\(\int (d+e x)^3 \log (c (a+\frac {b}{x})^p) \, dx\) [198]

Optimal result

Integrand size = 20, antiderivative size = 139 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {b e \left (6 a^2 d^2-4 a b d e+b^2 e^2\right ) p x}{4 a^3}+\frac {b e^2 (4 a d-b e) p x^2}{8 a^2}+\frac {b e^3 p x^3}{12 a}+\frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {d^4 p \log (x)}{4 e}-\frac {(a d-b e)^4 p \log (b+a x)}{4 a^4 e} \]

[Out]

1/4*b*e*(6*a^2*d^2-4*a*b*d*e+b^2*e^2)*p*x/a^3+1/8*b*e^2*(4*a*d-b*e)*p*x^2/a^2+1/12*b*e^3*p*x^3/a+1/4*(e*x+d)^4
*ln(c*(a+b/x)^p)/e+1/4*d^4*p*ln(x)/e-1/4*(a*d-b*e)^4*p*ln(a*x+b)/a^4/e

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2513, 528, 84} \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {p (a d-b e)^4 \log (a x+b)}{4 a^4 e}+\frac {b e^2 p x^2 (4 a d-b e)}{8 a^2}+\frac {b e p x \left (6 a^2 d^2-4 a b d e+b^2 e^2\right )}{4 a^3}+\frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {b e^3 p x^3}{12 a}+\frac {d^4 p \log (x)}{4 e} \]

[In]

Int[(d + e*x)^3*Log[c*(a + b/x)^p],x]

[Out]

(b*e*(6*a^2*d^2 - 4*a*b*d*e + b^2*e^2)*p*x)/(4*a^3) + (b*e^2*(4*a*d - b*e)*p*x^2)/(8*a^2) + (b*e^3*p*x^3)/(12*
a) + ((d + e*x)^4*Log[c*(a + b/x)^p])/(4*e) + (d^4*p*Log[x])/(4*e) - ((a*d - b*e)^4*p*Log[b + a*x])/(4*a^4*e)

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 528

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 2513

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[(f
+ g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n)^p])/(g*(r + 1))), x] - Dist[b*e*n*(p/(g*(r + 1))), Int[x^(n - 1)*((f
+ g*x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {(b p) \int \frac {(d+e x)^4}{\left (a+\frac {b}{x}\right ) x^2} \, dx}{4 e} \\ & = \frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {(b p) \int \frac {(d+e x)^4}{x (b+a x)} \, dx}{4 e} \\ & = \frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {(b p) \int \left (\frac {e^2 \left (6 a^2 d^2-4 a b d e+b^2 e^2\right )}{a^3}+\frac {d^4}{b x}+\frac {e^3 (4 a d-b e) x}{a^2}+\frac {e^4 x^2}{a}-\frac {(a d-b e)^4}{a^3 b (b+a x)}\right ) \, dx}{4 e} \\ & = \frac {b e \left (6 a^2 d^2-4 a b d e+b^2 e^2\right ) p x}{4 a^3}+\frac {b e^2 (4 a d-b e) p x^2}{8 a^2}+\frac {b e^3 p x^3}{12 a}+\frac {(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )}{4 e}+\frac {d^4 p \log (x)}{4 e}-\frac {(a d-b e)^4 p \log (b+a x)}{4 a^4 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.82 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {\frac {b e^2 p x \left (6 b^2 e^2-3 a b e (8 d+e x)+2 a^2 \left (18 d^2+6 d e x+e^2 x^2\right )\right )}{6 a^3}+(d+e x)^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+d^4 p \log (x)-\frac {(a d-b e)^4 p \log (b+a x)}{a^4}}{4 e} \]

[In]

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

[Out]

((b*e^2*p*x*(6*b^2*e^2 - 3*a*b*e*(8*d + e*x) + 2*a^2*(18*d^2 + 6*d*e*x + e^2*x^2)))/(6*a^3) + (d + e*x)^4*Log[
c*(a + b/x)^p] + d^4*p*Log[x] - ((a*d - b*e)^4*p*Log[b + a*x])/a^4)/(4*e)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.76

