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

Optimal. Leaf size=140 \[ -\frac {p (b d-a e)^4 \log (a+b x)}{4 b^4 e}-\frac {p x (b d-a e)^3}{4 b^3}-\frac {p (d+e x)^2 (b d-a e)^2}{8 b^2 e}+\frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e}-\frac {p (d+e x)^3 (b d-a e)}{12 b e}-\frac {p (d+e x)^4}{16 e} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2395, 43} \[ -\frac {p x (b d-a e)^3}{4 b^3}-\frac {p (d+e x)^2 (b d-a e)^2}{8 b^2 e}-\frac {p (b d-a e)^4 \log (a+b x)}{4 b^4 e}+\frac {(d+e x)^4 \log \left (c (a+b x)^p\right )}{4 e}-\frac {p (d+e x)^3 (b d-a e)}{12 b e}-\frac {p (d+e x)^4}{16 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Log[c*(a + b*x)^p],x]

[Out]

-((b*d - a*e)^3*p*x)/(4*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 43

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 2395

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*} \int (d+e x)^3 \log \left (c (a+b x)^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.21, size = 185, normalized size = 1.32 \[ -\frac {12 a^2 e p \left (a^2 e^2-4 a b d e+6 b^2 d^2\right ) \log (a+b x)+b p x \left (-12 a^3 e^3+6 a^2 b e^2 (8 d+e x)-4 a b^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+b^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )-12 b^3 \left (4 a d^3+b x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right ) \log \left (c (a+b x)^p\right )}{48 b^4} \]

Antiderivative was successfully verified.

