\(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{(a+b x)^4} \, dx\) [14]

Optimal result

Integrand size = 29, antiderivative size = 164 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=-\frac {p r}{9 b (a+b x)^3}-\frac {d q r}{6 b (b c-a d) (a+b x)^2}+\frac {d^2 q r}{3 b (b c-a d)^2 (a+b x)}+\frac {d^3 q r \log (a+b x)}{3 b (b c-a d)^3}-\frac {d^3 q r \log (c+d x)}{3 b (b c-a d)^3}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b (a+b x)^3} \]

[Out]

-1/9*p*r/b/(b*x+a)^3-1/6*d*q*r/b/(-a*d+b*c)/(b*x+a)^2+1/3*d^2*q*r/b/(-a*d+b*c)^2/(b*x+a)+1/3*d^3*q*r*ln(b*x+a)
/b/(-a*d+b*c)^3-1/3*d^3*q*r*ln(d*x+c)/b/(-a*d+b*c)^3-1/3*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/b/(b*x+a)^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2581, 32, 46} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {d^3 q r \log (a+b x)}{3 b (b c-a d)^3}-\frac {d^3 q r \log (c+d x)}{3 b (b c-a d)^3}+\frac {d^2 q r}{3 b (a+b x) (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b (a+b x)^3}-\frac {d q r}{6 b (a+b x)^2 (b c-a d)}-\frac {p r}{9 b (a+b x)^3} \]

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^4,x]

[Out]

-1/9*(p*r)/(b*(a + b*x)^3) - (d*q*r)/(6*b*(b*c - a*d)*(a + b*x)^2) + (d^2*q*r)/(3*b*(b*c - a*d)^2*(a + b*x)) +
(d^3*q*r*Log[a + b*x])/(3*b*(b*c - a*d)^3) - (d^3*q*r*Log[c + d*x])/(3*b*(b*c - a*d)^3) - Log[e*(f*(a + b*x)^
p*(c + d*x)^q)^r]/(3*b*(a + b*x)^3)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2581

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[(g + h*x)^(m + 1)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(h*(m + 1))), x] + (-Dist[b*p*(r/(h
*(m + 1))), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[d*q*(r/(h*(m + 1))), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b (a+b x)^3}+\frac {1}{3} (p r) \int \frac {1}{(a+b x)^4} \, dx+\frac {(d q r) \int \frac {1}{(a+b x)^3 (c+d x)} \, dx}{3 b} \\ & = -\frac {p r}{9 b (a+b x)^3}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b (a+b x)^3}+\frac {(d q r) \int \left (\frac {b}{(b c-a d) (a+b x)^3}-\frac {b d}{(b c-a d)^2 (a+b x)^2}+\frac {b d^2}{(b c-a d)^3 (a+b x)}-\frac {d^3}{(b c-a d)^3 (c+d x)}\right ) \, dx}{3 b} \\ & = -\frac {p r}{9 b (a+b x)^3}-\frac {d q r}{6 b (b c-a d) (a+b x)^2}+\frac {d^2 q r}{3 b (b c-a d)^2 (a+b x)}+\frac {d^3 q r \log (a+b x)}{3 b (b c-a d)^3}-\frac {d^3 q r \log (c+d x)}{3 b (b c-a d)^3}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b (a+b x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.86 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {r \left (\frac {-2 p+\frac {3 d q (a+b x)}{-b c+a d}+\frac {6 d^2 q (a+b x)^2}{(b c-a d)^2}}{6 (a+b x)^3}+\frac {d^3 q \log (a+b x)}{(b c-a d)^3}-\frac {d^3 q \log (c+d x)}{(b c-a d)^3}\right )-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3}}{3 b} \]

[In]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^4,x]

[Out]

(r*((-2*p + (3*d*q*(a + b*x))/(-(b*c) + a*d) + (6*d^2*q*(a + b*x)^2)/(b*c - a*d)^2)/(6*(a + b*x)^3) + (d^3*q*L
og[a + b*x])/(b*c - a*d)^3 - (d^3*q*Log[c + d*x])/(b*c - a*d)^3) - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b
*x)^3)/(3*b)

Maple [F]

\[\int \frac {\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right )}{\left (b x +a \right )^{4}}d x\]

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^4,x)

[Out]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^4,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (152) = 304\).

