3.1.21 \(\int x (d+e x)^3 (a+b \log (c x^n)) \, dx\) [21]

Optimal. Leaf size=122 \[ \frac {b d^4 n x}{5 e}+\frac {3}{20} b d^3 n x^2+\frac {1}{15} b d^2 e n x^3+\frac {1}{80} b d e^2 n x^4-\frac {b n (d+e x)^5}{25 e^2}+\frac {b d^5 n \log (x)}{20 e^2}-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

1/5*b*d^4*n*x/e+3/20*b*d^3*n*x^2+1/15*b*d^2*e*n*x^3+1/80*b*d*e^2*n*x^4-1/25*b*n*(e*x+d)^5/e^2+1/20*b*d^5*n*ln(
x)/e^2-1/20*(5*d*(e*x+d)^4/e^2-4*(e*x+d)^5/e^2)*(a+b*ln(c*x^n))

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Rubi [A]
time = 0.06, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {45, 2371, 12, 81} \begin {gather*} -\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {b d^5 n \log (x)}{20 e^2}+\frac {b d^4 n x}{5 e}+\frac {3}{20} b d^3 n x^2+\frac {1}{15} b d^2 e n x^3+\frac {1}{80} b d e^2 n x^4-\frac {b n (d+e x)^5}{25 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^4*n*x)/(5*e) + (3*b*d^3*n*x^2)/20 + (b*d^2*e*n*x^3)/15 + (b*d*e^2*n*x^4)/80 - (b*n*(d + e*x)^5)/(25*e^2)
+ (b*d^5*n*Log[x])/(20*e^2) - (((5*d*(d + e*x)^4)/e^2 - (4*(d + e*x)^5)/e^2)*(a + b*Log[c*x^n]))/20

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 2371

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int x (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {(d+e x)^4 (-d+4 e x)}{20 e^2 x} \, dx\\ &=-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \int \frac {(d+e x)^4 (-d+4 e x)}{x} \, dx}{20 e^2}\\ &=-\frac {b n (d+e x)^5}{25 e^2}-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b d n) \int \frac {(d+e x)^4}{x} \, dx}{20 e^2}\\ &=-\frac {b n (d+e x)^5}{25 e^2}-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b d n) \int \left (4 d^3 e+\frac {d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx}{20 e^2}\\ &=\frac {b d^4 n x}{5 e}+\frac {3}{20} b d^3 n x^2+\frac {1}{15} b d^2 e n x^3+\frac {1}{80} b d e^2 n x^4-\frac {b n (d+e x)^5}{25 e^2}+\frac {b d^5 n \log (x)}{20 e^2}-\frac {1}{20} \left (\frac {5 d (d+e x)^4}{e^2}-\frac {4 (d+e x)^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 130, normalized size = 1.07 \begin {gather*} -\frac {1}{4} b d^3 n x^2-\frac {1}{3} b d^2 e n x^3-\frac {3}{16} b d e^2 n x^4-\frac {1}{25} b e^3 n x^5+\frac {1}{2} d^3 x^2 \left (a+b \log \left (c x^n\right )\right )+d^2 e x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{4} d e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*d^3*n*x^2) - (b*d^2*e*n*x^3)/3 - (3*b*d*e^2*n*x^4)/16 - (b*e^3*n*x^5)/25 + (d^3*x^2*(a + b*Log[c*x^n])
)/2 + d^2*e*x^3*(a + b*Log[c*x^n]) + (3*d*e^2*x^4*(a + b*Log[c*x^n]))/4 + (e^3*x^5*(a + b*Log[c*x^n]))/5

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.12, size = 598, normalized size = 4.90

