3.3.18 \(\int (a g+b g x)^{-2-m} (c i+d i x)^m (A+B \log (e (\frac {a+b x}{c+d x})^n))^3 \, dx\) [218]

Optimal. Leaf size=309 \[ -\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m}}{(b c-a d) i^2 (1+m)^4 (c+d x)}-\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) i^2 (1+m)^3 (c+d x)}-\frac {3 B n (a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^2}{(b c-a d) i^2 (1+m)^2 (c+d x)}-\frac {(a+b x) (g (a+b x))^{-2-m} (i (c+d x))^{2+m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3}{(b c-a d) i^2 (1+m) (c+d x)} \]

[Out]

-6*B^3*n^3*(b*x+a)*(g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)/(-a*d+b*c)/i^2/(1+m)^4/(d*x+c)-6*B^2*n^2*(b*x+a)*(g*(b
*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(-a*d+b*c)/i^2/(1+m)^3/(d*x+c)-3*B*n*(b*x+a)*(
g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^2/(-a*d+b*c)/i^2/(1+m)^2/(d*x+c)-(b*x+a)*(
g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^3/(-a*d+b*c)/i^2/(1+m)/(d*x+c)

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Rubi [A]
time = 0.22, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {2563, 2342, 2341} \begin {gather*} -\frac {6 B^2 n^2 (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{i^2 (m+1)^3 (c+d x) (b c-a d)}-\frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^3}{i^2 (m+1) (c+d x) (b c-a d)}-\frac {3 B n (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )^2}{i^2 (m+1)^2 (c+d x) (b c-a d)}-\frac {6 B^3 n^3 (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2}}{i^2 (m+1)^4 (c+d x) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*B^3*n^3*(a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c + d*x))^(2 + m))/((b*c - a*d)*i^2*(1 + m)^4*(c + d*x)) - (6
*B^2*n^2*(a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c + d*x))^(2 + m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/((b*c
- a*d)*i^2*(1 + m)^3*(c + d*x)) - (3*B*n*(a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c + d*x))^(2 + m)*(A + B*Log[e*(
(a + b*x)/(c + d*x))^n])^2)/((b*c - a*d)*i^2*(1 + m)^2*(c + d*x)) - ((a + b*x)*(g*(a + b*x))^(-2 - m)*(i*(c +
d*x))^(2 + m)*(A + B*Log[e*((a + b*x)/(c + d*x))^n])^3)/((b*c - a*d)*i^2*(1 + m)*(c + d*x))

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 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] && GtQ[p, 0]

Rule 2563

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[d^2*((g*((a + b*x)/b))^m/(i^2*(b*c - a*d)*(i*((c + d*x)/d))^
m*((a + b*x)/(c + d*x))^m)), Subst[Int[x^m*(A + B*Log[e*x^n])^p, x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a,
b, c, d, e, f, g, h, i, A, B, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] &
& EqQ[m + q + 2, 0]

Rubi steps

\begin {align*} \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )^3 \, dx &=\int \left (A^3 (218 c+218 d x)^m (a g+b g x)^{-2-m}+3 A^2 B (218 c+218 d x)^m (a g+b g x)^{-2-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 A B^2 (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B^3 (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx\\ &=A^3 \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \, dx+\left (3 A^2 B\right ) \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+\left (3 A B^2\right ) \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+B^3 \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx\\ &=-\frac {A^3 (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m}}{218 (b c-a d) g (1+m)}-\frac {3 A^2 B (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{218 (b c-a d) g (1+m)}-\left (3 A^2 B\right ) \int \frac {218^m n (c+d x)^m (a g+b g x)^{-2-m}}{-1-m} \, dx+\left (3 A B^2\right ) \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+B^3 \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx\\ &=-\frac {A^3 (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m}}{218 (b c-a d) g (1+m)}-\frac {3 A^2 B (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{218 (b c-a d) g (1+m)}+\left (3 A B^2\right ) \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+B^3 \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+\frac {\left (3\ 218^m A^2 B n\right ) \int (c+d x)^m (a g+b g x)^{-2-m} \, dx}{1+m}\\ &=-\frac {3\ 218^m A^2 B n (c+d x)^{1+m} (a g+b g x)^{-1-m}}{(b c-a d) g (1+m)^2}-\frac {A^3 (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m}}{218 (b c-a d) g (1+m)}-\frac {3 A^2 B (218 c+218 d x)^{1+m} (a g+b g x)^{-1-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{218 (b c-a d) g (1+m)}+\left (3 A B^2\right ) \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx+B^3 \int (218 c+218 d x)^m (a g+b g x)^{-2-m} \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right ) \, dx\\ \end {align*}

