\(\int (a+b x) \log (e (f (a+b x)^p (c+d x)^q)^r) \, dx\) [10]

Optimal result

Integrand size = 27, antiderivative size = 116 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-\frac {1}{2} a p r x+\frac {(b c-a d) q r x}{2 d}-\frac {1}{4} b p r x^2-\frac {q r (a+b x)^2}{4 b}-\frac {(b c-a d)^2 q r \log (c+d x)}{2 b d^2}+\frac {(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2581, 45} \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-\frac {q r (b c-a d)^2 \log (c+d x)}{2 b d^2}+\frac {(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}+\frac {q r x (b c-a d)}{2 d}-\frac {q r (a+b x)^2}{4 b}-\frac {1}{2} a p r x-\frac {1}{4} b p r x^2 \]

[In]

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

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}-\frac {1}{2} (p r) \int (a+b x) \, dx-\frac {(d q r) \int \frac {(a+b x)^2}{c+d x} \, dx}{2 b} \\ & = -\frac {1}{2} a p r x-\frac {1}{4} b p r x^2+\frac {(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b}-\frac {(d q r) \int \left (-\frac {b (b c-a d)}{d^2}+\frac {b (a+b x)}{d}+\frac {(-b c+a d)^2}{d^2 (c+d x)}\right ) \, dx}{2 b} \\ & = -\frac {1}{2} a p r x+\frac {(b c-a d) q r x}{2 d}-\frac {1}{4} b p r x^2-\frac {q r (a+b x)^2}{4 b}-\frac {(b c-a d)^2 q r \log (c+d x)}{2 b d^2}+\frac {(a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {a^2 p r \log (a+b x)}{2 b}-\frac {2 c (b c-2 a d) q r \log (c+d x)+d x \left (r (-2 b c q+2 a d (p+2 q)+b d (p+q) x)-2 d (2 a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )\right )}{4 d^2} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(104)=208\).

Time = 12.56 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.63

method	result	size
parallelrisch	\(\frac {-x^{2} b^{2} d^{2} p r -x^{2} b^{2} d^{2} q r +6 \ln \left (b x +a \right ) a^{2} d^{2} p r +6 \ln \left (b x +a \right ) a b c d p r +4 \ln \left (d x +c \right ) a^{2} d^{2} q r +10 \ln \left (d x +c \right ) a b c d q r -2 \ln \left (d x +c \right ) b^{2} c^{2} q r +2 x^{2} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) b^{2} d^{2}-2 x a b \,d^{2} p r -4 x a b \,d^{2} q r +2 x \,b^{2} c d q r +4 x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a b \,d^{2}+2 a^{2} d^{2} p r +4 a^{2} q r \,d^{2}+3 a b c d p r +3 a b c d q r -2 b^{2} c^{2} q r -4 \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{2} d^{2}-6 \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a b c d}{4 b \,d^{2}}\)	\(305\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.70 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-\frac {{\left (b^{2} d^{2} p + b^{2} d^{2} q\right )} r x^{2} + 2 \, {\left (a b d^{2} p - {\left (b^{2} c d - 2 \, a b d^{2}\right )} q\right )} r x - 2 \, {\left (b^{2} d^{2} p r x^{2} + 2 \, a b d^{2} p r x + a^{2} d^{2} p r\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} q r x^{2} + 2 \, a b d^{2} q r x - {\left (b^{2} c^{2} - 2 \, a b c d\right )} q r\right )} \log \left (d x + c\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x\right )} \log \left (e\right ) - 2 \, {\left (b^{2} d^{2} r x^{2} + 2 \, a b d^{2} r x\right )} \log \left (f\right )}{4 \, b d^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (104) = 208\).

