3.2.100 \(\int \frac {1}{(a g+b g x)^3 (A+B \log (\frac {e (c+d x)}{a+b x}))^2} \, dx\) [200]

Optimal. Leaf size=159 \[ \frac {d e^{-\frac {A}{B}} \text {Ei}\left (\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{B}\right )}{B^2 (b c-a d)^2 e g^3}-\frac {2 b e^{-\frac {2 A}{B}} \text {Ei}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B^2 (b c-a d)^2 e^2 g^3}+\frac {c+d x}{B (b c-a d) g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )} \]

[Out]

d*Ei((A+B*ln(e*(d*x+c)/(b*x+a)))/B)/B^2/(-a*d+b*c)^2/e/exp(A/B)/g^3-2*b*Ei(2*(A+B*ln(e*(d*x+c)/(b*x+a)))/B)/B^
2/(-a*d+b*c)^2/e^2/exp(2*A/B)/g^3+(d*x+c)/B/(-a*d+b*c)/g^3/(b*x+a)^2/(A+B*ln(e*(d*x+c)/(b*x+a)))

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Rubi [A]
time = 0.16, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2552, 2357, 2367, 2336, 2209, 2346} \begin {gather*} -\frac {2 b e^{-\frac {2 A}{B}} \text {Ei}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{B^2 e^2 g^3 (b c-a d)^2}+\frac {d 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^3 (b c-a d)^2}+\frac {c+d x}{B g^3 (a+b x)^2 (b c-a d) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])^2),x]

[Out]

(d*ExpIntegralEi[(A + B*Log[(e*(c + d*x))/(a + b*x)])/B])/(B^2*(b*c - a*d)^2*e*E^(A/B)*g^3) - (2*b*ExpIntegral
Ei[(2*(A + B*Log[(e*(c + d*x))/(a + b*x)]))/B])/(B^2*(b*c - a*d)^2*e^2*E^((2*A)/B)*g^3) + (c + d*x)/(B*(b*c -
a*d)*g^3*(a + b*x)^2*(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 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 2346

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[x*(d + e*x)^q*((a
+ b*Log[c*x^n])^(p + 1)/(b*n*(p + 1))), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[d*(q/(b*n*(p + 1))), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 2367

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

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)^3 \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)^3 \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 = 0.31, size = 135, normalized size = 0.85 \begin {gather*} \frac {\frac {d e^{-\frac {A}{B}} \text {Ei}\left (\frac {A}{B}+\log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{e}-\frac {2 b e^{-\frac {2 A}{B}} \text {Ei}\left (\frac {2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}{B}\right )}{e^2}+\frac {B (b c-a d) (c+d x)}{(a+b x)^2 \left (A+B \log \left (\frac {e (c+d x)}{a+b x}\right )\right )}}{B^2 (b c-a d)^2 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a*g + b*g*x)^3*(A + B*Log[(e*(c + d*x))/(a + b*x)])^2),x]

[Out]

((d*ExpIntegralEi[A/B + Log[(e*(c + d*x))/(a + b*x)]])/(e*E^(A/B)) - (2*b*ExpIntegralEi[(2*(A + B*Log[(e*(c +
d*x))/(a + b*x)]))/B])/(e^2*E^((2*A)/B)) + (B*(b*c - a*d)*(c + d*x))/((a + b*x)^2*(A + B*Log[(e*(c + d*x))/(a
+ b*x)])))/(B^2*(b*c - a*d)^2*g^3)

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Maple [A]
time = 3.96, size = 268, normalized size = 1.69

method	result	size
derivativedivides	\(-\frac {\frac {b \left (-\frac {\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )^{2}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-2 \,{\mathrm e}^{-\frac {2 A}{B}} \expIntegral \left (1, -2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )\right )}{B^{2}}-\frac {e d \left (-\frac {\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )\right )}{B^{2}}}{e^{2} \left (a d -c b \right )^{2} g^{3}}\)	\(268\)
default	\(-\frac {\frac {b \left (-\frac {\left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )^{2}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-2 \,{\mathrm e}^{-\frac {2 A}{B}} \expIntegral \left (1, -2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )\right )}{B^{2}}-\frac {e d \left (-\frac {\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}}{\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )+\frac {A}{B}}-{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )\right )}{B^{2}}}{e^{2} \left (a d -c b \right )^{2} g^{3}}\)	\(268\)
risch	\(-\frac {d x +c}{\left (a d -c b \right ) B \left (b x +a \right )^{2} g^{3} \left (A +B \ln \left (\frac {e \left (d x +c \right )}{b x +a}\right )\right )}+\frac {b c d \,{\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \,g^{3} B^{2} \left (a d -c b \right )^{3}}-\frac {2 c \,b^{2} {\mathrm e}^{-\frac {2 A}{B}} \expIntegral \left (1, -2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{e^{2} g^{3} B^{2} \left (a d -c b \right )^{3}}-\frac {a \,d^{2} {\mathrm e}^{-\frac {A}{B}} \expIntegral \left (1, -\ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {A}{B}\right )}{e \,g^{3} B^{2} \left (a d -c b \right )^{3}}+\frac {2 d b a \,{\mathrm e}^{-\frac {2 A}{B}} \expIntegral \left (1, -2 \ln \left (\frac {d e}{b}-\frac {e \left (a d -c b \right )}{b \left (b x +a \right )}\right )-\frac {2 A}{B}\right )}{e^{2} g^{3} B^{2} \left (a d -c b \right )^{3}}\)	\(338\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*g*x+a*g)^3/(A+B*ln(e*(d*x+c)/(b*x+a)))^2,x,method=_RETURNVERBOSE)

