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

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

[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

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Rubi [A]
time = 0.05, 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 = {2442, 45} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 = 0.14, size = 185, normalized size = 1.32 \begin {gather*} -\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} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.72, size = 766, normalized size = 5.47

method	result	size
risch	\(\frac {e^{3} a p \,x^{3}}{12 b}-\frac {e^{3} a^{2} p \,x^{2}}{8 b^{2}}-d^{3} p x +\frac {d^{3} p a \ln \left (b x +a \right )}{b}+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 {e^{2} a^{2} d p x}{b^{2}}+\frac {3 e a \,d^{2} p x}{2 b}+\frac {e^{2} \ln \left (b x +a \right ) a^{3} d p}{b^{3}}-\frac {3 i e \pi \,d^{2} x^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{4}+\frac {i e^{3} \pi \,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 e^{3} \pi \,x^{4} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{8}-\frac {i e^{2} \pi d \,x^{3} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}-\frac {3 i e \pi \,d^{2} 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 ) \mathrm {csgn}\left (i c \right )}{4}+\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 {e^{3} \ln \left (c \right ) x^{4}}{4}+\ln \left (c \right ) d^{3} x -\frac {i \pi \,d^{3} x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}-\frac {i e^{3} \pi \,x^{4} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{8}-\frac {e^{3} p \,x^{4}}{16}-\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 ) \mathrm {csgn}\left (i c \right )}{2}-\frac {i e^{3} \pi \,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 ) \mathrm {csgn}\left (i c \right )}{8}+\frac {i e^{2} \pi d \,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 {i e^{2} \pi d \,x^{3} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}+\frac {3 i e \pi \,d^{2} 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 {3 i e \pi \,d^{2} x^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{4}+\frac {\left (e x +d \right )^{4} \ln \left (\left (b x +a \right )^{p}\right )}{4 e}-\frac {i e^{2} \pi d \,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 ) \mathrm {csgn}\left (i c \right )}{2}+\frac {i \pi \,d^{3} x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{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 d^{2} e p \,x^{2}}{4}-\frac {d \,e^{2} p \,x^{3}}{3}+\frac {e^{2} a d p \,x^{2}}{2 b}-\frac {3 e \ln \left (b x +a \right ) a^{2} d^{2} p}{2 b^{2}}\)	\(766\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 207, normalized size = 1.48 \begin {gather*} -\frac {1}{48} \, b p {\left (\frac {3 \, b^{3} x^{4} e^{3} + 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 (x^{4} e^{3} + 4 \, d x^{3} e^{2} + 6 \, d^{2} x^{2} e + 4 \, d^{3} x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (131) = 262\).
time = 0.36, size = 266, normalized size = 1.90 \begin {gather*} -\frac {48 \, b^{4} d^{3} p x + {\left (3 \, b^{4} p x^{4} - 4 \, a b^{3} p x^{3} + 6 \, a^{2} b^{2} p x^{2} - 12 \, a^{3} b p x\right )} e^{3} + 8 \, {\left (2 \, b^{4} d p x^{3} - 3 \, a b^{3} d p x^{2} + 6 \, a^{2} b^{2} d p x\right )} e^{2} + 36 \, {\left (b^{4} d^{2} p x^{2} - 2 \, a b^{3} d^{2} p x\right )} e - 12 \, {\left (4 \, b^{4} d^{3} p x + 4 \, a b^{3} d^{3} p + {\left (b^{4} p x^{4} - a^{4} p\right )} e^{3} + 4 \, {\left (b^{4} d p x^{3} + a^{3} b d p\right )} e^{2} + 6 \, {\left (b^{4} d^{2} p x^{2} - a^{2} b^{2} d^{2} p\right )} e\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} x^{4} e^{3} + 4 \, b^{4} d x^{3} e^{2} + 6 \, b^{4} d^{2} x^{2} e + 4 \, b^{4} d^{3} x\right )} \log \left (c\right )}{48 \, b^{4}} \end {gather*}

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*(48*b^4*d^3*p*x + (3*b^4*p*x^4 - 4*a*b^3*p*x^3 + 6*a^2*b^2*p*x^2 - 12*a^3*b*p*x)*e^3 + 8*(2*b^4*d*p*x^3
- 3*a*b^3*d*p*x^2 + 6*a^2*b^2*d*p*x)*e^2 + 36*(b^4*d^2*p*x^2 - 2*a*b^3*d^2*p*x)*e - 12*(4*b^4*d^3*p*x + 4*a*b^
3*d^3*p + (b^4*p*x^4 - a^4*p)*e^3 + 4*(b^4*d*p*x^3 + a^3*b*d*p)*e^2 + 6*(b^4*d^2*p*x^2 - a^2*b^2*d^2*p)*e)*log
(b*x + a) - 12*(b^4*x^4*e^3 + 4*b^4*d*x^3*e^2 + 6*b^4*d^2*x^2*e + 4*b^4*d^3*x)*log(c))/b^4

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (117) = 234\).
time = 0.95, size = 335, normalized size = 2.39 \begin {gather*} \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} \end {gather*}

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*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))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (131) = 262\).
time = 3.19, size = 558, normalized size = 3.99 \begin {gather*} \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 \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 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 \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} 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 \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}} \end {gather*}

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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Mupad [B]
time = 0.31, size = 208, normalized size = 1.49 \begin {gather*} \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} \end {gather*}

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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