\(\int (c g+d g x)^2 (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\) [31]

Optimal result

Integrand size = 33, antiderivative size = 124 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d)^2 g^2 n x}{3 b^2}-\frac {B (b c-a d) g^2 n (c+d x)^2}{6 b d}-\frac {B (b c-a d)^3 g^2 n \log (a+b x)}{3 b^3 d}+\frac {g^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2547, 21, 45} \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g^2 (c+d x)^3 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{3 d}-\frac {B g^2 n (b c-a d)^3 \log (a+b x)}{3 b^3 d}-\frac {B g^2 n x (b c-a d)^2}{3 b^2}-\frac {B g^2 n (c+d x)^2 (b c-a d)}{6 b d} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2547

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x
_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Dist[B*n*((b*c -
a*d)/(g*(m + 1))), Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, m
, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, -2]

Rubi steps \begin{align*} \text {integral}& = \frac {g^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d}-\frac {(B (b c-a d) n) \int \frac {(c g+d g x)^3}{(a+b x) (c+d x)} \, dx}{3 d g} \\ & = \frac {g^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d}-\frac {\left (B (b c-a d) g^2 n\right ) \int \frac {(c+d x)^2}{a+b x} \, dx}{3 d} \\ & = \frac {g^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d}-\frac {\left (B (b c-a d) g^2 n\right ) \int \left (\frac {d (b c-a d)}{b^2}+\frac {(b c-a d)^2}{b^2 (a+b x)}+\frac {d (c+d x)}{b}\right ) \, dx}{3 d} \\ & = -\frac {B (b c-a d)^2 g^2 n x}{3 b^2}-\frac {B (b c-a d) g^2 n (c+d x)^2}{6 b d}-\frac {B (b c-a d)^3 g^2 n \log (a+b x)}{3 b^3 d}+\frac {g^2 (c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.81 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g^2 \left (-\frac {B (b c-a d) n \left (2 b d (b c-a d) x+b^2 (c+d x)^2+2 (b c-a d)^2 \log (a+b x)\right )}{2 b^3}+(c+d x)^3 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )\right )}{3 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(462\) vs. \(2(116)=232\).

Time = 2.92 (sec) , antiderivative size = 463, normalized size of antiderivative = 3.73

method	result	size
parallelrisch	\(\frac {6 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c^{2} d \,g^{2} n -6 B \ln \left (b x +a \right ) a^{2} b c \,d^{2} g^{2} n^{2}-6 A \,b^{3} c^{3} g^{2} n +6 B x a \,b^{2} c \,d^{2} g^{2} n^{2}+6 B \ln \left (b x +a \right ) a \,b^{2} c^{2} d \,g^{2} n^{2}+2 B \,a^{3} d^{3} g^{2} n^{2}+4 B \,b^{3} c^{3} g^{2} n^{2}+6 B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c \,d^{2} g^{2} n -12 A a \,b^{2} c^{2} d \,g^{2} n +6 A \,x^{2} b^{3} c \,d^{2} g^{2} n +6 A x \,b^{3} c^{2} d \,g^{2} n +2 B \,x^{3} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} d^{3} g^{2} n +B \,x^{2} a \,b^{2} d^{3} g^{2} n^{2}-B \,x^{2} b^{3} c \,d^{2} g^{2} n^{2}-2 B x \,a^{2} b \,d^{3} g^{2} n^{2}-4 B x \,b^{3} c^{2} d \,g^{2} n^{2}-5 B \,a^{2} b c \,d^{2} g^{2} n^{2}-B a \,b^{2} c^{2} d \,g^{2} n^{2}+2 A \,x^{3} b^{3} d^{3} g^{2} n +2 B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{3} c^{3} g^{2} n +2 B \ln \left (b x +a \right ) a^{3} d^{3} g^{2} n^{2}-2 B \ln \left (b x +a \right ) b^{3} c^{3} g^{2} n^{2}}{6 b^{3} d n}\)	\(463\)

[In]

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

[Out]

