\(\int x^2 (d+e x^2)^2 (a+b \log (c x^n)) \, dx\) [190]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^2 n x^3-\frac {2}{25} b d e n x^5-\frac {1}{49} b e^2 n x^7+\frac {1}{105} \left (35 d^2 x^3+42 d e x^5+15 e^2 x^7\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/9*b*d^2*n*x^3-2/25*b*d*e*n*x^5-1/49*b*e^2*n*x^7+1/105*(15*e^2*x^7+42*d*e*x^5+35*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 = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {276, 2371} \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{105} \left (35 d^2 x^3+42 d e x^5+15 e^2 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b d^2 n x^3-\frac {2}{25} b d e n x^5-\frac {1}{49} b e^2 n x^7 \]

[In]

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

[Out]

-1/9*(b*d^2*n*x^3) - (2*b*d*e*n*x^5)/25 - (b*e^2*n*x^7)/49 + ((35*d^2*x^3 + 42*d*e*x^5 + 15*e^2*x^7)*(a + b*Lo
g[c*x^n]))/105

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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}{105} \left (35 d^2 x^3+42 d e x^5+15 e^2 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {d^2 x^2}{3}+\frac {2}{5} d e x^4+\frac {e^2 x^6}{7}\right ) \, dx \\ & = -\frac {1}{9} b d^2 n x^3-\frac {2}{25} b d e n x^5-\frac {1}{49} b e^2 n x^7+\frac {1}{105} \left (35 d^2 x^3+42 d e x^5+15 e^2 x^7\right ) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d^2 n x^3-\frac {2}{25} b d e n x^5-\frac {1}{49} b e^2 n x^7+\frac {1}{3} d^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {2}{5} d e x^5 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{7} e^2 x^7 \left (a+b \log \left (c x^n\right )\right ) \]

[In]

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

[Out]

-1/9*(b*d^2*n*x^3) - (2*b*d*e*n*x^5)/25 - (b*e^2*n*x^7)/49 + (d^2*x^3*(a + b*Log[c*x^n]))/3 + (2*d*e*x^5*(a +
b*Log[c*x^n]))/5 + (e^2*x^7*(a + b*Log[c*x^n]))/7

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\frac {x^{7} b \ln \left (c \,x^{n}\right ) e^{2}}{7}-\frac {b \,e^{2} n \,x^{7}}{49}+\frac {x^{7} a \,e^{2}}{7}+\frac {2 x^{5} \ln \left (c \,x^{n}\right ) b d e}{5}-\frac {2 b d e n \,x^{5}}{25}+\frac {2 a d e \,x^{5}}{5}+\frac {x^{3} b \ln \left (c \,x^{n}\right ) d^{2}}{3}-\frac {b \,d^{2} n \,x^{3}}{9}+\frac {a \,d^{2} x^{3}}{3}\)	\(101\)
risch	\(\frac {b \,x^{3} \left (15 e^{2} x^{4}+42 d e \,x^{2}+35 d^{2}\right ) \ln \left (x^{n}\right )}{105}+\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b \,e^{2} x^{7} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{14}+\frac {i \pi b d e \,x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{5}-\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}+\frac {\ln \left (c \right ) b \,e^{2} x^{7}}{7}-\frac {b \,e^{2} n \,x^{7}}{49}+\frac {x^{7} a \,e^{2}}{7}-\frac {i \pi b d e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{5}+\frac {i \pi b \,e^{2} x^{7} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{14}-\frac {i \pi b d e \,x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{5}-\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{6}+\frac {2 \ln \left (c \right ) b d e \,x^{5}}{5}-\frac {2 b d e n \,x^{5}}{25}+\frac {2 a d e \,x^{5}}{5}+\frac {i \pi b \,e^{2} x^{7} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{14}+\frac {i \pi b \,d^{2} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b d e \,x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{5}-\frac {i \pi b \,e^{2} x^{7} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{14}+\frac {\ln \left (c \right ) b \,d^{2} x^{3}}{3}-\frac {b \,d^{2} n \,x^{3}}{9}+\frac {a \,d^{2} x^{3}}{3}\)	\(434\)

[In]

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

[Out]

