\(\int \frac {(d+e x^2) (a+b \log (c x^n))}{x^6} \, dx\) [182]

Optimal result

Integrand size = 21, antiderivative size = 57 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b d n}{25 x^5}-\frac {b e n}{9 x^3}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{3 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2372, 12} \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {d \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {b d n}{25 x^5}-\frac {b e n}{9 x^3} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {a d}{5 x^5}-\frac {b d n}{25 x^5}-\frac {a e}{3 x^3}-\frac {b e n}{9 x^3}-\frac {b d \log \left (c x^n\right )}{5 x^5}-\frac {b e \log \left (c x^n\right )}{3 x^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(-\frac {75 b e \,x^{2} \ln \left (c \,x^{n}\right )+25 b e n \,x^{2}+75 a e \,x^{2}+45 b \ln \left (c \,x^{n}\right ) d +9 b d n +45 a d}{225 x^{5}}\)	\(54\)
risch	\(-\frac {b \left (5 e \,x^{2}+3 d \right ) \ln \left (x^{n}\right )}{15 x^{5}}-\frac {-75 i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) b e \,x^{2}+75 i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}+75 i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}-75 i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3} b e \,x^{2}+150 \ln \left (c \right ) b e \,x^{2}+50 b e n \,x^{2}+150 a e \,x^{2}-45 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+45 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+45 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-45 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+90 d b \ln \left (c \right )+18 b d n +90 a d}{450 x^{5}}\)	\(251\)

[In]

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

[Out]

-1/225/x^5*(75*b*e*x^2*ln(c*x^n)+25*b*e*n*x^2+75*a*e*x^2+45*b*ln(c*x^n)*d+9*b*d*n+45*a*d)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {9 \, b d n + 25 \, {\left (b e n + 3 \, a e\right )} x^{2} + 45 \, a d + 15 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \log \left (c\right ) + 15 \, {\left (5 \, b e n x^{2} + 3 \, b d n\right )} \log \left (x\right )}{225 \, x^{5}} \]

[In]

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

[Out]

-1/225*(9*b*d*n + 25*(b*e*n + 3*a*e)*x^2 + 45*a*d + 15*(5*b*e*x^2 + 3*b*d)*log(c) + 15*(5*b*e*n*x^2 + 3*b*d*n)
*log(x))/x^5

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=- \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} - \frac {b d n}{25 x^{5}} - \frac {b d \log {\left (c x^{n} \right )}}{5 x^{5}} - \frac {b e n}{9 x^{3}} - \frac {b e \log {\left (c x^{n} \right )}}{3 x^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b e n}{9 \, x^{3}} - \frac {b e \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a e}{3 \, x^{3}} - \frac {b d n}{25 \, x^{5}} - \frac {b d \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d}{5 \, x^{5}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {{\left (5 \, b e n x^{2} + 3 \, b d n\right )} \log \left (x\right )}{15 \, x^{5}} - \frac {25 \, b e n x^{2} + 75 \, b e x^{2} \log \left (c\right ) + 75 \, a e x^{2} + 9 \, b d n + 45 \, b d \log \left (c\right ) + 45 \, a d}{225 \, x^{5}} \]

[In]

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

[Out]

-1/15*(5*b*e*n*x^2 + 3*b*d*n)*log(x)/x^5 - 1/225*(25*b*e*n*x^2 + 75*b*e*x^2*log(c) + 75*a*e*x^2 + 9*b*d*n + 45
*b*d*log(c) + 45*a*d)/x^5

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {\left (5\,a\,e+\frac {5\,b\,e\,n}{3}\right )\,x^2+3\,a\,d+\frac {3\,b\,d\,n}{5}}{15\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^2}{3}+\frac {b\,d}{5}\right )}{x^5} \]

[In]

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

[Out]

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