\(\int x^2 (d+e x)^2 (a+b \log (c x^n)) \, dx\) [10]

Optimal result

Integrand size = 21, antiderivative size = 74 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^2 n x^3-\frac {1}{8} b d e n x^4-\frac {1}{25} b e^2 n x^5+\frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/9*b*d^2*n*x^3-1/8*b*d*e*n*x^4-1/25*b*e^2*n*x^5+1/30*(6*e^2*x^5+15*d*e*x^4+10*d^2*x^3)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {45, 2371, 12, 14} \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b d^2 n x^3-\frac {1}{8} b d e n x^4-\frac {1}{25} b e^2 n x^5 \]

[In]

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

[Out]

-1/9*(b*d^2*n*x^3) - (b*d*e*n*x^4)/8 - (b*e^2*n*x^5)/25 + ((10*d^2*x^3 + 15*d*e*x^4 + 6*e^2*x^5)*(a + b*Log[c*
x^n]))/30

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*} \text {integral}& = \frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{30} x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right ) \, dx \\ & = \frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{30} (b n) \int x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right ) \, dx \\ & = \frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{30} (b n) \int \left (10 d^2 x^2+15 d e x^3+6 e^2 x^4\right ) \, dx \\ & = -\frac {1}{9} b d^2 n x^3-\frac {1}{8} b d e n x^4-\frac {1}{25} b e^2 n x^5+\frac {1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^3 \left (60 a \left (10 d^2+15 d e x+6 e^2 x^2\right )-b n \left (200 d^2+225 d e x+72 e^2 x^2\right )+60 b \left (10 d^2+15 d e x+6 e^2 x^2\right ) \log \left (c x^n\right )\right )}{1800} \]

[In]

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

[Out]

(x^3*(60*a*(10*d^2 + 15*d*e*x + 6*e^2*x^2) - b*n*(200*d^2 + 225*d*e*x + 72*e^2*x^2) + 60*b*(10*d^2 + 15*d*e*x
+ 6*e^2*x^2)*Log[c*x^n]))/1800

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {x^{5} b \ln \left (c \,x^{n}\right ) e^{2}}{5}-\frac {b \,e^{2} n \,x^{5}}{25}+\frac {x^{5} a \,e^{2}}{5}+\frac {x^{4} b \ln \left (c \,x^{n}\right ) d e}{2}-\frac {b d e n \,x^{4}}{8}+\frac {x^{4} a d e}{2}+\frac {x^{3} b \ln \left (c \,x^{n}\right ) d^{2}}{3}-\frac {b \,d^{2} n \,x^{3}}{9}+\frac {a \,d^{2} x^{3}}{3}\)	\(101\)
risch	\(\frac {b \,x^{3} \left (6 e^{2} x^{2}+15 d e x +10 d^{2}\right ) \ln \left (x^{n}\right )}{30}+\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}+\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{6}+\frac {\ln \left (c \right ) b \,e^{2} x^{5}}{5}-\frac {b \,e^{2} n \,x^{5}}{25}+\frac {x^{5} a \,e^{2}}{5}+\frac {i \pi b d e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b d e \,x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{4}+\frac {\ln \left (c \right ) b d e \,x^{4}}{2}-\frac {b d e n \,x^{4}}{8}+\frac {x^{4} a d e}{2}-\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{10}+\frac {i \pi b d e \,x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {i \pi b \,e^{2} x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}-\frac {i \pi b d e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {\ln \left (c \right ) b \,d^{2} x^{3}}{3}-\frac {b \,d^{2} n \,x^{3}}{9}+\frac {a \,d^{2} x^{3}}{3}\)	\(432\)

[In]

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

[Out]

1/5*x^5*b*ln(c*x^n)*e^2-1/25*b*e^2*n*x^5+1/5*x^5*a*e^2+1/2*x^4*b*ln(c*x^n)*d*e-1/8*b*d*e*n*x^4+1/2*x^4*a*d*e+1
/3*x^3*b*ln(c*x^n)*d^2-1/9*b*d^2*n*x^3+1/3*a*d^2*x^3