1/7*x^7*b*ln(c*x^n)*e^2-1/49*b*e^2*n*x^7+1/7*x^7*a*e^2+2/5*x^5*ln(c*x^n)*b*d*e-2/25*b*d*e*n*x^5+2/5*a*d*e*x^5+
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 = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{49} \, {\left (b e^{2} n - 7 \, a e^{2}\right )} x^{7} - \frac {2}{25} \, {\left (b d e n - 5 \, a d e\right )} x^{5} - \frac {1}{9} \, {\left (b d^{2} n - 3 \, a d^{2}\right )} x^{3} + \frac {1}{105} \, {\left (15 \, b e^{2} x^{7} + 42 \, b d e x^{5} + 35 \, b d^{2} x^{3}\right )} \log \left (c\right ) + \frac {1}{105} \, {\left (15 \, b e^{2} n x^{7} + 42 \, b d e n x^{5} + 35 \, b d^{2} n x^{3}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/49*(b*e^2*n - 7*a*e^2)*x^7 - 2/25*(b*d*e*n - 5*a*d*e)*x^5 - 1/9*(b*d^2*n - 3*a*d^2)*x^3 + 1/105*(15*b*e^2*x
^7 + 42*b*d*e*x^5 + 35*b*d^2*x^3)*log(c) + 1/105*(15*b*e^2*n*x^7 + 42*b*d*e*n*x^5 + 35*b*d^2*n*x^3)*log(x)

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{2} x^{3}}{3} + \frac {2 a d e x^{5}}{5} + \frac {a e^{2} x^{7}}{7} - \frac {b d^{2} n x^{3}}{9} + \frac {b d^{2} x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {2 b d e n x^{5}}{25} + \frac {2 b d e x^{5} \log {\left (c x^{n} \right )}}{5} - \frac {b e^{2} n x^{7}}{49} + \frac {b e^{2} x^{7} \log {\left (c x^{n} \right )}}{7} \]

[In]

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

[Out]

a*d**2*x**3/3 + 2*a*d*e*x**5/5 + a*e**2*x**7/7 - b*d**2*n*x**3/9 + b*d**2*x**3*log(c*x**n)/3 - 2*b*d*e*n*x**5/
25 + 2*b*d*e*x**5*log(c*x**n)/5 - b*e**2*n*x**7/49 + b*e**2*x**7*log(c*x**n)/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{49} \, b e^{2} n x^{7} + \frac {1}{7} \, b e^{2} x^{7} \log \left (c x^{n}\right ) + \frac {1}{7} \, a e^{2} x^{7} - \frac {2}{25} \, b d e n x^{5} + \frac {2}{5} \, b d e x^{5} \log \left (c x^{n}\right ) + \frac {2}{5} \, a d e x^{5} - \frac {1}{9} \, b d^{2} n x^{3} + \frac {1}{3} \, b d^{2} x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d^{2} x^{3} \]

[In]

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

[Out]

-1/49*b*e^2*n*x^7 + 1/7*b*e^2*x^7*log(c*x^n) + 1/7*a*e^2*x^7 - 2/25*b*d*e*n*x^5 + 2/5*b*d*e*x^5*log(c*x^n) + 2
/5*a*d*e*x^5 - 1/9*b*d^2*n*x^3 + 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 = 0.36 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.66 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{7} \, b e^{2} n x^{7} \log \left (x\right ) - \frac {1}{49} \, b e^{2} n x^{7} + \frac {1}{7} \, b e^{2} x^{7} \log \left (c\right ) + \frac {1}{7} \, a e^{2} x^{7} + \frac {2}{5} \, b d e n x^{5} \log \left (x\right ) - \frac {2}{25} \, b d e n x^{5} + \frac {2}{5} \, b d e x^{5} \log \left (c\right ) + \frac {2}{5} \, a d e x^{5} + \frac {1}{3} \, b d^{2} n x^{3} \log \left (x\right ) - \frac {1}{9} \, b d^{2} n x^{3} + \frac {1}{3} \, b d^{2} x^{3} \log \left (c\right ) + \frac {1}{3} \, a d^{2} x^{3} \]

[In]

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

[Out]

1/7*b*e^2*n*x^7*log(x) - 1/49*b*e^2*n*x^7 + 1/7*b*e^2*x^7*log(c) + 1/7*a*e^2*x^7 + 2/5*b*d*e*n*x^5*log(x) - 2/
25*b*d*e*n*x^5 + 2/5*b*d*e*x^5*log(c) + 2/5*a*d*e*x^5 + 1/3*b*d^2*n*x^3*log(x) - 1/9*b*d^2*n*x^3 + 1/3*b*d^2*x
^3*log(c) + 1/3*a*d^2*x^3

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11 \[ \int x^2 \left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2\,x^3}{3}+\frac {2\,b\,d\,e\,x^5}{5}+\frac {b\,e^2\,x^7}{7}\right )+\frac {d^2\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e^2\,x^7\,\left (7\,a-b\,n\right )}{49}+\frac {2\,d\,e\,x^5\,\left (5\,a-b\,n\right )}{25} \]

[In]

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

[Out]

log(c*x^n)*((b*d^2*x^3)/3 + (b*e^2*x^7)/7 + (2*b*d*e*x^5)/5) + (d^2*x^3*(3*a - b*n))/9 + (e^2*x^7*(7*a - b*n))
/49 + (2*d*e*x^5*(5*a - b*n))/25