[Out]

-1/e^2/(a*d-b*c)^2/g^3*(b/B^2*(-(d*e/b-e*(a*d-b*c)/b/(b*x+a))^2/(ln(d*e/b-e*(a*d-b*c)/b/(b*x+a))+A/B)-2*exp(-2
*A/B)*Ei(1,-2*ln(d*e/b-e*(a*d-b*c)/b/(b*x+a))-2*A/B))-e*d/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)^3/(A+B*log(e*(d*x+c)/(b*x+a)))^2,x, algorithm="maxima")

[Out]

(d*x + c)/((a^2*b*c*g^3 - a^3*d*g^3)*A*B + (a^2*b*c*g^3 - a^3*d*g^3)*B^2 + ((b^3*c*g^3 - a*b^2*d*g^3)*A*B + (b
^3*c*g^3 - a*b^2*d*g^3)*B^2)*x^2 + 2*((a*b^2*c*g^3 - a^2*b*d*g^3)*A*B + (a*b^2*c*g^3 - a^2*b*d*g^3)*B^2)*x - (
(b^3*c*g^3 - a*b^2*d*g^3)*B^2*x^2 + 2*(a*b^2*c*g^3 - a^2*b*d*g^3)*B^2*x + (a^2*b*c*g^3 - a^3*d*g^3)*B^2)*log(b
*x + a) + ((b^3*c*g^3 - a*b^2*d*g^3)*B^2*x^2 + 2*(a*b^2*c*g^3 - a^2*b*d*g^3)*B^2*x + (a^2*b*c*g^3 - a^3*d*g^3)
*B^2)*log(d*x + c)) - integrate(-(b*d*x + 2*b*c - a*d)/(((b^4*c*g^3 - a*b^3*d*g^3)*A*B + (b^4*c*g^3 - a*b^3*d*
g^3)*B^2)*x^3 + (a^3*b*c*g^3 - a^4*d*g^3)*A*B + (a^3*b*c*g^3 - a^4*d*g^3)*B^2 + 3*((a*b^3*c*g^3 - a^2*b^2*d*g^
3)*A*B + (a*b^3*c*g^3 - a^2*b^2*d*g^3)*B^2)*x^2 + 3*((a^2*b^2*c*g^3 - a^3*b*d*g^3)*A*B + (a^2*b^2*c*g^3 - a^3*
b*d*g^3)*B^2)*x - ((b^4*c*g^3 - a*b^3*d*g^3)*B^2*x^3 + 3*(a*b^3*c*g^3 - a^2*b^2*d*g^3)*B^2*x^2 + 3*(a^2*b^2*c*
g^3 - a^3*b*d*g^3)*B^2*x + (a^3*b*c*g^3 - a^4*d*g^3)*B^2)*log(b*x + a) + ((b^4*c*g^3 - a*b^3*d*g^3)*B^2*x^3 +
3*(a*b^3*c*g^3 - a^2*b^2*d*g^3)*B^2*x^2 + 3*(a^2*b^2*c*g^3 - a^3*b*d*g^3)*B^2*x + (a^3*b*c*g^3 - a^4*d*g^3)*B^
2)*log(d*x + c)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (158) = 316\).
time = 0.36, size = 550, normalized size = 3.46 \begin {gather*} \frac {{\left (B b c^{2} - B a c d + {\left (B b c d - B a d^{2}\right )} x\right )} e^{\left (\frac {2 \, A}{B} + 2\right )} - 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b + {\left (B b^{3} x^{2} + 2 \, B a b^{2} x + B a^{2} b\right )} \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right )\right )} \operatorname {log\_integral}\left (\frac {{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} e^{\left (\frac {2 \, A}{B} + 2\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + {\left ({\left (B b^{2} d x^{2} + 2 \, B a b d x + B a^{2} d\right )} e^{\left (\frac {A}{B} + 1\right )} \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + {\left (A b^{2} d x^{2} + 2 \, A a b d x + A a^{2} d\right )} e^{\left (\frac {A}{B} + 1\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^{4} c^{2} - 2 \, B^{3} a b^{3} c d + B^{3} a^{2} b^{2} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (B^{3} a b^{3} c^{2} - 2 \, B^{3} a^{2} b^{2} c d + B^{3} a^{3} b d^{2}\right )} g^{3} x + {\left (B^{3} a^{2} b^{2} c^{2} - 2 \, B^{3} a^{3} b c d + B^{3} a^{4} d^{2}\right )} g^{3}\right )} e^{\left (\frac {2 \, A}{B} + 2\right )} \log \left (\frac {{\left (d x + c\right )} e}{b x + a}\right ) + {\left ({\left (A B^{2} b^{4} c^{2} - 2 \, A B^{2} a b^{3} c d + A B^{2} a^{2} b^{2} d^{2}\right )} g^{3} x^{2} + 2 \, {\left (A B^{2} a b^{3} c^{2} - 2 \, A B^{2} a^{2} b^{2} c d + A B^{2} a^{3} b d^{2}\right )} g^{3} x + {\left (A B^{2} a^{2} b^{2} c^{2} - 2 \, A B^{2} a^{3} b c d + A B^{2} a^{4} d^{2}\right )} g^{3}\right )} e^{\left (\frac {2 \, A}{B} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a)))^2,x, algorithm="fricas")