method	result	size
parts	\(\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e^{3} x^{4}}{4}+\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e^{2} d \,x^{3}+\frac {3 \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) e \,d^{2} x^{2}}{2}+\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) d^{3} x +\frac {\ln \left (c \left (a +\frac {b}{x}\right )^{p}\right ) d^{4}}{4 e}+\frac {p b \left (\frac {e^{2} \left (\frac {1}{3} a^{2} e^{2} x^{3}+2 a^{2} d e \,x^{2}-\frac {1}{2} a b \,e^{2} x^{2}+6 a^{2} d^{2} x -4 a b d e x +x \,e^{2} b^{2}\right )}{a^{3}}+\frac {d^{4} \ln \left (x \right )}{b}+\frac {\left (-a^{4} d^{4}+4 a^{3} b \,d^{3} e -6 a^{2} b^{2} d^{2} e^{2}+4 b^{3} d \,e^{3} a -b^{4} e^{4}\right ) \ln \left (a x +b \right )}{a^{4} b}\right )}{4 e}\)	\(244\)
parallelrisch	\(-\frac {-6 x^{4} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} e^{3}-24 x^{3} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} d \,e^{2}-2 x^{3} a^{3} b \,e^{3} p -36 x^{2} \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} d^{2} e -12 x^{2} a^{3} b d \,e^{2} p +3 x^{2} a^{2} b^{2} e^{3} p +24 \ln \left (x \right ) a^{3} b \,d^{3} p -48 \ln \left (a x +b \right ) a^{3} b \,d^{3} p +36 \ln \left (a x +b \right ) a^{2} b^{2} d^{2} e p -24 \ln \left (a x +b \right ) a \,b^{3} d \,e^{2} p +6 \ln \left (a x +b \right ) b^{4} e^{3} p -24 x \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{4} d^{3}-36 x \,a^{3} b \,d^{2} e p +24 x \,a^{2} b^{2} d \,e^{2} p -6 x a \,b^{3} e^{3} p +24 \ln \left (c \left (\frac {a x +b}{x}\right )^{p}\right ) a^{3} b \,d^{3}+36 a^{2} b^{2} d^{2} e p -24 a \,b^{3} d \,e^{2} p +6 b^{4} e^{3} p}{24 a^{4}}\)	\(321\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.72 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {2 \, a^{3} b e^{3} p x^{3} + 3 \, {\left (4 \, a^{3} b d e^{2} - a^{2} b^{2} e^{3}\right )} p x^{2} + 6 \, {\left (6 \, a^{3} b d^{2} e - 4 \, a^{2} b^{2} d e^{2} + a b^{3} e^{3}\right )} p x + 6 \, {\left (4 \, a^{3} b d^{3} - 6 \, a^{2} b^{2} d^{2} e + 4 \, a b^{3} d e^{2} - b^{4} e^{3}\right )} p \log \left (a x + b\right ) + 6 \, {\left (a^{4} e^{3} x^{4} + 4 \, a^{4} d e^{2} x^{3} + 6 \, a^{4} d^{2} e x^{2} + 4 \, a^{4} d^{3} x\right )} \log \left (c\right ) + 6 \, {\left (a^{4} e^{3} p x^{4} + 4 \, a^{4} d e^{2} p x^{3} + 6 \, a^{4} d^{2} e p x^{2} + 4 \, a^{4} d^{3} p x\right )} \log \left (\frac {a x + b}{x}\right )}{24 \, a^{4}} \]

[In]

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

[Out]

1/24*(2*a^3*b*e^3*p*x^3 + 3*(4*a^3*b*d*e^2 - a^2*b^2*e^3)*p*x^2 + 6*(6*a^3*b*d^2*e - 4*a^2*b^2*d*e^2 + a*b^3*e
^3)*p*x + 6*(4*a^3*b*d^3 - 6*a^2*b^2*d^2*e + 4*a*b^3*d*e^2 - b^4*e^3)*p*log(a*x + b) + 6*(a^4*e^3*x^4 + 4*a^4*
d*e^2*x^3 + 6*a^4*d^2*e*x^2 + 4*a^4*d^3*x)*log(c) + 6*(a^4*e^3*p*x^4 + 4*a^4*d*e^2*p*x^3 + 6*a^4*d^2*e*p*x^2 +
4*a^4*d^3*p*x)*log((a*x + b)/x))/a^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (128) = 256\).

