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

Optimal. Leaf size=120 \[ \frac{g^2 (a+b x)^3 \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{3 b}-\frac{2 B g^2 x (b c-a d)^2}{3 d^2}+\frac{2 B g^2 (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac{B g^2 (a+b x)^2 (b c-a d)}{3 b d} \]

[Out]

(-2*B*(b*c - a*d)^2*g^2*x)/(3*d^2) + (B*(b*c - a*d)*g^2*(a + b*x)^2)/(3*b*d) + (2*B*(b*c - a*d)^3*g^2*Log[c +
d*x])/(3*b*d^3) + (g^2*(a + b*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2]))/(3*b)

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Rubi [A]  time = 0.0783567, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {2525, 12, 43} \[ \frac{g^2 (a+b x)^3 \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{3 b}-\frac{2 B g^2 x (b c-a d)^2}{3 d^2}+\frac{2 B g^2 (b c-a d)^3 \log (c+d x)}{3 b d^3}+\frac{B g^2 (a+b x)^2 (b c-a d)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*B*(b*c - a*d)^2*g^2*x)/(3*d^2) + (B*(b*c - a*d)*g^2*(a + b*x)^2)/(3*b*d) + (2*B*(b*c - a*d)^3*g^2*Log[c +
d*x])/(3*b*d^3) + (g^2*(a + b*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2]))/(3*b)

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)^2 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right ) \, dx &=\frac{g^2 (a+b x)^3 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b}-\frac{B \int \frac{2 (-b c+a d) g^3 (a+b x)^2}{c+d x} \, dx}{3 b g}\\ &=\frac{g^2 (a+b x)^3 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b}+\frac{\left (2 B (b c-a d) g^2\right ) \int \frac{(a+b x)^2}{c+d x} \, dx}{3 b}\\ &=\frac{g^2 (a+b x)^3 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b}+\frac{\left (2 B (b c-a d) g^2\right ) \int \left (-\frac{b (b c-a d)}{d^2}+\frac{b (a+b x)}{d}+\frac{(-b c+a d)^2}{d^2 (c+d x)}\right ) \, dx}{3 b}\\ &=-\frac{2 B (b c-a d)^2 g^2 x}{3 d^2}+\frac{B (b c-a d) g^2 (a+b x)^2}{3 b d}+\frac{2 B (b c-a d)^3 g^2 \log (c+d x)}{3 b d^3}+\frac{g^2 (a+b x)^3 \left (A+B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )\right )}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.050639, size = 98, normalized size = 0.82 \[ \frac{g^2 \left (\frac{B (b c-a d) \left (d \left (a^2 d+4 a b d x+b^2 x (d x-2 c)\right )+2 (b c-a d)^2 \log (c+d x)\right )}{d^3}+(a+b x)^3 \left (B \log \left (\frac{e (c+d x)^2}{(a+b x)^2}\right )+A\right )\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g^2*((B*(b*c - a*d)*(d*(a^2*d + 4*a*b*d*x + b^2*x*(-2*c + d*x)) + 2*(b*c - a*d)^2*Log[c + d*x]))/d^3 + (a + b
*x)^3*(A + B*Log[(e*(c + d*x)^2)/(a + b*x)^2])))/(3*b)

