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

Optimal. Leaf size=137 \[ -\frac {(a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2} \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{i^2 (m+1) (c+d x) (b c-a d)}-\frac {B n (a+b x) (g (a+b x))^{-m-2} (i (c+d x))^{m+2}}{i^2 (m+1)^2 (c+d x) (b c-a d)} \]

[Out]

-B*n*(b*x+a)*(g*(b*x+a))^(-2-m)*(i*(d*x+c))^(2+m)/(-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))/(-a*d+b*c)/i^2/(1+m)/(d*x+c)

________________________________________________________________________________________

Rubi [A]  time = 0.62, antiderivative size = 170, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {6742, 37, 2554, 12} \[ -\frac {A (a g+b g x)^{-m-1} (c i+d i x)^{m+1}}{g i (m+1) (b c-a d)}-\frac {B (a g+b g x)^{-m-1} (c i+d i x)^{m+1} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{g i (m+1) (b c-a d)}-\frac {B n (a g+b g x)^{-m-1} (c i+d i x)^{m+1}}{g i (m+1)^2 (b c-a d)} \]

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]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int (220 c+220 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 ) \, dx &=\int \left (A (220 c+220 d x)^m (a g+b g x)^{-2-m}+B (220 c+220 d x)^m (a g+b g x)^{-2-m} \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \, dx\\ &=A \int (220 c+220 d x)^m (a g+b g x)^{-2-m} \, dx+B \int (220 c+220 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\\ &=-\frac {A (220 c+220 d x)^{1+m} (a g+b g x)^{-1-m}}{220 (b c-a d) g (1+m)}-\frac {B (220 c+220 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 )}{220 (b c-a d) g (1+m)}-B \int \frac {220^m n (c+d x)^m (a g+b g x)^{-2-m}}{-1-m} \, dx\\ &=-\frac {A (220 c+220 d x)^{1+m} (a g+b g x)^{-1-m}}{220 (b c-a d) g (1+m)}-\frac {B (220 c+220 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 )}{220 (b c-a d) g (1+m)}+\frac {\left (220^m B n\right ) \int (c+d x)^m (a g+b g x)^{-2-m} \, dx}{1+m}\\ &=-\frac {220^m B n (c+d x)^{1+m} (a g+b g x)^{-1-m}}{(b c-a d) g (1+m)^2}-\frac {A (220 c+220 d x)^{1+m} (a g+b g x)^{-1-m}}{220 (b c-a d) g (1+m)}-\frac {B (220 c+220 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 )}{220 (b c-a d) g (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.53, size = 78, normalized size = 0.57 \[ -\frac {(c+d x) (g (a+b x))^{-m-1} (i (c+d x))^m \left (B (m+1) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A m+A+B n\right )}{g (m+1)^2 (b c-a d)} \]

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]),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.05, size = 269, normalized size = 1.96 \[ -\frac {{\left (A a c m + B a c n + A a c + {\left (A b d m + B b d n + A b d\right )} x^{2} + {\left (A b c + A a d + {\left (A b c + A a d\right )} m + {\left (B b c + B a d\right )} n\right )} x + {\left (B a c m + B a c + {\left (B b d m + B b d\right )} x^{2} + {\left (B b c + B a d + {\left (B b c + B a d\right )} m\right )} x\right )} \log \relax (e) + {\left ({\left (B b d m + B b d\right )} n x^{2} + {\left (B b c + B a d + {\left (B b c + B a d\right )} m\right )} n x + {\left (B a c m + B a c\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )\right )} {\left (b g x + a g\right )}^{-m - 2} e^{\left (m \log \left (b g x + a g\right ) - m \log \left (\frac {b x + a}{d x + c}\right ) + m \log \left (\frac {i}{g}\right )\right )}}{{\left (b c - a d\right )} m^{2} + b c - a d + 2 \, {\left (b c - a d\right )} m} \]

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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 8.30, size = 0, normalized size = 0.00 \[ \int \left (B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )+A \right ) \left (b g x +a g \right )^{-m -2} \left (d i x +c i \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,i+d\,i\,x\right )}^m\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}{{\left (a\,g+b\,g\,x\right )}^{m+2}} \,d x \]

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)))/(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)))/(a*g + b*g*x)^(m + 2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)),x)

[Out]

Timed out

________________________________________________________________________________________