Time = 1.66 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.55 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\begin {cases} d^{3} x \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {3 d^{2} e x^{2} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{2} + d e^{2} x^{3} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )} + \frac {e^{3} x^{4} \log {\left (c \left (a + \frac {b}{x}\right )^{p} \right )}}{4} + \frac {b d^{3} p \log {\left (x + \frac {b}{a} \right )}}{a} + \frac {3 b d^{2} e p x}{2 a} + \frac {b d e^{2} p x^{2}}{2 a} + \frac {b e^{3} p x^{3}}{12 a} - \frac {3 b^{2} d^{2} e p \log {\left (x + \frac {b}{a} \right )}}{2 a^{2}} - \frac {b^{2} d e^{2} p x}{a^{2}} - \frac {b^{2} e^{3} p x^{2}}{8 a^{2}} + \frac {b^{3} d e^{2} p \log {\left (x + \frac {b}{a} \right )}}{a^{3}} + \frac {b^{3} e^{3} p x}{4 a^{3}} - \frac {b^{4} e^{3} p \log {\left (x + \frac {b}{a} \right )}}{4 a^{4}} & \text {for}\: a \neq 0 \\d^{3} p x + d^{3} x \log {\left (c \left (\frac {b}{x}\right )^{p} \right )} + \frac {3 d^{2} e p x^{2}}{4} + \frac {3 d^{2} e x^{2} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{2} + \frac {d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )} + \frac {e^{3} p x^{4}}{16} + \frac {e^{3} x^{4} \log {\left (c \left (\frac {b}{x}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((d**3*x*log(c*(a + b/x)**p) + 3*d**2*e*x**2*log(c*(a + b/x)**p)/2 + d*e**2*x**3*log(c*(a + b/x)**p)
+ e**3*x**4*log(c*(a + b/x)**p)/4 + b*d**3*p*log(x + b/a)/a + 3*b*d**2*e*p*x/(2*a) + b*d*e**2*p*x**2/(2*a) + b
*e**3*p*x**3/(12*a) - 3*b**2*d**2*e*p*log(x + b/a)/(2*a**2) - b**2*d*e**2*p*x/a**2 - b**2*e**3*p*x**2/(8*a**2)
+ b**3*d*e**2*p*log(x + b/a)/a**3 + b**3*e**3*p*x/(4*a**3) - b**4*e**3*p*log(x + b/a)/(4*a**4), Ne(a, 0)), (d
**3*p*x + d**3*x*log(c*(b/x)**p) + 3*d**2*e*p*x**2/4 + 3*d**2*e*x**2*log(c*(b/x)**p)/2 + d*e**2*p*x**3/3 + d*e
**2*x**3*log(c*(b/x)**p) + e**3*p*x**4/16 + e**3*x**4*log(c*(b/x)**p)/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.19 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=\frac {1}{24} \, b p {\left (\frac {2 \, a^{2} e^{3} x^{3} + 3 \, {\left (4 \, a^{2} d e^{2} - a b e^{3}\right )} x^{2} + 6 \, {\left (6 \, a^{2} d^{2} e - 4 \, a b d e^{2} + b^{2} e^{3}\right )} x}{a^{3}} + \frac {6 \, {\left (4 \, a^{3} d^{3} - 6 \, a^{2} b d^{2} e + 4 \, a b^{2} d e^{2} - b^{3} e^{3}\right )} \log \left (a x + b\right )}{a^{4}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) \]

[In]

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

[Out]

1/24*b*p*((2*a^2*e^3*x^3 + 3*(4*a^2*d*e^2 - a*b*e^3)*x^2 + 6*(6*a^2*d^2*e - 4*a*b*d*e^2 + b^2*e^3)*x)/a^3 + 6*
(4*a^3*d^3 - 6*a^2*b*d^2*e + 4*a*b^2*d*e^2 - b^3*e^3)*log(a*x + b)/a^4) + 1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e
*x^2 + 4*d^3*x)*log((a + b/x)^p*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 847 vs. \(2 (127) = 254\).