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Maple [B]  time = 0.241, size = 569, normalized size = 4.7 \begin{align*}{\frac{5\,{g}^{2}B{a}^{2}c}{3\,d}}+2\,{\frac{{g}^{2}B{a}^{2}c}{d}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) }+{\frac{{b}^{2}{g}^{2}Bc{x}^{2}}{3\,d}}-{\frac{2\,{b}^{2}{g}^{2}B{c}^{2}x}{3\,{d}^{2}}}-{\frac{2\,b{g}^{2}Ba{c}^{2}}{3\,{d}^{2}}}-{\frac{2\,{g}^{2}B{a}^{3}}{3\,b}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) }+{\frac{{b}^{2}A{x}^{3}{g}^{2}}{3}}+{\frac{{b}^{2}B{x}^{3}{g}^{2}}{3}\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ) }+bB\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ){x}^{2}a{g}^{2}+B\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ) x{a}^{2}{g}^{2}+{\frac{{g}^{2}B{a}^{3}}{3\,b}\ln \left ({\frac{e}{{b}^{2}} \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) ^{2}} \right ) }+{\frac{2\,{g}^{2}B{a}^{3}\ln \left ( \left ( bx+a \right ) ^{-1} \right ) }{3\,b}}+{\frac{2\,{b}^{2}{g}^{2}B{c}^{3}}{3\,{d}^{3}}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) }-{\frac{2\,{b}^{2}{g}^{2}B{c}^{3}\ln \left ( \left ( bx+a \right ) ^{-1} \right ) }{3\,{d}^{3}}}-2\,{\frac{b{g}^{2}Ba{c}^{2}}{{d}^{2}}\ln \left ({\frac{ad}{bx+a}}-{\frac{bc}{bx+a}}-d \right ) }+2\,{\frac{b{g}^{2}Ba\ln \left ( \left ( bx+a \right ) ^{-1} \right ){c}^{2}}{{d}^{2}}}-2\,{\frac{{g}^{2}B{a}^{2}\ln \left ( \left ( bx+a \right ) ^{-1} \right ) c}{d}}+2\,{\frac{b{g}^{2}Bacx}{d}}-{\frac{4\,Bx{a}^{2}{g}^{2}}{3}}-{\frac{bB{x}^{2}a{g}^{2}}{3}}+bA{x}^{2}a{g}^{2}+Ax{a}^{2}{g}^{2}+{\frac{A{a}^{3}{g}^{2}}{3\,b}}-{\frac{{g}^{2}B{a}^{3}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*g^2*B*a^2/d*c+2*g^2*B*a^2/d*ln(1/(b*x+a)*a*d-b*c/(b*x+a)-d)*c+1/3*b^2*g^2*B*c/d*x^2-2/3*b^2*g^2*B*c^2/d^2*
x-2/3*b*g^2*B*a/d^2*c^2-2/3/b*g^2*B*a^3*ln(1/(b*x+a)*a*d-b*c/(b*x+a)-d)+1/3*b^2*A*x^3*g^2+1/3*b^2*B*ln(e*(1/(b
*x+a)*a*d-b*c/(b*x+a)-d)^2/b^2)*x^3*g^2+b*B*ln(e*(1/(b*x+a)*a*d-b*c/(b*x+a)-d)^2/b^2)*x^2*a*g^2+B*ln(e*(1/(b*x
+a)*a*d-b*c/(b*x+a)-d)^2/b^2)*x*a^2*g^2+1/3/b*B*ln(e*(1/(b*x+a)*a*d-b*c/(b*x+a)-d)^2/b^2)*a^3*g^2+2/3/b*g^2*B*
a^3*ln(1/(b*x+a))+2/3*b^2*g^2*B*c^3/d^3*ln(1/(b*x+a)*a*d-b*c/(b*x+a)-d)-2/3*b^2*g^2*B*c^3/d^3*ln(1/(b*x+a))-2*
b*g^2*B*a/d^2*ln(1/(b*x+a)*a*d-b*c/(b*x+a)-d)*c^2+2*b*g^2*B*a/d^2*ln(1/(b*x+a))*c^2-2*g^2*B*a^2/d*ln(1/(b*x+a)
)*c+2*b*g^2*B*a/d*c*x-4/3*B*x*a^2*g^2-1/3*b*B*x^2*a*g^2+b*A*x^2*a*g^2+A*x*a^2*g^2+1/3/b*A*a^3*g^2-1/b*g^2*B*a^
3

