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

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Rubi [A]  time = 0.616366, 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 6742

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

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 12

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

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 = 0.508259, 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))

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Maple [F]  time = 4.586, size = 0, normalized size = 0. \begin{align*} \int \left ( bgx+ag \right ) ^{-2-m} \left ( dix+ci \right ) ^{m} \left ( A+B\ln \left ( e \left ({\frac{bx+a}{dx+c}} \right ) ^{n} \right ) \right ) \, dx \end{align*}

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)),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)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)

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Fricas [A]  time = 0.538503, size = 632, normalized size = 4.61 \begin{align*} -\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 \left (e\right ) +{\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} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)