method	result	size
risch	\(\frac {x^{5} a \,e^{3}}{5}+\frac {3 x^{4} a d \,e^{2}}{4}+x^{3} a \,d^{2} e +\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{2} e \,x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \pi b \,e^{3} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{10}-\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4}+\frac {i \pi b \,d^{2} e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {\ln \left (c \right ) b \,e^{3} x^{5}}{5}+\frac {\ln \left (c \right ) b \,d^{3} x^{2}}{2}-\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}-\frac {i \pi b \,e^{3} x^{5} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {b \,x^{2} \left (4 e^{3} x^{3}+15 d \,e^{2} x^{2}+20 d^{2} e x +10 d^{3}\right ) \ln \left (x^{n}\right )}{20}+\frac {a \,d^{3} x^{2}}{2}-\frac {i \pi b \,d^{2} e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2}-\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}+\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\ln \left (c \right ) b \,d^{2} e \,x^{3}+\frac {3 \ln \left (c \right ) b d \,e^{2} x^{4}}{4}-\frac {b \,d^{3} n \,x^{2}}{4}-\frac {b \,d^{2} e n \,x^{3}}{3}-\frac {3 b d \,e^{2} n \,x^{4}}{16}-\frac {b \,e^{3} n \,x^{5}}{25}-\frac {i \pi b \,d^{2} e \,x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i \pi b \,e^{3} x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,e^{3} x^{5} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{10}\)	\(598\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*a*e^3+3/4*x^4*a*d*e^2+x^3*a*d^2*e+1/5*ln(c)*b*e^3*x^5+1/2*ln(c)*b*d^3*x^2-1/4*I*Pi*b*d^3*x^2*csgn(I*c*
x^n)^3+1/20*b*x^2*(4*e^3*x^3+15*d*e^2*x^2+20*d^2*e*x+10*d^3)*ln(x^n)+1/2*a*d^3*x^2+3/8*I*Pi*b*d*e^2*x^4*csgn(I
*x^n)*csgn(I*c*x^n)^2+3/8*I*Pi*b*d*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*Pi*b*d^2*e*x^3*csgn(I*x^n)*csgn(I*c
*x^n)^2-1/10*I*Pi*b*e^3*x^5*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/2*I*Pi*b*d^2*e*x^3*csgn(I*c)*csgn(I*x^n)*csg
n(I*c*x^n)-3/8*I*Pi*b*d*e^2*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/4*I*Pi*b*d^3*x^2*csgn(I*c)*csgn(I*x^n)*c
sgn(I*c*x^n)+1/2*I*Pi*b*d^2*e*x^3*csgn(I*c)*csgn(I*c*x^n)^2-1/10*I*Pi*b*e^3*x^5*csgn(I*c*x^n)^3+ln(c)*b*d^2*e*
x^3+3/4*ln(c)*b*d*e^2*x^4-1/2*I*Pi*b*d^2*e*x^3*csgn(I*c*x^n)^3+1/10*I*Pi*b*e^3*x^5*csgn(I*c)*csgn(I*c*x^n)^2-1
/4*b*d^3*n*x^2-1/3*b*d^2*e*n*x^3-3/16*b*d*e^2*n*x^4-1/25*b*e^3*n*x^5+1/4*I*Pi*b*d^3*x^2*csgn(I*c)*csgn(I*c*x^n
)^2+1/4*I*Pi*b*d^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-3/8*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^3+1/10*I*Pi*b*e^3*x^5*cs
gn(I*x^n)*csgn(I*c*x^n)^2

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Maxima [A]
time = 0.27, size = 138, normalized size = 1.13 \begin {gather*} -\frac {1}{25} \, b n x^{5} e^{3} - \frac {3}{16} \, b d n x^{4} e^{2} - \frac {1}{3} \, b d^{2} n x^{3} e + \frac {1}{5} \, b x^{5} e^{3} \log \left (c x^{n}\right ) + \frac {3}{4} \, b d x^{4} e^{2} \log \left (c x^{n}\right ) + b d^{2} x^{3} e \log \left (c x^{n}\right ) - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{5} \, a x^{5} e^{3} + \frac {3}{4} \, a d x^{4} e^{2} + a d^{2} x^{3} e + \frac {1}{2} \, b d^{3} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{3} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*b*n*x^5*e^3 - 3/16*b*d*n*x^4*e^2 - 1/3*b*d^2*n*x^3*e + 1/5*b*x^5*e^3*log(c*x^n) + 3/4*b*d*x^4*e^2*log(c*
x^n) + b*d^2*x^3*e*log(c*x^n) - 1/4*b*d^3*n*x^2 + 1/5*a*x^5*e^3 + 3/4*a*d*x^4*e^2 + a*d^2*x^3*e + 1/2*b*d^3*x^
2*log(c*x^n) + 1/2*a*d^3*x^2

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Fricas [A]
time = 0.35, size = 157, normalized size = 1.29 \begin {gather*} -\frac {1}{25} \, {\left (b n - 5 \, a\right )} x^{5} e^{3} - \frac {3}{16} \, {\left (b d n - 4 \, a d\right )} x^{4} e^{2} - \frac {1}{3} \, {\left (b d^{2} n - 3 \, a d^{2}\right )} x^{3} e - \frac {1}{4} \, {\left (b d^{3} n - 2 \, a d^{3}\right )} x^{2} + \frac {1}{20} \, {\left (4 \, b x^{5} e^{3} + 15 \, b d x^{4} e^{2} + 20 \, b d^{2} x^{3} e + 10 \, b d^{3} x^{2}\right )} \log \left (c\right ) + \frac {1}{20} \, {\left (4 \, b n x^{5} e^{3} + 15 \, b d n x^{4} e^{2} + 20 \, b d^{2} n x^{3} e + 10 \, b d^{3} n x^{2}\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*(b*n - 5*a)*x^5*e^3 - 3/16*(b*d*n - 4*a*d)*x^4*e^2 - 1/3*(b*d^2*n - 3*a*d^2)*x^3*e - 1/4*(b*d^3*n - 2*a*
d^3)*x^2 + 1/20*(4*b*x^5*e^3 + 15*b*d*x^4*e^2 + 20*b*d^2*x^3*e + 10*b*d^3*x^2)*log(c) + 1/20*(4*b*n*x^5*e^3 +
15*b*d*n*x^4*e^2 + 20*b*d^2*n*x^3*e + 10*b*d^3*n*x^2)*log(x)

