3.201 \(\int \frac {(d+e x^2)^3 (a+b \log (c x^n))}{x^5} \, dx\)

Optimal. Leaf size=131 \[ -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{4 x^4}-\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{16 x^4}-\frac {3 b d^2 e n}{4 x^2}-\frac {3}{2} b d e^2 n \log ^2(x)-\frac {1}{4} b e^3 n x^2 \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 99, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {266, 43, 2334, 12, 14, 2301} \[ -\frac {1}{4} \left (\frac {6 d^2 e}{x^2}+\frac {d^3}{x^4}-12 d e^2 \log (x)-2 e^3 x^2\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {3 b d^2 e n}{4 x^2}-\frac {b d^3 n}{16 x^4}-\frac {3}{2} b d e^2 n \log ^2(x)-\frac {1}{4} b e^3 n x^2 \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*Log[c*x^n]))/x^5,x]

[Out]

-(b*d^3*n)/(16*x^4) - (3*b*d^2*e*n)/(4*x^2) - (b*e^3*n*x^2)/4 - (3*b*d*e^2*n*Log[x]^2)/2 - ((d^3/x^4 + (6*d^2*
e)/x^2 - 2*e^3*x^2 - 12*d*e^2*Log[x])*(a + b*Log[c*x^n]))/4

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 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 )^3 \left (a+b \log \left (c x^n\right )\right )}{x^5} \, dx &=-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {-d^3-6 d^2 e x^2+2 e^3 x^6+12 d e^2 x^4 \log (x)}{4 x^5} \, dx\\ &=-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \frac {-d^3-6 d^2 e x^2+2 e^3 x^6+12 d e^2 x^4 \log (x)}{x^5} \, dx\\ &=-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \left (\frac {-d^3-6 d^2 e x^2+2 e^3 x^6}{x^5}+\frac {12 d e^2 \log (x)}{x}\right ) \, dx\\ &=-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \frac {-d^3-6 d^2 e x^2+2 e^3 x^6}{x^5} \, dx-\left (3 b d e^2 n\right ) \int \frac {\log (x)}{x} \, dx\\ &=-\frac {3}{2} b d e^2 n \log ^2(x)-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \left (-\frac {d^3}{x^5}-\frac {6 d^2 e}{x^3}+2 e^3 x\right ) \, dx\\ &=-\frac {b d^3 n}{16 x^4}-\frac {3 b d^2 e n}{4 x^2}-\frac {1}{4} b e^3 n x^2-\frac {3}{2} b d e^2 n \log ^2(x)-\frac {1}{4} \left (\frac {d^3}{x^4}+\frac {6 d^2 e}{x^2}-2 e^3 x^2-12 d e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 115, normalized size = 0.88 \[ \frac {1}{16} \left (-\frac {4 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^4}-\frac {24 d^2 e \left (a+b \log \left (c x^n\right )\right )}{x^2}+\frac {24 d e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+8 e^3 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{x^4}-\frac {12 b d^2 e n}{x^2}-4 b e^3 n x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*Log[c*x^n]))/x^5,x]

[Out]

(-((b*d^3*n)/x^4) - (12*b*d^2*e*n)/x^2 - 4*b*e^3*n*x^2 - (4*d^3*(a + b*Log[c*x^n]))/x^4 - (24*d^2*e*(a + b*Log
[c*x^n]))/x^2 + 8*e^3*x^2*(a + b*Log[c*x^n]) + (24*d*e^2*(a + b*Log[c*x^n])^2)/(b*n))/16

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fricas [A]  time = 0.82, size = 157, normalized size = 1.20 \[ \frac {24 \, b d e^{2} n x^{4} \log \relax (x)^{2} - 4 \, {\left (b e^{3} n - 2 \, a e^{3}\right )} x^{6} - b d^{3} n - 4 \, a d^{3} - 12 \, {\left (b d^{2} e n + 2 \, a d^{2} e\right )} x^{2} + 4 \, {\left (2 \, b e^{3} x^{6} - 6 \, b d^{2} e x^{2} - b d^{3}\right )} \log \relax (c) + 4 \, {\left (2 \, b e^{3} n x^{6} + 12 \, b d e^{2} x^{4} \log \relax (c) + 12 \, a d e^{2} x^{4} - 6 \, b d^{2} e n x^{2} - b d^{3} n\right )} \log \relax (x)}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="fricas")

[Out]

1/16*(24*b*d*e^2*n*x^4*log(x)^2 - 4*(b*e^3*n - 2*a*e^3)*x^6 - b*d^3*n - 4*a*d^3 - 12*(b*d^2*e*n + 2*a*d^2*e)*x
^2 + 4*(2*b*e^3*x^6 - 6*b*d^2*e*x^2 - b*d^3)*log(c) + 4*(2*b*e^3*n*x^6 + 12*b*d*e^2*x^4*log(c) + 12*a*d*e^2*x^
4 - 6*b*d^2*e*n*x^2 - b*d^3*n)*log(x))/x^4

