3.2.13 \(\int \frac {1}{(a g+b g x)^3 (A+B \log (\frac {e (a+b x)}{c+d x}))} \, dx\) [113]

Optimal. Leaf size=107 \[ \frac {b e^2 e^{\frac {2 A}{B}} \text {Ei}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )}{B (b c-a d)^2 g^3}-\frac {d e e^{A/B} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )}{B (b c-a d)^2 g^3} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2550, 2395, 2346, 2209} \begin {gather*} \frac {b e^2 e^{\frac {2 A}{B}} \text {Ei}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )}{B g^3 (b c-a d)^2}-\frac {d e e^{A/B} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )}{B g^3 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2395

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

Rule 2550

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/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] && EqQ[n + mn, 0] && IGtQ[n, 0]
&& 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 g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx &=\int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )} \, dx\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 89, normalized size = 0.83 \begin {gather*} \frac {e e^{A/B} \left (b e e^{A/B} \text {Ei}\left (-\frac {2 \left (A+B \log \left (\frac {e (a+b x)}{c+d x}\right )\right )}{B}\right )-d \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)}{c+d x}\right )}{B}\right )\right )}{B (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*(a + b*x))/(c + d*x)])),x]

[Out]

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

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Maple [A]
time = 9.96, size = 117, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {e \left (\frac {d \,{\mathrm e}^{\frac {A}{B}} \expIntegral \left (1, \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{B}-\frac {b e \,{\mathrm e}^{\frac {2 A}{B}} \expIntegral \left (1, 2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{B}\right )}{\left (a d -c b \right )^{2} g^{3}}\)	\(117\)
default	\(\frac {e \left (\frac {d \,{\mathrm e}^{\frac {A}{B}} \expIntegral \left (1, \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{B}-\frac {b e \,{\mathrm e}^{\frac {2 A}{B}} \expIntegral \left (1, 2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{B}\right )}{\left (a d -c b \right )^{2} g^{3}}\)	\(117\)
risch	\(\frac {e d \,{\mathrm e}^{\frac {A}{B}} \expIntegral \left (1, \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {A}{B}\right )}{g^{3} \left (a d -c b \right )^{2} B}-\frac {e^{2} b \,{\mathrm e}^{\frac {2 A}{B}} \expIntegral \left (1, 2 \ln \left (\frac {b e}{d}+\frac {\left (a d -c b \right ) e}{d \left (d x +c \right )}\right )+\frac {2 A}{B}\right )}{g^{3} \left (a d -c b \right )^{2} B}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 122, normalized size = 1.14 \begin {gather*} -\frac {d 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 ) - b e^{\left (\frac {2 \, A}{B} + 2\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 (B b^{2} c^{2} - 2 \, B a b c d + B a^{2} d^{2}\right )} g^{3}} \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*(b*x+a)/(d*x+c))),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{A a^{3} + 3 A a^{2} b x + 3 A a b^{2} x^{2} + A b^{3} x^{3} + B a^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a^{2} b x \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + 3 B a b^{2} x^{2} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )} + B b^{3} x^{3} \log {\left (\frac {a e}{c + d x} + \frac {b e x}{c + d x} \right )}}\, dx}{g^{3}} \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*(b*x+a)/(d*x+c))),x)

[Out]

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

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

[Out]

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

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

[Out]

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

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