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

Optimal. Leaf size=86 \[ \frac{g (a+b x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}+\frac{B g n (b c-a d)^2 \log (c+d x)}{2 b d^2}-\frac{B g n x (b c-a d)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0614864, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2525, 12, 43} \[ \frac{g (a+b x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )}{2 b}+\frac{B g n (b c-a d)^2 \log (c+d x)}{2 b d^2}-\frac{B g n x (b c-a d)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 12

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

Rule 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0395174, size = 73, normalized size = 0.85 \[ \frac{g \left ((a+b x)^2 \left (B \log \left (e \left (\frac{a+b x}{c+d x}\right )^n\right )+A\right )+\frac{B n (a d-b c) ((a d-b c) \log (c+d x)+b d x)}{d^2}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.322, size = 0, normalized size = 0. \begin{align*} \int \left ( bgx+ag \right ) \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)*(A+B*ln(e*((b*x+a)/(d*x+c))^n)),x)

[Out]

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

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Maxima [A]  time = 1.13311, size = 211, normalized size = 2.45 \begin{align*} \frac{1}{2} \, B b 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 b g x^{2} - \frac{1}{2} \, B b 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 a g n{\left (\frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} + B a g x \log \left (e{\left (\frac{b x}{d x + c} + \frac{a}{d x + c}\right )}^{n}\right ) + A a g x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.948587, size = 360, normalized size = 4.19 \begin{align*} \frac{A b^{2} d^{2} g x^{2} + B a^{2} d^{2} g n \log \left (b x + a\right ) +{\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g n \log \left (d x + c\right ) +{\left (2 \, A a b d^{2} 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 a b d^{2} g x\right )} \log \left (e\right ) +{\left (B b^{2} d^{2} g n x^{2} + 2 \, B a b d^{2} g n x\right )} \log \left (\frac{b x + a}{d x + c}\right )}{2 \, b d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Timed out

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Giac [A]  time = 3.65069, size = 167, normalized size = 1.94 \begin{align*} \frac{B a^{2} g n \log \left (b x + a\right )}{2 \, b} + \frac{1}{2} \,{\left (A b g + B b g\right )} x^{2} + \frac{1}{2} \,{\left (B b g n x^{2} + 2 \, B a g n x\right )} \log \left (\frac{b x + a}{d x + c}\right ) - \frac{{\left (B b c g n - B a d g n - 2 \, A a d g - 2 \, B a d g\right )} x}{2 \, d} + \frac{{\left (B b c^{2} g n - 2 \, B a c d g n\right )} \log \left (d x + c\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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