3.1.11 \(\int x (d+e x)^2 (a+b \log (c x^n)) \, dx\) [11]

Optimal. Leaf size=74 \[ -\frac {1}{4} b d^2 n x^2-\frac {2}{9} b d e n x^3-\frac {1}{16} b e^2 n x^4+\frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/4*b*d^2*n*x^2-2/9*b*d*e*n*x^3-1/16*b*e^2*n*x^4+1/12*(3*e^2*x^4+8*d*e*x^3+6*d^2*x^2)*(a+b*ln(c*x^n))

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Rubi [A]
time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {45, 2371, 12, 14} \begin {gather*} \frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b d^2 n x^2-\frac {2}{9} b d e n x^3-\frac {1}{16} b e^2 n x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*d^2*n*x^2) - (2*b*d*e*n*x^3)/9 - (b*e^2*n*x^4)/16 + ((6*d^2*x^2 + 8*d*e*x^3 + 3*e^2*x^4)*(a + b*Log[c*
x^n]))/12

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 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 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)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{12} x \left (6 d^2+8 d e x+3 e^2 x^2\right ) \, dx\\ &=\frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} (b n) \int x \left (6 d^2+8 d e x+3 e^2 x^2\right ) \, dx\\ &=\frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} (b n) \int \left (6 d^2 x+8 d e x^2+3 e^2 x^3\right ) \, dx\\ &=-\frac {1}{4} b d^2 n x^2-\frac {2}{9} b d e n x^3-\frac {1}{16} b e^2 n x^4+\frac {1}{12} \left (6 d^2 x^2+8 d e x^3+3 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 81, normalized size = 1.09 \begin {gather*} \frac {1}{144} x^2 \left (12 a \left (6 d^2+8 d e x+3 e^2 x^2\right )-b n \left (36 d^2+32 d e x+9 e^2 x^2\right )+12 b \left (6 d^2+8 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(12*a*(6*d^2 + 8*d*e*x + 3*e^2*x^2) - b*n*(36*d^2 + 32*d*e*x + 9*e^2*x^2) + 12*b*(6*d^2 + 8*d*e*x + 3*e^2
*x^2)*Log[c*x^n]))/144

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.88, size = 432, normalized size = 5.84

method	result	size
risch	\(\frac {b \,x^{2} \left (3 e^{2} x^{2}+8 d e x +6 d^{2}\right ) \ln \left (x^{n}\right )}{12}-\frac {i \pi b \,d^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{3}+\frac {\ln \left (c \right ) b \,e^{2} x^{4}}{4}-\frac {b \,e^{2} n \,x^{4}}{16}+\frac {x^{4} a \,e^{2}}{4}-\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{3}-\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{3}+\frac {i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b d e \,x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{3}+\frac {2 \ln \left (c \right ) b d e \,x^{3}}{3}-\frac {2 b d e n \,x^{3}}{9}+\frac {2 x^{3} a d e}{3}-\frac {i \pi b \,d^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4}-\frac {i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}+\frac {i \pi b \,d^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {\ln \left (c \right ) b \,d^{2} x^{2}}{2}-\frac {b \,d^{2} n \,x^{2}}{4}+\frac {x^{2} a \,d^{2}}{2}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b*x^2*(3*e^2*x^2+8*d*e*x+6*d^2)*ln(x^n)-1/4*I*Pi*b*d^2*x^2*csgn(I*c*x^n)^3+1/8*I*Pi*b*e^2*x^4*csgn(I*x^n)
*csgn(I*c*x^n)^2+1/4*I*Pi*b*d^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*I*Pi*b*d*e*x^3*csgn(I*c)*csgn(I*c*x^n)^2+1
/4*ln(c)*b*e^2*x^4-1/16*b*e^2*n*x^4+1/4*x^4*a*e^2-1/3*I*Pi*b*d*e*x^3*csgn(I*c*x^n)^3-1/3*I*Pi*b*d*e*x^3*csgn(I
*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/8*I*Pi*b*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^2+1/3*I*Pi*b*d*e*x^3*csgn(I*x^n)*csgn
(I*c*x^n)^2+2/3*ln(c)*b*d*e*x^3-2/9*b*d*e*n*x^3+2/3*x^3*a*d*e-1/4*I*Pi*b*d^2*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*
c*x^n)-1/8*I*Pi*b*e^2*x^4*csgn(I*c*x^n)^3-1/8*I*Pi*b*e^2*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/4*I*Pi*b*d^
2*x^2*csgn(I*c)*csgn(I*c*x^n)^2+1/2*ln(c)*b*d^2*x^2-1/4*b*d^2*n*x^2+1/2*x^2*a*d^2

