3.9 \(\int x^3 (d+e x)^2 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=74 \[ \frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b d^2 n x^4-\frac {2}{25} b d e n x^5-\frac {1}{36} b e^2 n x^6 \]

[Out]

-1/16*b*d^2*n*x^4-2/25*b*d*e*n*x^5-1/36*b*e^2*n*x^6+1/60*(10*e^2*x^6+24*d*e*x^5+15*d^2*x^4)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {43, 2334, 12, 14} \[ \frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b d^2 n x^4-\frac {2}{25} b d e n x^5-\frac {1}{36} b e^2 n x^6 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + e*x)^2*(a + b*Log[c*x^n]),x]

[Out]

-(b*d^2*n*x^4)/16 - (2*b*d*e*n*x^5)/25 - (b*e^2*n*x^6)/36 + ((15*d^2*x^4 + 24*d*e*x^5 + 10*e^2*x^6)*(a + b*Log
[c*x^n]))/60

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2334

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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {align*} \int x^3 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{60} x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right ) \, dx\\ &=\frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} (b n) \int x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right ) \, dx\\ &=\frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{60} (b n) \int \left (15 d^2 x^3+24 d e x^4+10 e^2 x^5\right ) \, dx\\ &=-\frac {1}{16} b d^2 n x^4-\frac {2}{25} b d e n x^5-\frac {1}{36} b e^2 n x^6+\frac {1}{60} \left (15 d^2 x^4+24 d e x^5+10 e^2 x^6\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 81, normalized size = 1.09 \[ \frac {x^4 \left (60 a \left (15 d^2+24 d e x+10 e^2 x^2\right )+60 b \left (15 d^2+24 d e x+10 e^2 x^2\right ) \log \left (c x^n\right )-b n \left (225 d^2+288 d e x+100 e^2 x^2\right )\right )}{3600} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + e*x)^2*(a + b*Log[c*x^n]),x]

[Out]

(x^4*(60*a*(15*d^2 + 24*d*e*x + 10*e^2*x^2) - b*n*(225*d^2 + 288*d*e*x + 100*e^2*x^2) + 60*b*(15*d^2 + 24*d*e*
x + 10*e^2*x^2)*Log[c*x^n]))/3600

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fricas [A]  time = 0.44, size = 118, normalized size = 1.59 \[ -\frac {1}{36} \, {\left (b e^{2} n - 6 \, a e^{2}\right )} x^{6} - \frac {2}{25} \, {\left (b d e n - 5 \, a d e\right )} x^{5} - \frac {1}{16} \, {\left (b d^{2} n - 4 \, a d^{2}\right )} x^{4} + \frac {1}{60} \, {\left (10 \, b e^{2} x^{6} + 24 \, b d e x^{5} + 15 \, b d^{2} x^{4}\right )} \log \relax (c) + \frac {1}{60} \, {\left (10 \, b e^{2} n x^{6} + 24 \, b d e n x^{5} + 15 \, b d^{2} n x^{4}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

-1/36*(b*e^2*n - 6*a*e^2)*x^6 - 2/25*(b*d*e*n - 5*a*d*e)*x^5 - 1/16*(b*d^2*n - 4*a*d^2)*x^4 + 1/60*(10*b*e^2*x
^6 + 24*b*d*e*x^5 + 15*b*d^2*x^4)*log(c) + 1/60*(10*b*e^2*n*x^6 + 24*b*d*e*n*x^5 + 15*b*d^2*n*x^4)*log(x)

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giac [A]  time = 0.39, size = 123, normalized size = 1.66 \[ \frac {1}{6} \, b n x^{6} e^{2} \log \relax (x) + \frac {2}{5} \, b d n x^{5} e \log \relax (x) - \frac {1}{36} \, b n x^{6} e^{2} - \frac {2}{25} \, b d n x^{5} e + \frac {1}{6} \, b x^{6} e^{2} \log \relax (c) + \frac {2}{5} \, b d x^{5} e \log \relax (c) + \frac {1}{4} \, b d^{2} n x^{4} \log \relax (x) - \frac {1}{16} \, b d^{2} n x^{4} + \frac {1}{6} \, a x^{6} e^{2} + \frac {2}{5} \, a d x^{5} e + \frac {1}{4} \, b d^{2} x^{4} \log \relax (c) + \frac {1}{4} \, a d^{2} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

1/6*b*n*x^6*e^2*log(x) + 2/5*b*d*n*x^5*e*log(x) - 1/36*b*n*x^6*e^2 - 2/25*b*d*n*x^5*e + 1/6*b*x^6*e^2*log(c) +
2/5*b*d*x^5*e*log(c) + 1/4*b*d^2*n*x^4*log(x) - 1/16*b*d^2*n*x^4 + 1/6*a*x^6*e^2 + 2/5*a*d*x^5*e + 1/4*b*d^2*
x^4*log(c) + 1/4*a*d^2*x^4

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maple [C]  time = 0.22, size = 432, normalized size = 5.84 \[ -\frac {i \pi b \,e^{2} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{12}+\frac {i \pi b \,e^{2} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12}+\frac {i \pi b \,e^{2} x^{6} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12}-\frac {i \pi b \,e^{2} x^{6} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{12}-\frac {i \pi b d e \,x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{5}+\frac {i \pi b d e \,x^{5} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{5}+\frac {i \pi b d e \,x^{5} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{5}-\frac {i \pi b d e \,x^{5} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{5}-\frac {i \pi b \,d^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}+\frac {i \pi b \,d^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b \,d^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {b \,e^{2} n \,x^{6}}{36}+\frac {b \,e^{2} x^{6} \ln \relax (c )}{6}+\frac {a \,e^{2} x^{6}}{6}-\frac {2 b d e n \,x^{5}}{25}+\frac {2 b d e \,x^{5} \ln \relax (c )}{5}+\frac {2 a d e \,x^{5}}{5}-\frac {b \,d^{2} n \,x^{4}}{16}+\frac {b \,d^{2} x^{4} \ln \relax (c )}{4}+\frac {a \,d^{2} x^{4}}{4}+\frac {\left (10 e^{2} x^{2}+24 d e x +15 d^{2}\right ) b \,x^{4} \ln \left (x^{n}\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^2*(b*ln(c*x^n)+a),x)

