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

Optimal. Leaf size=36 \[ \frac{1}{n (b c-a d) \left (1-e \left (\frac{a+b x}{c+d x}\right )^n\right )} \]

[Out]

1/((b*c - a*d)*n*(1 - e*((a + b*x)/(c + d*x))^n))

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Rubi [A]  time = 0.366682, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 53, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {12, 6686} \[ \frac{1}{n (b c-a d) \left (1-e \left (\frac{a+b x}{c+d x}\right )^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/((b*c - a*d)*n*(1 - e*((a + b*x)/(c + d*x))^n))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.103027, size = 35, normalized size = 0.97 \[ \frac{1}{n (a d-b c) \left (e \left (\frac{a+b x}{c+d x}\right )^n-1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/((-(b*c) + a*d)*n*(-1 + e*((a + b*x)/(c + d*x))^n))

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Maple [A]  time = 0.171, size = 56, normalized size = 1.6 \begin{align*}{\frac{e}{n \left ( ad-bc \right ) }{{\rm e}^{n\ln \left ({\frac{bx+a}{dx+c}} \right ) }} \left ( e{{\rm e}^{n\ln \left ({\frac{bx+a}{dx+c}} \right ) }}-1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e/n/(a*d-b*c)*exp(n*ln((b*x+a)/(d*x+c)))/(e*exp(n*ln((b*x+a)/(d*x+c)))-1)

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Maxima [A]  time = 1.04494, size = 70, normalized size = 1.94 \begin{align*} -\frac{{\left (b x + a\right )}^{n} e}{{\left (b c e n - a d e n\right )}{\left (b x + a\right )}^{n} -{\left (b c n - a d n\right )}{\left (d x + c\right )}^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x + a)^n*e/((b*c*e*n - a*d*e*n)*(b*x + a)^n - (b*c*n - a*d*n)*(d*x + c)^n)

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Fricas [A]  time = 2.47833, size = 84, normalized size = 2.33 \begin{align*} -\frac{1}{{\left (b c - a d\right )} e n \left (\frac{b x + a}{d x + c}\right )^{n} -{\left (b c - a d\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e \left (\frac{b x + a}{d x + c}\right )^{n}}{{\left (b x + a\right )}{\left (d x + c\right )}{\left (e \left (\frac{b x + a}{d x + c}\right )^{n} - 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e*((b*x + a)/(d*x + c))^n/((b*x + a)*(d*x + c)*(e*((b*x + a)/(d*x + c))^n - 1)^2), x)