\(\int \log (e (f (a+b x)^p (c+d x)^q)^r) \, dx\) [29]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-((p+q) r x)+\frac {(b c-a d) q r \log (c+d x)}{b d}+\frac {(a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2579, 31, 8} \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {(a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b}+\frac {q r (b c-a d) \log (c+d x)}{b d}-(r x (p+q)) \]

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 31

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

Rule 2579

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[(a
+ b*x)*(Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s/b), x] + (Dist[q*r*s*((b*c - a*d)/b), Int[Log[e*(f*(a + b*x)^p
*(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] - Dist[r*s*(p + q), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1
), x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && NeQ[p + q, 0] && IGtQ[s, 0] &&
LtQ[s, 4]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {a p r \log (a+b x)}{b}+\frac {c q r \log (c+d x)}{d}+x \left (-((p+q) r)+\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00

method	result	size
default	\(\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) x -r \left (x p +x q -\frac {c q \ln \left (d x +c \right )}{d}-\frac {a p \ln \left (b x +a \right )}{b}\right )\)	\(61\)
parallelrisch	\(\frac {\ln \left (b x +a \right ) a d p q \,r^{2}-\ln \left (b x +a \right ) b c p q \,r^{2}-x b d p q \,r^{2}-x b d \,q^{2} r^{2}+x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) b d q r +a d p q \,r^{2}+a d \,q^{2} r^{2}+b c p q \,r^{2}+b c \,r^{2} q^{2}+\ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) b c q r}{b d q r}\)	\(152\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {b d r x \log \left (f\right ) + b d x \log \left (e\right ) - {\left (b d p + b d q\right )} r x + {\left (b d p r x + a d p r\right )} \log \left (b x + a\right ) + {\left (b d q r x + b c q r\right )} \log \left (d x + c\right )}{b d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (54) = 108\).

Time = 5.06 (sec) , antiderivative size = 184, normalized size of antiderivative = 3.02 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\begin {cases} x \log {\left (e \left (a^{p} c^{q} f\right )^{r} \right )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {c \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{d} - q r x + x \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )} & \text {for}\: b = 0 \\\frac {a \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{b} - p r x + x \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )} & \text {for}\: d = 0 \\- \frac {a q r \log {\left (c + d x \right )}}{b} + \frac {a \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{b} + \frac {c q r \log {\left (c + d x \right )}}{d} - p r x - q r x + x \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*log(e*(a**p*c**q*f)**r), Eq(b, 0) & Eq(d, 0)), (c*log(e*(a**p*f*(c + d*x)**q)**r)/d - q*r*x + x*l
og(e*(a**p*f*(c + d*x)**q)**r), Eq(b, 0)), (a*log(e*(c**q*f*(a + b*x)**p)**r)/b - p*r*x + x*log(e*(c**q*f*(a +
b*x)**p)**r), Eq(d, 0)), (-a*q*r*log(c + d*x)/b + a*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/b + c*q*r*log(c +
d*x)/d - p*r*x - q*r*x + x*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=x \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) - \frac {{\left (b f p {\left (\frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}}\right )} + d f q {\left (\frac {x}{d} - \frac {c \log \left (d x + c\right )}{d^{2}}\right )}\right )} r}{f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=p r x \log \left (b x + a\right ) + q r x \log \left (d x + c\right ) + \frac {a p r \log \left (b x + a\right )}{b} + \frac {c q r \log \left (-d x - c\right )}{d} - {\left (p r + q r - r \log \left (f\right ) - \log \left (e\right )\right )} x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.98 \[ \int \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=x\,\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )-p\,r\,x-q\,r\,x+\frac {a\,p\,r\,\ln \left (a+b\,x\right )}{b}+\frac {c\,q\,r\,\ln \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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