3.50 \(\int \frac{A+B \log (\frac{e (a+b x)}{c+d x})}{(c i+d i x)^3} \, dx\)

Optimal. Leaf size=144 \[ -\frac{B \log \left (\frac{e (a+b x)}{c+d x}\right )+A}{2 d i^3 (c+d x)^2}+\frac{b^2 B \log (a+b x)}{2 d i^3 (b c-a d)^2}-\frac{b^2 B \log (c+d x)}{2 d i^3 (b c-a d)^2}+\frac{b B}{2 d i^3 (c+d x) (b c-a d)}+\frac{B}{4 d i^3 (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0993907, antiderivative size = 144, 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, 44} \[ -\frac{B \log \left (\frac{e (a+b x)}{c+d x}\right )+A}{2 d i^3 (c+d x)^2}+\frac{b^2 B \log (a+b x)}{2 d i^3 (b c-a d)^2}-\frac{b^2 B \log (c+d x)}{2 d i^3 (b c-a d)^2}+\frac{b B}{2 d i^3 (c+d x) (b c-a d)}+\frac{B}{4 d i^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(c*i + d*i*x)^3,x]

[Out]

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

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 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])

Rubi steps

\begin{align*} \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(50 c+50 d x)^3} \, dx &=-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{250000 d (c+d x)^2}+\frac{B \int \frac{b c-a d}{2500 (a+b x) (c+d x)^3} \, dx}{100 d}\\ &=-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{250000 d (c+d x)^2}+\frac{(B (b c-a d)) \int \frac{1}{(a+b x) (c+d x)^3} \, dx}{250000 d}\\ &=-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{250000 d (c+d x)^2}+\frac{(B (b c-a d)) \int \left (\frac{b^3}{(b c-a d)^3 (a+b x)}-\frac{d}{(b c-a d) (c+d x)^3}-\frac{b d}{(b c-a d)^2 (c+d x)^2}-\frac{b^2 d}{(b c-a d)^3 (c+d x)}\right ) \, dx}{250000 d}\\ &=\frac{B}{500000 d (c+d x)^2}+\frac{b B}{250000 d (b c-a d) (c+d x)}+\frac{b^2 B \log (a+b x)}{250000 d (b c-a d)^2}-\frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{250000 d (c+d x)^2}-\frac{b^2 B \log (c+d x)}{250000 d (b c-a d)^2}\\ \end{align*}

Mathematica [A]  time = 0.11675, size = 111, normalized size = 0.77 \[ \frac{\frac{B \left (2 b^2 (c+d x)^2 \log (a+b x)+(b c-a d) (-a d+3 b c+2 b d x)-2 b^2 (c+d x)^2 \log (c+d x)\right )}{(b c-a d)^2}-2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{4 d i^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Log[(e*(a + b*x))/(c + d*x)])/(c*i + d*i*x)^3,x]

[Out]

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

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Maple [B]  time = 0.052, size = 746, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*ln(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^3,x)

[Out]

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

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Maxima [A]  time = 1.19764, size = 344, normalized size = 2.39 \begin{align*} \frac{1}{4} \, B{\left (\frac{2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} i^{3} x^{2} + 2 \,{\left (b c^{2} d^{2} - a c d^{3}\right )} i^{3} x +{\left (b c^{3} d - a c^{2} d^{2}\right )} i^{3}} - \frac{2 \, \log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right )}{d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}} + \frac{2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}} - \frac{2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}}\right )} - \frac{A}{2 \,{\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.486076, size = 455, normalized size = 3.16 \begin{align*} -\frac{{\left (2 \, A - 3 \, B\right )} b^{2} c^{2} - 4 \,{\left (A - B\right )} a b c d +{\left (2 \, A - B\right )} a^{2} d^{2} - 2 \,{\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \,{\left (B b^{2} d^{2} x^{2} + 2 \, B b^{2} c d x + 2 \, B a b c d - B a^{2} d^{2}\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{4 \,{\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} i^{3} x^{2} + 2 \,{\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} i^{3} x +{\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} i^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 2.99029, size = 422, normalized size = 2.93 \begin{align*} - \frac{B b^{2} \log{\left (x + \frac{- \frac{B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} + \frac{3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} - \frac{3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d + \frac{B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} + \frac{B b^{2} \log{\left (x + \frac{\frac{B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} - \frac{3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} + \frac{3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d - \frac{B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} - \frac{B \log{\left (\frac{e \left (a + b x\right )}{c + d x} \right )}}{2 c^{2} d i^{3} + 4 c d^{2} i^{3} x + 2 d^{3} i^{3} x^{2}} - \frac{2 A a d - 2 A b c - B a d + 3 B b c + 2 B b d x}{4 a c^{2} d^{2} i^{3} - 4 b c^{3} d i^{3} + x^{2} \left (4 a d^{4} i^{3} - 4 b c d^{3} i^{3}\right ) + x \left (8 a c d^{3} i^{3} - 8 b c^{2} d^{2} i^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*ln(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)**3,x)

[Out]

-B*b**2*log(x + (-B*a**3*b**2*d**3/(a*d - b*c)**2 + 3*B*a**2*b**3*c*d**2/(a*d - b*c)**2 - 3*B*a*b**4*c**2*d/(a
*d - b*c)**2 + B*a*b**2*d + B*b**5*c**3/(a*d - b*c)**2 + B*b**3*c)/(2*B*b**3*d))/(2*d*i**3*(a*d - b*c)**2) + B
*b**2*log(x + (B*a**3*b**2*d**3/(a*d - b*c)**2 - 3*B*a**2*b**3*c*d**2/(a*d - b*c)**2 + 3*B*a*b**4*c**2*d/(a*d
- b*c)**2 + B*a*b**2*d - B*b**5*c**3/(a*d - b*c)**2 + B*b**3*c)/(2*B*b**3*d))/(2*d*i**3*(a*d - b*c)**2) - B*lo
g(e*(a + b*x)/(c + d*x))/(2*c**2*d*i**3 + 4*c*d**2*i**3*x + 2*d**3*i**3*x**2) - (2*A*a*d - 2*A*b*c - B*a*d + 3
*B*b*c + 2*B*b*d*x)/(4*a*c**2*d**2*i**3 - 4*b*c**3*d*i**3 + x**2*(4*a*d**4*i**3 - 4*b*c*d**3*i**3) + x*(8*a*c*
d**3*i**3 - 8*b*c**2*d**2*i**3))

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Giac [A]  time = 1.31786, size = 282, normalized size = 1.96 \begin{align*} -\frac{B b^{2} \log \left (b x + a\right )}{2 \,{\left (b^{2} c^{2} d i - 2 \, a b c d^{2} i + a^{2} d^{3} i\right )}} + \frac{B b^{2} \log \left (d x + c\right )}{2 \,{\left (b^{2} c^{2} d i - 2 \, a b c d^{2} i + a^{2} d^{3} i\right )}} - \frac{B i \log \left (\frac{b x + a}{d x + c}\right )}{2 \,{\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} + \frac{2 \, B b d i x - 2 \, A b c i + B b c i + 2 \, A a d i + B a d i}{4 \,{\left (b c d^{3} x^{2} - a d^{4} x^{2} + 2 \, b c^{2} d^{2} x - 2 \, a c d^{3} x + b c^{3} d - a c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*log(e*(b*x+a)/(d*x+c)))/(d*i*x+c*i)^3,x, algorithm="giac")

[Out]

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