\(\int (d+e x)^3 \log (c (a+b x)^p) \, dx\) [176]

Optimal result

Integrand size = 18, antiderivative size = 140 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=-\frac {(b d-a e)^3 p x}{4 b^3}-\frac {(b d-a e)^2 p (d+e x)^2}{8 b^2 e}-\frac {(b d-a e) p (d+e x)^3}{12 b e}-\frac {p (d+e x)^4}{16 e}-\frac {(b d-a e)^4 p \log (a+b x)}{4 b^4 e}+\frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e} \]

[Out]

-1/4*(-a*e+b*d)^3*p*x/b^3-1/8*(-a*e+b*d)^2*p*(e*x+d)^2/b^2/e-1/12*(-a*e+b*d)*p*(e*x+d)^3/b/e-1/16*p*(e*x+d)^4/
e-1/4*(-a*e+b*d)^4*p*ln(b*x+a)/b^4/e+1/4*(e*x+d)^4*ln(c*(b*x+a)^p)/e

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2442, 45} \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=-\frac {p (b d-a e)^4 \log (a+b x)}{4 b^4 e}-\frac {p x (b d-a e)^3}{4 b^3}-\frac {p (d+e x)^2 (b d-a e)^2}{8 b^2 e}+\frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e}-\frac {p (d+e x)^3 (b d-a e)}{12 b e}-\frac {p (d+e x)^4}{16 e} \]

[In]

Int[(d + e*x)^3*Log[c*(a + b*x)^p],x]

[Out]

-1/4*((b*d - a*e)^3*p*x)/b^3 - ((b*d - a*e)^2*p*(d + e*x)^2)/(8*b^2*e) - ((b*d - a*e)*p*(d + e*x)^3)/(12*b*e)
- (p*(d + e*x)^4)/(16*e) - ((b*d - a*e)^4*p*Log[a + b*x])/(4*b^4*e) + ((d + e*x)^4*Log[c*(a + b*x)^p])/(4*e)

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 2442

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

Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e}-\frac {(b p) \int \frac {(d+e x)^4}{a+b x} \, dx}{4 e} \\ & = \frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e}-\frac {(b p) \int \left (\frac {e (b d-a e)^3}{b^4}+\frac {(b d-a e)^4}{b^4 (a+b x)}+\frac {e (b d-a e)^2 (d+e x)}{b^3}+\frac {e (b d-a e) (d+e x)^2}{b^2}+\frac {e (d+e x)^3}{b}\right ) \, dx}{4 e} \\ & = -\frac {(b d-a e)^3 p x}{4 b^3}-\frac {(b d-a e)^2 p (d+e x)^2}{8 b^2 e}-\frac {(b d-a e) p (d+e x)^3}{12 b e}-\frac {p (d+e x)^4}{16 e}-\frac {(b d-a e)^4 p \log (a+b x)}{4 b^4 e}+\frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.32 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=-\frac {b p x \left (-12 a^3 e^3+6 a^2 b e^2 (8 d+e x)-4 a b^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+b^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )+12 a^2 e \left (6 b^2 d^2-4 a b d e+a^2 e^2\right ) p \log (a+b x)-12 b^3 \left (4 a d^3+b x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right ) \log \left (c (a+b x)^p\right )}{48 b^4} \]

[In]

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

[Out]

-1/48*(b*p*x*(-12*a^3*e^3 + 6*a^2*b*e^2*(8*d + e*x) - 4*a*b^2*e*(18*d^2 + 6*d*e*x + e^2*x^2) + b^3*(48*d^3 + 3
6*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3)) + 12*a^2*e*(6*b^2*d^2 - 4*a*b*d*e + a^2*e^2)*p*Log[a + b*x] - 12*b^3*(4
*a*d^3 + b*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3))*Log[c*(a + b*x)^p])/b^4

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(283\) vs. \(2(128)=256\).

