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

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Rubi [A]  time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 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]

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 (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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.87, size = 174, normalized size = 1.67 \[ \frac {A b^{2} d^{2} g x^{2} + 2 \, {\left (A b^{2} d^{2} f - {\left (B b^{2} c d - B a b d^{2}\right )} g\right )} x + 2 \, {\left (2 \, B a b d^{2} f - B a^{2} d^{2} g\right )} \log \left (b x + a\right ) - 2 \, {\left (2 \, B b^{2} c d f - B b^{2} c^{2} g\right )} \log \left (d x + c\right ) + {\left (B b^{2} d^{2} g x^{2} + 2 \, B b^{2} d^{2} f x\right )} \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right )}{2 \, b^{2} d^{2}} \]

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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giac [A]  time = 1.11, size = 145, normalized size = 1.39 \[ \frac {1}{2} \, {\left (A g + B g\right )} x^{2} + \frac {1}{2} \, {\left (B g x^{2} + 2 \, B f x\right )} \log \left (\frac {b^{2} x^{2} + 2 \, a b x + a^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac {{\left (A b d f + B b d f - B b c g + B a d g\right )} x}{b d} + \frac {{\left (2 \, B a b f - B a^{2} g\right )} \log \left (b x + a\right )}{b^{2}} - \frac {{\left (2 \, B c d f - B c^{2} g\right )} \log \left (-d x - c\right )}{d^{2}} \]

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

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maple [B]  time = 0.08, size = 656, normalized size = 6.31 \[ -\frac {2 B \,a^{2} c g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) b}+\frac {2 B \,a^{2} d f \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) b}+\frac {4 B a \,c^{2} g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) d}-\frac {4 B a c f \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{a d -b c}-\frac {2 B b \,c^{3} g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) d^{2}}+\frac {2 B b \,c^{2} f \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{\left (a d -b c \right ) d}+\frac {B g \,x^{2} \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{2}+\frac {A g \,x^{2}}{2}+B f x \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )+A f x +\frac {B \,a^{2} g \ln \left (\frac {1}{d x +c}\right )}{b^{2}}-\frac {B \,a^{2} g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b^{2}}+\frac {2 B a c g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{b d}-\frac {2 B a f \ln \left (\frac {1}{d x +c}\right )}{b}+\frac {B a g x}{b}-\frac {B \,c^{2} g \ln \left (\frac {1}{d x +c}\right )}{d^{2}}-\frac {B \,c^{2} g \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{2 d^{2}}-\frac {B \,c^{2} g \ln \left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )}{d^{2}}+\frac {2 B c f \ln \left (\frac {1}{d x +c}\right )}{d}+\frac {B c f \ln \left (\frac {\left (\frac {a d}{d x +c}-\frac {b c}{d x +c}+b \right )^{2} e}{d^{2}}\right )}{d}-\frac {B c g x}{d}-\frac {A \,c^{2} g}{2 d^{2}}+\frac {A c f}{d}+\frac {B a c g}{b d}-\frac {B \,c^{2} g}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.76, size = 246, normalized size = 2.37 \[ \frac {1}{2} \, A g x^{2} + {\left (x \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + \frac {2 \, a \log \left (b x + a\right )}{b} - \frac {2 \, c \log \left (d x + c\right )}{d}\right )} B f + \frac {1}{2} \, {\left (x^{2} \log \left (\frac {b^{2} e x^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {2 \, a b e x}{d^{2} x^{2} + 2 \, c d x + c^{2}} + \frac {a^{2} e}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - \frac {2 \, a^{2} \log \left (b x + a\right )}{b^{2}} + \frac {2 \, c^{2} \log \left (d x + c\right )}{d^{2}} - \frac {2 \, {\left (b c - a d\right )} x}{b d}\right )} B g + A f x \]

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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mupad [B]  time = 4.50, size = 133, normalized size = 1.28 \[ \ln \left (\frac {e\,{\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2}\right )\,\left (\frac {B\,g\,x^2}{2}+B\,f\,x\right )+x\,\left (\frac {A\,a\,d\,g+A\,b\,c\,g+A\,b\,d\,f+B\,a\,d\,g-B\,b\,c\,g}{b\,d}-\frac {A\,g\,\left (a\,d+b\,c\right )}{b\,d}\right )+\frac {A\,g\,x^2}{2}-\frac {B\,a\,\ln \left (a+b\,x\right )\,\left (a\,g-2\,b\,f\right )}{b^2}+\frac {B\,c\,\ln \left (c+d\,x\right )\,\left (c\,g-2\,d\,f\right )}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.79, size = 314, normalized size = 3.02 \[ \frac {A g x^{2}}{2} - \frac {B a \left (a g - 2 b f\right ) \log {\left (x + \frac {B a^{2} c d g + \frac {B a^{2} d^{2} \left (a g - 2 b f\right )}{b} + B a b c^{2} g - 4 B a b c d f - B a c d \left (a g - 2 b f\right )}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{b^{2}} + \frac {B c \left (c g - 2 d f\right ) \log {\left (x + \frac {B a^{2} c d g + B a b c^{2} g - 4 B a b c d f - B a b c \left (c g - 2 d f\right ) + \frac {B b^{2} c^{2} \left (c g - 2 d f\right )}{d}}{B a^{2} d^{2} g - 2 B a b d^{2} f + B b^{2} c^{2} g - 2 B b^{2} c d f} \right )}}{d^{2}} + x \left (A f + \frac {B a g}{b} - \frac {B c g}{d}\right ) + \left (B f x + \frac {B g x^{2}}{2}\right ) \log {\left (\frac {e \left (a + b x\right )^{2}}{\left (c + d x\right )^{2}} \right )} \]

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

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