Time = 0.34 (sec) , antiderivative size = 580, normalized size of antiderivative = 3.54 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {6 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )} q r x^{2} - 3 \, {\left (b^{3} c^{2} d - 6 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} q r x - 6 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} r \log \left (f\right ) - {\left (2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} p + 3 \, {\left (a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, a^{3} d^{3}\right )} q\right )} r + 6 \, {\left (b^{3} d^{3} q r x^{3} + 3 \, a b^{2} d^{3} q r x^{2} + 3 \, a^{2} b d^{3} q r x + {\left (a^{3} d^{3} q - {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} p\right )} r\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} d^{3} q r x^{3} + 3 \, a b^{2} d^{3} q r x^{2} + 3 \, a^{2} b d^{3} q r x + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2}\right )} q r\right )} \log \left (d x + c\right ) - 6 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (e\right )}{18 \, {\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3} + {\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{3} + 3 \, {\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x\right )}} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^4,x, algorithm="fricas")

[Out]

1/18*(6*(b^3*c*d^2 - a*b^2*d^3)*q*r*x^2 - 3*(b^3*c^2*d - 6*a*b^2*c*d^2 + 5*a^2*b*d^3)*q*r*x - 6*(b^3*c^3 - 3*a
*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*r*log(f) - (2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*p + 3*
(a*b^2*c^2*d - 4*a^2*b*c*d^2 + 3*a^3*d^3)*q)*r + 6*(b^3*d^3*q*r*x^3 + 3*a*b^2*d^3*q*r*x^2 + 3*a^2*b*d^3*q*r*x
+ (a^3*d^3*q - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*p)*r)*log(b*x + a) - 6*(b^3*d^3*q*r*x^3 + 3
*a*b^2*d^3*q*r*x^2 + 3*a^2*b*d^3*q*r*x + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2)*q*r)*log(d*x + c) - 6*(b^3*
c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(e))/(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*
b*d^3 + (b^7*c^3 - 3*a*b^6*c^2*d + 3*a^2*b^5*c*d^2 - a^3*b^4*d^3)*x^3 + 3*(a*b^6*c^3 - 3*a^2*b^5*c^2*d + 3*a^3
*b^4*c*d^2 - a^4*b^3*d^3)*x^2 + 3*(a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\text {Timed out} \]

[In]

integrate(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/(b*x+a)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.76 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {{\left (3 \, {\left (\frac {2 \, d^{2} \log \left (b x + a\right )}{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}} - \frac {2 \, d^{2} \log \left (d x + c\right )}{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}} + \frac {2 \, b d x - b c + 3 \, a d}{a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x}\right )} d f q - \frac {2 \, b f p}{b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b}\right )} r}{18 \, b f} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{3 \, {\left (b x + a\right )}^{3} b} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^4,x, algorithm="maxima")

[Out]

1/18*(3*(2*d^2*log(b*x + a)/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) - 2*d^2*log(d*x + c)/(b^3*c^3
- 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) + (2*b*d*x - b*c + 3*a*d)/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2 + (b
^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^2 + 2*(a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x))*d*f*q - 2*b*f*p/(b^4*x
^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b))*r/(b*f) - 1/3*log(((b*x + a)^p*(d*x + c)^q*f)^r*e)/((b*x + a)^3*b)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (152) = 304\).

Time = 0.28 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.90 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {d^{3} q r \log \left (b x + a\right )}{3 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}} - \frac {d^{3} q r \log \left (d x + c\right )}{3 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )}} - \frac {p r \log \left (b x + a\right )}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {q r \log \left (d x + c\right )}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} + \frac {6 \, b^{2} d^{2} q r x^{2} - 3 \, b^{2} c d q r x + 15 \, a b d^{2} q r x - 2 \, b^{2} c^{2} p r + 4 \, a b c d p r - 2 \, a^{2} d^{2} p r - 3 \, a b c d q r + 9 \, a^{2} d^{2} q r - 6 \, b^{2} c^{2} r \log \left (f\right ) + 12 \, a b c d r \log \left (f\right ) - 6 \, a^{2} d^{2} r \log \left (f\right ) - 6 \, b^{2} c^{2} \log \left (e\right ) + 12 \, a b c d \log \left (e\right ) - 6 \, a^{2} d^{2} \log \left (e\right )}{18 \, {\left (b^{6} c^{2} x^{3} - 2 \, a b^{5} c d x^{3} + a^{2} b^{4} d^{2} x^{3} + 3 \, a b^{5} c^{2} x^{2} - 6 \, a^{2} b^{4} c d x^{2} + 3 \, a^{3} b^{3} d^{2} x^{2} + 3 \, a^{2} b^{4} c^{2} x - 6 \, a^{3} b^{3} c d x + 3 \, a^{4} b^{2} d^{2} x + a^{3} b^{3} c^{2} - 2 \, a^{4} b^{2} c d + a^{5} b d^{2}\right )}} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^4,x, algorithm="giac")