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{5} - \frac {1}{8} \, {\left (b d e n - 4 \, a d e\right )} x^{4} - \frac {1}{9} \, {\left (b d^{2} n - 3 \, a d^{2}\right )} x^{3} + \frac {1}{30} \, {\left (6 \, b e^{2} x^{5} + 15 \, b d e x^{4} + 10 \, b d^{2} x^{3}\right )} \log \left (c\right ) + \frac {1}{30} \, {\left (6 \, b e^{2} n x^{5} + 15 \, b d e n x^{4} + 10 \, b d^{2} n x^{3}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/25*(b*e^2*n - 5*a*e^2)*x^5 - 1/8*(b*d*e*n - 4*a*d*e)*x^4 - 1/9*(b*d^2*n - 3*a*d^2)*x^3 + 1/30*(6*b*e^2*x^5
+ 15*b*d*e*x^4 + 10*b*d^2*x^3)*log(c) + 1/30*(6*b*e^2*n*x^5 + 15*b*d*e*n*x^4 + 10*b*d^2*n*x^3)*log(x)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{2} x^{3}}{3} + \frac {a d e x^{4}}{2} + \frac {a e^{2} x^{5}}{5} - \frac {b d^{2} n x^{3}}{9} + \frac {b d^{2} x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {b d e n x^{4}}{8} + \frac {b d e x^{4} \log {\left (c x^{n} \right )}}{2} - \frac {b e^{2} n x^{5}}{25} + \frac {b e^{2} x^{5} \log {\left (c x^{n} \right )}}{5} \]

[In]

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

[Out]

a*d**2*x**3/3 + a*d*e*x**4/2 + a*e**2*x**5/5 - b*d**2*n*x**3/9 + b*d**2*x**3*log(c*x**n)/3 - b*d*e*n*x**4/8 +
b*d*e*x**4*log(c*x**n)/2 - b*e**2*n*x**5/25 + b*e**2*x**5*log(c*x**n)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, b e^{2} n x^{5} + \frac {1}{5} \, b e^{2} x^{5} \log \left (c x^{n}\right ) - \frac {1}{8} \, b d e n x^{4} + \frac {1}{5} \, a e^{2} x^{5} + \frac {1}{2} \, b d e x^{4} \log \left (c x^{n}\right ) - \frac {1}{9} \, b d^{2} n x^{3} + \frac {1}{2} \, a d e x^{4} + \frac {1}{3} \, b d^{2} x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d^{2} x^{3} \]

[In]

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

[Out]

-1/25*b*e^2*n*x^5 + 1/5*b*e^2*x^5*log(c*x^n) - 1/8*b*d*e*n*x^4 + 1/5*a*e^2*x^5 + 1/2*b*d*e*x^4*log(c*x^n) - 1/
9*b*d^2*n*x^3 + 1/2*a*d*e*x^4 + 1/3*b*d^2*x^3*log(c*x^n) + 1/3*a*d^2*x^3

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.66 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, b e^{2} n x^{5} \log \left (x\right ) - \frac {1}{25} \, b e^{2} n x^{5} + \frac {1}{5} \, b e^{2} x^{5} \log \left (c\right ) + \frac {1}{2} \, b d e n x^{4} \log \left (x\right ) - \frac {1}{8} \, b d e n x^{4} + \frac {1}{5} \, a e^{2} x^{5} + \frac {1}{2} \, b d e x^{4} \log \left (c\right ) + \frac {1}{3} \, b d^{2} n x^{3} \log \left (x\right ) - \frac {1}{9} \, b d^{2} n x^{3} + \frac {1}{2} \, a d e x^{4} + \frac {1}{3} \, b d^{2} x^{3} \log \left (c\right ) + \frac {1}{3} \, a d^{2} x^{3} \]

[In]

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

[Out]

1/5*b*e^2*n*x^5*log(x) - 1/25*b*e^2*n*x^5 + 1/5*b*e^2*x^5*log(c) + 1/2*b*d*e*n*x^4*log(x) - 1/8*b*d*e*n*x^4 +
1/5*a*e^2*x^5 + 1/2*b*d*e*x^4*log(c) + 1/3*b*d^2*n*x^3*log(x) - 1/9*b*d^2*n*x^3 + 1/2*a*d*e*x^4 + 1/3*b*d^2*x^
3*log(c) + 1/3*a*d^2*x^3

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int x^2 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^3}{3}+\frac {b\,d\,e\,x^4}{2}+\frac {b\,e^2\,x^5}{5}\right )+\frac {d^2\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e^2\,x^5\,\left (5\,a-b\,n\right )}{25}+\frac {d\,e\,x^4\,\left (4\,a-b\,n\right )}{8} \]

[In]

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

[Out]

log(c*x^n)*((b*d^2*x^3)/3 + (b*e^2*x^5)/5 + (b*d*e*x^4)/2) + (d^2*x^3*(3*a - b*n))/9 + (e^2*x^5*(5*a - b*n))/2
5 + (d*e*x^4*(4*a - b*n))/8