\(\int (a+b x)^2 \log (e (f (a+b x)^p (c+d x)^q)^r) \, dx\) [9]

Optimal result

Integrand size = 29, antiderivative size = 143 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-\frac {(b c-a d)^2 q r x}{3 d^2}+\frac {(b c-a d) q r (a+b x)^2}{6 b d}-\frac {p r (a+b x)^3}{9 b}-\frac {q r (a+b x)^3}{9 b}+\frac {(b c-a d)^3 q r \log (c+d x)}{3 b d^3}+\frac {(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 143, 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, 45} \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {q r (b c-a d)^3 \log (c+d x)}{3 b d^3}-\frac {q r x (b c-a d)^2}{3 d^2}+\frac {(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b}+\frac {q r (a+b x)^2 (b c-a d)}{6 b d}-\frac {p r (a+b x)^3}{9 b}-\frac {q r (a+b x)^3}{9 b} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {-\frac {r \left (6 b d (b c-a d)^2 (p+3 q) x-3 b^2 (b c-a d) (2 p+3 q) (c+d x)^2+2 b^3 (p+q) (c+d x)^3-6 (b c-a d)^3 q \log (c+d x)\right )}{6 d^3}+(a+b x)^3 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{3 b} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(131)=262\).

Time = 70.99 (sec) , antiderivative size = 451, normalized size of antiderivative = 3.15

method	result	size
parallelrisch	\(\frac {12 a^{2} b c \,d^{2} p r +9 a^{2} b c \,d^{2} q r +6 x^{3} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) b^{3} d^{3}-18 \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{3} d^{3}+6 c^{3} q r \,b^{3}+6 a^{3} d^{3} p r +18 a^{3} d^{3} q r -15 a \,c^{2} d q r \,b^{2}-2 x^{3} b^{3} d^{3} p r -2 x^{3} b^{3} d^{3} q r +18 x^{2} \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a \,b^{2} d^{3}+18 x \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{2} b \,d^{3}-36 \ln \left (e \left (f \left (b x +a \right )^{p} \left (d x +c \right )^{q}\right )^{r}\right ) a^{2} b c \,d^{2}+24 \ln \left (b x +a \right ) a^{3} d^{3} p r +18 \ln \left (d x +c \right ) a^{3} d^{3} q r +6 \ln \left (d x +c \right ) b^{3} c^{3} q r -6 x^{2} a \,b^{2} d^{3} p r -9 x^{2} a \,b^{2} d^{3} q r +3 x^{2} b^{3} c \,d^{2} q r -6 x \,a^{2} b \,d^{3} p r -18 x \,a^{2} b \,d^{3} q r -6 x \,b^{3} c^{2} d q r +18 x a \,b^{2} c \,d^{2} q r +36 \ln \left (b x +a \right ) a^{2} b c \,d^{2} p r +54 \ln \left (d x +c \right ) a^{2} b c \,d^{2} q r -18 \ln \left (d x +c \right ) a \,b^{2} c^{2} d q r}{18 b \,d^{3}}\)	\(451\)

[In]

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

[Out]

1/18*(12*a^2*b*c*d^2*p*r+9*a^2*b*c*d^2*q*r+6*x^3*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*b^3*d^3-18*ln(e*(f*(b*x+a)^p*
(d*x+c)^q)^r)*a^3*d^3+6*c^3*q*r*b^3+6*a^3*d^3*p*r+18*a^3*d^3*q*r-15*a*c^2*d*q*r*b^2-2*x^3*b^3*d^3*p*r-2*x^3*b^
3*d^3*q*r+18*x^2*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a*b^2*d^3+18*x*ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^2*b*d^3-36*l
n(e*(f*(b*x+a)^p*(d*x+c)^q)^r)*a^2*b*c*d^2+24*ln(b*x+a)*a^3*d^3*p*r+18*ln(d*x+c)*a^3*d^3*q*r+6*ln(d*x+c)*b^3*c
^3*q*r-6*x^2*a*b^2*d^3*p*r-9*x^2*a*b^2*d^3*q*r+3*x^2*b^3*c*d^2*q*r-6*x*a^2*b*d^3*p*r-18*x*a^2*b*d^3*q*r-6*x*b^
3*c^2*d*q*r+18*x*a*b^2*c*d^2*q*r+36*ln(b*x+a)*a^2*b*c*d^2*p*r+54*ln(d*x+c)*a^2*b*c*d^2*q*r-18*ln(d*x+c)*a*b^2*
c^2*d*q*r)/b/d^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (131) = 262\).