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Maxima [B]  time = 1.33054, size = 589, normalized size = 4.91 \begin{align*} \frac{1}{3} \, A b^{2} g^{2} x^{3} + A a b g^{2} x^{2} +{\left (x \log \left (\frac{d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{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^{2} g^{2} +{\left (x^{2} \log \left (\frac{d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{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 a b g^{2} + \frac{1}{3} \,{\left (x^{3} \log \left (\frac{d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac{c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) - \frac{2 \, a^{3} \log \left (b x + a\right )}{b^{3}} + \frac{2 \, c^{3} \log \left (d x + c\right )}{d^{3}} + \frac{{\left (b^{2} c d - a b d^{2}\right )} x^{2} - 2 \,{\left (b^{2} c^{2} - a^{2} d^{2}\right )} x}{b^{2} d^{2}}\right )} B b^{2} g^{2} + A a^{2} g^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^2*(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.11278, size = 508, normalized size = 4.23 \begin{align*} \frac{A b^{3} d^{3} g^{2} x^{3} - 2 \, B a^{3} d^{3} g^{2} \log \left (b x + a\right ) +{\left (B b^{3} c d^{2} +{\left (3 \, A - B\right )} a b^{2} d^{3}\right )} g^{2} x^{2} -{\left (2 \, B b^{3} c^{2} d - 6 \, B a b^{2} c d^{2} -{\left (3 \, A - 4 \, B\right )} a^{2} b d^{3}\right )} g^{2} x + 2 \,{\left (B b^{3} c^{3} - 3 \, B a b^{2} c^{2} d + 3 \, B a^{2} b c d^{2}\right )} g^{2} \log \left (d x + c\right ) +{\left (B b^{3} d^{3} g^{2} x^{3} + 3 \, B a b^{2} d^{3} g^{2} x^{2} + 3 \, B a^{2} b d^{3} g^{2} x\right )} \log \left (\frac{d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{3 \, b d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^2*(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="fricas")

[Out]

1/3*(A*b^3*d^3*g^2*x^3 - 2*B*a^3*d^3*g^2*log(b*x + a) + (B*b^3*c*d^2 + (3*A - B)*a*b^2*d^3)*g^2*x^2 - (2*B*b^3
*c^2*d - 6*B*a*b^2*c*d^2 - (3*A - 4*B)*a^2*b*d^3)*g^2*x + 2*(B*b^3*c^3 - 3*B*a*b^2*c^2*d + 3*B*a^2*b*c*d^2)*g^
2*log(d*x + c) + (B*b^3*d^3*g^2*x^3 + 3*B*a*b^2*d^3*g^2*x^2 + 3*B*a^2*b*d^3*g^2*x)*log((d^2*e*x^2 + 2*c*d*e*x
+ c^2*e)/(b^2*x^2 + 2*a*b*x + a^2)))/(b*d^3)

