - 2 \, a b c^{3} d^{3} + a^{2} c^{2} d^{4}\right )} g^{4} x + {\left (b^{2} c^{5} d - 2 \, a b c^{4} d^{2} + a^{2} c^{3} d^{3}\right )} g^{4}} + \frac {6 \, b^{3} \log \left (b x + a\right )}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} g^{4}} - \frac {6 \, b^{3} \log \left (d x + c\right )}{{\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - a^{3} d^{4}\right )} g^{4}}\right )} - \frac {B \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right )}{3 \, {\left (d^{4} g^{4} x^{3} + 3 \, c d^{3} g^{4} x^{2} + 3 \, c^{2} d^{2} g^{4} x + c^{3} d g^{4}\right )}} - \frac {A}{3 \, {\left (d^{4} g^{4} x^{3} + 3 \, c d^{3} g^{4} x^{2} + 3 \, c^{2} d^{2} g^{4} x + c^{3} d g^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*g*x+c*g)^4,x, algorithm="maxima")

[Out]

1/18*B*n*((6*b^2*d^2*x^2 + 11*b^2*c^2 - 7*a*b*c*d + 2*a^2*d^2 + 3*(5*b^2*c*d - a*b*d^2)*x)/((b^2*c^2*d^4 - 2*a
*b*c*d^5 + a^2*d^6)*g^4*x^3 + 3*(b^2*c^3*d^3 - 2*a*b*c^2*d^4 + a^2*c*d^5)*g^4*x^2 + 3*(b^2*c^4*d^2 - 2*a*b*c^3
*d^3 + a^2*c^2*d^4)*g^4*x + (b^2*c^5*d - 2*a*b*c^4*d^2 + a^2*c^3*d^3)*g^4) + 6*b^3*log(b*x + a)/((b^3*c^3*d -
3*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*g^4) - 6*b^3*log(d*x + c)/((b^3*c^3*d - 3*a*b^2*c^2*d^2 + 3*a^2*b*c
*d^3 - a^3*d^4)*g^4)) - 1/3*B*log(e*(b*x/(d*x + c) + a/(d*x + c))^n)/(d^4*g^4*x^3 + 3*c*d^3*g^4*x^2 + 3*c^2*d^
2*g^4*x + c^3*d*g^4) - 1/3*A/(d^4*g^4*x^3 + 3*c*d^3*g^4*x^2 + 3*c^2*d^2*g^4*x + c^3*d*g^4)

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mupad [B]  time = 4.72, size = 349, normalized size = 1.91 \[ \frac {B\,a^2\,d\,n}{9\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {A\,a^2\,d}{3\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {A\,b^2\,c^2}{3\,d\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{3\,d\,g^4\,{\left (c+d\,x\right )}^3}+\frac {2\,A\,a\,b\,c}{3\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}+\frac {B\,b^2\,d\,n\,x^2}{3\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {7\,B\,a\,b\,c\,n}{18\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}+\frac {11\,B\,b^2\,c^2\,n}{18\,d\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}+\frac {5\,B\,b^2\,c\,n\,x}{6\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}-\frac {B\,a\,b\,d\,n\,x}{6\,g^4\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^3}+\frac {B\,b^3\,n\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{3\,d\,g^4\,{\left (a\,d-b\,c\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*log(e*((a + b*x)/(c + d*x))^n))/(c*g + d*g*x)^4,x)

[Out]

(B*a^2*d*n)/(9*g^4*(a*d - b*c)^2*(c + d*x)^3) - (A*a^2*d)/(3*g^4*(a*d - b*c)^2*(c + d*x)^3) - (A*b^2*c^2)/(3*d
*g^4*(a*d - b*c)^2*(c + d*x)^3) - (B*log(e*((a + b*x)/(c + d*x))^n))/(3*d*g^4*(c + d*x)^3) + (B*b^3*n*atan((a*
d*1i + b*c*1i + b*d*x*2i)/(a*d - b*c))*2i)/(3*d*g^4*(a*d - b*c)^3) + (2*A*a*b*c)/(3*g^4*(a*d - b*c)^2*(c + d*x
)^3) + (B*b^2*d*n*x^2)/(3*g^4*(a*d - b*c)^2*(c + d*x)^3) - (7*B*a*b*c*n)/(18*g^4*(a*d - b*c)^2*(c + d*x)^3) +
(11*B*b^2*c^2*n)/(18*d*g^4*(a*d - b*c)^2*(c + d*x)^3) + (5*B*b^2*c*n*x)/(6*g^4*(a*d - b*c)^2*(c + d*x)^3) - (B
*a*b*d*n*x)/(6*g^4*(a*d - b*c)^2*(c + d*x)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*((b*x+a)/(d*x+c))**n))/(d*g*x+c*g)**4,x)

[Out]

Timed out

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