1/6*(6*B*x*ln(e*((b*x+a)/(d*x+c))^n)*b^3*c^2*d*g^2*n-6*B*ln(b*x+a)*a^2*b*c*d^2*g^2*n^2-6*A*b^3*c^3*g^2*n+6*B*x
*a*b^2*c*d^2*g^2*n^2+6*B*ln(b*x+a)*a*b^2*c^2*d*g^2*n^2+2*B*a^3*d^3*g^2*n^2+4*B*b^3*c^3*g^2*n^2+6*B*x^2*ln(e*((
b*x+a)/(d*x+c))^n)*b^3*c*d^2*g^2*n-12*A*a*b^2*c^2*d*g^2*n+6*A*x^2*b^3*c*d^2*g^2*n+6*A*x*b^3*c^2*d*g^2*n+2*B*x^
3*ln(e*((b*x+a)/(d*x+c))^n)*b^3*d^3*g^2*n+B*x^2*a*b^2*d^3*g^2*n^2-B*x^2*b^3*c*d^2*g^2*n^2-2*B*x*a^2*b*d^3*g^2*
n^2-4*B*x*b^3*c^2*d*g^2*n^2-5*B*a^2*b*c*d^2*g^2*n^2-B*a*b^2*c^2*d*g^2*n^2+2*A*x^3*b^3*d^3*g^2*n+2*B*ln(e*((b*x
+a)/(d*x+c))^n)*b^3*c^3*g^2*n+2*B*ln(b*x+a)*a^3*d^3*g^2*n^2-2*B*ln(b*x+a)*b^3*c^3*g^2*n^2)/b^3/d/n

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (116) = 232\).

Time = 0.29 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.40 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {2 \, A b^{3} d^{3} g^{2} x^{3} - 2 \, B b^{3} c^{3} g^{2} n \log \left (d x + c\right ) + 2 \, {\left (3 \, B a b^{2} c^{2} d - 3 \, B a^{2} b c d^{2} + B a^{3} d^{3}\right )} g^{2} n \log \left (b x + a\right ) + {\left (6 \, A b^{3} c d^{2} g^{2} - {\left (B b^{3} c d^{2} - B a b^{2} d^{3}\right )} g^{2} n\right )} x^{2} + 2 \, {\left (3 \, A b^{3} c^{2} d g^{2} - {\left (2 \, B b^{3} c^{2} d - 3 \, B a b^{2} c d^{2} + B a^{2} b d^{3}\right )} g^{2} n\right )} x + 2 \, {\left (B b^{3} d^{3} g^{2} x^{3} + 3 \, B b^{3} c d^{2} g^{2} x^{2} + 3 \, B b^{3} c^{2} d g^{2} x\right )} \log \left (e\right ) + 2 \, {\left (B b^{3} d^{3} g^{2} n x^{3} + 3 \, B b^{3} c d^{2} g^{2} n x^{2} + 3 \, B b^{3} c^{2} d g^{2} n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{6 \, b^{3} d} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (116) = 232\).

Time = 0.19 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.49 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{3} \, B d^{2} g^{2} x^{3} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{3} \, A d^{2} g^{2} x^{3} + B c d g^{2} x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c d g^{2} x^{2} + \frac {1}{6} \, B d^{2} g^{2} n {\left (\frac {2 \, a^{3} \log \left (b x + a\right )}{b^{3}} - \frac {2 \, c^{3} \log \left (d x + c\right )}{d^{3}} - \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} - B c d g^{2} n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B c^{2} g^{2} n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B c^{2} g^{2} x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c^{2} g^{2} x \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 990 vs. \(2 (116) = 232\).