Time = 1.36 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.03

method	result	size
parts	\(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{3} x^{4}}{4}+\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} d \,x^{3}+\frac {3 \ln \left (c \left (b x +a \right )^{p}\right ) e \,d^{2} x^{2}}{2}+d^{3} \ln \left (c \left (b x +a \right )^{p}\right ) x +\frac {\ln \left (c \left (b x +a \right )^{p}\right ) d^{4}}{4 e}-\frac {p b \left (-\frac {e \left (-\frac {b^{3} e^{3} x^{4}}{4}+\frac {\left (\left (a e -2 b d \right ) e^{2} b^{2}-2 b^{3} d \,e^{2}\right ) x^{3}}{3}+\frac {\left (2 \left (a e -2 b d \right ) d e \,b^{2}-b e \left (a^{2} e^{2}-2 a d e b +2 b^{2} d^{2}\right )\right ) x^{2}}{2}+x \left (a e -2 b d \right ) \left (a^{2} e^{2}-2 a d e b +2 b^{2} d^{2}\right )\right )}{b^{4}}+\frac {\left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \ln \left (b x +a \right )}{b^{5}}\right )}{4 e}\)	\(284\)
parallelrisch	\(-\frac {-12 x^{4} \ln \left (c \left (b x +a \right )^{p}\right ) a \,b^{4} e^{3}+3 x^{4} a \,b^{4} e^{3} p -48 x^{3} \ln \left (c \left (b x +a \right )^{p}\right ) a \,b^{4} d \,e^{2}-4 x^{3} a^{2} b^{3} e^{3} p +16 x^{3} a \,b^{4} d \,e^{2} p -72 x^{2} \ln \left (c \left (b x +a \right )^{p}\right ) a \,b^{4} d^{2} e +6 x^{2} a^{3} b^{2} e^{3} p -24 x^{2} a^{2} b^{3} d \,e^{2} p +36 x^{2} a \,b^{4} d^{2} e p +12 \ln \left (b x +a \right ) a^{5} e^{3} p -48 \ln \left (b x +a \right ) a^{4} b d \,e^{2} p +72 \ln \left (b x +a \right ) a^{3} b^{2} d^{2} e p -96 \ln \left (b x +a \right ) a^{2} b^{3} d^{3} p -48 x \ln \left (c \left (b x +a \right )^{p}\right ) a \,b^{4} d^{3}-12 x \,a^{4} b \,e^{3} p +48 x \,a^{3} b^{2} d \,e^{2} p -72 x \,a^{2} b^{3} d^{2} e p +48 x a \,b^{4} d^{3} p +48 \ln \left (c \left (b x +a \right )^{p}\right ) a^{2} b^{3} d^{3}}{48 b^{4} a}\)	\(325\)
risch	\(\frac {i e^{2} \pi d \,x^{3} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i e^{2} \pi d \,x^{3} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi \,d^{3} x \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}+e^{2} \ln \left (c \right ) d \,x^{3}+\frac {3 e \ln \left (c \right ) d^{2} x^{2}}{2}-\frac {\ln \left (b x +a \right ) d^{4} p}{4 e}-\frac {i e^{3} \pi \,x^{4} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{8}-\frac {i \pi \,d^{3} x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {3 i e \pi \,d^{2} x^{2} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4}-\frac {i e^{3} \pi \,x^{4} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{8}+\frac {\left (e x +d \right )^{4} \ln \left (\left (b x +a \right )^{p}\right )}{4 e}-d^{3} p x +\frac {i \pi \,d^{3} x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {e^{2} \ln \left (b x +a \right ) a^{3} d p}{b^{3}}-\frac {3 e \ln \left (b x +a \right ) a^{2} d^{2} p}{2 b^{2}}+\frac {e^{3} \ln \left (c \right ) x^{4}}{4}+\ln \left (c \right ) d^{3} x +\frac {i \pi \,d^{3} x \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {i e^{3} \pi \,x^{4} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {i e^{3} \pi \,x^{4} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{8}-\frac {i e^{2} \pi d \,x^{3} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}-\frac {3 i e \pi \,d^{2} x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{4}+\frac {d^{3} p a \ln \left (b x +a \right )}{b}+\frac {e^{2} a d p \,x^{2}}{2 b}-\frac {e^{2} a^{2} d p x}{b^{2}}+\frac {3 e a \,d^{2} p x}{2 b}-\frac {i e^{2} \pi d \,x^{3} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {3 i e \pi \,d^{2} x^{2} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}+\frac {e^{3} a p \,x^{3}}{12 b}-\frac {e^{3} a^{2} p \,x^{2}}{8 b^{2}}+\frac {e^{3} a^{3} p x}{4 b^{3}}-\frac {e^{3} \ln \left (b x +a \right ) a^{4} p}{4 b^{4}}+\frac {3 i e \pi \,d^{2} x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{4}-\frac {d \,e^{2} p \,x^{3}}{3}-\frac {3 d^{2} e p \,x^{2}}{4}-\frac {e^{3} p \,x^{4}}{16}\)	\(766\)

