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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2525, 12, 43} \[ \frac {g (a+b x)^2 \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )}{2 b}-\frac {B g (b c-a d)^2 \log (c+d x)}{b d^2}+\frac {B g x (b c-a d)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 72, normalized size = 0.92 \[ \frac {g \left ((a+b x)^2 \left (B \log \left (\frac {e (c+d x)^2}{(a+b x)^2}\right )+A\right )-\frac {2 B (a d-b c) ((a d-b c) \log (c+d x)+b d x)}{d^2}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 149, normalized size = 1.91 \[ \frac {A b^{2} d^{2} g x^{2} - 2 \, B a^{2} d^{2} g \log \left (b x + a\right ) + 2 \, {\left (B b^{2} c d + {\left (A - B\right )} a b d^{2}\right )} g x - 2 \, {\left (B b^{2} c^{2} - 2 \, B a b c d\right )} g \log \left (d x + c\right ) + {\left (B b^{2} d^{2} g x^{2} + 2 \, B a b d^{2} g x\right )} \log \left (\frac {d^{2} e x^{2} + 2 \, c d e x + c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}{2 \, b d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.90, size = 128, normalized size = 1.64 \[ -\frac {B a^{2} g \log \left (b x + a\right )}{b} + \frac {1}{2} \, {\left (A b g + B b g\right )} x^{2} + \frac {1}{2} \, {\left (B b g x^{2} + 2 \, B a g x\right )} \log \left (\frac {d^{2} x^{2} + 2 \, c d x + c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + \frac {{\left (B b c g + A a d g\right )} x}{d} - \frac {{\left (B b c^{2} g - 2 \, B a c d g\right )} \log \left (-d x - c\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.06, size = 340, normalized size = 4.36 \[ \frac {B b g \,x^{2} \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )}{2}+\frac {A b g \,x^{2}}{2}+B a g x \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )+A a g x +\frac {B \,a^{2} g \ln \left (\frac {1}{b x +a}\right )}{b}+\frac {B \,a^{2} g \ln \left (\frac {\left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )^{2} e}{b^{2}}\right )}{2 b}-\frac {B \,a^{2} g \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{b}-\frac {2 B a c g \ln \left (\frac {1}{b x +a}\right )}{d}+\frac {2 B a c g \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{d}-B a g x +\frac {B b \,c^{2} g \ln \left (\frac {1}{b x +a}\right )}{d^{2}}-\frac {B b \,c^{2} g \ln \left (\frac {a d}{b x +a}-\frac {b c}{b x +a}-d \right )}{d^{2}}+\frac {B b c g x}{d}+\frac {A \,a^{2} g}{2 b}-\frac {B \,a^{2} g}{b}+\frac {B a c g}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.44, size = 250, normalized size = 3.21 \[ \frac {1}{2} \, A b g x^{2} + {\left (x \log \left (\frac {d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) - \frac {2 \, a \log \left (b x + a\right )}{b} + \frac {2 \, c \log \left (d x + c\right )}{d}\right )} B a g + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {d^{2} e x^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {2 \, c d e x}{b^{2} x^{2} + 2 \, a b x + a^{2}} + \frac {c^{2} e}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right ) + \frac {2 \, a^{2} \log \left (b x + a\right )}{b^{2}} - \frac {2 \, c^{2} \log \left (d x + c\right )}{d^{2}} + \frac {2 \, {\left (b c - a d\right )} x}{b d}\right )} B b g + A a g x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*A*b*g*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*g + 1/2*(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*b*g + A*a*g*x

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mupad [B]  time = 4.38, size = 120, normalized size = 1.54 \[ x\,\left (\frac {g\,\left (2\,A\,a\,d+A\,b\,c-B\,a\,d+B\,b\,c\right )}{d}-\frac {A\,g\,\left (a\,d+b\,c\right )}{d}\right )+\ln \left (\frac {e\,{\left (c+d\,x\right )}^2}{{\left (a+b\,x\right )}^2}\right )\,\left (\frac {B\,b\,g\,x^2}{2}+B\,a\,g\,x\right )+\frac {A\,b\,g\,x^2}{2}-\frac {B\,a^2\,g\,\ln \left (a+b\,x\right )}{b}+\frac {B\,c\,g\,\ln \left (c+d\,x\right )\,\left (2\,a\,d-b\,c\right )}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.92, size = 250, normalized size = 3.21 \[ \frac {A b g x^{2}}{2} - \frac {B a^{2} g \log {\left (x + \frac {\frac {B a^{3} d^{2} g}{b} + 2 B a^{2} c d g - B a b c^{2} g}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{b} + \frac {B c g \left (2 a d - b c\right ) \log {\left (x + \frac {3 B a^{2} c d g - B a b c^{2} g - B a c g \left (2 a d - b c\right ) + \frac {B b c^{2} g \left (2 a d - b c\right )}{d}}{B a^{2} d^{2} g + 2 B a b c d g - B b^{2} c^{2} g} \right )}}{d^{2}} + x \left (A a g - B a g + \frac {B b c g}{d}\right ) + \left (B a g x + \frac {B b g x^{2}}{2}\right ) \log {\left (\frac {e \left (c + d x\right )^{2}}{\left (a + b x\right )^{2}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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