3.188 \(\int \frac {(d+e x^2)^2 (a+b \log (c x^n))}{x^5} \, dx\)

Optimal. Leaf size=90 \[ -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {d e \left (a+b \log \left (c x^n\right )\right )}{x^2}+e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {b d^2 n}{16 x^4}-\frac {b d e n}{2 x^2}-\frac {1}{2} b e^2 n \log ^2(x) \]

[Out]

-1/16*b*d^2*n/x^4-1/2*b*d*e*n/x^2-1/2*b*e^2*n*ln(x)^2-1/4*d^2*(a+b*ln(c*x^n))/x^4-d*e*(a+b*ln(c*x^n))/x^2+e^2*
ln(x)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.09, antiderivative size = 73, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {266, 43, 2334, 14, 2301} \[ -\frac {1}{4} \left (\frac {d^2}{x^4}+\frac {4 d e}{x^2}-4 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b d^2 n}{16 x^4}-\frac {b d e n}{2 x^2}-\frac {1}{2} b e^2 n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^5,x]

[Out]

-(b*d^2*n)/(16*x^4) - (b*d*e*n)/(2*x^2) - (b*e^2*n*Log[x]^2)/2 - ((d^2/x^4 + (4*d*e)/x^2 - 4*e^2*Log[x])*(a +
b*Log[c*x^n]))/4

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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^5} \, dx &=-\frac {1}{4} \left (\frac {d^2}{x^4}+\frac {4 d e}{x^2}-4 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (-\frac {d \left (d+4 e x^2\right )}{4 x^5}+\frac {e^2 \log (x)}{x}\right ) \, dx\\ &=-\frac {1}{4} \left (\frac {d^2}{x^4}+\frac {4 d e}{x^2}-4 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} (b d n) \int \frac {d+4 e x^2}{x^5} \, dx-\left (b e^2 n\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {1}{2} b e^2 n \log ^2(x)-\frac {1}{4} \left (\frac {d^2}{x^4}+\frac {4 d e}{x^2}-4 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} (b d n) \int \left (\frac {d}{x^5}+\frac {4 e}{x^3}\right ) \, dx\\ &=-\frac {b d^2 n}{16 x^4}-\frac {b d e n}{2 x^2}-\frac {1}{2} b e^2 n \log ^2(x)-\frac {1}{4} \left (\frac {d^2}{x^4}+\frac {4 d e}{x^2}-4 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.91 \[ \frac {1}{16} \left (-\frac {4 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^4}-\frac {16 d e \left (a+b \log \left (c x^n\right )\right )}{x^2}+\frac {8 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac {b d^2 n}{x^4}-\frac {8 b d e n}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*Log[c*x^n]))/x^5,x]

[Out]

(-((b*d^2*n)/x^4) - (8*b*d*e*n)/x^2 - (4*d^2*(a + b*Log[c*x^n]))/x^4 - (16*d*e*(a + b*Log[c*x^n]))/x^2 + (8*e^
2*(a + b*Log[c*x^n])^2)/(b*n))/16

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fricas [A]  time = 0.59, size = 108, normalized size = 1.20 \[ \frac {8 \, b e^{2} n x^{4} \log \relax (x)^{2} - b d^{2} n - 4 \, a d^{2} - 8 \, {\left (b d e n + 2 \, a d e\right )} x^{2} - 4 \, {\left (4 \, b d e x^{2} + b d^{2}\right )} \log \relax (c) + 4 \, {\left (4 \, b e^{2} x^{4} \log \relax (c) + 4 \, a e^{2} x^{4} - 4 \, b d e n x^{2} - b d^{2} n\right )} \log \relax (x)}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^5,x, algorithm="fricas")

[Out]

1/16*(8*b*e^2*n*x^4*log(x)^2 - b*d^2*n - 4*a*d^2 - 8*(b*d*e*n + 2*a*d*e)*x^2 - 4*(4*b*d*e*x^2 + b*d^2)*log(c)
+ 4*(4*b*e^2*x^4*log(c) + 4*a*e^2*x^4 - 4*b*d*e*n*x^2 - b*d^2*n)*log(x))/x^4

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giac [A]  time = 0.31, size = 113, normalized size = 1.26 \[ \frac {8 \, b n x^{4} e^{2} \log \relax (x)^{2} + 16 \, b x^{4} e^{2} \log \relax (c) \log \relax (x) + 16 \, a x^{4} e^{2} \log \relax (x) - 16 \, b d n x^{2} e \log \relax (x) - 8 \, b d n x^{2} e - 16 \, b d x^{2} e \log \relax (c) - 16 \, a d x^{2} e - 4 \, b d^{2} n \log \relax (x) - b d^{2} n - 4 \, b d^{2} \log \relax (c) - 4 \, a d^{2}}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^5,x, algorithm="giac")

[Out]

