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

Optimal. Leaf size=45 \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{2 B n (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6686} \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{2 B n (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2/(2*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 {A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )}{(a+b x) (c+d x)} \, dx &=\frac {\left (A+B \log \left (e (a+b x)^n (c+d x)^{-n}\right )\right )^2}{2 B (b c-a d) n}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 43, normalized size = 0.96 \[ \frac {\left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )^2}{2 (b B c n-a B d n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.13, size = 72, normalized size = 1.60 \[ \frac {B n \log \left (b x + a\right )^{2} + B n \log \left (d x + c\right )^{2} + 2 \, {\left (B \log \relax (e) + A\right )} \log \left (b x + a\right ) - 2 \, {\left (B n \log \left (b x + a\right ) + B \log \relax (e) + A\right )} \log \left (d x + c\right )}{2 \, {\left (b c - a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.61, size = 1152, normalized size = 25.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.20, size = 151, normalized size = 3.36 \[ B {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} \log \left (\frac {{\left (b x + a\right )}^{n} e}{{\left (d x + c\right )}^{n}}\right ) + A {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} - \frac {{\left (e n \log \left (b x + a\right )^{2} - 2 \, e n \log \left (b x + a\right ) \log \left (d x + c\right ) + e n \log \left (d x + c\right )^{2}\right )} B}{2 \, {\left (b c - a d\right )} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.67, size = 71, normalized size = 1.58 \[ -\frac {B\,{\ln \left (\frac {e\,{\left (a+b\,x\right )}^n}{{\left (c+d\,x\right )}^n}\right )}^2-A\,n\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{2\,n\,\left (a\,d-b\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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