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Mathematica [A]
time = 4.38, size = 206, normalized size = 0.67 \begin {gather*} -\frac {(g (a+b x))^{-1-m} (c+d x) (i (c+d x))^m \left (A^3 (1+m)^3+3 A^2 B (1+m)^2 n+6 A B^2 (1+m) n^2+6 B^3 n^3+3 B (1+m) \left (A^2 (1+m)^2+2 A B (1+m) n+2 B^2 n^2\right ) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+3 B^2 (1+m)^2 (A+A m+B n) \log ^2\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+B^3 (1+m)^3 \log ^3\left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(b c-a d) g (1+m)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((g*(a + b*x))^(-1 - m)*(c + d*x)*(i*(c + d*x))^m*(A^3*(1 + m)^3 + 3*A^2*B*(1 + m)^2*n + 6*A*B^2*(1 + m)*n^2
+ 6*B^3*n^3 + 3*B*(1 + m)*(A^2*(1 + m)^2 + 2*A*B*(1 + m)*n + 2*B^2*n^2)*Log[e*((a + b*x)/(c + d*x))^n] + 3*B^
2*(1 + m)^2*(A + A*m + B*n)*Log[e*((a + b*x)/(c + d*x))^n]^2 + B^3*(1 + m)^3*Log[e*((a + b*x)/(c + d*x))^n]^3)
)/((b*c - a*d)*g*(1 + m)^4))

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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \left (b g x +a g \right )^{-2-m} \left (d i x +c i \right )^{m} \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*g*x+a*g)^(-2-m)*(d*i*x+c*i)^m*(A+B*ln(e*((b*x+a)/(d*x+c))^n))^3,x)

[Out]