Time = 0.33 (sec) , antiderivative size = 847, normalized size of antiderivative = 6.09 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=-\frac {\frac {6 \, {\left (4 \, a^{3} b^{2} d^{3} p - 6 \, a^{2} b^{3} d^{2} e p + 4 \, a b^{4} d e^{2} p - b^{5} e^{3} p - \frac {12 \, {\left (a x + b\right )} a^{2} b^{2} d^{3} p}{x} + \frac {12 \, {\left (a x + b\right )} a b^{3} d^{2} e p}{x} - \frac {4 \, {\left (a x + b\right )} b^{4} d e^{2} p}{x} + \frac {12 \, {\left (a x + b\right )}^{2} a b^{2} d^{3} p}{x^{2}} - \frac {6 \, {\left (a x + b\right )}^{2} b^{3} d^{2} e p}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} b^{2} d^{3} p}{x^{3}}\right )} \log \left (\frac {a x + b}{x}\right )}{a^{4} - \frac {4 \, {\left (a x + b\right )} a^{3}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{2}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a}{x^{3}} + \frac {{\left (a x + b\right )}^{4}}{x^{4}}} + \frac {36 \, a^{5} b^{3} d^{2} e p - 36 \, a^{4} b^{4} d e^{2} p + 11 \, a^{3} b^{5} e^{3} p + 24 \, a^{6} b^{2} d^{3} \log \left (c\right ) - 36 \, a^{5} b^{3} d^{2} e \log \left (c\right ) + 24 \, a^{4} b^{4} d e^{2} \log \left (c\right ) - 6 \, a^{3} b^{5} e^{3} \log \left (c\right ) - \frac {108 \, {\left (a x + b\right )} a^{4} b^{3} d^{2} e p}{x} + \frac {96 \, {\left (a x + b\right )} a^{3} b^{4} d e^{2} p}{x} - \frac {26 \, {\left (a x + b\right )} a^{2} b^{5} e^{3} p}{x} - \frac {72 \, {\left (a x + b\right )} a^{5} b^{2} d^{3} \log \left (c\right )}{x} + \frac {72 \, {\left (a x + b\right )} a^{4} b^{3} d^{2} e \log \left (c\right )}{x} - \frac {24 \, {\left (a x + b\right )} a^{3} b^{4} d e^{2} \log \left (c\right )}{x} + \frac {108 \, {\left (a x + b\right )}^{2} a^{3} b^{3} d^{2} e p}{x^{2}} - \frac {84 \, {\left (a x + b\right )}^{2} a^{2} b^{4} d e^{2} p}{x^{2}} + \frac {21 \, {\left (a x + b\right )}^{2} a b^{5} e^{3} p}{x^{2}} + \frac {72 \, {\left (a x + b\right )}^{2} a^{4} b^{2} d^{3} \log \left (c\right )}{x^{2}} - \frac {36 \, {\left (a x + b\right )}^{2} a^{3} b^{3} d^{2} e \log \left (c\right )}{x^{2}} - \frac {36 \, {\left (a x + b\right )}^{3} a^{2} b^{3} d^{2} e p}{x^{3}} + \frac {24 \, {\left (a x + b\right )}^{3} a b^{4} d e^{2} p}{x^{3}} - \frac {6 \, {\left (a x + b\right )}^{3} b^{5} e^{3} p}{x^{3}} - \frac {24 \, {\left (a x + b\right )}^{3} a^{3} b^{2} d^{3} \log \left (c\right )}{x^{3}}}{a^{7} - \frac {4 \, {\left (a x + b\right )} a^{6}}{x} + \frac {6 \, {\left (a x + b\right )}^{2} a^{5}}{x^{2}} - \frac {4 \, {\left (a x + b\right )}^{3} a^{4}}{x^{3}} + \frac {{\left (a x + b\right )}^{4} a^{3}}{x^{4}}} + \frac {6 \, {\left (4 \, a^{3} b^{2} d^{3} p - 6 \, a^{2} b^{3} d^{2} e p + 4 \, a b^{4} d e^{2} p - b^{5} e^{3} p\right )} \log \left (-a + \frac {a x + b}{x}\right )}{a^{4}} - \frac {6 \, {\left (4 \, a^{3} b^{2} d^{3} p - 6 \, a^{2} b^{3} d^{2} e p + 4 \, a b^{4} d e^{2} p - b^{5} e^{3} p\right )} \log \left (\frac {a x + b}{x}\right )}{a^{4}}}{24 \, b} \]

[In]

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

[Out]