[In]

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

[Out]

1/4*ln(c*(b*x+a)^p)*e^3*x^4+ln(c*(b*x+a)^p)*e^2*d*x^3+3/2*ln(c*(b*x+a)^p)*e*d^2*x^2+d^3*ln(c*(b*x+a)^p)*x+1/4*
ln(c*(b*x+a)^p)/e*d^4-1/4/e*p*b*(-e/b^4*(-1/4*b^3*e^3*x^4+1/3*((a*e-2*b*d)*e^2*b^2-2*b^3*d*e^2)*x^3+1/2*(2*(a*
e-2*b*d)*d*e*b^2-b*e*(a^2*e^2-2*a*b*d*e+2*b^2*d^2))*x^2+x*(a*e-2*b*d)*(a^2*e^2-2*a*b*d*e+2*b^2*d^2))+(a^4*e^4-
4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e+b^4*d^4)/b^5*ln(b*x+a))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (128) = 256\).

Time = 0.34 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.92 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=-\frac {3 \, b^{4} e^{3} p x^{4} + 4 \, {\left (4 \, b^{4} d e^{2} - a b^{3} e^{3}\right )} p x^{3} + 6 \, {\left (6 \, b^{4} d^{2} e - 4 \, a b^{3} d e^{2} + a^{2} b^{2} e^{3}\right )} p x^{2} + 12 \, {\left (4 \, b^{4} d^{3} - 6 \, a b^{3} d^{2} e + 4 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} p x - 12 \, {\left (b^{4} e^{3} p x^{4} + 4 \, b^{4} d e^{2} p x^{3} + 6 \, b^{4} d^{2} e p x^{2} + 4 \, b^{4} d^{3} p x + {\left (4 \, a b^{3} d^{3} - 6 \, a^{2} b^{2} d^{2} e + 4 \, a^{3} b d e^{2} - a^{4} e^{3}\right )} p\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} e^{3} x^{4} + 4 \, b^{4} d e^{2} x^{3} + 6 \, b^{4} d^{2} e x^{2} + 4 \, b^{4} d^{3} x\right )} \log \left (c\right )}{48 \, b^{4}} \]

[In]

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

[Out]

-1/48*(3*b^4*e^3*p*x^4 + 4*(4*b^4*d*e^2 - a*b^3*e^3)*p*x^3 + 6*(6*b^4*d^2*e - 4*a*b^3*d*e^2 + a^2*b^2*e^3)*p*x
^2 + 12*(4*b^4*d^3 - 6*a*b^3*d^2*e + 4*a^2*b^2*d*e^2 - a^3*b*e^3)*p*x - 12*(b^4*e^3*p*x^4 + 4*b^4*d*e^2*p*x^3
+ 6*b^4*d^2*e*p*x^2 + 4*b^4*d^3*p*x + (4*a*b^3*d^3 - 6*a^2*b^2*d^2*e + 4*a^3*b*d*e^2 - a^4*e^3)*p)*log(b*x + a
) - 12*(b^4*e^3*x^4 + 4*b^4*d*e^2*x^3 + 6*b^4*d^2*e*x^2 + 4*b^4*d^3*x)*log(c))/b^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (117) = 234\).

Time = 0.93 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.39 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=\begin {cases} - \frac {a^{4} e^{3} \log {\left (c \left (a + b x\right )^{p} \right )}}{4 b^{4}} + \frac {a^{3} d e^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{b^{3}} + \frac {a^{3} e^{3} p x}{4 b^{3}} - \frac {3 a^{2} d^{2} e \log {\left (c \left (a + b x\right )^{p} \right )}}{2 b^{2}} - \frac {a^{2} d e^{2} p x}{b^{2}} - \frac {a^{2} e^{3} p x^{2}}{8 b^{2}} + \frac {a d^{3} \log {\left (c \left (a + b x\right )^{p} \right )}}{b} + \frac {3 a d^{2} e p x}{2 b} + \frac {a d e^{2} p x^{2}}{2 b} + \frac {a e^{3} p x^{3}}{12 b} - d^{3} p x + d^{3} x \log {\left (c \left (a + b x\right )^{p} \right )} - \frac {3 d^{2} e p x^{2}}{4} + \frac {3 d^{2} e x^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{2} - \frac {d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\left (c \left (a + b x\right )^{p} \right )} - \frac {e^{3} p x^{4}}{16} + \frac {e^{3} x^{4} \log {\left (c \left (a + b x\right )^{p} \right )}}{4} & \text {for}\: b \neq 0 \\\left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.53 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=-\frac {1}{48} \, b p {\left (\frac {3 \, b^{3} e^{3} x^{4} + 4 \, {\left (4 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{3} + 6 \, {\left (6 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + 12 \, {\left (4 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 4 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} x}{b^{4}} - \frac {12 \, {\left (4 \, a b^{3} d^{3} - 6 \, a^{2} b^{2} d^{2} e + 4 \, a^{3} b d e^{2} - a^{4} e^{3}\right )} \log \left (b x + a\right )}{b^{5}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[In]