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

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2059 vs. \(2 (303) = 606\).
time = 0.43, size = 2059, normalized size = 6.66 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*B^3*a*c*n^3 + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*c*m^3 + 3*(A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*c*m^2 + 3*(A^3
+ 3*A^2*B + 3*A*B^2 + B^3)*a*c*m + ((B^3*b*d*m^3 + 3*B^3*b*d*m^2 + 3*B^3*b*d*m + B^3*b*d)*n^3*x^2 + (B^3*b*c
+ B^3*a*d + (B^3*b*c + B^3*a*d)*m^3 + 3*(B^3*b*c + B^3*a*d)*m^2 + 3*(B^3*b*c + B^3*a*d)*m)*n^3*x + (B^3*a*c*m^
3 + 3*B^3*a*c*m^2 + 3*B^3*a*c*m + B^3*a*c)*n^3)*log((b*x + a)/(d*x + c))^3 + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a
*c + 6*((A*B^2 + B^3)*a*c*m + (A*B^2 + B^3)*a*c)*n^2 + (6*B^3*b*d*n^3 + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*b*d*m^
3 + 3*(A^3 + 3*A^2*B + 3*A*B^2 + B^3)*b*d*m^2 + 3*(A^3 + 3*A^2*B + 3*A*B^2 + B^3)*b*d*m + (A^3 + 3*A^2*B + 3*A
*B^2 + B^3)*b*d + 6*((A*B^2 + B^3)*b*d*m + (A*B^2 + B^3)*b*d)*n^2 + 3*((A^2*B + 2*A*B^2 + B^3)*b*d*m^2 + 2*(A^
2*B + 2*A*B^2 + B^3)*b*d*m + (A^2*B + 2*A*B^2 + B^3)*b*d)*n)*x^2 + 3*((B^3*a*c*m^2 + 2*B^3*a*c*m + B^3*a*c)*n^
3 + ((A*B^2 + B^3)*a*c*m^3 + 3*(A*B^2 + B^3)*a*c*m^2 + 3*(A*B^2 + B^3)*a*c*m + (A*B^2 + B^3)*a*c)*n^2 + ((B^3*
b*d*m^2 + 2*B^3*b*d*m + B^3*b*d)*n^3 + ((A*B^2 + B^3)*b*d*m^3 + 3*(A*B^2 + B^3)*b*d*m^2 + 3*(A*B^2 + B^3)*b*d*
m + (A*B^2 + B^3)*b*d)*n^2)*x^2 + ((B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m^2 + 2*(B^3*b*c + B^3*a*d)*m)*n^3
+ (((A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d)*m^3 + (A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d + 3*((A*B^2 + B^3)*b*c
+ (A*B^2 + B^3)*a*d)*m^2 + 3*((A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d)*m)*n^2)*x)*log((b*x + a)/(d*x + c))^2 +
3*((A^2*B + 2*A*B^2 + B^3)*a*c*m^2 + 2*(A^2*B + 2*A*B^2 + B^3)*a*c*m + (A^2*B + 2*A*B^2 + B^3)*a*c)*n + (((A^3
+ 3*A^2*B + 3*A*B^2 + B^3)*b*c + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*d)*m^3 + 6*(B^3*b*c + B^3*a*d)*n^3 + (A^3
+ 3*A^2*B + 3*A*B^2 + B^3)*b*c + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*d + 3*((A^3 + 3*A^2*B + 3*A*B^2 + B^3)*b*c
+ (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*d)*m^2 + 6*((A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d + ((A*B^2 + B^3)*b*c + (
A*B^2 + B^3)*a*d)*m)*n^2 + 3*((A^3 + 3*A^2*B + 3*A*B^2 + B^3)*b*c + (A^3 + 3*A^2*B + 3*A*B^2 + B^3)*a*d)*m + 3
*((A^2*B + 2*A*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2 + B^3)*a*d + ((A^2*B + 2*A*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2
+ B^3)*a*d)*m^2 + 2*((A^2*B + 2*A*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2 + B^3)*a*d)*m)*n)*x + 3*(2*(B^3*a*c*m + B^
3*a*c)*n^3 + 2*((A*B^2 + B^3)*a*c*m^2 + 2*(A*B^2 + B^3)*a*c*m + (A*B^2 + B^3)*a*c)*n^2 + (2*(B^3*b*d*m + B^3*b
*d)*n^3 + 2*((A*B^2 + B^3)*b*d*m^2 + 2*(A*B^2 + B^3)*b*d*m + (A*B^2 + B^3)*b*d)*n^2 + ((A^2*B + 2*A*B^2 + B^3)
*b*d*m^3 + 3*(A^2*B + 2*A*B^2 + B^3)*b*d*m^2 + 3*(A^2*B + 2*A*B^2 + B^3)*b*d*m + (A^2*B + 2*A*B^2 + B^3)*b*d)*
n)*x^2 + ((A^2*B + 2*A*B^2 + B^3)*a*c*m^3 + 3*(A^2*B + 2*A*B^2 + B^3)*a*c*m^2 + 3*(A^2*B + 2*A*B^2 + B^3)*a*c*
m + (A^2*B + 2*A*B^2 + B^3)*a*c)*n + (2*(B^3*b*c + B^3*a*d + (B^3*b*c + B^3*a*d)*m)*n^3 + 2*((A*B^2 + B^3)*b*c
+ (A*B^2 + B^3)*a*d + ((A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d)*m^2 + 2*((A*B^2 + B^3)*b*c + (A*B^2 + B^3)*a*d)
*m)*n^2 + (((A^2*B + 2*A*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2 + B^3)*a*d)*m^3 + (A^2*B + 2*A*B^2 + B^3)*b*c + (A^
2*B + 2*A*B^2 + B^3)*a*d + 3*((A^2*B + 2*A*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2 + B^3)*a*d)*m^2 + 3*((A^2*B + 2*A
*B^2 + B^3)*b*c + (A^2*B + 2*A*B^2 + B^3)*a*d)*m)*n)*x)*log((b*x + a)/(d*x + c)))*(I*d*x + I*c)^m*e^(-(m + 2)*
log(I*d*x + I*c) - (m + 2)*log(-I*g) - (m + 2)*log((b*x + a)/(d*x + c)))/((b*c - a*d)*m^4 + 4*(b*c - a*d)*m^3
+ 6*(b*c - a*d)*m^2 + b*c - a*d + 4*(b*c - a*d)*m)

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,i+d\,i\,x\right )}^m\,{\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}^3}{{\left (a\,g+b\,g\,x\right )}^{m+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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