\(\int \frac {\log (e (f (a+b x)^p (c+d x)^q)^r)}{(a+b x)^2} \, dx\) [12]

Optimal result

Integrand size = 29, antiderivative size = 95 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=-\frac {p r}{b (a+b x)}+\frac {d q r \log (a+b x)}{b (b c-a d)}-\frac {d q r \log (c+d x)}{b (b c-a d)}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2581, 32, 36, 31} \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)}+\frac {d q r \log (a+b x)}{b (b c-a d)}-\frac {d q r \log (c+d x)}{b (b c-a d)}-\frac {p r}{b (a+b x)} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 )}{b (a+b x)}+(p r) \int \frac {1}{(a+b x)^2} \, dx+\frac {(d q r) \int \frac {1}{(a+b x) (c+d x)} \, dx}{b} \\ & = -\frac {p r}{b (a+b x)}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)}+\frac {(d q r) \int \frac {1}{a+b x} \, dx}{b c-a d}-\frac {\left (d^2 q r\right ) \int \frac {1}{c+d x} \, dx}{b (b c-a d)} \\ & = -\frac {p r}{b (a+b x)}+\frac {d q r \log (a+b x)}{b (b c-a d)}-\frac {d q r \log (c+d x)}{b (b c-a d)}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=\frac {r \left (-\frac {p}{a+b x}+\frac {d q \log (a+b x)}{b c-a d}-\frac {d q \log (c+d x)}{b c-a d}\right )}{b}-\frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 31.53 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.78

method	result	size
parallelrisch	\(-\frac {\ln \left (b x +a \right ) x \,b^{3} d^{2} q r -\ln \left (d x +c \right ) x \,b^{3} d^{2} q r +\ln \left (b x +a \right ) a \,b^{2} d^{2} q r -\ln \left (d x +c \right ) a \,b^{2} d^{2} q r +a \,b^{2} d^{2} p r -b^{3} c d p r +\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a \,b^{2} d^{2}-\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) b^{3} c d}{\left (a d -c b \right ) \left (b x +a \right ) b^{3} d}\)	\(169\)

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.26 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=-\frac {{\left (b c - a d\right )} p r + {\left (b c - a d\right )} r \log \left (f\right ) - {\left (b d q r x + {\left (a d q - {\left (b c - a d\right )} p\right )} r\right )} \log \left (b x + a\right ) + {\left (b d q r x + b c q r\right )} \log \left (d x + c\right ) + {\left (b c - a d\right )} \log \left (e\right )}{a b^{2} c - a^{2} b d + {\left (b^{3} c - a b^{2} d\right )} x} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=\text {Exception raised: NotImplementedError} \]

[In]

integrate(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/(b*x+a)**2,x)

[Out]

Exception raised: NotImplementedError >> no valid subset found

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=\frac {{\left (d f q {\left (\frac {\log \left (b x + a\right )}{b c - a d} - \frac {\log \left (d x + c\right )}{b c - a d}\right )} - \frac {b f p}{b^{2} x + a b}\right )} r}{b f} - \frac {\log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right )}{{\left (b x + a\right )} b} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.19 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=\frac {d q r \log \left (b x + a\right )}{b^{2} c - a b d} - \frac {d q r \log \left (d x + c\right )}{b^{2} c - a b d} - \frac {p r \log \left (b x + a\right )}{b^{2} x + a b} - \frac {q r \log \left (d x + c\right )}{b^{2} x + a b} - \frac {p r + r \log \left (f\right ) + \log \left (e\right )}{b^{2} x + a b} \]

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{(a+b x)^2} \, dx=-\frac {\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (x+\frac {a}{b}\right )}{{\left (a+b\,x\right )}^2}-\frac {p\,r}{x\,b^2+a\,b}+\frac {d\,q\,r\,\mathrm {atan}\left (\frac {b\,c\,2{}\mathrm {i}+b\,d\,x\,2{}\mathrm {i}}{a\,d-b\,c}+1{}\mathrm {i}\right )\,2{}\mathrm {i}}{b\,\left (a\,d-b\,c\right )} \]

[In]

int(log(e*(f*(a + b*x)^p*(c + d*x)^q)^r)/(a + b*x)^2,x)

[Out]

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