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

[Out]

-1/48*b*p*((3*b^3*e^3*x^4 + 4*(4*b^3*d*e^2 - a*b^2*e^3)*x^3 + 6*(6*b^3*d^2*e - 4*a*b^2*d*e^2 + a^2*b*e^3)*x^2
+ 12*(4*b^3*d^3 - 6*a*b^2*d^2*e + 4*a^2*b*d*e^2 - a^3*e^3)*x)/b^4 - 12*(4*a*b^3*d^3 - 6*a^2*b^2*d^2*e + 4*a^3*
b*d*e^2 - a^4*e^3)*log(b*x + a)/b^5) + 1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)*log((b*x + a)^p*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (128) = 256\).

Time = 0.31 (sec) , antiderivative size = 573, normalized size of antiderivative = 4.09 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=\frac {{\left (b x + a\right )} d^{3} p \log \left (b x + a\right )}{b} + \frac {3 \, {\left (b x + a\right )}^{2} d^{2} e p \log \left (b x + a\right )}{2 \, b^{2}} - \frac {3 \, {\left (b x + a\right )} a d^{2} e p \log \left (b x + a\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{3} d e^{2} p \log \left (b x + a\right )}{b^{3}} - \frac {3 \, {\left (b x + a\right )}^{2} a d e^{2} p \log \left (b x + a\right )}{b^{3}} + \frac {3 \, {\left (b x + a\right )} a^{2} d e^{2} p \log \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )}^{4} e^{3} p \log \left (b x + a\right )}{4 \, b^{4}} - \frac {{\left (b x + a\right )}^{3} a e^{3} p \log \left (b x + a\right )}{b^{4}} + \frac {3 \, {\left (b x + a\right )}^{2} a^{2} e^{3} p \log \left (b x + a\right )}{2 \, b^{4}} - \frac {{\left (b x + a\right )} a^{3} e^{3} p \log \left (b x + a\right )}{b^{4}} - \frac {{\left (b x + a\right )} d^{3} p}{b} - \frac {3 \, {\left (b x + a\right )}^{2} d^{2} e p}{4 \, b^{2}} + \frac {3 \, {\left (b x + a\right )} a d^{2} e p}{b^{2}} - \frac {{\left (b x + a\right )}^{3} d e^{2} p}{3 \, b^{3}} + \frac {3 \, {\left (b x + a\right )}^{2} a d e^{2} p}{2 \, b^{3}} - \frac {3 \, {\left (b x + a\right )} a^{2} d e^{2} p}{b^{3}} - \frac {{\left (b x + a\right )}^{4} e^{3} p}{16 \, b^{4}} + \frac {{\left (b x + a\right )}^{3} a e^{3} p}{3 \, b^{4}} - \frac {3 \, {\left (b x + a\right )}^{2} a^{2} e^{3} p}{4 \, b^{4}} + \frac {{\left (b x + a\right )} a^{3} e^{3} p}{b^{4}} + \frac {{\left (b x + a\right )} d^{3} \log \left (c\right )}{b} + \frac {3 \, {\left (b x + a\right )}^{2} d^{2} e \log \left (c\right )}{2 \, b^{2}} - \frac {3 \, {\left (b x + a\right )} a d^{2} e \log \left (c\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{3} d e^{2} \log \left (c\right )}{b^{3}} - \frac {3 \, {\left (b x + a\right )}^{2} a d e^{2} \log \left (c\right )}{b^{3}} + \frac {3 \, {\left (b x + a\right )} a^{2} d e^{2} \log \left (c\right )}{b^{3}} + \frac {{\left (b x + a\right )}^{4} e^{3} \log \left (c\right )}{4 \, b^{4}} - \frac {{\left (b x + a\right )}^{3} a e^{3} \log \left (c\right )}{b^{4}} + \frac {3 \, {\left (b x + a\right )}^{2} a^{2} e^{3} \log \left (c\right )}{2 \, b^{4}} - \frac {{\left (b x + a\right )} a^{3} e^{3} \log \left (c\right )}{b^{4}} \]

