\(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{(a+b x)^3} \, dx\) [13]

Optimal result

Integrand size = 29, antiderivative size = 135 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=-\frac {p r}{4 b (a+b x)^2}-\frac {d q r}{2 b (b c-a d) (a+b x)}-\frac {d^2 q r \log (a+b x)}{2 b (b c-a d)^2}+\frac {d^2 q r \log (c+d x)}{2 b (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 135, 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} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=-\frac {d^2 q r \log (a+b x)}{2 b (b c-a d)^2}+\frac {d^2 q r \log (c+d x)}{2 b (b c-a d)^2}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b (a+b x)^2}-\frac {d q r}{2 b (a+b x) (b c-a d)}-\frac {p r}{4 b (a+b x)^2} \]

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^3,x]

[Out]

-1/4*(p*r)/(b*(a + b*x)^2) - (d*q*r)/(2*b*(b*c - a*d)*(a + b*x)) - (d^2*q*r*Log[a + b*x])/(2*b*(b*c - a*d)^2)
+ (d^2*q*r*Log[c + d*x])/(2*b*(b*c - a*d)^2) - Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(2*b*(a + b*x)^2)

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.86 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=\frac {r \left (-\frac {p-\frac {2 d q (a+b x)}{-b c+a d}}{2 (a+b x)^2}-\frac {d^2 q \log (a+b x)}{(b c-a d)^2}+\frac {d^2 q \log (c+d x)}{(b c-a d)^2}\right )-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2}}{2 b} \]

[In]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^3,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(766\) vs. \(2(125)=250\).

Time = 150.01 (sec) , antiderivative size = 767, normalized size of antiderivative = 5.68

method	result	size
parallelrisch	\(-\frac {2 \ln \left (b x +a \right ) x^{2} a^{4} b^{2} c \,d^{2} p r +2 \ln \left (b x +a \right ) x^{2} a^{4} b^{2} c \,d^{2} q r -4 \ln \left (b x +a \right ) x^{2} a^{3} b^{3} c^{2} d p r -4 \ln \left (d x +c \right ) x^{2} a^{3} b^{3} c^{2} d q r +4 \ln \left (b x +a \right ) x \,a^{5} b c \,d^{2} p r +4 \ln \left (b x +a \right ) x \,a^{5} b c \,d^{2} q r -8 \ln \left (b x +a \right ) x \,a^{4} b^{2} c^{2} d p r -8 \ln \left (d x +c \right ) x \,a^{4} b^{2} c^{2} d q r -4 \ln \left (b x +a \right ) a^{5} b \,c^{2} d p r -4 \ln \left (d x +c \right ) a^{5} b \,c^{2} d q r -x^{2} a^{4} b^{2} c \,d^{2} p r +2 x^{2} a^{4} b^{2} c \,d^{2} q r +2 x^{2} a^{3} b^{3} c^{2} d p r -2 x^{2} a^{3} b^{3} c^{2} d q r -2 x \,a^{5} b c \,d^{2} p r +2 x \,a^{5} b c \,d^{2} q r +4 x \,a^{4} b^{2} c^{2} d p r -2 x \,a^{4} b^{2} c^{2} d q r -2 x^{2} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{2} b^{4} c^{3}-4 x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{3} b^{3} c^{3}+2 \ln \left (b x +a \right ) x^{2} a^{2} b^{4} c^{3} p r +2 \ln \left (d x +c \right ) x^{2} a^{2} b^{4} c^{3} q r +4 \ln \left (b x +a \right ) x \,a^{3} b^{3} c^{3} p r +4 \ln \left (d x +c \right ) x \,a^{3} b^{3} c^{3} q r +2 \ln \left (b x +a \right ) a^{6} c \,d^{2} p r +2 \ln \left (b x +a \right ) a^{6} c \,d^{2} q r +2 \ln \left (b x +a \right ) a^{4} b^{2} c^{3} p r +2 \ln \left (d x +c \right ) a^{4} b^{2} c^{3} q r -x^{2} a^{2} b^{4} c^{3} p r -2 x^{2} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{4} b^{2} c \,d^{2}+4 x^{2} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{3} b^{3} c^{2} d -2 x \,a^{3} b^{3} c^{3} p r -4 x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{5} b c \,d^{2}+8 x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{4} b^{2} c^{2} d}{4 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) b \left (b x +a \right )^{2} c \,a^{4}}\)	\(767\)