-1/24*(6*(4*a^3*b^2*d^3*p - 6*a^2*b^3*d^2*e*p + 4*a*b^4*d*e^2*p - b^5*e^3*p - 12*(a*x + b)*a^2*b^2*d^3*p/x + 1
2*(a*x + b)*a*b^3*d^2*e*p/x - 4*(a*x + b)*b^4*d*e^2*p/x + 12*(a*x + b)^2*a*b^2*d^3*p/x^2 - 6*(a*x + b)^2*b^3*d
^2*e*p/x^2 - 4*(a*x + b)^3*b^2*d^3*p/x^3)*log((a*x + b)/x)/(a^4 - 4*(a*x + b)*a^3/x + 6*(a*x + b)^2*a^2/x^2 -
4*(a*x + b)^3*a/x^3 + (a*x + b)^4/x^4) + (36*a^5*b^3*d^2*e*p - 36*a^4*b^4*d*e^2*p + 11*a^3*b^5*e^3*p + 24*a^6*
b^2*d^3*log(c) - 36*a^5*b^3*d^2*e*log(c) + 24*a^4*b^4*d*e^2*log(c) - 6*a^3*b^5*e^3*log(c) - 108*(a*x + b)*a^4*
b^3*d^2*e*p/x + 96*(a*x + b)*a^3*b^4*d*e^2*p/x - 26*(a*x + b)*a^2*b^5*e^3*p/x - 72*(a*x + b)*a^5*b^2*d^3*log(c
)/x + 72*(a*x + b)*a^4*b^3*d^2*e*log(c)/x - 24*(a*x + b)*a^3*b^4*d*e^2*log(c)/x + 108*(a*x + b)^2*a^3*b^3*d^2*
e*p/x^2 - 84*(a*x + b)^2*a^2*b^4*d*e^2*p/x^2 + 21*(a*x + b)^2*a*b^5*e^3*p/x^2 + 72*(a*x + b)^2*a^4*b^2*d^3*log
(c)/x^2 - 36*(a*x + b)^2*a^3*b^3*d^2*e*log(c)/x^2 - 36*(a*x + b)^3*a^2*b^3*d^2*e*p/x^3 + 24*(a*x + b)^3*a*b^4*
d*e^2*p/x^3 - 6*(a*x + b)^3*b^5*e^3*p/x^3 - 24*(a*x + b)^3*a^3*b^2*d^3*log(c)/x^3)/(a^7 - 4*(a*x + b)*a^6/x +
6*(a*x + b)^2*a^5/x^2 - 4*(a*x + b)^3*a^4/x^3 + (a*x + b)^4*a^3/x^4) + 6*(4*a^3*b^2*d^3*p - 6*a^2*b^3*d^2*e*p
+ 4*a*b^4*d*e^2*p - b^5*e^3*p)*log(-a + (a*x + b)/x)/a^4 - 6*(4*a^3*b^2*d^3*p - 6*a^2*b^3*d^2*e*p + 4*a*b^4*d*
e^2*p - b^5*e^3*p)*log((a*x + b)/x)/a^4)/b

Mupad [B] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.32 \[ \int (d+e x)^3 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx=x\,\left (\frac {b\,\left (\frac {b^2\,e^3\,p}{4\,a^2}-\frac {b\,d\,e^2\,p}{a}\right )}{a}+\frac {3\,b\,d^2\,e\,p}{2\,a}\right )+\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )\,\left (d^3\,x+\frac {3\,d^2\,e\,x^2}{2}+d\,e^2\,x^3+\frac {e^3\,x^4}{4}\right )-x^2\,\left (\frac {b^2\,e^3\,p}{8\,a^2}-\frac {b\,d\,e^2\,p}{2\,a}\right )-\frac {\ln \left (b+a\,x\right )\,\left (-4\,p\,a^3\,b\,d^3+6\,p\,a^2\,b^2\,d^2\,e-4\,p\,a\,b^3\,d\,e^2+p\,b^4\,e^3\right )}{4\,a^4}+\frac {b\,e^3\,p\,x^3}{12\,a} \]

[In]

int(log(c*(a + b/x)^p)*(d + e*x)^3,x)

[Out]

x*((b*((b^2*e^3*p)/(4*a^2) - (b*d*e^2*p)/a))/a + (3*b*d^2*e*p)/(2*a)) + log(c*(a + b/x)^p)*(d^3*x + (e^3*x^4)/
4 + (3*d^2*e*x^2)/2 + d*e^2*x^3) - x^2*((b^2*e^3*p)/(8*a^2) - (b*d*e^2*p)/(2*a)) - (log(b + a*x)*(b^4*e^3*p -
4*a^3*b*d^3*p - 4*a*b^3*d*e^2*p + 6*a^2*b^2*d^2*e*p))/(4*a^4) + (b*e^3*p*x^3)/(12*a)