[In]

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

[Out]

(b*x + a)*d^3*p*log(b*x + a)/b + 3/2*(b*x + a)^2*d^2*e*p*log(b*x + a)/b^2 - 3*(b*x + a)*a*d^2*e*p*log(b*x + a)
/b^2 + (b*x + a)^3*d*e^2*p*log(b*x + a)/b^3 - 3*(b*x + a)^2*a*d*e^2*p*log(b*x + a)/b^3 + 3*(b*x + a)*a^2*d*e^2
*p*log(b*x + a)/b^3 + 1/4*(b*x + a)^4*e^3*p*log(b*x + a)/b^4 - (b*x + a)^3*a*e^3*p*log(b*x + a)/b^4 + 3/2*(b*x
+ a)^2*a^2*e^3*p*log(b*x + a)/b^4 - (b*x + a)*a^3*e^3*p*log(b*x + a)/b^4 - (b*x + a)*d^3*p/b - 3/4*(b*x + a)^
2*d^2*e*p/b^2 + 3*(b*x + a)*a*d^2*e*p/b^2 - 1/3*(b*x + a)^3*d*e^2*p/b^3 + 3/2*(b*x + a)^2*a*d*e^2*p/b^3 - 3*(b
*x + a)*a^2*d*e^2*p/b^3 - 1/16*(b*x + a)^4*e^3*p/b^4 + 1/3*(b*x + a)^3*a*e^3*p/b^4 - 3/4*(b*x + a)^2*a^2*e^3*p
/b^4 + (b*x + a)*a^3*e^3*p/b^4 + (b*x + a)*d^3*log(c)/b + 3/2*(b*x + a)^2*d^2*e*log(c)/b^2 - 3*(b*x + a)*a*d^2
*e*log(c)/b^2 + (b*x + a)^3*d*e^2*log(c)/b^3 - 3*(b*x + a)^2*a*d*e^2*log(c)/b^3 + 3*(b*x + a)*a^2*d*e^2*log(c)
/b^3 + 1/4*(b*x + a)^4*e^3*log(c)/b^4 - (b*x + a)^3*a*e^3*log(c)/b^4 + 3/2*(b*x + a)^2*a^2*e^3*log(c)/b^4 - (b
*x + a)*a^3*e^3*log(c)/b^4

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.49 \[ \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx=\ln \left (c\,{\left (a+b\,x\right )}^p\right )\,\left (d^3\,x+\frac {3\,d^2\,e\,x^2}{2}+d\,e^2\,x^3+\frac {e^3\,x^4}{4}\right )+x^2\,\left (\frac {a\,\left (d\,e^2\,p-\frac {a\,e^3\,p}{4\,b}\right )}{2\,b}-\frac {3\,d^2\,e\,p}{4}\right )-x\,\left (d^3\,p+\frac {a\,\left (\frac {a\,\left (d\,e^2\,p-\frac {a\,e^3\,p}{4\,b}\right )}{b}-\frac {3\,d^2\,e\,p}{2}\right )}{b}\right )-x^3\,\left (\frac {d\,e^2\,p}{3}-\frac {a\,e^3\,p}{12\,b}\right )-\frac {e^3\,p\,x^4}{16}-\frac {\ln \left (a+b\,x\right )\,\left (p\,a^4\,e^3-4\,p\,a^3\,b\,d\,e^2+6\,p\,a^2\,b^2\,d^2\,e-4\,p\,a\,b^3\,d^3\right )}{4\,b^4} \]

[In]

int(log(c*(a + b*x)^p)*(d + e*x)^3,x)

[Out]

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*((a*(d*e^2*p - (a*e^3*p)/(4*b)))/
(2*b) - (3*d^2*e*p)/4) - x*(d^3*p + (a*((a*(d*e^2*p - (a*e^3*p)/(4*b)))/b - (3*d^2*e*p)/2))/b) - x^3*((d*e^2*p
)/3 - (a*e^3*p)/(12*b)) - (e^3*p*x^4)/16 - (log(a + b*x)*(a^4*e^3*p - 4*a*b^3*d^3*p - 4*a^3*b*d*e^2*p + 6*a^2*
b^2*d^2*e*p))/(4*b^4)