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Sympy [B]  time = 4.15602, size = 527, normalized size = 4.39 \begin{align*} \frac{A b^{2} g^{2} x^{3}}{3} - \frac{2 B a^{3} g^{2} \log{\left (x + \frac{\frac{2 B a^{4} d^{3} g^{2}}{b} + 6 B a^{3} c d^{2} g^{2} - 6 B a^{2} b c^{2} d g^{2} + 2 B a b^{2} c^{3} g^{2}}{2 B a^{3} d^{3} g^{2} + 6 B a^{2} b c d^{2} g^{2} - 6 B a b^{2} c^{2} d g^{2} + 2 B b^{3} c^{3} g^{2}} \right )}}{3 b} + \frac{2 B c g^{2} \left (3 a^{2} d^{2} - 3 a b c d + b^{2} c^{2}\right ) \log{\left (x + \frac{8 B a^{3} c d^{2} g^{2} - 6 B a^{2} b c^{2} d g^{2} + 2 B a b^{2} c^{3} g^{2} - 2 B a c g^{2} \left (3 a^{2} d^{2} - 3 a b c d + b^{2} c^{2}\right ) + \frac{2 B b c^{2} g^{2} \left (3 a^{2} d^{2} - 3 a b c d + b^{2} c^{2}\right )}{d}}{2 B a^{3} d^{3} g^{2} + 6 B a^{2} b c d^{2} g^{2} - 6 B a b^{2} c^{2} d g^{2} + 2 B b^{3} c^{3} g^{2}} \right )}}{3 d^{3}} + \left (B a^{2} g^{2} x + B a b g^{2} x^{2} + \frac{B b^{2} g^{2} x^{3}}{3}\right ) \log{\left (\frac{e \left (c + d x\right )^{2}}{\left (a + b x\right )^{2}} \right )} + \frac{x^{2} \left (3 A a b d g^{2} - B a b d g^{2} + B b^{2} c g^{2}\right )}{3 d} + \frac{x \left (3 A a^{2} d^{2} g^{2} - 4 B a^{2} d^{2} g^{2} + 6 B a b c d g^{2} - 2 B b^{2} c^{2} g^{2}\right )}{3 d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*b**2*g**2*x**3/3 - 2*B*a**3*g**2*log(x + (2*B*a**4*d**3*g**2/b + 6*B*a**3*c*d**2*g**2 - 6*B*a**2*b*c**2*d*g*
*2 + 2*B*a*b**2*c**3*g**2)/(2*B*a**3*d**3*g**2 + 6*B*a**2*b*c*d**2*g**2 - 6*B*a*b**2*c**2*d*g**2 + 2*B*b**3*c*
*3*g**2))/(3*b) + 2*B*c*g**2*(3*a**2*d**2 - 3*a*b*c*d + b**2*c**2)*log(x + (8*B*a**3*c*d**2*g**2 - 6*B*a**2*b*
c**2*d*g**2 + 2*B*a*b**2*c**3*g**2 - 2*B*a*c*g**2*(3*a**2*d**2 - 3*a*b*c*d + b**2*c**2) + 2*B*b*c**2*g**2*(3*a
**2*d**2 - 3*a*b*c*d + b**2*c**2)/d)/(2*B*a**3*d**3*g**2 + 6*B*a**2*b*c*d**2*g**2 - 6*B*a*b**2*c**2*d*g**2 + 2
*B*b**3*c**3*g**2))/(3*d**3) + (B*a**2*g**2*x + B*a*b*g**2*x**2 + B*b**2*g**2*x**3/3)*log(e*(c + d*x)**2/(a +
b*x)**2) + x**2*(3*A*a*b*d*g**2 - B*a*b*d*g**2 + B*b**2*c*g**2)/(3*d) + x*(3*A*a**2*d**2*g**2 - 4*B*a**2*d**2*
g**2 + 6*B*a*b*c*d*g**2 - 2*B*b**2*c**2*g**2)/(3*d**2)

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Giac [B]  time = 4.72178, size = 335, normalized size = 2.79 \begin{align*} -\frac{2 \, B a^{3} g^{2} \log \left (b x + a\right )}{3 \, b} + \frac{1}{3} \,{\left (A b^{2} g^{2} + B b^{2} g^{2}\right )} x^{3} + \frac{{\left (B b^{2} c g^{2} + 3 \, A a b d g^{2} + 2 \, B a b d g^{2}\right )} x^{2}}{3 \, d} + \frac{1}{3} \,{\left (B b^{2} g^{2} x^{3} + 3 \, B a b g^{2} x^{2} + 3 \, B a^{2} g^{2} x\right )} \log \left (\frac{d^{2} x^{2} + 2 \, c d x + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) - \frac{{\left (2 \, B b^{2} c^{2} g^{2} - 6 \, B a b c d g^{2} - 3 \, A a^{2} d^{2} g^{2} + B a^{2} d^{2} g^{2}\right )} x}{3 \, d^{2}} + \frac{2 \,{\left (B b^{2} c^{3} g^{2} - 3 \, B a b c^{2} d g^{2} + 3 \, B a^{2} c d^{2} g^{2}\right )} \log \left (d x + c\right )}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*g*x+a*g)^2*(A+B*log(e*(d*x+c)^2/(b*x+a)^2)),x, algorithm="giac")

[Out]

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