3.265 \(\int (f+g x) (A+B \log (\frac{e (a+b x)^2}{(c+d x)^2})) \, dx\)

Optimal. Leaf size=104 \[ \frac{(f+g x)^2 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 g}-\frac{B (b f-a g)^2 \log (a+b x)}{b^2 g}-\frac{B g x (b c-a d)}{b d}+\frac{B (d f-c g)^2 \log (c+d x)}{d^2 g} \]

[Out]

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

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Rubi [A]  time = 0.0870162, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2525, 12, 72} \[ \frac{(f+g x)^2 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 g}-\frac{B (b f-a g)^2 \log (a+b x)}{b^2 g}-\frac{B g x (b c-a d)}{b d}+\frac{B (d f-c g)^2 \log (c+d x)}{d^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.104722, size = 118, normalized size = 1.13 \[ \frac{b \left (d \left (2 B g^2 x (a d-b c)+A b d (f+g x)^2\right )+b B d^2 (f+g x)^2 \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+2 b B (d f-c g)^2 \log (c+d x)\right )-2 B d^2 (b f-a g)^2 \log (a+b x)}{2 b^2 d^2 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.244, size = 656, normalized size = 6.3 \begin{align*}{\frac{Bgac}{bd}}-4\,{\frac{Bacf}{ad-bc}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{A{c}^{2}g}{2\,{d}^{2}}}+{\frac{Acf}{d}}-{\frac{B{c}^{2}g}{{d}^{2}}}+B\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) xf+{\frac{Bg{x}^{2}}{2}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }-2\,{\frac{B\ln \left ( \left ( dx+c \right ) ^{-1} \right ) af}{b}}+{\frac{Bcf}{d}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }+2\,{\frac{B\ln \left ( \left ( dx+c \right ) ^{-1} \right ) cf}{d}}-{\frac{B{c}^{2}g}{{d}^{2}}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{B{c}^{2}g}{2\,{d}^{2}}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }-{\frac{B\ln \left ( \left ( dx+c \right ) ^{-1} \right ){c}^{2}g}{{d}^{2}}}-{\frac{Bcgx}{d}}+2\,{\frac{Bgac}{bd}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+2\,{\frac{B{c}^{2}bf}{d \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+4\,{\frac{Ba{c}^{2}g}{d \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+2\,{\frac{dB{a}^{2}f}{b \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+{\frac{Bgax}{b}}-2\,{\frac{B{c}^{3}bg}{{d}^{2} \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-2\,{\frac{B{a}^{2}cg}{b \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{Bg{a}^{2}}{{b}^{2}}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+{\frac{Bg\ln \left ( \left ( dx+c \right ) ^{-1} \right ){a}^{2}}{{b}^{2}}}+{\frac{A{x}^{2}g}{2}}+Afx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(A+B*ln(e*(b*x+a)^2/(d*x+c)^2)),x)

[Out]

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

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Maxima [B]  time = 1.16359, size = 332, normalized size = 3.19 \begin{align*} \frac{1}{2} \, A g x^{2} +{\left (x \log \left (\frac{b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac{2 \, a \log \left (b x + a\right )}{b} - \frac{2 \, c \log \left (d x + c\right )}{d}\right )} B f + \frac{1}{2} \,{\left (x^{2} \log \left (\frac{b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac{a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac{2 \, a^{2} \log \left (b x + a\right )}{b^{2}} + \frac{2 \, c^{2} \log \left (d x + c\right )}{d^{2}} - \frac{2 \,{\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)^2/(d*x+c)^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.04507, size = 373, normalized size = 3.59 \begin{align*} \frac{A b^{2} d^{2} g x^{2} + 2 \,{\left (A b^{2} d^{2} f -{\left (B b^{2} c d - B a b d^{2}\right )} g\right )} x + 2 \,{\left (2 \, B a b d^{2} f - B a^{2} d^{2} g\right )} \log \left (b x + a\right ) - 2 \,{\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^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\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)^2/(d*x+c)^2)),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*(2*B*a*b*d^2*f - B*a^2*d^2*g)*log(b*x
+ a) - 2*(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^2*e*x^2 + 2*
a*b*e*x + a^2*e)/(d^2*x^2 + 2*c*d*x + c^2)))/(b^2*d^2)

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Sympy [B]  time = 4.26136, size = 321, normalized size = 3.09 \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 )}}{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 )}}{d^{2}} + \left (B f x + \frac{B g x^{2}}{2}\right ) \log{\left (\frac{e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} + \frac{x \left (A b d f + B a d g - B b c g\right )}{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)**2/(d*x+c)**2)),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))/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))/d**2 + (B*f*x + B*g*x**2/2)*log(e*
(a + b*x)**2/(c + d*x)**2) + x*(A*b*d*f + B*a*d*g - B*b*c*g)/(b*d)

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Giac [A]  time = 2.10539, size = 196, normalized size = 1.88 \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^{2} x^{2} + 2 \, a b x + a^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac{{\left (A b d f + B b d f - B b c g + B a d g\right )} x}{b d} + \frac{{\left (2 \, B a b f - B a^{2} g\right )} \log \left (b x + a\right )}{b^{2}} - \frac{{\left (2 \, B c d f - B c^{2} g\right )} \log \left (-d x - c\right )}{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)^2/(d*x+c)^2)),x, algorithm="giac")

[Out]

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