Time = 20.45 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.80 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\begin {cases} a x \log {\left (e \left (a^{p} c^{q} f\right )^{r} \right )} & \text {for}\: b = 0 \wedge d = 0 \\a \left (\frac {c \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{d} - q r x + x \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}\right ) & \text {for}\: b = 0 \\\frac {a^{2} \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{2 b} - \frac {a p r x}{2} + a x \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )} - \frac {b p r x^{2}}{4} + \frac {b x^{2} \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{2} & \text {for}\: d = 0 \\- \frac {a^{2} q r \log {\left (\frac {c}{d} + x \right )}}{2 b} + \frac {a^{2} \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{2 b} + \frac {a c q r \log {\left (\frac {c}{d} + x \right )}}{d} - \frac {a p r x}{2} - a q r x + a x \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )} - \frac {b c^{2} q r \log {\left (\frac {c}{d} + x \right )}}{2 d^{2}} + \frac {b c q r x}{2 d} - \frac {b p r x^{2}}{4} - \frac {b q r x^{2}}{4} + \frac {b x^{2} \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x*log(e*(a**p*c**q*f)**r), Eq(b, 0) & Eq(d, 0)), (a*(c*log(e*(a**p*f*(c + d*x)**q)**r)/d - q*r*x
+ x*log(e*(a**p*f*(c + d*x)**q)**r)), Eq(b, 0)), (a**2*log(e*(c**q*f*(a + b*x)**p)**r)/(2*b) - a*p*r*x/2 + a*x
*log(e*(c**q*f*(a + b*x)**p)**r) - b*p*r*x**2/4 + b*x**2*log(e*(c**q*f*(a + b*x)**p)**r)/2, Eq(d, 0)), (-a**2*
q*r*log(c/d + x)/(2*b) + a**2*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(2*b) + a*c*q*r*log(c/d + x)/d - a*p*r*x
/2 - a*q*r*x + a*x*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r) - b*c**2*q*r*log(c/d + x)/(2*d**2) + b*c*q*r*x/(2*d
) - b*p*r*x**2/4 - b*q*r*x**2/4 + b*x**2*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac {{\left (\frac {2 \, a^{2} f p \log \left (b x + a\right )}{b} - \frac {b d f {\left (p + q\right )} x^{2} + 2 \, {\left (a d f {\left (p + 2 \, q\right )} - b c f q\right )} x}{d} - \frac {2 \, {\left (b c^{2} f q - 2 \, a c d f q\right )} \log \left (d x + c\right )}{d^{2}}\right )} r}{4 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.31 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {a^{2} p r \log \left (b x + a\right )}{2 \, b} - \frac {1}{4} \, {\left (b p r + b q r - 2 \, b r \log \left (f\right ) - 2 \, b \log \left (e\right )\right )} x^{2} + \frac {1}{2} \, {\left (b p r x^{2} + 2 \, a p r x\right )} \log \left (b x + a\right ) + \frac {1}{2} \, {\left (b q r x^{2} + 2 \, a q r x\right )} \log \left (d x + c\right ) - \frac {{\left (a d p r - b c q r + 2 \, a d q r - 2 \, a d r \log \left (f\right ) - 2 \, a d \log \left (e\right )\right )} x}{2 \, d} - \frac {{\left (b c^{2} q r - 2 \, a c d q r\right )} \log \left (-d x - c\right )}{2 \, d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.10 \[ \int (a+b x) \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (\frac {b\,x^2}{2}+a\,x\right )-x\,\left (\frac {r\,\left (2\,a\,d\,p+b\,c\,p+3\,a\,d\,q\right )}{2\,d}-\frac {r\,\left (p+q\right )\,\left (2\,a\,d+2\,b\,c\right )}{4\,d}\right )-\frac {\ln \left (c+d\,x\right )\,\left (b\,c^2\,q\,r-2\,a\,c\,d\,q\,r\right )}{2\,d^2}-\frac {b\,r\,x^2\,\left (p+q\right )}{4}+\frac {a^2\,p\,r\,\ln \left (a+b\,x\right )}{2\,b} \]

[In]

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

[Out]

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