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

Optimal. Leaf size=45 \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^4}{4 B n (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6686} \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^4}{4 B n (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.96 \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^4}{4 (b B c n-a B d n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.80, size = 375, normalized size = 8.33 \[ \frac {B^{3} n^{3} \log \left (b x + a\right )^{4} + B^{3} n^{3} \log \left (d x + c\right )^{4} + 4 \, {\left (B^{3} n^{2} \log \relax (e) + A B^{2} n^{2}\right )} \log \left (b x + a\right )^{3} - 4 \, {\left (B^{3} n^{3} \log \left (b x + a\right ) + B^{3} n^{2} \log \relax (e) + A B^{2} n^{2}\right )} \log \left (d x + c\right )^{3} + 6 \, {\left (B^{3} n \log \relax (e)^{2} + 2 \, A B^{2} n \log \relax (e) + A^{2} B n\right )} \log \left (b x + a\right )^{2} + 6 \, {\left (B^{3} n^{3} \log \left (b x + a\right )^{2} + B^{3} n \log \relax (e)^{2} + 2 \, A B^{2} n \log \relax (e) + A^{2} B n + 2 \, {\left (B^{3} n^{2} \log \relax (e) + A B^{2} n^{2}\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )^{2} + 4 \, {\left (B^{3} \log \relax (e)^{3} + 3 \, A B^{2} \log \relax (e)^{2} + 3 \, A^{2} B \log \relax (e) + A^{3}\right )} \log \left (b x + a\right ) - 4 \, {\left (B^{3} n^{3} \log \left (b x + a\right )^{3} + B^{3} \log \relax (e)^{3} + 3 \, A B^{2} \log \relax (e)^{2} + 3 \, A^{2} B \log \relax (e) + A^{3} + 3 \, {\left (B^{3} n^{2} \log \relax (e) + A B^{2} n^{2}\right )} \log \left (b x + a\right )^{2} + 3 \, {\left (B^{3} n \log \relax (e)^{2} + 2 \, A B^{2} n \log \relax (e) + A^{2} B n\right )} \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{4 \, {\left (b c - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

integrate((B*log((b*x + a)^n*e/(d*x + c)^n) + A)^3/((b*x + a)*(d*x + c)), x)

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maple [C]  time = 35.35, size = 64288, normalized size = 1428.62 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [B]  time = 1.78, size = 766, normalized size = 17.02 \[ B^{3} {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{3} + 3 \, A B^{2} {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{2} + 3 \, A^{2} B {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) - \frac {1}{4} \, B^{3} {\left (\frac {6 \, {\left (e n \log \left (b x + a\right )^{2} - 2 \, e n \log \left (b x + a\right ) \log \left (d x + c\right ) + e n \log \left (d x + c\right )^{2}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )^{2}}{{\left (b c - a d\right )} e} - \frac {\frac {4 \, {\left (e^{2} n^{2} \log \left (b x + a\right )^{3} - 3 \, e^{2} n^{2} \log \left (b x + a\right )^{2} \log \left (d x + c\right ) + 3 \, e^{2} n^{2} \log \left (b x + a\right ) \log \left (d x + c\right )^{2} - e^{2} n^{2} \log \left (d x + c\right )^{3}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{{\left (b c - a d\right )} e} - \frac {e^{3} n^{3} \log \left (b x + a\right )^{4} - 4 \, e^{3} n^{3} \log \left (b x + a\right )^{3} \log \left (d x + c\right ) + 6 \, e^{3} n^{3} \log \left (b x + a\right )^{2} \log \left (d x + c\right )^{2} - 4 \, e^{3} n^{3} \log \left (b x + a\right ) \log \left (d x + c\right )^{3} + e^{3} n^{3} \log \left (d x + c\right )^{4}}{{\left (b c - a d\right )} e^{2}}}{e}\right )} + A^{3} {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} - A B^{2} {\left (\frac {3 \, {\left (e n \log \left (b x + a\right )^{2} - 2 \, e n \log \left (b x + a\right ) \log \left (d x + c\right ) + e n \log \left (d x + c\right )^{2}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right )}{{\left (b c - a d\right )} e} - \frac {e^{2} n^{2} \log \left (b x + a\right )^{3} - 3 \, e^{2} n^{2} \log \left (b x + a\right )^{2} \log \left (d x + c\right ) + 3 \, e^{2} n^{2} \log \left (b x + a\right ) \log \left (d x + c\right )^{2} - e^{2} n^{2} \log \left (d x + c\right )^{3}}{{\left (b c - a d\right )} e^{2}}\right )} - \frac {3 \, {\left (e n \log \left (b x + a\right )^{2} - 2 \, e n \log \left (b x + a\right ) \log \left (d x + c\right ) + e n \log \left (d x + c\right )^{2}\right )} A^{2} B}{2 \, {\left (b c - a d\right )} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^3/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.75, size = 141, normalized size = 3.13 \[ -\frac {\frac {3\,A^2\,B\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2}{2}+A\,B^2\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^3+\frac {B^3\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^4}{4}}{n\,\left (a\,d-b\,c\right )}+\frac {A^3\,\mathrm {atan}\left (\frac {a\,d\,1{}\mathrm {i}+b\,c\,1{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}\right )\,2{}\mathrm {i}}{a\,d-b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**n/((d*x+c)**n)))**3/(b*x+a)/(d*x+c),x)

[Out]

Timed out

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