[Out]

((B*b*c^2 - B*a*c*d + (B*b*c*d - B*a*d^2)*x)*e^(2*A/B + 2) - 2*(A*b^3*x^2 + 2*A*a*b^2*x + A*a^2*b + (B*b^3*x^2
+ 2*B*a*b^2*x + B*a^2*b)*log((d*x + c)*e/(b*x + a)))*log_integral((d^2*x^2 + 2*c*d*x + c^2)*e^(2*A/B + 2)/(b^
2*x^2 + 2*a*b*x + a^2)) + ((B*b^2*d*x^2 + 2*B*a*b*d*x + B*a^2*d)*e^(A/B + 1)*log((d*x + c)*e/(b*x + a)) + (A*b
^2*d*x^2 + 2*A*a*b*d*x + A*a^2*d)*e^(A/B + 1))*log_integral((d*x + c)*e^(A/B + 1)/(b*x + a)))/(((B^3*b^4*c^2 -
2*B^3*a*b^3*c*d + B^3*a^2*b^2*d^2)*g^3*x^2 + 2*(B^3*a*b^3*c^2 - 2*B^3*a^2*b^2*c*d + B^3*a^3*b*d^2)*g^3*x + (B
^3*a^2*b^2*c^2 - 2*B^3*a^3*b*c*d + B^3*a^4*d^2)*g^3)*e^(2*A/B + 2)*log((d*x + c)*e/(b*x + a)) + ((A*B^2*b^4*c^
2 - 2*A*B^2*a*b^3*c*d + A*B^2*a^2*b^2*d^2)*g^3*x^2 + 2*(A*B^2*a*b^3*c^2 - 2*A*B^2*a^2*b^2*c*d + A*B^2*a^3*b*d^
2)*g^3*x + (A*B^2*a^2*b^2*c^2 - 2*A*B^2*a^3*b*c*d + A*B^2*a^4*d^2)*g^3)*e^(2*A/B + 2))

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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*g*x+a*g)**3/(A+B*ln(e*(d*x+c)/(b*x+a)))**2,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (158) = 316\).
time = 5.14, size = 317, normalized size = 1.99 \begin {gather*} {\left (\frac {d {\rm Ei}\left (\frac {A}{B} + \log \left (\frac {d x e + c e}{b x + a}\right )\right ) e^{\left (-\frac {A}{B} + 1\right )}}{B^{2} b c g^{3} e - B^{2} a d g^{3} e} - \frac {2 \, b {\rm Ei}\left (\frac {2 \, A}{B} + 2 \, \log \left (\frac {d x e + c e}{b x + a}\right )\right ) e^{\left (-\frac {2 \, A}{B}\right )}}{B^{2} b c g^{3} e - B^{2} a d g^{3} e} - \frac {\frac {{\left (d x e + c e\right )} d e}{b x + a} - \frac {{\left (d x e + c e\right )}^{2} b}{{\left (b x + a\right )}^{2}}}{B^{2} b c g^{3} e \log \left (\frac {d x e + c e}{b x + a}\right ) - B^{2} a d g^{3} e \log \left (\frac {d x e + c e}{b x + a}\right ) + A B b c g^{3} e - A B a d g^{3} e}\right )} {\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*g*x+a*g)^3/(A+B*log(e*(d*x+c)/(b*x+a)))^2,x, algorithm="giac")

[Out]

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

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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 )}^3\,{\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)^3*(A + B*log((e*(c + d*x))/(a + b*x)))^2),x)

[Out]

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

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