3.2.99 \(\int \frac {1}{(a g+b g x)^2 (A+B \log (\frac {e (c+d x)}{a+b x}))^2} \, dx\) [199]

Optimal. Leaf size=104 \[ -\frac {e^{-\frac {A}{B}} \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B^2 (b c-a d) e g^2}+\frac {c+d x}{B (b c-a d) g^2 (a+b x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2552, 2334, 2336, 2209} \begin {gather*} \frac {c+d x}{B g^2 (a+b x) (b c-a d) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )}-\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^2 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)])^2),x]

[Out]

-(ExpIntegralEi[(A + B*Log[(e*(c + d*x))/(a + b*x)])/B]/(B^2*(b*c - a*d)*e*E^(A/B)*g^2)) + (c + d*x)/(B*(b*c -
a*d)*g^2*(a + b*x)*(A + B*Log[(e*(c + d*x))/(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 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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 )^2} \, 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 )^2} \, dx\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 88, normalized size = 0.85 \begin {gather*} \frac {\frac {e^{-\frac {A}{B}} \text {Ei}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{e}-\frac {B (c+d x)}{(a+b x) \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}}{B^2 (-b c+a d) 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)])^2),x]

[Out]

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

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Maple [A]
time = 2.72, size = 138, normalized size = 1.33

method	result	size
risch	\(-\frac {d x +c}{\left (a d -c b \right ) B \left (b x +a \right ) g^{2} \left (A +B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )\right )}-\frac {{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{g^{2} B^{2} e \left (a d -c b \right )}\)	\(121\)
derivativedivides	\(\frac {-\frac {\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B^{2}}\)	\(138\)
default	\(\frac {-\frac {\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \left (a d -c b \right ) g^{2} B^{2}}\)	\(138\)

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)))^2,x,method=_RETURNVERBOSE)

[Out]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 205, normalized size = 1.97 \begin {gather*} \frac {{\left (B d x + B c\right )} e^{\left (\frac {A}{B} + 1\right )} - {\left (A b x + A a + {\left (B b x + B a\right )} \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right )\right )} \operatorname {log\_integral}\left (\frac {{\left (d x + c\right )} e^{\left (\frac {A}{B} + 1\right )}}{b x + a}\right )}{{\left ({\left (B^{3} b^{2} c - B^{3} a b d\right )} g^{2} x + {\left (B^{3} a b c - B^{3} a^{2} d\right )} g^{2}\right )} e^{\left (\frac {A}{B} + 1\right )} \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + {\left ({\left (A B^{2} b^{2} c - A B^{2} a b d\right )} g^{2} x + {\left (A B^{2} a b c - A B^{2} a^{2} d\right )} g^{2}\right )} e^{\left (\frac {A}{B} + 1\right )}} \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)))^2,x, algorithm="fricas")

[Out]

((B*d*x + B*c)*e^(A/B + 1) - (A*b*x + A*a + (B*b*x + B*a)*log((d*x + c)*e/(b*x + a)))*log_integral((d*x + c)*e
^(A/B + 1)/(b*x + a)))/(((B^3*b^2*c - B^3*a*b*d)*g^2*x + (B^3*a*b*c - B^3*a^2*d)*g^2)*e^(A/B + 1)*log((d*x + c
)*e/(b*x + a)) + ((A*B^2*b^2*c - A*B^2*a*b*d)*g^2*x + (A*B^2*a*b*c - A*B^2*a^2*d)*g^2)*e^(A/B + 1))

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

[Out]

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

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Giac [A]
time = 4.46, size = 152, normalized size = 1.46 \begin {gather*} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} {\left (\frac {d x e + c e}{{\left (B^{2} g^{2} \log \left (\frac {d x e + c e}{b x + a}\right ) + A B g^{2}\right )} {\left (b x + a\right )}} - \frac {{\rm Ei}\left (\frac {A}{B} + \log \left (\frac {d x e + c e}{b x + a}\right )\right ) e^{\left (-\frac {A}{B}\right )}}{B^{2} g^{2}}\right )} \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)))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \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 )}^2} \,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)))^2),x)

[Out]

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

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