3.3.46 \(\int \frac {1}{(a+b x)^2 \log (e (\frac {a+b x}{c+d x})^n)} \, dx\) [246]

Optimal. Leaf size=71 \[ \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \text {Ei}\left (-\frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2549, 2347, 2209} \begin {gather*} \frac {(c+d x) \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} \text {Ei}\left (-\frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{n (a+b x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2549

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/b)^m, Subst[Int[x^m*((A + B*Log[e*x^n])^p/(b - d*x)^(m + 2)), x]
, x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && NeQ[b*c - a*d, 0] && IntegersQ[m,
p] && EqQ[b*f - a*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 71, normalized size = 1.00 \begin {gather*} \frac {\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )^{\frac {1}{n}} (c+d x) \text {Ei}\left (-\frac {\log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{n}\right )}{(b c-a d) n (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right )^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 38, normalized size = 0.54 \begin {gather*} \frac {e^{\frac {1}{n}} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (-\frac {1}{n}\right )}}{b x + a}\right )}{{\left (b c - a d\right )} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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