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]

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

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Rubi [A]  time = 0.10, antiderivative size = 144, 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, 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 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 (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*}

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Mathematica [A]  time = 0.12, 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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fricas [A]  time = 0.92, size = 221, normalized size = 1.53 \[ -\frac {{\left (2 \, A - 3 \, B\right )} b^{2} c^{2} - 4 \, {\left (A - B\right )} a b c d + {\left (2 \, A - B\right )} a^{2} d^{2} - 2 \, {\left (B b^{2} c d - B a b d^{2}\right )} x - 2 \, {\left (B b^{2} d^{2} x^{2} + 2 \, B b^{2} c d x + 2 \, B a b c d - B a^{2} d^{2}\right )} \log \left (\frac {b e x + a e}{d x + c}\right )}{4 \, {\left ({\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} i^{3} x^{2} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} i^{3} x + {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3}\right )} i^{3}\right )}} \]

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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giac [A]  time = 0.94, size = 254, normalized size = 1.76 \[ \frac {{\left (\frac {4 \, {\left (b x e + a e\right )} B b i e \log \left (\frac {b x e + a e}{d x + c}\right )}{d x + c} + \frac {4 \, {\left (b x e + a e\right )} A b i e}{d x + c} - \frac {4 \, {\left (b x e + a e\right )} B b i e}{d x + c} - \frac {2 \, {\left (b x e + a e\right )}^{2} B d i \log \left (\frac {b x e + a e}{d x + c}\right )}{{\left (d x + c\right )}^{2}} - \frac {2 \, {\left (b x e + a e\right )}^{2} A d i}{{\left (d x + c\right )}^{2}} + \frac {{\left (b x e + a e\right )}^{2} B d i}{{\left (d x + c\right )}^{2}}\right )} {\left (\frac {b c}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}} - \frac {a d}{{\left (b c e - a d e\right )} {\left (b c - a d\right )}}\right )}}{4 \, {\left (b c e - a d e\right )}} \]

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]

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

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maple [B]  time = 0.05, size = 746, normalized size = 5.18 \[ -\frac {B \,a^{3} d^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}+\frac {3 B \,a^{2} b c d \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}-\frac {3 B a \,b^{2} c^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}+\frac {B \,b^{3} c^{3} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} d \,i^{3}}-\frac {A \,a^{3} d^{2}}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}+\frac {3 A \,a^{2} b c d}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}-\frac {3 A a \,b^{2} c^{2}}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}+\frac {A \,b^{3} c^{3}}{2 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} d \,i^{3}}+\frac {B \,a^{3} d^{2}}{4 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}-\frac {3 B \,a^{2} b c d}{4 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}+\frac {3 B a \,b^{2} c^{2}}{4 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} i^{3}}-\frac {B \,b^{3} c^{3}}{4 \left (a d -b c \right )^{3} \left (d x +c \right )^{2} d \,i^{3}}-\frac {B \,a^{2} b d}{2 \left (a d -b c \right )^{3} \left (d x +c \right ) i^{3}}+\frac {B a \,b^{2} c}{\left (a d -b c \right )^{3} \left (d x +c \right ) i^{3}}+\frac {B a \,b^{2} \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} i^{3}}-\frac {B \,b^{3} c^{2}}{2 \left (a d -b c \right )^{3} \left (d x +c \right ) d \,i^{3}}-\frac {B \,b^{3} c \ln \left (\frac {b e}{d}+\frac {\left (a d -b c \right ) e}{\left (d x +c \right ) d}\right )}{2 \left (a d -b c \right )^{3} d \,i^{3}}+\frac {A a \,b^{2}}{2 \left (a d -b c \right )^{3} i^{3}}-\frac {A \,b^{3} c}{2 \left (a d -b c \right )^{3} d \,i^{3}}-\frac {3 B a \,b^{2}}{4 \left (a d -b c \right )^{3} i^{3}}+\frac {3 B \,b^{3} c}{4 \left (a d -b c \right )^{3} d \,i^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 255, normalized size = 1.77 \[ \frac {1}{4} \, B {\left (\frac {2 \, b d x + 3 \, b c - a d}{{\left (b c d^{3} - a d^{4}\right )} i^{3} x^{2} + 2 \, {\left (b c^{2} d^{2} - a c d^{3}\right )} i^{3} x + {\left (b c^{3} d - a c^{2} d^{2}\right )} i^{3}} - \frac {2 \, \log \left (\frac {b e x}{d x + c} + \frac {a e}{d x + c}\right )}{d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}} + \frac {2 \, b^{2} \log \left (b x + a\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}} - \frac {2 \, b^{2} \log \left (d x + c\right )}{{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} i^{3}}\right )} - \frac {A}{2 \, {\left (d^{3} i^{3} x^{2} + 2 \, c d^{2} i^{3} x + c^{2} d i^{3}\right )}} \]

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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mupad [B]  time = 5.43, size = 208, normalized size = 1.44 \[ \frac {B\,b^2\,\mathrm {atanh}\left (\frac {2\,a^2\,d^3\,i^3-2\,b^2\,c^2\,d\,i^3}{2\,d\,i^3\,{\left (a\,d-b\,c\right )}^2}+\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{d\,i^3\,{\left (a\,d-b\,c\right )}^2}-\frac {B\,\ln \left (\frac {e\,\left (a+b\,x\right )}{c+d\,x}\right )}{2\,d^2\,i^3\,\left (2\,c\,x+d\,x^2+\frac {c^2}{d}\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-B\,a\,d+3\,B\,b\,c}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,x}{a\,d-b\,c}}{2\,c^2\,d\,i^3+4\,c\,d^2\,i^3\,x+2\,d^3\,i^3\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 2.58, size = 422, normalized size = 2.93 \[ - \frac {B b^{2} \log {\left (x + \frac {- \frac {B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} + \frac {3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} - \frac {3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d + \frac {B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} + \frac {B b^{2} \log {\left (x + \frac {\frac {B a^{3} b^{2} d^{3}}{\left (a d - b c\right )^{2}} - \frac {3 B a^{2} b^{3} c d^{2}}{\left (a d - b c\right )^{2}} + \frac {3 B a b^{4} c^{2} d}{\left (a d - b c\right )^{2}} + B a b^{2} d - \frac {B b^{5} c^{3}}{\left (a d - b c\right )^{2}} + B b^{3} c}{2 B b^{3} d} \right )}}{2 d i^{3} \left (a d - b c\right )^{2}} - \frac {B \log {\left (\frac {e \left (a + b x\right )}{c + d x} \right )}}{2 c^{2} d i^{3} + 4 c d^{2} i^{3} x + 2 d^{3} i^{3} x^{2}} + \frac {- 2 A a d + 2 A b c + B a d - 3 B b c - 2 B b d x}{4 a c^{2} d^{2} i^{3} - 4 b c^{3} d i^{3} + x^{2} \left (4 a d^{4} i^{3} - 4 b c d^{3} i^{3}\right ) + x \left (8 a c d^{3} i^{3} - 8 b c^{2} d^{2} i^{3}\right )} \]

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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