[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

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fricas [B]  time = 0.45, size = 269, normalized size = 1.92 \[ -\frac {3 \, b^{4} e^{3} p x^{4} + 4 \, {\left (4 \, b^{4} d e^{2} - a b^{3} e^{3}\right )} p x^{3} + 6 \, {\left (6 \, b^{4} d^{2} e - 4 \, a b^{3} d e^{2} + a^{2} b^{2} e^{3}\right )} p x^{2} + 12 \, {\left (4 \, b^{4} d^{3} - 6 \, a b^{3} d^{2} e + 4 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} p x - 12 \, {\left (b^{4} e^{3} p x^{4} + 4 \, b^{4} d e^{2} p x^{3} + 6 \, b^{4} d^{2} e p x^{2} + 4 \, b^{4} d^{3} p x + {\left (4 \, a b^{3} d^{3} - 6 \, a^{2} b^{2} d^{2} e + 4 \, a^{3} b d e^{2} - a^{4} e^{3}\right )} p\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} e^{3} x^{4} + 4 \, b^{4} d e^{2} x^{3} + 6 \, b^{4} d^{2} e x^{2} + 4 \, b^{4} d^{3} x\right )} \log \relax (c)}{48 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.20, size = 558, normalized size = 3.99 \[ \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} p e \log \left (b x + a\right )}{2 \, b^{2}} - \frac {3 \, {\left (b x + a\right )} a d^{2} p e \log \left (b x + a\right )}{b^{2}} - \frac {{\left (b x + a\right )} d^{3} p}{b} - \frac {3 \, {\left (b x + a\right )}^{2} d^{2} p e}{4 \, b^{2}} + \frac {3 \, {\left (b x + a\right )} a d^{2} p e}{b^{2}} + \frac {{\left (b x + a\right )}^{3} d p e^{2} \log \left (b x + a\right )}{b^{3}} - \frac {3 \, {\left (b x + a\right )}^{2} a d p e^{2} \log \left (b x + a\right )}{b^{3}} + \frac {3 \, {\left (b x + a\right )} a^{2} d p e^{2} \log \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )} d^{3} \log \relax (c)}{b} + \frac {3 \, {\left (b x + a\right )}^{2} d^{2} e \log \relax (c)}{2 \, b^{2}} - \frac {3 \, {\left (b x + a\right )} a d^{2} e \log \relax (c)}{b^{2}} - \frac {{\left (b x + a\right )}^{3} d p e^{2}}{3 \, b^{3}} + \frac {3 \, {\left (b x + a\right )}^{2} a d p e^{2}}{2 \, b^{3}} - \frac {3 \, {\left (b x + a\right )} a^{2} d p e^{2}}{b^{3}} + \frac {{\left (b x + a\right )}^{4} p e^{3} \log \left (b x + a\right )}{4 \, b^{4}} - \frac {{\left (b x + a\right )}^{3} a p e^{3} \log \left (b x + a\right )}{b^{4}} + \frac {3 \, {\left (b x + a\right )}^{2} a^{2} p e^{3} \log \left (b x + a\right )}{2 \, b^{4}} - \frac {{\left (b x + a\right )} a^{3} p e^{3} \log \left (b x + a\right )}{b^{4}} + \frac {{\left (b x + a\right )}^{3} d e^{2} \log \relax (c)}{b^{3}} - \frac {3 \, {\left (b x + a\right )}^{2} a d e^{2} \log \relax (c)}{b^{3}} + \frac {3 \, {\left (b x + a\right )} a^{2} d e^{2} \log \relax (c)}{b^{3}} - \frac {{\left (b x + a\right )}^{4} p e^{3}}{16 \, b^{4}} + \frac {{\left (b x + a\right )}^{3} a p e^{3}}{3 \, b^{4}} - \frac {3 \, {\left (b x + a\right )}^{2} a^{2} p e^{3}}{4 \, b^{4}} + \frac {{\left (b x + a\right )} a^{3} p e^{3}}{b^{4}} + \frac {{\left (b x + a\right )}^{4} e^{3} \log \relax (c)}{4 \, b^{4}} - \frac {{\left (b x + a\right )}^{3} a e^{3} \log \relax (c)}{b^{4}} + \frac {3 \, {\left (b x + a\right )}^{2} a^{2} e^{3} \log \relax (c)}{2 \, b^{4}} - \frac {{\left (b x + a\right )} a^{3} e^{3} \log \relax (c)}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [C]  time = 0.54, size = 766, normalized size = 5.47 \[ d \,e^{2} x^{3} \ln \relax (c )+\frac {3 d^{2} e \,x^{2} \ln \relax (c )}{2}-\frac {d^{4} p \ln \left (b x +a \right )}{4 e}+\frac {e^{3} x^{4} \ln \relax (c )}{4}+d^{3} x \ln \relax (c )+\frac {\left (e x +d \right )^{4} \ln \left (\left (b x +a \right )^{p}\right )}{4 e}-\frac {e^{3} p \,x^{4}}{16}-\frac {3 d^{2} e p \,x^{2}}{4}+\frac {a \,e^{3} p \,x^{3}}{12 b}-\frac {a^{2} e^{3} p \,x^{2}}{8 b^{2}}+\frac {a^{3} e^{3} p x}{4 b^{3}}-d^{3} p x -\frac {a^{4} e^{3} p \ln \left (b x +a \right )}{4 b^{4}}+\frac {a \,d^{3} p \ln \left (b x +a \right )}{b}-\frac {d \,e^{2} p \,x^{3}}{3}+\frac {a^{3} d \,e^{2} p \ln \left (b x +a \right )}{b^{3}}-\frac {3 a^{2} d^{2} e p \ln \left (b x +a \right )}{2 b^{2}}+\frac {i \pi \,e^{3} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{8}+\frac {i \pi \,e^{3} x^{4} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{8}-\frac {i \pi d \,e^{2} x^{3} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}-\frac {3 i \pi \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{4}+\frac {i \pi \,d^{3} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {i \pi \,d^{3} x \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi d \,e^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{2}-\frac {3 i \pi \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{4}+\frac {a d \,e^{2} p \,x^{2}}{2 b}-\frac {a^{2} d \,e^{2} p x}{b^{2}}+\frac {3 a \,d^{2} e p x}{2 b}-\frac {i \pi \,e^{3} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{8}-\frac {i \pi \,e^{3} x^{4} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{8}-\frac {i \pi \,d^{3} x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi d \,e^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {i \pi d \,e^{2} x^{3} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {3 i \pi \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4}+\frac {3 i \pi \,d^{2} e \,x^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4}-\frac {i \pi \,d^{3} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 214, normalized size = 1.53 \[ -\frac {1}{48} \, b p {\left (\frac {3 \, b^{3} e^{3} x^{4} + 4 \, {\left (4 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{3} + 6 \, {\left (6 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} + 12 \, {\left (4 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 4 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} x}{b^{4}} - \frac {12 \, {\left (4 \, a b^{3} d^{3} - 6 \, a^{2} b^{2} d^{2} e + 4 \, a^{3} b d e^{2} - a^{4} e^{3}\right )} \log \left (b x + a\right )}{b^{5}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.31, size = 208, normalized size = 1.49 \[ \ln \left (c\,{\left (a+b\,x\right )}^p\right )\,\left (d^3\,x+\frac {3\,d^2\,e\,x^2}{2}+d\,e^2\,x^3+\frac {e^3\,x^4}{4}\right )+x^2\,\left (\frac {a\,\left (d\,e^2\,p-\frac {a\,e^3\,p}{4\,b}\right )}{2\,b}-\frac {3\,d^2\,e\,p}{4}\right )-x\,\left (d^3\,p+\frac {a\,\left (\frac {a\,\left (d\,e^2\,p-\frac {a\,e^3\,p}{4\,b}\right )}{b}-\frac {3\,d^2\,e\,p}{2}\right )}{b}\right )-x^3\,\left (\frac {d\,e^2\,p}{3}-\frac {a\,e^3\,p}{12\,b}\right )-\frac {e^3\,p\,x^4}{16}-\frac {\ln \left (a+b\,x\right )\,\left (p\,a^4\,e^3-4\,p\,a^3\,b\,d\,e^2+6\,p\,a^2\,b^2\,d^2\,e-4\,p\,a\,b^3\,d^3\right )}{4\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 6.40, size = 369, normalized size = 2.64 \[ \begin {cases} - \frac {a^{4} e^{3} p \log {\left (a + b x \right )}}{4 b^{4}} + \frac {a^{3} d e^{2} p \log {\left (a + b x \right )}}{b^{3}} + \frac {a^{3} e^{3} p x}{4 b^{3}} - \frac {3 a^{2} d^{2} e p \log {\left (a + b x \right )}}{2 b^{2}} - \frac {a^{2} d e^{2} p x}{b^{2}} - \frac {a^{2} e^{3} p x^{2}}{8 b^{2}} + \frac {a d^{3} p \log {\left (a + b x \right )}}{b} + \frac {3 a d^{2} e p x}{2 b} + \frac {a d e^{2} p x^{2}}{2 b} + \frac {a e^{3} p x^{3}}{12 b} + d^{3} p x \log {\left (a + b x \right )} - d^{3} p x + d^{3} x \log {\relax (c )} + \frac {3 d^{2} e p x^{2} \log {\left (a + b x \right )}}{2} - \frac {3 d^{2} e p x^{2}}{4} + \frac {3 d^{2} e x^{2} \log {\relax (c )}}{2} + d e^{2} p x^{3} \log {\left (a + b x \right )} - \frac {d e^{2} p x^{3}}{3} + d e^{2} x^{3} \log {\relax (c )} + \frac {e^{3} p x^{4} \log {\left (a + b x \right )}}{4} - \frac {e^{3} p x^{4}}{16} + \frac {e^{3} x^{4} \log {\relax (c )}}{4} & \text {for}\: b \neq 0 \\\left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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