3.2.46 \(\int \frac {1}{(a g+b g x)^3 (A+B \log (\frac {e (a+b x)^2}{(c+d x)^2}))^2} \, dx\) [146]

Optimal. Leaf size=266 \[ \frac {d e^{\frac {A}{2 B}} \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} (c+d x) \text {Ei}\left (\frac {-A-B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{2 B}\right )}{4 B^2 (b c-a d)^2 g^3 (a+b x)}-\frac {b e e^{A/B} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{B}\right )}{2 B^2 (b c-a d)^2 g^3}+\frac {d (c+d x)}{2 B (b c-a d)^2 g^3 (a+b x) \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )}-\frac {b (c+d x)^2}{2 B (b c-a d)^2 g^3 (a+b x)^2 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 263, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2550, 2395, 2343, 2347, 2209} \begin {gather*} \frac {d e^{\frac {A}{2 B}} (c+d x) \sqrt {\frac {e (a+b x)^2}{(c+d x)^2}} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{2 B}\right )}{4 B^2 g^3 (a+b x) (b c-a d)^2}-\frac {b e e^{A/B} \text {Ei}\left (-\frac {A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )}{B}\right )}{2 B^2 g^3 (b c-a d)^2}-\frac {b (c+d x)^2}{2 B g^3 (a+b x)^2 (b c-a d)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )}+\frac {d (c+d x)}{2 B g^3 (a+b x) (b c-a d)^2 \left (B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )+A\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*e*E^(A/B)*ExpIntegralEi[-((A + B*Log[(e*(a + b*x)^2)/(c + d*x)^2])/B)])/(B^2*(b*c - a*d)^2*g^3) + (d*E
^(A/(2*B))*Sqrt[(e*(a + b*x)^2)/(c + d*x)^2]*(c + d*x)*ExpIntegralEi[-1/2*(A + B*Log[(e*(a + b*x)^2)/(c + d*x)
^2])/B])/(4*B^2*(b*c - a*d)^2*g^3*(a + b*x)) + (d*(c + d*x))/(2*B*(b*c - a*d)^2*g^3*(a + b*x)*(A + B*Log[(e*(a
+ b*x)^2)/(c + d*x)^2])) - (b*(c + d*x)^2)/(2*B*(b*c - a*d)^2*g^3*(a + b*x)^2*(A + B*Log[(e*(a + b*x)^2)/(c +
d*x)^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 2343

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

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 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)^2}{(c+d x)^2}\right )\right )^2} \, dx &=\int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx\\ \end {align*}

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Mathematica [F]
time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(a g+b g x)^3 \left (A+B \log \left (\frac {e (a+b x)^2}{(c+d x)^2}\right )\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 1.14, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b g x +a g \right )^{3} \left (A +B \ln \left (\frac {e \left (b x +a \right )^{2}}{\left (d x +c \right )^{2}}\right )\right )^{2}}\, dx\]

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)^2/(d*x+c)^2))^2,x)

[Out]

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

[Out]

-1/2*(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 + 2*((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) - 2*((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(1/2*(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 + 2*((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) - 2*((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 [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)^2/(d*x+c)^2))^2,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \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 )}^2}{{\left (c+d\,x\right )}^2}\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*(a + b*x)^2)/(c + d*x)^2))^2),x)

[Out]

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

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