Time = 0.29 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.27 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=-\frac {2 \, {\left (b^{3} d^{3} p + b^{3} d^{3} q\right )} r x^{3} + 3 \, {\left (2 \, a b^{2} d^{3} p - {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} q\right )} r x^{2} + 6 \, {\left (a^{2} b d^{3} p + {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} q\right )} r x - 6 \, {\left (b^{3} d^{3} p r x^{3} + 3 \, a b^{2} d^{3} p r x^{2} + 3 \, a^{2} b d^{3} p r x + a^{3} d^{3} p r\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} d^{3} q r x^{3} + 3 \, a b^{2} d^{3} q r x^{2} + 3 \, a^{2} b d^{3} q r x + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2}\right )} q r\right )} \log \left (d x + c\right ) - 6 \, {\left (b^{3} d^{3} x^{3} + 3 \, a b^{2} d^{3} x^{2} + 3 \, a^{2} b d^{3} x\right )} \log \left (e\right ) - 6 \, {\left (b^{3} d^{3} r x^{3} + 3 \, a b^{2} d^{3} r x^{2} + 3 \, a^{2} b d^{3} r x\right )} \log \left (f\right )}{18 \, b d^{3}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (124) = 248\).

Time = 132.78 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.41 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\begin {cases} a^{2} x \log {\left (e \left (a^{p} c^{q} f\right )^{r} \right )} & \text {for}\: b = 0 \wedge d = 0 \\a^{2} \left (\frac {c \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}}{d} - q r x + x \log {\left (e \left (a^{p} f \left (c + d x\right )^{q}\right )^{r} \right )}\right ) & \text {for}\: b = 0 \\\frac {a^{3} \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{3 b} - \frac {a^{2} p r x}{3} + a^{2} x \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )} - \frac {a b p r x^{2}}{3} + a b x^{2} \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )} - \frac {b^{2} p r x^{3}}{9} + \frac {b^{2} x^{3} \log {\left (e \left (c^{q} f \left (a + b x\right )^{p}\right )^{r} \right )}}{3} & \text {for}\: d = 0 \\- \frac {a^{3} q r \log {\left (\frac {c}{d} + x \right )}}{3 b} + \frac {a^{3} \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{3 b} + \frac {a^{2} c q r \log {\left (\frac {c}{d} + x \right )}}{d} - \frac {a^{2} p r x}{3} - a^{2} q r x + a^{2} x \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )} - \frac {a b c^{2} q r \log {\left (\frac {c}{d} + x \right )}}{d^{2}} + \frac {a b c q r x}{d} - \frac {a b p r x^{2}}{3} - \frac {a b q r x^{2}}{2} + a b x^{2} \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )} + \frac {b^{2} c^{3} q r \log {\left (\frac {c}{d} + x \right )}}{3 d^{3}} - \frac {b^{2} c^{2} q r x}{3 d^{2}} + \frac {b^{2} c q r x^{2}}{6 d} - \frac {b^{2} p r x^{3}}{9} - \frac {b^{2} q r x^{3}}{9} + \frac {b^{2} x^{3} \log {\left (e \left (f \left (a + b x\right )^{p} \left (c + d x\right )^{q}\right )^{r} \right )}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x*log(e*(a**p*c**q*f)**r), Eq(b, 0) & Eq(d, 0)), (a**2*(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**3*log(e*(c**q*f*(a + b*x)**p)**r)/(3*b) - a**2*p*r*
x/3 + a**2*x*log(e*(c**q*f*(a + b*x)**p)**r) - a*b*p*r*x**2/3 + a*b*x**2*log(e*(c**q*f*(a + b*x)**p)**r) - b**
2*p*r*x**3/9 + b**2*x**3*log(e*(c**q*f*(a + b*x)**p)**r)/3, Eq(d, 0)), (-a**3*q*r*log(c/d + x)/(3*b) + a**3*lo
g(e*(f*(a + b*x)**p*(c + d*x)**q)**r)/(3*b) + a**2*c*q*r*log(c/d + x)/d - a**2*p*r*x/3 - a**2*q*r*x + a**2*x*l
og(e*(f*(a + b*x)**p*(c + d*x)**q)**r) - a*b*c**2*q*r*log(c/d + x)/d**2 + a*b*c*q*r*x/d - a*b*p*r*x**2/3 - a*b
*q*r*x**2/2 + a*b*x**2*log(e*(f*(a + b*x)**p*(c + d*x)**q)**r) + b**2*c**3*q*r*log(c/d + x)/(3*d**3) - b**2*c*
*2*q*r*x/(3*d**2) + b**2*c*q*r*x**2/(6*d) - b**2*p*r*x**3/9 - b**2*q*r*x**3/9 + b**2*x**3*log(e*(f*(a + b*x)**
p*(c + d*x)**q)**r)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.36 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {1}{3} \, {\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} \log \left (\left ({\left (b x + a\right )}^{p} {\left (d x + c\right )}^{q} f\right )^{r} e\right ) + \frac {{\left (\frac {6 \, a^{3} f p \log \left (b x + a\right )}{b} - \frac {2 \, b^{2} d^{2} f {\left (p + q\right )} x^{3} + 3 \, {\left (a b d^{2} f {\left (2 \, p + 3 \, q\right )} - b^{2} c d f q\right )} x^{2} + 6 \, {\left (a^{2} d^{2} f {\left (p + 3 \, q\right )} + b^{2} c^{2} f q - 3 \, a b c d f q\right )} x}{d^{2}} + \frac {6 \, {\left (b^{2} c^{3} f q - 3 \, a b c^{2} d f q + 3 \, a^{2} c d^{2} f q\right )} \log \left (d x + c\right )}{d^{3}}\right )} r}{18 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (131) = 262\).

