3.2.54 \(\int \frac {A+B \log (e (\frac {a+b x}{c+d x})^n)}{(c i+d i x)^3} \, dx\) [154]

Optimal. Leaf size=151 \[ \frac {B n}{4 d i^3 (c+d x)^2}+\frac {b B n}{2 d (b c-a d) i^3 (c+d x)}+\frac {b^2 B n \log (a+b x)}{2 d (b c-a d)^2 i^3}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{2 d i^3 (c+d x)^2}-\frac {b^2 B n \log (c+d x)}{2 d (b c-a d)^2 i^3} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2547, 21, 46} \begin {gather*} -\frac {B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A}{2 d i^3 (c+d x)^2}+\frac {b^2 B n \log (a+b x)}{2 d i^3 (b c-a d)^2}-\frac {b^2 B n \log (c+d x)}{2 d i^3 (b c-a d)^2}+\frac {b B n}{2 d i^3 (c+d x) (b c-a d)}+\frac {B n}{4 d i^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2547

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))*((f_.) + (g_.)*(x_))^(m_.), x
_Symbol] :> Simp[(f + g*x)^(m + 1)*((A + B*Log[e*((a + b*x)/(c + d*x))^n])/(g*(m + 1))), x] - Dist[B*n*((b*c -
a*d)/(g*(m + 1))), Int[(f + g*x)^(m + 1)/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, m
, n}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, -2]

Rubi steps

\begin {align*} \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(154 c+154 d x)^3} \, dx &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7304528 d (c+d x)^2}+\frac {(B n) \int \frac {b c-a d}{23716 (a+b x) (c+d x)^3} \, dx}{308 d}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7304528 d (c+d x)^2}+\frac {(B (b c-a d) n) \int \frac {1}{(a+b x) (c+d x)^3} \, dx}{7304528 d}\\ &=-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7304528 d (c+d x)^2}+\frac {(B (b c-a d) n) \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}{7304528 d}\\ &=\frac {B n}{14609056 d (c+d x)^2}+\frac {b B n}{7304528 d (b c-a d) (c+d x)}+\frac {b^2 B n \log (a+b x)}{7304528 d (b c-a d)^2}-\frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{7304528 d (c+d x)^2}-\frac {b^2 B n \log (c+d x)}{7304528 d (b c-a d)^2}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 115, normalized size = 0.76 \begin {gather*} \frac {-2 \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )+\frac {B n \left ((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)\right )}{(b c-a d)^2}}{4 d i^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A + B*Log[e*((a + b*x)/(c + d*x))^n]) + (B*n*((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 [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )}{\left (d i x +c i \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 235, normalized size = 1.56 \begin {gather*} -{\left (\frac {b^{2} \log \left (b x + a\right )}{2 i \, b^{2} c^{2} d - 4 i \, a b c d^{2} + 2 i \, a^{2} d^{3}} - \frac {b^{2} \log \left (d x + c\right )}{2 i \, b^{2} c^{2} d - 4 i \, a b c d^{2} + 2 i \, a^{2} d^{3}} - \frac {2 \, b d x + 3 \, b c - a d}{-4 i \, b c^{3} d + 4 i \, a c^{2} d^{2} - 4 \, {\left (i \, b c d^{3} - i \, a d^{4}\right )} x^{2} - 8 \, {\left (i \, b c^{2} d^{2} - i \, a c d^{3}\right )} x}\right )} B n + \frac {B \log \left ({\left (\frac {b x}{d x + c} + \frac {a}{d x + c}\right )}^{n} e\right )}{2 i \, d^{3} x^{2} + 4 i \, c d^{2} x + 2 i \, c^{2} d} + \frac {A}{2 i \, d^{3} x^{2} + 4 i \, c d^{2} x + 2 i \, c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*log(b*x + a)/(2*I*b^2*c^2*d - 4*I*a*b*c*d^2 + 2*I*a^2*d^3) - b^2*log(d*x + c)/(2*I*b^2*c^2*d - 4*I*a*b*c
*d^2 + 2*I*a^2*d^3) - (2*b*d*x + 3*b*c - a*d)/(-4*I*b*c^3*d + 4*I*a*c^2*d^2 - 4*(I*b*c*d^3 - I*a*d^4)*x^2 - 8*
(I*b*c^2*d^2 - I*a*c*d^3)*x))*B*n + B*log((b*x/(d*x + c) + a/(d*x + c))^n*e)/(2*I*d^3*x^2 + 4*I*c*d^2*x + 2*I*
c^2*d) + A/(2*I*d^3*x^2 + 4*I*c*d^2*x + 2*I*c^2*d)

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Fricas [A]
time = 0.40, size = 248, normalized size = 1.64 \begin {gather*} -\frac {2 \, {\left (i \, A + i \, B\right )} b^{2} c^{2} + 4 \, {\left (-i \, A - i \, B\right )} a b c d + 2 \, {\left (i \, A + i \, B\right )} a^{2} d^{2} + 2 \, {\left (-i \, B b^{2} c d + i \, B a b d^{2}\right )} n x - {\left (3 i \, B b^{2} c^{2} - 4 i \, B a b c d + i \, B a^{2} d^{2}\right )} n + 2 \, {\left (-i \, B b^{2} d^{2} n x^{2} - 2 i \, B b^{2} c d n x + {\left (-2 i \, B a b c d + i \, B a^{2} d^{2}\right )} n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{4 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + {\left (b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}\right )} x^{2} + 2 \, {\left (b^{2} c^{3} d^{2} - 2 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 2.58, size = 181, normalized size = 1.20 \begin {gather*} -\frac {1}{4} \, {\left (2 \, {\left (-\frac {2 i \, {\left (b x + a\right )} B b n}{{\left (b c - a d\right )} {\left (d x + c\right )}} + \frac {i \, {\left (b x + a\right )}^{2} B d n}{{\left (b c - a d\right )} {\left (d x + c\right )}^{2}}\right )} \log \left (\frac {b x + a}{d x + c}\right ) + \frac {{\left (-i \, B d n + 2 i \, A d + 2 i \, B d\right )} {\left (b x + a\right )}^{2}}{{\left (b c - a d\right )} {\left (d x + c\right )}^{2}} + \frac {4 \, {\left (i \, B b n - i \, A b - i \, B b\right )} {\left (b x + a\right )}}{{\left (b c - a d\right )} {\left (d x + c\right )}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.97, size = 221, normalized size = 1.46 \begin {gather*} \frac {B\,b^2\,n\,\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 (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )}{2\,d\,\left (c^2\,i^3+2\,c\,d\,i^3\,x+d^2\,i^3\,x^2\right )}-\frac {\frac {2\,A\,a\,d-2\,A\,b\,c-B\,a\,d\,n+3\,B\,b\,c\,n}{2\,\left (a\,d-b\,c\right )}+\frac {B\,b\,d\,n\,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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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