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4*(2*ln(b*x+a)*x^2*a^4*b^2*c*d^2*p*r+2*ln(b*x+a)*x^2*a^4*b^2*c*d^2*q*r-4*ln(b*x+a)*x^2*a^3*b^3*c^2*d*p*r-4*
ln(d*x+c)*x^2*a^3*b^3*c^2*d*q*r+4*ln(b*x+a)*x*a^5*b*c*d^2*p*r+4*ln(b*x+a)*x*a^5*b*c*d^2*q*r-8*ln(b*x+a)*x*a^4*
b^2*c^2*d*p*r-8*ln(d*x+c)*x*a^4*b^2*c^2*d*q*r-4*ln(b*x+a)*a^5*b*c^2*d*p*r-4*ln(d*x+c)*a^5*b*c^2*d*q*r-x^2*a^4*
b^2*c*d^2*p*r+2*x^2*a^4*b^2*c*d^2*q*r+2*x^2*a^3*b^3*c^2*d*p*r-2*x^2*a^3*b^3*c^2*d*q*r-2*x*a^5*b*c*d^2*p*r+2*x*
a^5*b*c*d^2*q*r+4*x*a^4*b^2*c^2*d*p*r-2*x*a^4*b^2*c^2*d*q*r-2*x^2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^2*b^4*c^3-
4*x*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^3*b^3*c^3+2*ln(b*x+a)*x^2*a^2*b^4*c^3*p*r+2*ln(d*x+c)*x^2*a^2*b^4*c^3*q*
r+4*ln(b*x+a)*x*a^3*b^3*c^3*p*r+4*ln(d*x+c)*x*a^3*b^3*c^3*q*r+2*ln(b*x+a)*a^6*c*d^2*p*r+2*ln(b*x+a)*a^6*c*d^2*
q*r+2*ln(b*x+a)*a^4*b^2*c^3*p*r+2*ln(d*x+c)*a^4*b^2*c^3*q*r-x^2*a^2*b^4*c^3*p*r-2*x^2*ln(e*(f*(b*x+a)^p*(d*x+c
)^q)^r)*a^4*b^2*c*d^2+4*x^2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^3*b^3*c^2*d-2*x*a^3*b^3*c^3*p*r-4*x*ln(e*(f*(b*x
+a)^p*(d*x+c)^q)^r)*a^5*b*c*d^2+8*x*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^4*b^2*c^2*d)/(a^2*d^2-2*a*b*c*d+b^2*c^2)
/b/(b*x+a)^2/c/a^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (125) = 250\).

Time = 0.32 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.39 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=-\frac {2 \, {\left (b^{2} c d - a b d^{2}\right )} q r x + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} r \log \left (f\right ) + {\left ({\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} p + 2 \, {\left (a b c d - a^{2} d^{2}\right )} q\right )} r + 2 \, {\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x + {\left (a^{2} d^{2} q + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} p\right )} r\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x - {\left (b^{2} c^{2} - 2 \, a b c d\right )} q r\right )} \log \left (d x + c\right ) + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (e\right )}{4 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{2} + 2 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x\right )}} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^3,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=\text {Timed out} \]

[In]

integrate(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/(b*x+a)**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.22 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=-\frac {{\left (2 \, d f q {\left (\frac {d \log \left (b x + a\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {d \log \left (d x + c\right )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} + \frac {1}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x}\right )} + \frac {b f p}{b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b}\right )} r}{4 \, b f} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{2 \, {\left (b x + a\right )}^{2} b} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.85 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=-\frac {d^{2} q r \log \left (b x + a\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} + \frac {d^{2} q r \log \left (d x + c\right )}{2 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )}} - \frac {p r \log \left (b x + a\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {q r \log \left (d x + c\right )}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} - \frac {2 \, b d q r x + b c p r - a d p r + 2 \, a d q r + 2 \, b c r \log \left (f\right ) - 2 \, a d r \log \left (f\right ) + 2 \, b c \log \left (e\right ) - 2 \, a d \log \left (e\right )}{4 \, {\left (b^{4} c x^{2} - a b^{3} d x^{2} + 2 \, a b^{3} c x - 2 \, a^{2} b^{2} d x + a^{2} b^{2} c - a^{3} b d\right )}} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.43 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.35 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^3} \, dx=\frac {\frac {b\,c\,p\,r-a\,d\,p\,r+2\,a\,d\,q\,r}{2\,\left (a\,d-b\,c\right )}+\frac {b\,d\,q\,r\,x}{a\,d-b\,c}}{2\,a^2\,b+4\,a\,b^2\,x+2\,b^3\,x^2}-\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )}{{\left (a+b\,x\right )}^3}+\frac {d^2\,q\,r\,\mathrm {atanh}\left (\frac {2\,b^3\,c^2-2\,a^2\,b\,d^2}{2\,b\,{\left (a\,d-b\,c\right )}^2}-\frac {2\,b\,d\,x}{a\,d-b\,c}\right )}{b\,{\left (a\,d-b\,c\right )}^2} \]

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(a + b*x)^3,x)

[Out]

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