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

Optimal. Leaf size=130 \[ d^3 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {3}{2} d^2 e x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{4} d e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{6} e^3 x^6 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b d^3 n \log ^2(x)-\frac {3}{4} b d^2 e n x^2-\frac {3}{16} b d e^2 n x^4-\frac {1}{36} b e^3 n x^6 \]

[Out]

-3/4*b*d^2*e*n*x^2-3/16*b*d*e^2*n*x^4-1/36*b*e^3*n*x^6-1/2*b*d^3*n*ln(x)^2+3/2*d^2*e*x^2*(a+b*ln(c*x^n))+3/4*d
*e^2*x^4*(a+b*ln(c*x^n))+1/6*e^3*x^6*(a+b*ln(c*x^n))+d^3*ln(x)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.10, antiderivative size = 100, normalized size of antiderivative = 0.77, 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} \[ \frac {1}{12} \left (18 d^2 e x^2+12 d^3 \log (x)+9 d e^2 x^4+2 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {3}{4} b d^2 e n x^2-\frac {1}{2} b d^3 n \log ^2(x)-\frac {3}{16} b d e^2 n x^4-\frac {1}{36} b e^3 n x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*d^2*e*n*x^2)/4 - (3*b*d*e^2*n*x^4)/16 - (b*e^3*n*x^6)/36 - (b*d^3*n*Log[x]^2)/2 + ((18*d^2*e*x^2 + 9*d*e
^2*x^4 + 2*e^3*x^6 + 12*d^3*Log[x])*(a + b*Log[c*x^n]))/12

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} \, dx &=\frac {1}{12} \left (18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {1}{12} e x \left (18 d^2+9 d e x^2+2 e^2 x^4\right )+\frac {d^3 \log (x)}{x}\right ) \, dx\\ &=\frac {1}{12} \left (18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\left (b d^3 n\right ) \int \frac {\log (x)}{x} \, dx-\frac {1}{12} (b e n) \int x \left (18 d^2+9 d e x^2+2 e^2 x^4\right ) \, dx\\ &=-\frac {1}{2} b d^3 n \log ^2(x)+\frac {1}{12} \left (18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} (b e n) \int \left (18 d^2 x+9 d e x^3+2 e^2 x^5\right ) \, dx\\ &=-\frac {3}{4} b d^2 e n x^2-\frac {3}{16} b d e^2 n x^4-\frac {1}{36} b e^3 n x^6-\frac {1}{2} b d^3 n \log ^2(x)+\frac {1}{12} \left (18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \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 = 116, normalized size = 0.89 \[ \frac {1}{144} \left (\frac {72 d^3 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+216 d^2 e x^2 \left (a+b \log \left (c x^n\right )\right )+108 d e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+24 e^3 x^6 \left (a+b \log \left (c x^n\right )\right )-108 b d^2 e n x^2-27 b d e^2 n x^4-4 b e^3 n x^6\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-108*b*d^2*e*n*x^2 - 27*b*d*e^2*n*x^4 - 4*b*e^3*n*x^6 + 216*d^2*e*x^2*(a + b*Log[c*x^n]) + 108*d*e^2*x^4*(a +
b*Log[c*x^n]) + 24*e^3*x^6*(a + b*Log[c*x^n]) + (72*d^3*(a + b*Log[c*x^n])^2)/(b*n))/144

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fricas [A]  time = 0.77, size = 155, normalized size = 1.19 \[ -\frac {1}{36} \, {\left (b e^{3} n - 6 \, a e^{3}\right )} x^{6} + \frac {1}{2} \, b d^{3} n \log \relax (x)^{2} - \frac {3}{16} \, {\left (b d e^{2} n - 4 \, a d e^{2}\right )} x^{4} - \frac {3}{4} \, {\left (b d^{2} e n - 2 \, a d^{2} e\right )} x^{2} + \frac {1}{12} \, {\left (2 \, b e^{3} x^{6} + 9 \, b d e^{2} x^{4} + 18 \, b d^{2} e x^{2}\right )} \log \relax (c) + \frac {1}{12} \, {\left (2 \, b e^{3} n x^{6} + 9 \, b d e^{2} n x^{4} + 18 \, b d^{2} e n x^{2} + 12 \, b d^{3} \log \relax (c) + 12 \, a d^{3}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(b*e^3*n - 6*a*e^3)*x^6 + 1/2*b*d^3*n*log(x)^2 - 3/16*(b*d*e^2*n - 4*a*d*e^2)*x^4 - 3/4*(b*d^2*e*n - 2*a
*d^2*e)*x^2 + 1/12*(2*b*e^3*x^6 + 9*b*d*e^2*x^4 + 18*b*d^2*e*x^2)*log(c) + 1/12*(2*b*e^3*n*x^6 + 9*b*d*e^2*n*x
^4 + 18*b*d^2*e*n*x^2 + 12*b*d^3*log(c) + 12*a*d^3)*log(x)

