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

Optimal. Leaf size=91 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac {2 d e \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x} \]

[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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 0.79, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {270, 2334, 12, 14} \[ -\frac {1}{15} \left (\frac {3 d^2}{x^5}+\frac {10 d e}{x^3}+\frac {15 e^2}{x}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b d^2 n}{25 x^5}-\frac {2 b d e n}{9 x^3}-\frac {b e^2 n}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^2*n)/(25*x^5) - (2*b*d*e*n)/(9*x^3) - (b*e^2*n)/x - (((3*d^2)/x^5 + (10*d*e)/x^3 + (15*e^2)/x)*(a + b*Lo
g[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 270

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

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Mathematica [A]  time = 0.04, size = 86, normalized size = 0.95 \[ -\frac {15 a \left (3 d^2+10 d e x^2+15 e^2 x^4\right )+15 b \left (3 d^2+10 d e x^2+15 e^2 x^4\right ) \log \left (c x^n\right )+b n \left (9 d^2+50 d e x^2+225 e^2 x^4\right )}{225 x^5} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.90, size = 111, normalized size = 1.22 \[ -\frac {225 \, {\left (b e^{2} n + a e^{2}\right )} x^{4} + 9 \, b d^{2} n + 45 \, a d^{2} + 50 \, {\left (b d e n + 3 \, a d e\right )} x^{2} + 15 \, {\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \relax (c) + 15 \, {\left (15 \, b e^{2} n x^{4} + 10 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \relax (x)}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.26, size = 116, normalized size = 1.27 \[ -\frac {225 \, b n x^{4} e^{2} \log \relax (x) + 225 \, b n x^{4} e^{2} + 225 \, b x^{4} e^{2} \log \relax (c) + 150 \, b d n x^{2} e \log \relax (x) + 225 \, a x^{4} e^{2} + 50 \, b d n x^{2} e + 150 \, b d x^{2} e \log \relax (c) + 150 \, a d x^{2} e + 45 \, b d^{2} n \log \relax (x) + 9 \, b d^{2} n + 45 \, b d^{2} \log \relax (c) + 45 \, a d^{2}}{225 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.17, size = 419, normalized size = 4.60 \[ -\frac {\left (15 e^{2} x^{4}+10 d e \,x^{2}+3 d^{2}\right ) b \ln \left (x^{n}\right )}{15 x^{5}}-\frac {-225 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+225 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+225 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-225 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-150 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+150 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+150 i \pi b d e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-150 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+450 b \,e^{2} n \,x^{4}+450 b \,e^{2} x^{4} \ln \relax (c )+450 a \,e^{2} x^{4}-45 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+45 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+45 i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-45 i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+100 b d e n \,x^{2}+300 b d e \,x^{2} \ln \relax (c )+300 a d e \,x^{2}+18 b \,d^{2} n +90 b \,d^{2} \ln \relax (c )+90 a \,d^{2}}{450 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*b*(15*e^2*x^4+10*d*e*x^2+3*d^2)/x^5*ln(x^n)-1/450*(-150*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^3+150*I*Pi*b*d*e*x^
2*csgn(I*x^n)*csgn(I*c*x^n)^2+45*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-45*I*Pi*b*d^2*csgn(I*c*x^n)^3+450*b*e^
2*x^4*ln(c)+450*b*e^2*n*x^4+450*a*e^2*x^4-225*I*Pi*b*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-45*I*Pi*b*d^2
*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+225*I*Pi*b*e^2*x^4*csgn(I*c*x^n)^2*csgn(I*c)-150*I*Pi*b*d*e*x^2*csgn(I*x^
n)*csgn(I*c*x^n)*csgn(I*c)+300*b*d*e*x^2*ln(c)+100*b*d*e*n*x^2+300*a*d*e*x^2-225*I*Pi*b*e^2*x^4*csgn(I*c*x^n)^
3+150*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^2*csgn(I*c)+225*I*Pi*b*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+45*I*Pi*b*d^2*cs
gn(I*c*x^n)^2*csgn(I*c)+90*b*d^2*ln(c)+18*b*d^2*n+90*a*d^2)/x^5

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maxima [A]  time = 0.47, size = 100, normalized size = 1.10 \[ -\frac {b e^{2} n}{x} - \frac {b e^{2} \log \left (c x^{n}\right )}{x} - \frac {a e^{2}}{x} - \frac {2 \, b d e n}{9 \, x^{3}} - \frac {2 \, b d e \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {b d^{2} n}{25 \, x^{5}} - \frac {b d^{2} \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d^{2}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.51, size = 88, normalized size = 0.97 \[ -\frac {x^4\,\left (15\,a\,e^2+15\,b\,e^2\,n\right )+x^2\,\left (10\,a\,d\,e+\frac {10\,b\,d\,e\,n}{3}\right )+3\,a\,d^2+\frac {3\,b\,d^2\,n}{5}}{15\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{5}+\frac {2\,b\,d\,e\,x^2}{3}+b\,e^2\,x^4\right )}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 4.21, size = 146, normalized size = 1.60 \[ - \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} - \frac {b d^{2} n \log {\relax (x )}}{5 x^{5}} - \frac {b d^{2} n}{25 x^{5}} - \frac {b d^{2} \log {\relax (c )}}{5 x^{5}} - \frac {2 b d e n \log {\relax (x )}}{3 x^{3}} - \frac {2 b d e n}{9 x^{3}} - \frac {2 b d e \log {\relax (c )}}{3 x^{3}} - \frac {b e^{2} n \log {\relax (x )}}{x} - \frac {b e^{2} n}{x} - \frac {b e^{2} \log {\relax (c )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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