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giac [A]  time = 0.29, size = 162, normalized size = 1.24 \[ \frac {8 \, b n x^{6} e^{3} \log \relax (x) + 24 \, b d n x^{4} e^{2} \log \relax (x)^{2} - 4 \, b n x^{6} e^{3} + 8 \, b x^{6} e^{3} \log \relax (c) + 48 \, b d x^{4} e^{2} \log \relax (c) \log \relax (x) + 8 \, a x^{6} e^{3} + 48 \, a d x^{4} e^{2} \log \relax (x) - 24 \, b d^{2} n x^{2} e \log \relax (x) - 12 \, b d^{2} n x^{2} e - 24 \, b d^{2} x^{2} e \log \relax (c) - 24 \, a d^{2} x^{2} e - 4 \, b d^{3} n \log \relax (x) - b d^{3} n - 4 \, b d^{3} \log \relax (c) - 4 \, a d^{3}}{16 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="giac")

[Out]

1/16*(8*b*n*x^6*e^3*log(x) + 24*b*d*n*x^4*e^2*log(x)^2 - 4*b*n*x^6*e^3 + 8*b*x^6*e^3*log(c) + 48*b*d*x^4*e^2*l
og(c)*log(x) + 8*a*x^6*e^3 + 48*a*d*x^4*e^2*log(x) - 24*b*d^2*n*x^2*e*log(x) - 12*b*d^2*n*x^2*e - 24*b*d^2*x^2
*e*log(c) - 24*a*d^2*x^2*e - 4*b*d^3*n*log(x) - b*d^3*n - 4*b*d^3*log(c) - 4*a*d^3)/x^4

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maple [C]  time = 0.33, size = 602, normalized size = 4.60 \[ -\frac {\left (-2 e^{3} x^{6}-12 d \,e^{2} x^{4} \ln \relax (x )+6 d^{2} e \,x^{2}+d^{3}\right ) b \ln \left (x^{n}\right )}{4 x^{4}}-\frac {-8 a \,e^{3} x^{6}+4 a \,d^{3}-8 b \,e^{3} x^{6} \ln \relax (c )+b \,d^{3} n +4 b \,d^{3} \ln \relax (c )+24 a \,d^{2} e \,x^{2}+24 i \pi b d \,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 )-2 i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+24 b \,d^{2} e \,x^{2} \ln \relax (c )+4 b \,e^{3} n \,x^{6}-48 a d \,e^{2} x^{4} \ln \relax (x )-4 i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-4 i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+12 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+12 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+4 i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )-24 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )-12 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+24 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )+2 i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+2 i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+4 i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-48 b d \,e^{2} x^{4} \ln \relax (c ) \ln \relax (x )+24 b d \,e^{2} n \,x^{4} \ln \relax (x )^{2}-24 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )+12 b \,d^{2} e n \,x^{2}-12 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{16 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(b*ln(c*x^n)+a)/x^5,x)

[Out]

-1/4*b*(-2*e^3*x^6-12*d*e^2*ln(x)*x^4+6*d^2*e*x^2+d^3)/x^4*ln(x^n)-1/16*(24*I*ln(x)*Pi*b*d*e^2*csgn(I*x^n)*csg
n(I*c*x^n)*csgn(I*c)*x^4-8*a*e^3*x^6+4*a*d^3-8*b*e^3*x^6*ln(c)+b*d^3*n+4*b*d^3*ln(c)+24*a*d^2*e*x^2+24*b*d^2*e
*x^2*ln(c)-2*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+4*b*e^3*n*x^6-24*I*ln(x)*Pi*b*d*e^2*csgn(I*x^n)*cs
gn(I*c*x^n)^2*x^4-24*I*ln(x)*Pi*b*d*e^2*csgn(I*c*x^n)^2*csgn(I*c)*x^4-48*ln(x)*a*d*e^2*x^4-12*I*Pi*b*d^2*e*x^2
*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+2*I*Pi*b*d^3*csgn(I*c*x^n)^2*csgn(I*c)+4*I*Pi*b*e^3*x^6*csgn(I*c*x^n)^3+2
*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*Pi*b*d^3*csgn(I*c*x^n)^3+12*I*Pi*b*d^2*e*x^2*csgn(I*c*x^n)^2*csgn(
I*c)+4*I*Pi*b*e^3*x^6*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+12*I*Pi*b*d^2*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+24*I
*ln(x)*Pi*b*d*e^2*csgn(I*c*x^n)^3*x^4-48*ln(x)*ln(c)*b*d*e^2*x^4+24*b*d*e^2*n*ln(x)^2*x^4-4*I*Pi*b*e^3*x^6*csg
n(I*x^n)*csgn(I*c*x^n)^2-4*I*Pi*b*e^3*x^6*csgn(I*c*x^n)^2*csgn(I*c)+12*b*d^2*e*n*x^2-12*I*Pi*b*d^2*e*x^2*csgn(
I*c*x^n)^3)/x^4

