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

Optimal. Leaf size=78 \[ \frac{g (a+b x)^2 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 b}+\frac{B g (b c-a d)^2 \log (c+d x)}{b d^2}-\frac{B g x (b c-a d)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0568225, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2525, 12, 43} \[ \frac{g (a+b x)^2 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )}{2 b}+\frac{B g (b c-a d)^2 \log (c+d x)}{b d^2}-\frac{B g x (b c-a d)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0360573, size = 72, normalized size = 0.92 \[ \frac{g \left ((a+b x)^2 \left (B \log \left (\frac{e (a+b x)^2}{(c+d x)^2}\right )+A\right )+\frac{2 B (a d-b c) ((a d-b c) \log (c+d x)+b d x)}{d^2}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.235, size = 560, normalized size = 7.2 \begin{align*} -{\frac{gBbcx}{d}}-{\frac{gBb{c}^{2}}{2\,{d}^{2}}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }+{\frac{gBb{x}^{2}}{2}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }+B\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) xag-{\frac{gBb{c}^{2}}{{d}^{2}}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-{\frac{Bg{a}^{2}}{b}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+{\frac{Bgac}{d}}-{\frac{gBb{c}^{2}}{{d}^{2}}}-{\frac{Bg\ln \left ( \left ( dx+c \right ) ^{-1} \right ){c}^{2}b}{{d}^{2}}}-{\frac{Bg\ln \left ( \left ( dx+c \right ) ^{-1} \right ){a}^{2}}{b}}+gBax+gAax-{\frac{Ag{c}^{2}b}{2\,{d}^{2}}}+{\frac{Ag{x}^{2}b}{2}}+2\,{\frac{Bg\ln \left ( \left ( dx+c \right ) ^{-1} \right ) ac}{d}}+2\,{\frac{Bgac}{d}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+{\frac{Bgac}{d}\ln \left ({\frac{e}{{d}^{2}} \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) ^{2}} \right ) }+{\frac{Agac}{d}}+2\,{\frac{dBg{a}^{3}}{b \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-6\,{\frac{Bg{a}^{2}c}{ad-bc}\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }+6\,{\frac{gBa{c}^{2}b}{d \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) }-2\,{\frac{Bg{c}^{3}{b}^{2}}{{d}^{2} \left ( ad-bc \right ) }\ln \left ({\frac{ad}{dx+c}}-{\frac{bc}{dx+c}}+b \right ) } \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)^2/(d*x+c)^2)),x)

[Out]

-1/d*g*B*b*c*x-1/2/d^2*B*g*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*b*c^2+1/2*B*g*b*ln(e*(1/(d*x+c)*a*d-b*c/(
d*x+c)+b)^2/d^2)*x^2+B*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*x*a*g-1/d^2*B*g*b*ln(1/(d*x+c)*a*d-b*c/(d*x+c
)+b)*c^2-B*g/b*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^2+1/d*B*g*a*c-1/d^2*B*g*c^2*b-1/d^2*B*g*ln(1/(d*x+c))*c^2*b-B
*g/b*ln(1/(d*x+c))*a^2+g*B*a*x+g*A*a*x-1/2/d^2*A*g*c^2*b+1/2*A*g*x^2*b+2/d*B*g*ln(1/(d*x+c))*a*c+2/d*B*g*ln(1/
(d*x+c)*a*d-b*c/(d*x+c)+b)*a*c+1/d*B*ln(e*(1/(d*x+c)*a*d-b*c/(d*x+c)+b)^2/d^2)*a*c*g+1/d*A*g*a*c+2*d*B*g/b/(a*
d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^3-6*B*g/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a^2*c+6/d*B*g/(a*d-
b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*a*c^2*b-2/d^2*B*g/(a*d-b*c)*ln(1/(d*x+c)*a*d-b*c/(d*x+c)+b)*c^3*b^2

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Maxima [B]  time = 1.22796, size = 338, normalized size = 4.33 \begin{align*} \frac{1}{2} \, A b 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 a g + \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 b g + 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)^2/(d*x+c)^2)),x, algorithm="maxima")

[Out]

1/2*A*b*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*a*g + 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*b*g + A*a*g*x

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Fricas [A]  time = 1.08462, size = 329, normalized size = 4.22 \begin{align*} \frac{A b^{2} d^{2} g x^{2} + 2 \, B a^{2} d^{2} g \log \left (b x + a\right ) - 2 \,{\left (B b^{2} c d -{\left (A + B\right )} a b d^{2}\right )} g x + 2 \,{\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g \log \left (d x + c\right ) +{\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g 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 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)^2/(d*x+c)^2)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 2.46477, size = 253, normalized size = 3.24 \begin{align*} \frac{A b g x^{2}}{2} + \frac{B a^{2} g \log{\left (x + \frac{\frac{B a^{3} d^{2} g}{b} + 2 B a^{2} c d g - B a b c^{2} g}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{b} - \frac{B c g \left (2 a d - b c\right ) \log{\left (x + \frac{3 B a^{2} c d g - B a b c^{2} g - B a c g \left (2 a d - b c\right ) + \frac{B b c^{2} g \left (2 a d - b c\right )}{d}}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{d^{2}} + \left (B a g x + \frac{B 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 a d g + B a d g - B b c g\right )}{d} \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)**2/(d*x+c)**2)),x)

[Out]

A*b*g*x**2/2 + B*a**2*g*log(x + (B*a**3*d**2*g/b + 2*B*a**2*c*d*g - B*a*b*c**2*g)/(B*a**2*d**2*g + 2*B*a*b*c*d
*g - B*b**2*c**2*g))/b - B*c*g*(2*a*d - b*c)*log(x + (3*B*a**2*c*d*g - B*a*b*c**2*g - B*a*c*g*(2*a*d - b*c) +
B*b*c**2*g*(2*a*d - b*c)/d)/(B*a**2*d**2*g + 2*B*a*b*c*d*g - B*b**2*c**2*g))/d**2 + (B*a*g*x + B*b*g*x**2/2)*l
og(e*(a + b*x)**2/(c + d*x)**2) + x*(A*a*d*g + B*a*d*g - B*b*c*g)/d

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Giac [A]  time = 1.74159, size = 177, normalized size = 2.27 \begin{align*} \frac{B a^{2} g \log \left (b x + a\right )}{b} + \frac{1}{2} \,{\left (A b g + B b g\right )} x^{2} + \frac{1}{2} \,{\left (B b g x^{2} + 2 \, B a g 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 (B b c g - A a d g - 2 \, B a d g\right )} x}{d} + \frac{{\left (B b c^{2} g - 2 \, B a c d g\right )} \log \left (d x + c\right )}{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)^2/(d*x+c)^2)),x, algorithm="giac")

[Out]

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