1/16*(8*b*n*x^4*e^2*log(x)^2 + 16*b*x^4*e^2*log(c)*log(x) + 16*a*x^4*e^2*log(x) - 16*b*d*n*x^2*e*log(x) - 8*b*
d*n*x^2*e - 16*b*d*x^2*e*log(c) - 16*a*d*x^2*e - 4*b*d^2*n*log(x) - b*d^2*n - 4*b*d^2*log(c) - 4*a*d^2)/x^4

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maple [C]  time = 0.25, size = 434, normalized size = 4.82 \[ -\frac {\left (-4 e^{2} x^{4} \ln \relax (x )+4 d e \,x^{2}+d^{2}\right ) b \ln \left (x^{n}\right )}{4 x^{4}}-\frac {8 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )-8 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )-8 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )+8 i \pi b \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )+8 b \,e^{2} n \,x^{4} \ln \relax (x )^{2}-16 b \,e^{2} x^{4} \ln \relax (c ) \ln \relax (x )-16 a \,e^{2} x^{4} \ln \relax (x )-8 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+8 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+8 i \pi b d e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-8 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-2 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+2 i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+2 i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+8 b d e n \,x^{2}+16 b d e \,x^{2} \ln \relax (c )+16 a d e \,x^{2}+b \,d^{2} n +4 b \,d^{2} \ln \relax (c )+4 a \,d^{2}}{16 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(b*ln(c*x^n)+a)/x^5,x)

[Out]

-1/4*b*(-4*e^2*ln(x)*x^4+4*d*e*x^2+d^2)/x^4*ln(x^n)-1/16*(8*I*ln(x)*Pi*b*e^2*csgn(I*c*x^n)^3*x^4-8*I*ln(x)*Pi*
b*e^2*csgn(I*c*x^n)^2*csgn(I*c)*x^4+2*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)-2*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x
^n)*csgn(I*c)-8*I*Pi*b*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+8*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^2*csgn(I*c)+
2*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+8*I*Pi*b*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+8*b*e^2*n*ln(x)^2*x^4-16
*ln(x)*ln(c)*b*e^2*x^4+8*I*ln(x)*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^4-2*I*Pi*b*d^2*csgn(I*c*x^n)^3
-8*I*Pi*b*d*e*x^2*csgn(I*c*x^n)^3-8*I*ln(x)*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2*x^4-16*ln(x)*a*e^2*x^4+16*b*d
*e*x^2*ln(c)+8*b*d*e*n*x^2+16*a*d*e*x^2+4*b*d^2*ln(c)+b*d^2*n+4*a*d^2)/x^4

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maxima [A]  time = 0.52, size = 90, normalized size = 1.00 \[ \frac {b e^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + a e^{2} \log \relax (x) - \frac {b d e n}{2 \, x^{2}} - \frac {b d e \log \left (c x^{n}\right )}{x^{2}} - \frac {a d e}{x^{2}} - \frac {b d^{2} n}{16 \, x^{4}} - \frac {b d^{2} \log \left (c x^{n}\right )}{4 \, x^{4}} - \frac {a d^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*log(c*x^n))/x^5,x, algorithm="maxima")

[Out]

1/2*b*e^2*log(c*x^n)^2/n + a*e^2*log(x) - 1/2*b*d*e*n/x^2 - b*d*e*log(c*x^n)/x^2 - a*d*e/x^2 - 1/16*b*d^2*n/x^
4 - 1/4*b*d^2*log(c*x^n)/x^4 - 1/4*a*d^2/x^4

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mupad [B]  time = 3.57, size = 102, normalized size = 1.13 \[ \ln \relax (x)\,\left (a\,e^2+\frac {3\,b\,e^2\,n}{4}\right )-\frac {x^2\,\left (4\,a\,d\,e+2\,b\,d\,e\,n\right )+a\,d^2+\frac {b\,d^2\,n}{4}}{4\,x^4}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{4}+b\,d\,e\,x^2+\frac {3\,b\,e^2\,x^4}{4}\right )}{x^4}+\frac {b\,e^2\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)^2*(a + b*log(c*x^n)))/x^5,x)

[Out]

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

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sympy [A]  time = 6.17, size = 105, normalized size = 1.17 \[ - \frac {a d^{2}}{4 x^{4}} - \frac {a d e}{x^{2}} + a e^{2} \log {\relax (x )} + b d^{2} \left (- \frac {n}{16 x^{4}} - \frac {\log {\left (c x^{n} \right )}}{4 x^{4}}\right ) + 2 b d e \left (- \frac {n}{4 x^{2}} - \frac {\log {\left (c x^{n} \right )}}{2 x^{2}}\right ) - b e^{2} \left (\begin {cases} - \log {\relax (c )} \log {\relax (x )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{2}}{2 n} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*ln(c*x**n))/x**5,x)

[Out]

-a*d**2/(4*x**4) - a*d*e/x**2 + a*e**2*log(x) + b*d**2*(-n/(16*x**4) - log(c*x**n)/(4*x**4)) + 2*b*d*e*(-n/(4*
x**2) - log(c*x**n)/(2*x**2)) - b*e**2*Piecewise((-log(c)*log(x), Eq(n, 0)), (-log(c*x**n)**2/(2*n), True))

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