3.232 \(\int \frac{1}{(a+b x) (c+d x) (A+B \log (e (a+b x)^n (c+d x)^{-n}))^2} \, dx\)

Optimal. Leaf size=43 \[ -\frac{1}{B n (b c-a d) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )} \]

[Out]

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

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Rubi [A]  time = 0.121829, antiderivative size = 43, normalized size of antiderivative = 1., 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} \[ -\frac{1}{B n (b c-a d) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2),x]

[Out]

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

Mathematica [A]  time = 0.0203908, size = 41, normalized size = 0.95 \[ -\frac{1}{(b B c n-a B d n) \left (B \log \left (e (a+b x)^n (c+d x)^{-n}\right )+A\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)*(A + B*Log[(e*(a + b*x)^n)/(c + d*x)^n])^2),x]

[Out]

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

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Maple [C]  time = 0.399, size = 366, normalized size = 8.5 \begin{align*} 2\,{\frac{1}{Bn \left ( ad-bc \right ) } \left ( 2\,A+2\,B\ln \left ( e \right ) +2\,B\ln \left ( \left ( bx+a \right ) ^{n} \right ) -2\,B\ln \left ( \left ( dx+c \right ) ^{n} \right ) -iB\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ){\it csgn} \left ({\frac{i}{ \left ( dx+c \right ) ^{n}}} \right ){\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) +iB\pi \,{\it csgn} \left ( i \left ( bx+a \right ) ^{n} \right ) \left ({\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{2}+iB\pi \,{\it csgn} \left ({\frac{i}{ \left ( dx+c \right ) ^{n}}} \right ) \left ({\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{2}-iB\pi \, \left ({\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{3}-iB\pi \,{\it csgn} \left ( ie \right ){\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ){\it csgn} \left ({\frac{ie \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) +iB\pi \,{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ({\frac{ie \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{2}+iB\pi \,{\it csgn} \left ({\frac{i \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \left ({\it csgn} \left ({\frac{ie \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{2}-iB\pi \, \left ({\it csgn} \left ({\frac{ie \left ( bx+a \right ) ^{n}}{ \left ( dx+c \right ) ^{n}}} \right ) \right ) ^{3} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.80463, size = 109, normalized size = 2.53 \begin{align*} -\frac{1}{{\left (b c n - a d n\right )} B^{2} \log \left ({\left (b x + a\right )}^{n}\right ) -{\left (b c n - a d n\right )} B^{2} \log \left ({\left (d x + c\right )}^{n}\right ) +{\left (b c n - a d n\right )} A B +{\left (b c n \log \left (e\right ) - a d n \log \left (e\right )\right )} B^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.507261, size = 185, normalized size = 4.3 \begin{align*} -\frac{1}{{\left (B^{2} b c - B^{2} a d\right )} n^{2} \log \left (b x + a\right ) -{\left (B^{2} b c - B^{2} a d\right )} n^{2} \log \left (d x + c\right ) +{\left (B^{2} b c - B^{2} a d\right )} n \log \left (e\right ) +{\left (A B b c - A B a d\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.29514, size = 128, normalized size = 2.98 \begin{align*} -\frac{1}{B^{2} b c n^{2} \log \left (b x + a\right ) - B^{2} a d n^{2} \log \left (b x + a\right ) - B^{2} b c n^{2} \log \left (d x + c\right ) + B^{2} a d n^{2} \log \left (d x + c\right ) + A B b c n + B^{2} b c n - A B a d n - B^{2} a d n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(A+B*log(e*(b*x+a)^n/((d*x+c)^n)))^2,x, algorithm="giac")

[Out]

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