\(\int \frac {(d+e x^2)^2 (a+b \log (c x^n))}{x^6} \, dx\) [194]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {276, 2372, 12, 14} \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x} \]

[In]

Int[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^6,x]

[Out]

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

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 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 2372

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {15 a \left (3 d^2+10 d e x^2+15 e^2 x^4\right )+b n \left (9 d^2+50 d e x^2+225 e^2 x^4\right )+15 b \left (3 d^2+10 d e x^2+15 e^2 x^4\right ) \log \left (c x^n\right )}{225 x^5} \]

[In]

Integrate[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-1/225*(15*a*(3*d^2 + 10*d*e*x^2 + 15*e^2*x^4) + b*n*(9*d^2 + 50*d*e*x^2 + 225*e^2*x^4) + 15*b*(3*d^2 + 10*d*e
*x^2 + 15*e^2*x^4)*Log[c*x^n])/x^5

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(-\frac {225 x^{4} b \ln \left (c \,x^{n}\right ) e^{2}+225 b \,e^{2} n \,x^{4}+225 x^{4} a \,e^{2}+150 b \ln \left (c \,x^{n}\right ) d e \,x^{2}+50 b d e n \,x^{2}+150 a d e \,x^{2}+45 b \ln \left (c \,x^{n}\right ) d^{2}+9 b \,d^{2} n +45 a \,d^{2}}{225 x^{5}}\)	\(97\)
risch	\(-\frac {b \left (15 e^{2} x^{4}+10 d e \,x^{2}+3 d^{2}\right ) \ln \left (x^{n}\right )}{15 x^{5}}-\frac {-150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+45 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+450 \ln \left (c \right ) b \,e^{2} x^{4}+450 b \,e^{2} n \,x^{4}+450 x^{4} a \,e^{2}-45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+300 e \ln \left (c \right ) b d \,x^{2}+100 b d e n \,x^{2}+300 a d e \,x^{2}+150 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+45 i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+225 i \pi b \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+90 d^{2} b \ln \left (c \right )+18 b \,d^{2} n +90 a \,d^{2}}{450 x^{5}}\)	\(419\)

[In]

int((e*x^2+d)^2*(a+b*ln(c*x^n))/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/225/x^5*(225*x^4*b*ln(c*x^n)*e^2+225*b*e^2*n*x^4+225*x^4*a*e^2+150*b*ln(c*x^n)*d*e*x^2+50*b*d*e*n*x^2+150*a
*d*e*x^2+45*b*ln(c*x^n)*d^2+9*b*d^2*n+45*a*d^2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {225 \, {\left (b e^{2} n + a e^{2}\right )} x^{4} + 9 \, b d^{2} n + 45 \, a d^{2} + 50 \, {\left (b d e n + 3 \, a d e\right )} x^{2} + 15 \, {\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (15 \, b e^{2} n x^{4} + 10 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )}{225 \, x^{5}} \]

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^6,x, algorithm="fricas")

[Out]

-1/225*(225*(b*e^2*n + a*e^2)*x^4 + 9*b*d^2*n + 45*a*d^2 + 50*(b*d*e*n + 3*a*d*e)*x^2 + 15*(15*b*e^2*x^4 + 10*
b*d*e*x^2 + 3*b*d^2)*log(c) + 15*(15*b*e^2*n*x^4 + 10*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))/x^5

Sympy [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=- \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} - \frac {b d^{2} n}{25 x^{5}} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{5 x^{5}} - \frac {2 b d e n}{9 x^{3}} - \frac {2 b d e \log {\left (c x^{n} \right )}}{3 x^{3}} - \frac {b e^{2} n}{x} - \frac {b e^{2} \log {\left (c x^{n} \right )}}{x} \]

[In]

integrate((e*x**2+d)**2*(a+b*ln(c*x**n))/x**6,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b e^{2} n}{x} - \frac {b e^{2} \log \left (c x^{n}\right )}{x} - \frac {a e^{2}}{x} - \frac {2 \, b d e n}{9 \, x^{3}} - \frac {2 \, b d e \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {b d^{2} n}{25 \, x^{5}} - \frac {b d^{2} \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d^{2}}{5 \, x^{5}} \]

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^6,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {{\left (15 \, b e^{2} n x^{4} + 10 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )}{15 \, x^{5}} - \frac {225 \, b e^{2} n x^{4} + 225 \, b e^{2} x^{4} \log \left (c\right ) + 225 \, a e^{2} x^{4} + 50 \, b d e n x^{2} + 150 \, b d e x^{2} \log \left (c\right ) + 150 \, a d e x^{2} + 9 \, b d^{2} n + 45 \, b d^{2} \log \left (c\right ) + 45 \, a d^{2}}{225 \, x^{5}} \]

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^6,x, algorithm="giac")

[Out]

-1/15*(15*b*e^2*n*x^4 + 10*b*d*e*n*x^2 + 3*b*d^2*n)*log(x)/x^5 - 1/225*(225*b*e^2*n*x^4 + 225*b*e^2*x^4*log(c)
+ 225*a*e^2*x^4 + 50*b*d*e*n*x^2 + 150*b*d*e*x^2*log(c) + 150*a*d*e*x^2 + 9*b*d^2*n + 45*b*d^2*log(c) + 45*a*
d^2)/x^5

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int \frac {\left (d+e x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {x^4\,\left (15\,a\,e^2+15\,b\,e^2\,n\right )+x^2\,\left (10\,a\,d\,e+\frac {10\,b\,d\,e\,n}{3}\right )+3\,a\,d^2+\frac {3\,b\,d^2\,n}{5}}{15\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{5}+\frac {2\,b\,d\,e\,x^2}{3}+b\,e^2\,x^4\right )}{x^5} \]

[In]

int(((d + e*x^2)^2*(a + b*log(c*x^n)))/x^6,x)

[Out]

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