Time = 1.13 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.92 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=\frac {a^{3} p r \log \left (b x + a\right )}{3 \, b} - \frac {1}{9} \, {\left (b^{2} p r + b^{2} q r - 3 \, b^{2} r \log \left (f\right ) - 3 \, b^{2} \log \left (e\right )\right )} x^{3} - \frac {{\left (2 \, a b d p r - b^{2} c q r + 3 \, a b d q r - 6 \, a b d r \log \left (f\right ) - 6 \, a b d \log \left (e\right )\right )} x^{2}}{6 \, d} + \frac {1}{3} \, {\left (b^{2} p r x^{3} + 3 \, a b p r x^{2} + 3 \, a^{2} p r x\right )} \log \left (b x + a\right ) + \frac {1}{3} \, {\left (b^{2} q r x^{3} + 3 \, a b q r x^{2} + 3 \, a^{2} q r x\right )} \log \left (d x + c\right ) - \frac {{\left (a^{2} d^{2} p r + b^{2} c^{2} q r - 3 \, a b c d q r + 3 \, a^{2} d^{2} q r - 3 \, a^{2} d^{2} r \log \left (f\right ) - 3 \, a^{2} d^{2} \log \left (e\right )\right )} x}{3 \, d^{2}} + \frac {{\left (b^{2} c^{3} q r - 3 \, a b c^{2} d q r + 3 \, a^{2} c d^{2} q r\right )} \log \left (-d x - c\right )}{3 \, d^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.78 \[ \int (a+b x)^2 \log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right ) \, dx=x\,\left (\frac {\left (\frac {b\,r\,\left (3\,a\,d\,p+b\,c\,p+4\,a\,d\,q\right )}{3\,d}-\frac {b\,r\,\left (p+q\right )\,\left (3\,a\,d+3\,b\,c\right )}{9\,d}\right )\,\left (3\,a\,d+3\,b\,c\right )}{3\,b\,d}-\frac {a\,r\,\left (a\,d\,p+b\,c\,p+2\,a\,d\,q\right )}{d}+\frac {a\,b\,c\,r\,\left (p+q\right )}{3\,d}\right )-x^2\,\left (\frac {b\,r\,\left (3\,a\,d\,p+b\,c\,p+4\,a\,d\,q\right )}{6\,d}-\frac {b\,r\,\left (p+q\right )\,\left (3\,a\,d+3\,b\,c\right )}{18\,d}\right )+\ln \left (e\,{\left (f\,{\left (a+b\,x\right )}^p\,{\left (c+d\,x\right )}^q\right )}^r\right )\,\left (a^2\,x+a\,b\,x^2+\frac {b^2\,x^3}{3}\right )+\frac {\ln \left (c+d\,x\right )\,\left (3\,q\,r\,a^2\,c\,d^2-3\,q\,r\,a\,b\,c^2\,d+q\,r\,b^2\,c^3\right )}{3\,d^3}-\frac {b^2\,r\,x^3\,\left (p+q\right )}{9}+\frac {a^3\,p\,r\,\ln \left (a+b\,x\right )}{3\,b} \]

[In]

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

[Out]

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