Time = 0.63 (sec) , antiderivative size = 990, normalized size of antiderivative = 7.98 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{6} \, {\left (\frac {2 \, {\left (B b^{4} c^{4} g^{2} n - 4 \, B a b^{3} c^{3} d g^{2} n + 6 \, B a^{2} b^{2} c^{2} d^{2} g^{2} n - 4 \, B a^{3} b c d^{3} g^{2} n + B a^{4} d^{4} g^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{3} d - \frac {3 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {3 \, {\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{3} d^{4}}{{\left (d x + c\right )}^{3}}} - \frac {3 \, B b^{6} c^{4} g^{2} n - 12 \, B a b^{5} c^{3} d g^{2} n - \frac {5 \, {\left (b x + a\right )} B b^{5} c^{4} d g^{2} n}{d x + c} + 18 \, B a^{2} b^{4} c^{2} d^{2} g^{2} n + \frac {20 \, {\left (b x + a\right )} B a b^{4} c^{3} d^{2} g^{2} n}{d x + c} + \frac {2 \, {\left (b x + a\right )}^{2} B b^{4} c^{4} d^{2} g^{2} n}{{\left (d x + c\right )}^{2}} - 12 \, B a^{3} b^{3} c d^{3} g^{2} n - \frac {30 \, {\left (b x + a\right )} B a^{2} b^{3} c^{2} d^{3} g^{2} n}{d x + c} - \frac {8 \, {\left (b x + a\right )}^{2} B a b^{3} c^{3} d^{3} g^{2} n}{{\left (d x + c\right )}^{2}} + 3 \, B a^{4} b^{2} d^{4} g^{2} n + \frac {20 \, {\left (b x + a\right )} B a^{3} b^{2} c d^{4} g^{2} n}{d x + c} + \frac {12 \, {\left (b x + a\right )}^{2} B a^{2} b^{2} c^{2} d^{4} g^{2} n}{{\left (d x + c\right )}^{2}} - \frac {5 \, {\left (b x + a\right )} B a^{4} b d^{5} g^{2} n}{d x + c} - \frac {8 \, {\left (b x + a\right )}^{2} B a^{3} b c d^{5} g^{2} n}{{\left (d x + c\right )}^{2}} + \frac {2 \, {\left (b x + a\right )}^{2} B a^{4} d^{6} g^{2} n}{{\left (d x + c\right )}^{2}} - 2 \, B b^{6} c^{4} g^{2} \log \left (e\right ) + 8 \, B a b^{5} c^{3} d g^{2} \log \left (e\right ) - 12 \, B a^{2} b^{4} c^{2} d^{2} g^{2} \log \left (e\right ) + 8 \, B a^{3} b^{3} c d^{3} g^{2} \log \left (e\right ) - 2 \, B a^{4} b^{2} d^{4} g^{2} \log \left (e\right ) - 2 \, A b^{6} c^{4} g^{2} + 8 \, A a b^{5} c^{3} d g^{2} - 12 \, A a^{2} b^{4} c^{2} d^{2} g^{2} + 8 \, A a^{3} b^{3} c d^{3} g^{2} - 2 \, A a^{4} b^{2} d^{4} g^{2}}{b^{5} d - \frac {3 \, {\left (b x + a\right )} b^{4} d^{2}}{d x + c} + \frac {3 \, {\left (b x + a\right )}^{2} b^{3} d^{3}}{{\left (d x + c\right )}^{2}} - \frac {{\left (b x + a\right )}^{3} b^{2} d^{4}}{{\left (d x + c\right )}^{3}}} + \frac {2 \, {\left (B b^{4} c^{4} g^{2} n - 4 \, B a b^{3} c^{3} d g^{2} n + 6 \, B a^{2} b^{2} c^{2} d^{2} g^{2} n - 4 \, B a^{3} b c d^{3} g^{2} n + B a^{4} d^{4} g^{2} n\right )} \log \left (b - \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{3} d} - \frac {2 \, {\left (B b^{4} c^{4} g^{2} n - 4 \, B a b^{3} c^{3} d g^{2} n + 6 \, B a^{2} b^{2} c^{2} d^{2} g^{2} n - 4 \, B a^{3} b c d^{3} g^{2} n + B a^{4} d^{4} g^{2} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{3} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[In]

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

[Out]

