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

Optimal. Leaf size=109 \[ \frac{(f+g x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 g}-\frac{B (b f-a g)^2 \log (a+b x)}{2 b^2 g}-\frac{B g x (b c-a d)}{2 b d}+\frac{B (d f-c g)^2 \log (c+d x)}{2 d^2 g} \]

[Out]

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

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Rubi [A]  time = 0.0975082, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2525, 12, 72} \[ \frac{(f+g x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 g}-\frac{B (b f-a g)^2 \log (a+b x)}{2 b^2 g}-\frac{B g x (b c-a d)}{2 b d}+\frac{B (d f-c g)^2 \log (c+d x)}{2 d^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [A]  time = 0.107345, size = 114, normalized size = 1.05 \[ \frac{b \left (d \left (B g^2 x (a d-b c)+A b d (f+g x)^2\right )+b B d^2 (f+g x)^2 \log \left (\frac{e (a+b x)}{c+d x}\right )+b B (d f-c g)^2 \log (c+d x)\right )-B d^2 (b f-a g)^2 \log (a+b x)}{2 b^2 d^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.174, size = 1809, normalized size = 16.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.19729, size = 189, normalized size = 1.73 \begin{align*} \frac{1}{2} \, A g x^{2} +{\left (x \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right ) + \frac{a \log \left (b x + a\right )}{b} - \frac{c \log \left (d x + c\right )}{d}\right )} B f + \frac{1}{2} \,{\left (x^{2} \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right ) - \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 g + A f x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.35456, size = 321, normalized size = 2.94 \begin{align*} \frac{A b^{2} d^{2} g x^{2} +{\left (2 \, A b^{2} d^{2} f -{\left (B b^{2} c d - B a b d^{2}\right )} g\right )} x +{\left (2 \, B a b d^{2} f - B a^{2} d^{2} g\right )} \log \left (b x + a\right ) -{\left (2 \, B b^{2} c d f - B b^{2} c^{2} g\right )} \log \left (d x + c\right ) +{\left (B b^{2} d^{2} g x^{2} + 2 \, B b^{2} d^{2} f x\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{2 \, b^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 4.78579, size = 325, normalized size = 2.98 \begin{align*} \frac{A g x^{2}}{2} - \frac{B a \left (a g - 2 b f\right ) \log{\left (x + \frac{B a^{2} c d g + \frac{B a^{2} d^{2} \left (a g - 2 b f\right )}{b} + B a b c^{2} g - 4 B a b c d f - B a c d \left (a g - 2 b f\right )}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{2 b^{2}} + \frac{B c \left (c g - 2 d f\right ) \log{\left (x + \frac{B a^{2} c d g + B a b c^{2} g - 4 B a b c d f - B a b c \left (c g - 2 d f\right ) + \frac{B b^{2} c^{2} \left (c g - 2 d f\right )}{d}}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{2 d^{2}} + \left (B f x + \frac{B g x^{2}}{2}\right ) \log{\left (\frac{e \left (a + b x\right )}{c + d x} \right )} + \frac{x \left (2 A b d f + B a d g - B b c g\right )}{2 b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.52862, size = 171, normalized size = 1.57 \begin{align*} \frac{1}{2} \,{\left (A g + B g\right )} x^{2} + \frac{1}{2} \,{\left (B g x^{2} + 2 \, B f x\right )} \log \left (\frac{b x + a}{d x + c}\right ) + \frac{{\left (2 \, A b d f + 2 \, B b d f - B b c g + B a d g\right )} x}{2 \, b d} + \frac{{\left (2 \, B a b f - B a^{2} g\right )} \log \left (b x + a\right )}{2 \, b^{2}} - \frac{{\left (2 \, B c d f - B c^{2} g\right )} \log \left (-d x - c\right )}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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