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Maxima [A]
time = 0.26, size = 100, normalized size = 1.35 \begin {gather*} -\frac {1}{16} \, b n x^{4} e^{2} - \frac {2}{9} \, b d n x^{3} e + \frac {1}{4} \, b x^{4} e^{2} \log \left (c x^{n}\right ) + \frac {2}{3} \, b d x^{3} e \log \left (c x^{n}\right ) - \frac {1}{4} \, b d^{2} n x^{2} + \frac {1}{4} \, a x^{4} e^{2} + \frac {2}{3} \, a d x^{3} e + \frac {1}{2} \, b d^{2} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*b*n*x^4*e^2 - 2/9*b*d*n*x^3*e + 1/4*b*x^4*e^2*log(c*x^n) + 2/3*b*d*x^3*e*log(c*x^n) - 1/4*b*d^2*n*x^2 +
1/4*a*x^4*e^2 + 2/3*a*d*x^3*e + 1/2*b*d^2*x^2*log(c*x^n) + 1/2*a*d^2*x^2

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Fricas [A]
time = 0.37, size = 114, normalized size = 1.54 \begin {gather*} -\frac {1}{16} \, {\left (b n - 4 \, a\right )} x^{4} e^{2} - \frac {2}{9} \, {\left (b d n - 3 \, a d\right )} x^{3} e - \frac {1}{4} \, {\left (b d^{2} n - 2 \, a d^{2}\right )} x^{2} + \frac {1}{12} \, {\left (3 \, b x^{4} e^{2} + 8 \, b d x^{3} e + 6 \, b d^{2} x^{2}\right )} \log \left (c\right ) + \frac {1}{12} \, {\left (3 \, b n x^{4} e^{2} + 8 \, b d n x^{3} e + 6 \, b d^{2} 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)^2*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

-1/16*(b*n - 4*a)*x^4*e^2 - 2/9*(b*d*n - 3*a*d)*x^3*e - 1/4*(b*d^2*n - 2*a*d^2)*x^2 + 1/12*(3*b*x^4*e^2 + 8*b*
d*x^3*e + 6*b*d^2*x^2)*log(c) + 1/12*(3*b*n*x^4*e^2 + 8*b*d*n*x^3*e + 6*b*d^2*n*x^2)*log(x)

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Sympy [A]
time = 0.28, size = 121, normalized size = 1.64 \begin {gather*} \frac {a d^{2} x^{2}}{2} + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{4}}{4} - \frac {b d^{2} n x^{2}}{4} + \frac {b d^{2} x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {2 b d e n x^{3}}{9} + \frac {2 b d e x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {b e^{2} n x^{4}}{16} + \frac {b e^{2} x^{4} \log {\left (c x^{n} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*x**2/2 + 2*a*d*e*x**3/3 + a*e**2*x**4/4 - b*d**2*n*x**2/4 + b*d**2*x**2*log(c*x**n)/2 - 2*b*d*e*n*x**3/
9 + 2*b*d*e*x**3*log(c*x**n)/3 - b*e**2*n*x**4/16 + b*e**2*x**4*log(c*x**n)/4

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Giac [A]
time = 2.46, size = 123, normalized size = 1.66 \begin {gather*} \frac {1}{4} \, b n x^{4} e^{2} \log \left (x\right ) + \frac {2}{3} \, b d n x^{3} e \log \left (x\right ) - \frac {1}{16} \, b n x^{4} e^{2} - \frac {2}{9} \, b d n x^{3} e + \frac {1}{4} \, b x^{4} e^{2} \log \left (c\right ) + \frac {2}{3} \, b d x^{3} e \log \left (c\right ) + \frac {1}{2} \, b d^{2} n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d^{2} n x^{2} + \frac {1}{4} \, a x^{4} e^{2} + \frac {2}{3} \, a d x^{3} e + \frac {1}{2} \, b d^{2} x^{2} \log \left (c\right ) + \frac {1}{2} \, a d^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*n*x^4*e^2*log(x) + 2/3*b*d*n*x^3*e*log(x) - 1/16*b*n*x^4*e^2 - 2/9*b*d*n*x^3*e + 1/4*b*x^4*e^2*log(c) +
2/3*b*d*x^3*e*log(c) + 1/2*b*d^2*n*x^2*log(x) - 1/4*b*d^2*n*x^2 + 1/4*a*x^4*e^2 + 2/3*a*d*x^3*e + 1/2*b*d^2*x^
2*log(c) + 1/2*a*d^2*x^2

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Mupad [B]
time = 3.63, size = 82, normalized size = 1.11 \begin {gather*} \ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^2}{2}+\frac {2\,b\,d\,e\,x^3}{3}+\frac {b\,e^2\,x^4}{4}\right )+\frac {d^2\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e^2\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {2\,d\,e\,x^3\,\left (3\,a-b\,n\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^n)*((b*d^2*x^2)/2 + (b*e^2*x^4)/4 + (2*b*d*e*x^3)/3) + (d^2*x^2*(2*a - b*n))/4 + (e^2*x^4*(4*a - b*n))
/16 + (2*d*e*x^3*(3*a - b*n))/9

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