1/6*(2*(B*b^4*c^4*g^2*n - 4*B*a*b^3*c^3*d*g^2*n + 6*B*a^2*b^2*c^2*d^2*g^2*n - 4*B*a^3*b*c*d^3*g^2*n + B*a^4*d^
4*g^2*n)*log((b*x + a)/(d*x + c))/(b^3*d - 3*(b*x + a)*b^2*d^2/(d*x + c) + 3*(b*x + a)^2*b*d^3/(d*x + c)^2 - (
b*x + a)^3*d^4/(d*x + c)^3) - (3*B*b^6*c^4*g^2*n - 12*B*a*b^5*c^3*d*g^2*n - 5*(b*x + a)*B*b^5*c^4*d*g^2*n/(d*x
+ c) + 18*B*a^2*b^4*c^2*d^2*g^2*n + 20*(b*x + a)*B*a*b^4*c^3*d^2*g^2*n/(d*x + c) + 2*(b*x + a)^2*B*b^4*c^4*d^
2*g^2*n/(d*x + c)^2 - 12*B*a^3*b^3*c*d^3*g^2*n - 30*(b*x + a)*B*a^2*b^3*c^2*d^3*g^2*n/(d*x + c) - 8*(b*x + a)^
2*B*a*b^3*c^3*d^3*g^2*n/(d*x + c)^2 + 3*B*a^4*b^2*d^4*g^2*n + 20*(b*x + a)*B*a^3*b^2*c*d^4*g^2*n/(d*x + c) + 1
2*(b*x + a)^2*B*a^2*b^2*c^2*d^4*g^2*n/(d*x + c)^2 - 5*(b*x + a)*B*a^4*b*d^5*g^2*n/(d*x + c) - 8*(b*x + a)^2*B*
a^3*b*c*d^5*g^2*n/(d*x + c)^2 + 2*(b*x + a)^2*B*a^4*d^6*g^2*n/(d*x + c)^2 - 2*B*b^6*c^4*g^2*log(e) + 8*B*a*b^5
*c^3*d*g^2*log(e) - 12*B*a^2*b^4*c^2*d^2*g^2*log(e) + 8*B*a^3*b^3*c*d^3*g^2*log(e) - 2*B*a^4*b^2*d^4*g^2*log(e
) - 2*A*b^6*c^4*g^2 + 8*A*a*b^5*c^3*d*g^2 - 12*A*a^2*b^4*c^2*d^2*g^2 + 8*A*a^3*b^3*c*d^3*g^2 - 2*A*a^4*b^2*d^4
*g^2)/(b^5*d - 3*(b*x + a)*b^4*d^2/(d*x + c) + 3*(b*x + a)^2*b^3*d^3/(d*x + c)^2 - (b*x + a)^3*b^2*d^4/(d*x +
c)^3) + 2*(B*b^4*c^4*g^2*n - 4*B*a*b^3*c^3*d*g^2*n + 6*B*a^2*b^2*c^2*d^2*g^2*n - 4*B*a^3*b*c*d^3*g^2*n + B*a^4
*d^4*g^2*n)*log(b - (b*x + a)*d/(d*x + c))/(b^3*d) - 2*(B*b^4*c^4*g^2*n - 4*B*a*b^3*c^3*d*g^2*n + 6*B*a^2*b^2*
c^2*d^2*g^2*n - 4*B*a^3*b*c*d^3*g^2*n + B*a^4*d^4*g^2*n)*log((b*x + a)/(d*x + c))/(b^3*d))*(b*c/(b*c - a*d)^2
- a*d/(b*c - a*d)^2)

Mupad [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.44 \[ \int (c g+d g x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (B\,c^2\,g^2\,x+B\,c\,d\,g^2\,x^2+\frac {B\,d^2\,g^2\,x^3}{3}\right )-x\,\left (\frac {\left (3\,a\,d+3\,b\,c\right )\,\left (\frac {d\,g^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{3\,b}-\frac {A\,d\,g^2\,\left (3\,a\,d+3\,b\,c\right )}{3\,b}\right )}{3\,b\,d}-\frac {c\,g^2\,\left (3\,A\,a\,d+3\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{b}+\frac {A\,a\,c\,d\,g^2}{b}\right )+x^2\,\left (\frac {d\,g^2\,\left (3\,A\,a\,d+9\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{6\,b}-\frac {A\,d\,g^2\,\left (3\,a\,d+3\,b\,c\right )}{6\,b}\right )+\frac {\ln \left (a+b\,x\right )\,\left (B\,n\,a^3\,d^2\,g^2-3\,B\,n\,a^2\,b\,c\,d\,g^2+3\,B\,n\,a\,b^2\,c^2\,g^2\right )}{3\,b^3}+\frac {A\,d^2\,g^2\,x^3}{3}-\frac {B\,c^3\,g^2\,n\,\ln \left (c+d\,x\right )}{3\,d} \]

[In]

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

[Out]

log(e*((a + b*x)/(c + d*x))^n)*((B*d^2*g^2*x^3)/3 + B*c^2*g^2*x + B*c*d*g^2*x^2) - x*(((3*a*d + 3*b*c)*((d*g^2
*(3*A*a*d + 9*A*b*c + B*a*d*n - B*b*c*n))/(3*b) - (A*d*g^2*(3*a*d + 3*b*c))/(3*b)))/(3*b*d) - (c*g^2*(3*A*a*d
+ 3*A*b*c + B*a*d*n - B*b*c*n))/b + (A*a*c*d*g^2)/b) + x^2*((d*g^2*(3*A*a*d + 9*A*b*c + B*a*d*n - B*b*c*n))/(6
*b) - (A*d*g^2*(3*a*d + 3*b*c))/(6*b)) + (log(a + b*x)*(B*a^3*d^2*g^2*n + 3*B*a*b^2*c^2*g^2*n - 3*B*a^2*b*c*d*
g^2*n))/(3*b^3) + (A*d^2*g^2*x^3)/3 - (B*c^3*g^2*n*log(c + d*x))/(3*d)