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maxima [A]  time = 0.50, size = 133, normalized size = 1.02 \[ -\frac {1}{4} \, b e^{3} n x^{2} + \frac {1}{2} \, b e^{3} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a e^{3} x^{2} + \frac {3 \, b d e^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d e^{2} \log \relax (x) - \frac {3 \, b d^{2} e n}{4 \, x^{2}} - \frac {3 \, b d^{2} e \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {3 \, a d^{2} e}{2 \, x^{2}} - \frac {b d^{3} n}{16 \, x^{4}} - \frac {b d^{3} \log \left (c x^{n}\right )}{4 \, x^{4}} - \frac {a d^{3}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="maxima")

[Out]

-1/4*b*e^3*n*x^2 + 1/2*b*e^3*x^2*log(c*x^n) + 1/2*a*e^3*x^2 + 3/2*b*d*e^2*log(c*x^n)^2/n + 3*a*d*e^2*log(x) -
3/4*b*d^2*e*n/x^2 - 3/2*b*d^2*e*log(c*x^n)/x^2 - 3/2*a*d^2*e/x^2 - 1/16*b*d^3*n/x^4 - 1/4*b*d^3*log(c*x^n)/x^4
- 1/4*a*d^3/x^4

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mupad [B]  time = 3.66, size = 149, normalized size = 1.14 \[ \ln \relax (x)\,\left (3\,a\,d\,e^2+\frac {9\,b\,d\,e^2\,n}{4}\right )-\ln \left (c\,x^n\right )\,\left (\frac {\frac {b\,d^3}{4}+\frac {3\,b\,d^2\,e\,x^2}{2}+\frac {9\,b\,d\,e^2\,x^4}{4}+b\,e^3\,x^6}{x^4}-\frac {3\,b\,e^3\,x^2}{2}\right )-\frac {a\,d^3+x^2\,\left (6\,a\,d^2\,e+3\,b\,d^2\,e\,n\right )+\frac {b\,d^3\,n}{4}}{4\,x^4}+\frac {e^3\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {3\,b\,d\,e^2\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*(3*a*d*e^2 + (9*b*d*e^2*n)/4) - log(c*x^n)*(((b*d^3)/4 + b*e^3*x^6 + (3*b*d^2*e*x^2)/2 + (9*b*d*e^2*x^4
)/4)/x^4 - (3*b*e^3*x^2)/2) - (a*d^3 + x^2*(6*a*d^2*e + 3*b*d^2*e*n) + (b*d^3*n)/4)/(4*x^4) + (e^3*x^2*(2*a -
b*n))/4 + (3*b*d*e^2*log(c*x^n)^2)/(2*n)

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sympy [A]  time = 6.92, size = 209, normalized size = 1.60 \[ - \frac {a d^{3}}{4 x^{4}} - \frac {3 a d^{2} e}{2 x^{2}} + 3 a d e^{2} \log {\relax (x )} + \frac {a e^{3} x^{2}}{2} - \frac {b d^{3} n \log {\relax (x )}}{4 x^{4}} - \frac {b d^{3} n}{16 x^{4}} - \frac {b d^{3} \log {\relax (c )}}{4 x^{4}} - \frac {3 b d^{2} e n \log {\relax (x )}}{2 x^{2}} - \frac {3 b d^{2} e n}{4 x^{2}} - \frac {3 b d^{2} e \log {\relax (c )}}{2 x^{2}} + \frac {3 b d e^{2} n \log {\relax (x )}^{2}}{2} + 3 b d e^{2} \log {\relax (c )} \log {\relax (x )} + \frac {b e^{3} n x^{2} \log {\relax (x )}}{2} - \frac {b e^{3} n x^{2}}{4} + \frac {b e^{3} x^{2} \log {\relax (c )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*ln(c*x**n))/x**5,x)

[Out]

-a*d**3/(4*x**4) - 3*a*d**2*e/(2*x**2) + 3*a*d*e**2*log(x) + a*e**3*x**2/2 - b*d**3*n*log(x)/(4*x**4) - b*d**3
*n/(16*x**4) - b*d**3*log(c)/(4*x**4) - 3*b*d**2*e*n*log(x)/(2*x**2) - 3*b*d**2*e*n/(4*x**2) - 3*b*d**2*e*log(
c)/(2*x**2) + 3*b*d*e**2*n*log(x)**2/2 + 3*b*d*e**2*log(c)*log(x) + b*e**3*n*x**2*log(x)/2 - b*e**3*n*x**2/4 +
b*e**3*x**2*log(c)/2

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