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

Optimal. Leaf size=64 \[ -\frac {A-B}{b g^2 (a+b x)}-\frac {B (c+d x) \log \left (\frac {e (c+d x)}{a+b x}\right )}{g^2 (a+b x) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 101, normalized size of antiderivative = 1.58, 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, 44} \[ -\frac {B \log \left (\frac {e (c+d x)}{a+b x}\right )+A}{b g^2 (a+b x)}+\frac {B d \log (a+b x)}{b g^2 (b c-a d)}-\frac {B d \log (c+d x)}{b g^2 (b c-a d)}+\frac {B}{b g^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 \frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{(a g+b g x)^2} \, dx &=-\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{b g^2 (a+b x)}+\frac {B \int \frac {-b c+a d}{g (a+b x)^2 (c+d x)} \, dx}{b g}\\ &=-\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{b g^2 (a+b x)}-\frac {(B (b c-a d)) \int \frac {1}{(a+b x)^2 (c+d x)} \, dx}{b g^2}\\ &=-\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{b g^2 (a+b x)}-\frac {(B (b c-a d)) \int \left (\frac {b}{(b c-a d) (a+b x)^2}-\frac {b d}{(b c-a d)^2 (a+b x)}+\frac {d^2}{(b c-a d)^2 (c+d x)}\right ) \, dx}{b g^2}\\ &=\frac {B}{b g^2 (a+b x)}+\frac {B d \log (a+b x)}{b (b c-a d) g^2}-\frac {B d \log (c+d x)}{b (b c-a d) g^2}-\frac {A+B \log \left (\frac {e (c+d x)}{a+b x}\right )}{b g^2 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 86, normalized size = 1.34 \[ \frac {-(b c-a d) \left (B \log \left (\frac {e (c+d x)}{a+b x}\right )+A-B\right )-B d (a+b x) \log (c+d x)+B d (a+b x) \log (a+b x)}{b g^2 (a+b x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.73, size = 87, normalized size = 1.36 \[ -\frac {{\left (A - B\right )} b c - {\left (A - B\right )} a d + {\left (B b d x + B b c\right )} \log \left (\frac {d e x + c e}{b x + a}\right )}{{\left (b^{3} c - a b^{2} d\right )} g^{2} x + {\left (a b^{2} c - a^{2} b d\right )} g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.90, size = 126, normalized size = 1.97 \[ -{\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )} {\left (\frac {{\left (d x e + c e\right )} B \log \left (\frac {d x e + c e}{b x + a}\right )}{{\left (b x + a\right )} g^{2}} + \frac {{\left (d x e + c e\right )} {\left (A - B\right )}}{{\left (b x + a\right )} g^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 520, normalized size = 8.12 \[ -\frac {B \,a^{2} d^{2} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{\left (a d -b c \right )^{2} \left (b x +a \right ) b \,g^{2}}+\frac {2 B a c d \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}-\frac {B b \,c^{2} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}-\frac {A \,a^{2} d^{2}}{\left (a d -b c \right )^{2} \left (b x +a \right ) b \,g^{2}}+\frac {2 A a c d}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}-\frac {A b \,c^{2}}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}+\frac {B \,a^{2} d^{2}}{\left (a d -b c \right )^{2} \left (b x +a \right ) b \,g^{2}}+\frac {B a \,d^{2} \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{\left (a d -b c \right )^{2} b \,g^{2}}-\frac {2 B a c d}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}+\frac {B b \,c^{2}}{\left (a d -b c \right )^{2} \left (b x +a \right ) g^{2}}-\frac {B c d \ln \left (\frac {d e}{b}-\frac {\left (a d -b c \right ) e}{\left (b x +a \right ) b}\right )}{\left (a d -b c \right )^{2} g^{2}}+\frac {A a \,d^{2}}{\left (a d -b c \right )^{2} b \,g^{2}}-\frac {A c d}{\left (a d -b c \right )^{2} g^{2}}-\frac {B a \,d^{2}}{\left (a d -b c \right )^{2} b \,g^{2}}+\frac {B c d}{\left (a d -b c \right )^{2} g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.07, size = 134, normalized size = 2.09 \[ -B {\left (\frac {\log \left (\frac {d e x}{b x + a} + \frac {c e}{b x + a}\right )}{b^{2} g^{2} x + a b g^{2}} - \frac {1}{b^{2} g^{2} x + a b g^{2}} - \frac {d \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} g^{2}} + \frac {d \log \left (d x + c\right )}{{\left (b^{2} c - a b d\right )} g^{2}}\right )} - \frac {A}{b^{2} g^{2} x + a b g^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.01, size = 106, normalized size = 1.66 \[ -\frac {A-B}{x\,b^2\,g^2+a\,b\,g^2}-\frac {B\,\ln \left (\frac {e\,\left (c+d\,x\right )}{a+b\,x}\right )}{b^2\,g^2\,\left (x+\frac {a}{b}\right )}+\frac {B\,d\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{b\,g^2\,\left (a\,d-b\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.54, size = 231, normalized size = 3.61 \[ - \frac {B \log {\left (\frac {e \left (c + d x\right )}{a + b x} \right )}}{a b g^{2} + b^{2} g^{2} x} + \frac {B d \log {\left (x + \frac {- \frac {B a^{2} d^{3}}{a d - b c} + \frac {2 B a b c d^{2}}{a d - b c} + B a d^{2} - \frac {B b^{2} c^{2} d}{a d - b c} + B b c d}{2 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} - \frac {B d \log {\left (x + \frac {\frac {B a^{2} d^{3}}{a d - b c} - \frac {2 B a b c d^{2}}{a d - b c} + B a d^{2} + \frac {B b^{2} c^{2} d}{a d - b c} + B b c d}{2 B b d^{2}} \right )}}{b g^{2} \left (a d - b c\right )} + \frac {- A + B}{a b g^{2} + b^{2} g^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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