[Out]

1/60*b*x^4*(10*e^2*x^2+24*d*e*x+15*d^2)*ln(x^n)-1/5*I*Pi*b*d*e*x^5*csgn(I*c*x^n)^3-1/12*I*Pi*b*e^2*x^6*csgn(I*
c*x^n)^3-1/12*I*Pi*b*e^2*x^6*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/8*I*Pi*b*d^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)*
csgn(I*c)+1/6*ln(c)*b*e^2*x^6-1/36*b*e^2*n*x^6+1/6*a*e^2*x^6-1/5*I*Pi*b*d*e*x^5*csgn(I*x^n)*csgn(I*c*x^n)*csgn
(I*c)+1/12*I*Pi*b*e^2*x^6*csgn(I*c*x^n)^2*csgn(I*c)+1/12*I*Pi*b*e^2*x^6*csgn(I*x^n)*csgn(I*c*x^n)^2+1/8*I*Pi*b
*d^2*x^4*csgn(I*c*x^n)^2*csgn(I*c)+2/5*ln(c)*b*d*e*x^5-2/25*b*d*e*n*x^5+2/5*a*d*e*x^5-1/8*I*Pi*b*d^2*x^4*csgn(
I*c*x^n)^3+1/8*I*Pi*b*d^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+1/5*I*Pi*b*d*e*x^5*csgn(I*x^n)*csgn(I*c*x^n)^2+1/5*I
*Pi*b*d*e*x^5*csgn(I*c*x^n)^2*csgn(I*c)+1/4*ln(c)*b*d^2*x^4-1/16*b*d^2*n*x^4+1/4*a*d^2*x^4

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maxima [A]  time = 0.67, size = 100, normalized size = 1.35 \[ -\frac {1}{36} \, b e^{2} n x^{6} + \frac {1}{6} \, b e^{2} x^{6} \log \left (c x^{n}\right ) - \frac {2}{25} \, b d e n x^{5} + \frac {1}{6} \, a e^{2} x^{6} + \frac {2}{5} \, b d e x^{5} \log \left (c x^{n}\right ) - \frac {1}{16} \, b d^{2} n x^{4} + \frac {2}{5} \, a d e x^{5} + \frac {1}{4} \, b d^{2} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{2} x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^2*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-1/36*b*e^2*n*x^6 + 1/6*b*e^2*x^6*log(c*x^n) - 2/25*b*d*e*n*x^5 + 1/6*a*e^2*x^6 + 2/5*b*d*e*x^5*log(c*x^n) - 1
/16*b*d^2*n*x^4 + 2/5*a*d*e*x^5 + 1/4*b*d^2*x^4*log(c*x^n) + 1/4*a*d^2*x^4

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mupad [B]  time = 3.69, size = 82, normalized size = 1.11 \[ \ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^4}{4}+\frac {2\,b\,d\,e\,x^5}{5}+\frac {b\,e^2\,x^6}{6}\right )+\frac {d^2\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^2\,x^6\,\left (6\,a-b\,n\right )}{36}+\frac {2\,d\,e\,x^5\,\left (5\,a-b\,n\right )}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*log(c*x^n))*(d + e*x)^2,x)

[Out]

log(c*x^n)*((b*d^2*x^4)/4 + (b*e^2*x^6)/6 + (2*b*d*e*x^5)/5) + (d^2*x^4*(4*a - b*n))/16 + (e^2*x^6*(6*a - b*n)
)/36 + (2*d*e*x^5*(5*a - b*n))/25

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sympy [B]  time = 3.92, size = 158, normalized size = 2.14 \[ \frac {a d^{2} x^{4}}{4} + \frac {2 a d e x^{5}}{5} + \frac {a e^{2} x^{6}}{6} + \frac {b d^{2} n x^{4} \log {\relax (x )}}{4} - \frac {b d^{2} n x^{4}}{16} + \frac {b d^{2} x^{4} \log {\relax (c )}}{4} + \frac {2 b d e n x^{5} \log {\relax (x )}}{5} - \frac {2 b d e n x^{5}}{25} + \frac {2 b d e x^{5} \log {\relax (c )}}{5} + \frac {b e^{2} n x^{6} \log {\relax (x )}}{6} - \frac {b e^{2} n x^{6}}{36} + \frac {b e^{2} x^{6} \log {\relax (c )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x+d)**2*(a+b*ln(c*x**n)),x)

[Out]

a*d**2*x**4/4 + 2*a*d*e*x**5/5 + a*e**2*x**6/6 + b*d**2*n*x**4*log(x)/4 - b*d**2*n*x**4/16 + b*d**2*x**4*log(c
)/4 + 2*b*d*e*n*x**5*log(x)/5 - 2*b*d*e*n*x**5/25 + 2*b*d*e*x**5*log(c)/5 + b*e**2*n*x**6*log(x)/6 - b*e**2*n*
x**6/36 + b*e**2*x**6*log(c)/6

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