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giac [A]  time = 0.27, size = 158, normalized size = 1.22 \[ \frac {1}{6} \, b n x^{6} e^{3} \log \relax (x) - \frac {1}{36} \, b n x^{6} e^{3} + \frac {1}{6} \, b x^{6} e^{3} \log \relax (c) + \frac {3}{4} \, b d n x^{4} e^{2} \log \relax (x) + \frac {1}{6} \, a x^{6} e^{3} - \frac {3}{16} \, b d n x^{4} e^{2} + \frac {3}{4} \, b d x^{4} e^{2} \log \relax (c) + \frac {3}{2} \, b d^{2} n x^{2} e \log \relax (x) + \frac {3}{4} \, a d x^{4} e^{2} - \frac {3}{4} \, b d^{2} n x^{2} e + \frac {3}{2} \, b d^{2} x^{2} e \log \relax (c) + \frac {1}{2} \, b d^{3} n \log \relax (x)^{2} + \frac {3}{2} \, a d^{2} x^{2} e + b d^{3} \log \relax (c) \log \relax (x) + a d^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*n*x^6*e^3*log(x) - 1/36*b*n*x^6*e^3 + 1/6*b*x^6*e^3*log(c) + 3/4*b*d*n*x^4*e^2*log(x) + 1/6*a*x^6*e^3 -
3/16*b*d*n*x^4*e^2 + 3/4*b*d*x^4*e^2*log(c) + 3/2*b*d^2*n*x^2*e*log(x) + 3/4*a*d*x^4*e^2 - 3/4*b*d^2*n*x^2*e +
3/2*b*d^2*x^2*e*log(c) + 1/2*b*d^3*n*log(x)^2 + 3/2*a*d^2*x^2*e + b*d^3*log(c)*log(x) + a*d^3*log(x)

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maple [C]  time = 0.30, size = 595, normalized size = 4.58 \[ \frac {a \,e^{3} x^{6}}{6}+\frac {3 b d \,e^{2} x^{4} \ln \relax (c )}{4}+\frac {3 a d \,e^{2} x^{4}}{4}+\left (\frac {b \,e^{3} x^{6}}{6}+\frac {3 b d \,e^{2} x^{4}}{4}+\frac {3 b \,d^{2} e \,x^{2}}{2}+b \,d^{3} \ln \relax (x )\right ) \ln \left (x^{n}\right )+\frac {b \,e^{3} x^{6} \ln \relax (c )}{6}+\frac {3 a \,d^{2} e \,x^{2}}{2}+b \,d^{3} \ln \relax (c ) \ln \relax (x )+a \,d^{3} \ln \relax (x )+\frac {3 b \,d^{2} e \,x^{2} \ln \relax (c )}{2}-\frac {b \,e^{3} n \,x^{6}}{36}+\frac {i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12}+\frac {i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{12}-\frac {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 ) \ln \relax (x )}{2}+\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {3 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {3 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}-\frac {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 )}{12}-\frac {b \,d^{3} n \ln \relax (x )^{2}}{2}-\frac {i \pi b \,e^{3} x^{6} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{12}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2}-\frac {3 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 )}{4}-\frac {3 b d \,e^{2} n \,x^{4}}{16}-\frac {3 b \,d^{2} e n \,x^{2}}{4}-\frac {3 i \pi b d \,e^{2} x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {3 i \pi b \,d^{2} e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}-\frac {3 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 )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*ln(x)*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3/4*I*Pi*b*d^2*e*x^2*csgn(I*c*x^n)^2*csgn(I*c)+1/6*a
*e^3*x^6+3/4*b*d*e^2*x^4*ln(c)+3/4*a*d*e^2*x^4+(1/6*b*e^3*x^6+3/4*b*d*e^2*x^4+3/2*b*d^2*e*x^2+b*d^3*ln(x))*ln(
x^n)+1/6*b*e^3*x^6*ln(c)+3/2*a*d^2*e*x^2+b*d^3*ln(c)*ln(x)+a*d^3*ln(x)-3/8*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^3-3/
4*I*Pi*b*d^2*e*x^2*csgn(I*c*x^n)^3-1/12*I*Pi*b*e^3*x^6*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3/8*I*Pi*b*d*e^2*x^
4*csgn(I*x^n)*csgn(I*c*x^n)^2+3/2*ln(c)*b*d^2*e*x^2-1/36*b*e^3*n*x^6+3/4*I*Pi*b*d^2*e*x^2*csgn(I*x^n)*csgn(I*c
*x^n)^2-3/4*I*Pi*b*d^2*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3/8*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^2*csgn(I*c
)-3/8*I*Pi*b*d*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/2*I*ln(x)*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2+1/
2*I*ln(x)*Pi*b*d^3*csgn(I*c*x^n)^2*csgn(I*c)+1/12*I*Pi*b*e^3*x^6*csgn(I*x^n)*csgn(I*c*x^n)^2+1/12*I*Pi*b*e^3*x
^6*csgn(I*c*x^n)^2*csgn(I*c)-1/12*I*Pi*b*e^3*x^6*csgn(I*c*x^n)^3-1/2*I*ln(x)*Pi*b*d^3*csgn(I*c*x^n)^3-1/2*b*d^
3*n*ln(x)^2-3/16*b*d*e^2*n*x^4-3/4*b*d^2*e*n*x^2

