\(\int (c g+d g x) (A+B \log (e (\frac {a+b x}{c+d x})^n)) \, dx\) [32]

Optimal result

Integrand size = 31, antiderivative size = 86 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=-\frac {B (b c-a d) g n x}{2 b}-\frac {B (b c-a d)^2 g n \log (a+b x)}{2 b^2 d}+\frac {g (c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2547, 21, 45} \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g (c+d x)^2 \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{2 d}-\frac {B g n (b c-a d)^2 \log (a+b x)}{2 b^2 d}-\frac {B g n x (b c-a d)}{2 b} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {g \left (-\frac {B (b c-a d) n (b d x+(b c-a d) \log (a+b x))}{b^2}+(c+d x)^2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )\right )}{2 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(249\) vs. \(2(80)=160\).

Time = 1.07 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.91

method	result	size
parallelrisch	\(\frac {B \,x^{2} \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} d^{2} g n +A \,x^{2} b^{2} d^{2} g n -B \ln \left (b x +a \right ) a^{2} d^{2} g \,n^{2}+2 B \ln \left (b x +a \right ) a b c d g \,n^{2}-B \ln \left (b x +a \right ) b^{2} c^{2} g \,n^{2}+2 B x \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} c d g n +B x a b \,d^{2} g \,n^{2}-B x \,b^{2} c d g \,n^{2}+2 A x \,b^{2} c d g n +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right ) b^{2} c^{2} g n -B \,a^{2} d^{2} g \,n^{2}+B \,b^{2} c^{2} g \,n^{2}-3 A a b c d g n -2 A \,b^{2} c^{2} g n}{2 b^{2} n d}\)	\(250\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (80) = 160\).

Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.88 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {A b^{2} d^{2} g x^{2} - B b^{2} c^{2} g n \log \left (d x + c\right ) + {\left (2 \, B a b c d - B a^{2} d^{2}\right )} g n \log \left (b x + a\right ) + {\left (2 \, A b^{2} c d g - {\left (B b^{2} c d - B a b d^{2}\right )} g n\right )} x + {\left (B b^{2} d^{2} g x^{2} + 2 \, B b^{2} c d g x\right )} \log \left (e\right ) + {\left (B b^{2} d^{2} g n x^{2} + 2 \, B b^{2} c d g n x\right )} \log \left (\frac {b x + a}{d x + c}\right )}{2 \, b^{2} d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (73) = 146\).

Time = 55.24 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.44 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\begin {cases} c g x \left (A + B \log {\left (e \left (\frac {a}{c}\right )^{n} \right )}\right ) & \text {for}\: b = 0 \wedge d = 0 \\A c g x + \frac {A d g x^{2}}{2} + \frac {B c^{2} g \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}}{2 d} + \frac {B c g n x}{2} + B c g x \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )} + \frac {B d g n x^{2}}{4} + \frac {B d g x^{2} \log {\left (e \left (\frac {a}{c + d x}\right )^{n} \right )}}{2} & \text {for}\: b = 0 \\c g \left (A x + \frac {B a \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}}{b} - B n x + B x \log {\left (e \left (\frac {a}{c} + \frac {b x}{c}\right )^{n} \right )}\right ) & \text {for}\: d = 0 \\A c g x + \frac {A d g x^{2}}{2} - \frac {B a^{2} d g n \log {\left (\frac {c}{d} + x \right )}}{2 b^{2}} - \frac {B a^{2} d g \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2 b^{2}} + \frac {B a c g n \log {\left (\frac {c}{d} + x \right )}}{b} + \frac {B a c g \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{b} + \frac {B a d g n x}{2 b} - \frac {B c^{2} g n \log {\left (\frac {c}{d} + x \right )}}{2 d} - \frac {B c g n x}{2} + B c g x \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )} + \frac {B d g x^{2} \log {\left (e \left (\frac {a}{c + d x} + \frac {b x}{c + d x}\right )^{n} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((c*g*x*(A + B*log(e*(a/c)**n)), Eq(b, 0) & Eq(d, 0)), (A*c*g*x + A*d*g*x**2/2 + B*c**2*g*log(e*(a/(c
+ d*x))**n)/(2*d) + B*c*g*n*x/2 + B*c*g*x*log(e*(a/(c + d*x))**n) + B*d*g*n*x**2/4 + B*d*g*x**2*log(e*(a/(c +
d*x))**n)/2, Eq(b, 0)), (c*g*(A*x + B*a*log(e*(a/c + b*x/c)**n)/b - B*n*x + B*x*log(e*(a/c + b*x/c)**n)), Eq(
d, 0)), (A*c*g*x + A*d*g*x**2/2 - B*a**2*d*g*n*log(c/d + x)/(2*b**2) - B*a**2*d*g*log(e*(a/(c + d*x) + b*x/(c
+ d*x))**n)/(2*b**2) + B*a*c*g*n*log(c/d + x)/b + B*a*c*g*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/b + B*a*d*g*
n*x/(2*b) - B*c**2*g*n*log(c/d + x)/(2*d) - B*c*g*n*x/2 + B*c*g*x*log(e*(a/(c + d*x) + b*x/(c + d*x))**n) + B*
d*g*x**2*log(e*(a/(c + d*x) + b*x/(c + d*x))**n)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.81 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{2} \, B d g x^{2} \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + \frac {1}{2} \, A d g x^{2} - \frac {1}{2} \, B d g n {\left (\frac {a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {{\left (b c - a d\right )} x}{b d}\right )} + B c g n {\left (\frac {a \log \left (b x + a\right )}{b} - \frac {c \log \left (d x + c\right )}{d}\right )} + B c g x \log \left (e {\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n}\right ) + A c g x \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (80) = 160\).

Time = 0.44 (sec) , antiderivative size = 580, normalized size of antiderivative = 6.74 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=\frac {1}{2} \, {\left (\frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d - \frac {2 \, {\left (b x + a\right )} b d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{3}}{{\left (d x + c\right )}^{2}}} - \frac {B b^{4} c^{3} g n - 3 \, B a b^{3} c^{2} d g n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d g n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g n}{d x + c} - B a^{3} b d^{3} g n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} g n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} g n}{d x + c} - B b^{4} c^{3} g \log \left (e\right ) + 3 \, B a b^{3} c^{2} d g \log \left (e\right ) - 3 \, B a^{2} b^{2} c d^{2} g \log \left (e\right ) + B a^{3} b d^{3} g \log \left (e\right ) - A b^{4} c^{3} g + 3 \, A a b^{3} c^{2} d g - 3 \, A a^{2} b^{2} c d^{2} g + A a^{3} b d^{3} g}{b^{3} d - \frac {2 \, {\left (b x + a\right )} b^{2} d^{2}}{d x + c} + \frac {{\left (b x + a\right )}^{2} b d^{3}}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b^{2} d} - \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int (c g+d g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx=x\,\left (\frac {g\,\left (2\,A\,a\,d+4\,A\,b\,c+B\,a\,d\,n-B\,b\,c\,n\right )}{2\,b}-\frac {A\,g\,\left (2\,a\,d+2\,b\,c\right )}{2\,b}\right )+\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\,\left (\frac {B\,d\,g\,x^2}{2}+B\,c\,g\,x\right )-\frac {\ln \left (a+b\,x\right )\,\left (B\,a^2\,d\,g\,n-2\,B\,a\,b\,c\,g\,n\right )}{2\,b^2}+\frac {A\,d\,g\,x^2}{2}-\frac {B\,c^2\,g\,n\,\ln \left (c+d\,x\right )}{2\,d} \]

[In]

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

[Out]

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