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

Optimal. Leaf size=57 \[ -\frac {d \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {b d n}{25 x^5}-\frac {b e n}{9 x^3} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 48, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2334, 12} \[ -\frac {1}{15} \left (\frac {3 d}{x^5}+\frac {5 e}{x^3}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b d n}{25 x^5}-\frac {b e n}{9 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*n)/(25*x^5) - (b*e*n)/(9*x^3) - (((3*d)/x^5 + (5*e)/x^3)*(a + b*Log[c*x^n]))/15

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 2334

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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 69, normalized size = 1.21 \[ -\frac {a d}{5 x^5}-\frac {a e}{3 x^3}-\frac {b d \log \left (c x^n\right )}{5 x^5}-\frac {b e \log \left (c x^n\right )}{3 x^3}-\frac {b d n}{25 x^5}-\frac {b e n}{9 x^3} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.45, size = 63, normalized size = 1.11 \[ -\frac {9 \, b d n + 25 \, {\left (b e n + 3 \, a e\right )} x^{2} + 45 \, a d + 15 \, {\left (5 \, b e x^{2} + 3 \, b d\right )} \log \relax (c) + 15 \, {\left (5 \, b e n x^{2} + 3 \, b d n\right )} \log \relax (x)}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.28, size = 66, normalized size = 1.16 \[ -\frac {75 \, b n x^{2} e \log \relax (x) + 25 \, b n x^{2} e + 75 \, b x^{2} e \log \relax (c) + 75 \, a x^{2} e + 45 \, b d n \log \relax (x) + 9 \, b d n + 45 \, b d \log \relax (c) + 45 \, a d}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 251, normalized size = 4.40 \[ -\frac {\left (5 e \,x^{2}+3 d \right ) b \ln \left (x^{n}\right )}{15 x^{5}}-\frac {-75 i \pi b e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+75 i \pi b e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+75 i \pi b e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-75 i \pi b e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-45 i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+45 i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+45 i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-45 i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+50 b e n \,x^{2}+150 b e \,x^{2} \ln \relax (c )+150 a e \,x^{2}+18 b d n +90 b d \ln \relax (c )+90 a d}{450 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*b*(5*e*x^2+3*d)/x^5*ln(x^n)-1/450*(75*I*Pi*b*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-75*I*Pi*b*e*x^2*csgn(I*x^
n)*csgn(I*c*x^n)*csgn(I*c)-75*I*Pi*b*e*x^2*csgn(I*c*x^n)^3+75*I*Pi*b*e*x^2*csgn(I*c*x^n)^2*csgn(I*c)+150*b*e*x
^2*ln(c)+50*b*e*n*x^2+150*a*e*x^2+45*I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n)^2-45*I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n
)*csgn(I*c)-45*I*Pi*b*d*csgn(I*c*x^n)^3+45*I*Pi*b*d*csgn(I*c*x^n)^2*csgn(I*c)+90*b*d*ln(c)+18*b*d*n+90*a*d)/x^
5

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maxima [A]  time = 0.48, size = 57, normalized size = 1.00 \[ -\frac {b e n}{9 \, x^{3}} - \frac {b e \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a e}{3 \, x^{3}} - \frac {b d n}{25 \, x^{5}} - \frac {b d \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.61, size = 53, normalized size = 0.93 \[ -\frac {\left (5\,a\,e+\frac {5\,b\,e\,n}{3}\right )\,x^2+3\,a\,d+\frac {3\,b\,d\,n}{5}}{15\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^2}{3}+\frac {b\,d}{5}\right )}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 3.91, size = 88, normalized size = 1.54 \[ - \frac {a d}{5 x^{5}} - \frac {a e}{3 x^{3}} - \frac {b d n \log {\relax (x )}}{5 x^{5}} - \frac {b d n}{25 x^{5}} - \frac {b d \log {\relax (c )}}{5 x^{5}} - \frac {b e n \log {\relax (x )}}{3 x^{3}} - \frac {b e n}{9 x^{3}} - \frac {b e \log {\relax (c )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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