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maxima [A]  time = 0.49, size = 133, normalized size = 1.02 \[ -\frac {1}{36} \, b e^{3} n x^{6} + \frac {1}{6} \, b e^{3} x^{6} \log \left (c x^{n}\right ) + \frac {1}{6} \, a e^{3} x^{6} - \frac {3}{16} \, b d e^{2} n x^{4} + \frac {3}{4} \, b d e^{2} x^{4} \log \left (c x^{n}\right ) + \frac {3}{4} \, a d e^{2} x^{4} - \frac {3}{4} \, b d^{2} e n x^{2} + \frac {3}{2} \, b d^{2} e x^{2} \log \left (c x^{n}\right ) + \frac {3}{2} \, a d^{2} e x^{2} + \frac {b d^{3} \log \left (c x^{n}\right )^{2}}{2 \, n} + a d^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*b*e^3*n*x^6 + 1/6*b*e^3*x^6*log(c*x^n) + 1/6*a*e^3*x^6 - 3/16*b*d*e^2*n*x^4 + 3/4*b*d*e^2*x^4*log(c*x^n)
+ 3/4*a*d*e^2*x^4 - 3/4*b*d^2*e*n*x^2 + 3/2*b*d^2*e*x^2*log(c*x^n) + 3/2*a*d^2*e*x^2 + 1/2*b*d^3*log(c*x^n)^2
/n + a*d^3*log(x)

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mupad [B]  time = 3.67, size = 112, normalized size = 0.86 \[ \ln \left (c\,x^n\right )\,\left (\frac {3\,b\,d^2\,e\,x^2}{2}+\frac {3\,b\,d\,e^2\,x^4}{4}+\frac {b\,e^3\,x^6}{6}\right )+\frac {e^3\,x^6\,\left (6\,a-b\,n\right )}{36}+a\,d^3\,\ln \relax (x)+\frac {b\,d^3\,{\ln \left (c\,x^n\right )}^2}{2\,n}+\frac {3\,d^2\,e\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {3\,d\,e^2\,x^4\,\left (4\,a-b\,n\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^n)*((b*e^3*x^6)/6 + (3*b*d^2*e*x^2)/2 + (3*b*d*e^2*x^4)/4) + (e^3*x^6*(6*a - b*n))/36 + a*d^3*log(x) +
(b*d^3*log(c*x^n)^2)/(2*n) + (3*d^2*e*x^2*(2*a - b*n))/4 + (3*d*e^2*x^4*(4*a - b*n))/16

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sympy [A]  time = 6.66, size = 212, normalized size = 1.63 \[ a d^{3} \log {\relax (x )} + \frac {3 a d^{2} e x^{2}}{2} + \frac {3 a d e^{2} x^{4}}{4} + \frac {a e^{3} x^{6}}{6} + \frac {b d^{3} n \log {\relax (x )}^{2}}{2} + b d^{3} \log {\relax (c )} \log {\relax (x )} + \frac {3 b d^{2} e n x^{2} \log {\relax (x )}}{2} - \frac {3 b d^{2} e n x^{2}}{4} + \frac {3 b d^{2} e x^{2} \log {\relax (c )}}{2} + \frac {3 b d e^{2} n x^{4} \log {\relax (x )}}{4} - \frac {3 b d e^{2} n x^{4}}{16} + \frac {3 b d e^{2} x^{4} \log {\relax (c )}}{4} + \frac {b e^{3} n x^{6} \log {\relax (x )}}{6} - \frac {b e^{3} n x^{6}}{36} + \frac {b e^{3} x^{6} \log {\relax (c )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*log(x) + 3*a*d**2*e*x**2/2 + 3*a*d*e**2*x**4/4 + a*e**3*x**6/6 + b*d**3*n*log(x)**2/2 + b*d**3*log(c)*l
og(x) + 3*b*d**2*e*n*x**2*log(x)/2 - 3*b*d**2*e*n*x**2/4 + 3*b*d**2*e*x**2*log(c)/2 + 3*b*d*e**2*n*x**4*log(x)
/4 - 3*b*d*e**2*n*x**4/16 + 3*b*d*e**2*x**4*log(c)/4 + b*e**3*n*x**6*log(x)/6 - b*e**3*n*x**6/36 + b*e**3*x**6
*log(c)/6

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