[Out]

1/3*d^3*q*r*log(b*x + a)/(b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) - 1/3*d^3*q*r*log(d*x + c)/(b
^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3) - 1/3*p*r*log(b*x + a)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^
2*x + a^3*b) - 1/3*q*r*log(d*x + c)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) + 1/18*(6*b^2*d^2*q*r*x^2 -
3*b^2*c*d*q*r*x + 15*a*b*d^2*q*r*x - 2*b^2*c^2*p*r + 4*a*b*c*d*p*r - 2*a^2*d^2*p*r - 3*a*b*c*d*q*r + 9*a^2*d^2
*q*r - 6*b^2*c^2*r*log(f) + 12*a*b*c*d*r*log(f) - 6*a^2*d^2*r*log(f) - 6*b^2*c^2*log(e) + 12*a*b*c*d*log(e) -
6*a^2*d^2*log(e))/(b^6*c^2*x^3 - 2*a*b^5*c*d*x^3 + a^2*b^4*d^2*x^3 + 3*a*b^5*c^2*x^2 - 6*a^2*b^4*c*d*x^2 + 3*a
^3*b^3*d^2*x^2 + 3*a^2*b^4*c^2*x - 6*a^3*b^3*c*d*x + 3*a^4*b^2*d^2*x + a^3*b^3*c^2 - 2*a^4*b^2*c*d + a^5*b*d^2
)

Mupad [B] (verification not implemented)

Time = 3.96 (sec) , antiderivative size = 346, normalized size of antiderivative = 2.11 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^4} \, dx=\frac {\frac {x\,\left (5\,a\,b\,d^2\,q\,r-b^2\,c\,d\,q\,r\right )}{2\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}-\frac {2\,a^2\,d^2\,p\,r+2\,b^2\,c^2\,p\,r-9\,a^2\,d^2\,q\,r-4\,a\,b\,c\,d\,p\,r+3\,a\,b\,c\,d\,q\,r}{6\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}+\frac {b^2\,d^2\,q\,r\,x^2}{a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}}{3\,a^3\,b+9\,a^2\,b^2\,x+9\,a\,b^3\,x^2+3\,b^4\,x^3}-\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {x}{3}+\frac {a}{3\,b}\right )}{{\left (a+b\,x\right )}^4}-\frac {2\,d^3\,q\,r\,\mathrm {atanh}\left (\frac {3\,a^3\,b\,d^3-3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+3\,b^4\,c^3}{3\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {2\,b\,d\,x\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{{\left (a\,d-b\,c\right )}^3}\right )}{3\,b\,{\left (a\,d-b\,c\right )}^3} \]

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(a + b*x)^4,x)

[Out]

((x*(5*a*b*d^2*q*r - b^2*c*d*q*r))/(2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) - (2*a^2*d^2*p*r + 2*b^2*c^2*p*r - 9*a^
2*d^2*q*r - 4*a*b*c*d*p*r + 3*a*b*c*d*q*r)/(6*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)) + (b^2*d^2*q*r*x^2)/(a^2*d^2 +
b^2*c^2 - 2*a*b*c*d))/(3*a^3*b + 3*b^4*x^3 + 9*a^2*b^2*x + 9*a*b^3*x^2) - (log(e*(f*(a + b*x)^p*(c + d*x)^q)^r
)*(x/3 + a/(3*b)))/(a + b*x)^4 - (2*d^3*q*r*atanh((3*b^4*c^3 + 3*a^3*b*d^3 - 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)/
(3*b*(a*d - b*c)^3) + (2*b*d*x*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(a*d - b*c)^3))/(3*b*(a*d - b*c)^3)