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Sympy [A]
time = 0.42, size = 167, normalized size = 1.37 \begin {gather*} \frac {a d^{3} x^{2}}{2} + a d^{2} e x^{3} + \frac {3 a d e^{2} x^{4}}{4} + \frac {a e^{3} x^{5}}{5} - \frac {b d^{3} n x^{2}}{4} + \frac {b d^{3} x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {b d^{2} e n x^{3}}{3} + b d^{2} e x^{3} \log {\left (c x^{n} \right )} - \frac {3 b d e^{2} n x^{4}}{16} + \frac {3 b d e^{2} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b e^{3} n x^{5}}{25} + \frac {b e^{3} x^{5} \log {\left (c x^{n} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*x**2/2 + a*d**2*e*x**3 + 3*a*d*e**2*x**4/4 + a*e**3*x**5/5 - b*d**3*n*x**2/4 + b*d**3*x**2*log(c*x**n)/
2 - b*d**2*e*n*x**3/3 + b*d**2*e*x**3*log(c*x**n) - 3*b*d*e**2*n*x**4/16 + 3*b*d*e**2*x**4*log(c*x**n)/4 - b*e
**3*n*x**5/25 + b*e**3*x**5*log(c*x**n)/5

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Giac [A]
time = 2.22, size = 170, normalized size = 1.39 \begin {gather*} \frac {1}{5} \, b n x^{5} e^{3} \log \left (x\right ) + \frac {3}{4} \, b d n x^{4} e^{2} \log \left (x\right ) + b d^{2} n x^{3} e \log \left (x\right ) - \frac {1}{25} \, b n x^{5} e^{3} - \frac {3}{16} \, b d n x^{4} e^{2} - \frac {1}{3} \, b d^{2} n x^{3} e + \frac {1}{5} \, b x^{5} e^{3} \log \left (c\right ) + \frac {3}{4} \, b d x^{4} e^{2} \log \left (c\right ) + b d^{2} x^{3} e \log \left (c\right ) + \frac {1}{2} \, b d^{3} n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{5} \, a x^{5} e^{3} + \frac {3}{4} \, a d x^{4} e^{2} + a d^{2} x^{3} e + \frac {1}{2} \, b d^{3} x^{2} \log \left (c\right ) + \frac {1}{2} \, a d^{3} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*n*x^5*e^3*log(x) + 3/4*b*d*n*x^4*e^2*log(x) + b*d^2*n*x^3*e*log(x) - 1/25*b*n*x^5*e^3 - 3/16*b*d*n*x^4*e
^2 - 1/3*b*d^2*n*x^3*e + 1/5*b*x^5*e^3*log(c) + 3/4*b*d*x^4*e^2*log(c) + b*d^2*x^3*e*log(c) + 1/2*b*d^3*n*x^2*
log(x) - 1/4*b*d^3*n*x^2 + 1/5*a*x^5*e^3 + 3/4*a*d*x^4*e^2 + a*d^2*x^3*e + 1/2*b*d^3*x^2*log(c) + 1/2*a*d^3*x^
2

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Mupad [B]
time = 3.61, size = 112, normalized size = 0.92 \begin {gather*} \ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^2}{2}+b\,d^2\,e\,x^3+\frac {3\,b\,d\,e^2\,x^4}{4}+\frac {b\,e^3\,x^5}{5}\right )+\frac {d^3\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e^3\,x^5\,\left (5\,a-b\,n\right )}{25}+\frac {d^2\,e\,x^3\,\left (3\,a-b\,n\right )}{3}+\frac {3\,d\,e^2\,x^4\,\left (4\,a-b\,n\right )}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*log(c*x^n))*(d + e*x)^3,x)

[Out]

log(c*x^n)*((b*d^3*x^2)/2 + (b*e^3*x^5)/5 + b*d^2*e*x^3 + (3*b*d*e^2*x^4)/4) + (d^3*x^2*(2*a - b*n))/4 + (e^3*
x^5*(5*a - b*n))/25 + (d^2*e*x^3*(3*a - b*n))/3 + (3*d*e^2*x^4*(4*a - b*n))/16

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