3.2.94 \(\int \frac {1}{(a g+b g x)^2 (A+B \log (\frac {e (c+d x)}{a+b x}))} \, dx\) [194]

Optimal. Leaf size=53 \[ -\frac {e^{-\frac {A}{B}} \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B (b c-a d) e g^2} \]

[Out]

-Ei((A+B*ln(e*(d*x+c)/(b*x+a)))/B)/B/(-a*d+b*c)/e/exp(A/B)/g^2

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Rubi [A]
time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2552, 2336, 2209} \begin {gather*} -\frac {e^{-\frac {A}{B}} \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B e g^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2336

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

Rule 2552

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(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] && EqQ[n + mn, 0] && IGtQ[n, 0] && NeQ
[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 50, normalized size = 0.94 \begin {gather*} \frac {e^{-\frac {A}{B}} \text {Ei}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B (-b c+a d) e g^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 4.18, size = 69, normalized size = 1.30

method	result	size
derivativedivides	\(-\frac {{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B}\)	\(69\)
default	\(-\frac {{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B}\)	\(69\)
risch	\(-\frac {{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*g*x+a*g)^2/(A+B*ln(e*(d*x+c)/(b*x+a))),x,method=_RETURNVERBOSE)

[Out]

-1/e/(a*d-b*c)/g^2/B*exp(-A/B)*Ei(1,-ln(d*e/b-e*(a*d-b*c)/b/(b*x+a))-A/B)

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

[Out]

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

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Fricas [A]
time = 0.37, size = 48, normalized size = 0.91 \begin {gather*} -\frac {e^{\left (-\frac {A}{B} - 1\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (\frac {A}{B} + 1\right )}}{b x + a}\right )}{{\left (B b c - B a d\right )} g^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{A a^{2} + 2 A a b x + A b^{2} x^{2} + B a^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + 2 B a b x \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )} + B b^{2} x^{2} \log {\left (\frac {c e}{a + b x} + \frac {d e x}{a + b x} \right )}}\, dx}{g^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (a\,g+b\,g\,x\right )}^2\,\left (A+B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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