3.198 \(\int x (d+e x^2)^3 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=91 \[ \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {b d^4 n \log (x)}{8 e}-\frac {1}{4} b d^3 n x^2-\frac {3}{16} b d^2 e n x^4-\frac {1}{12} b d e^2 n x^6-\frac {1}{64} b e^3 n x^8 \]

[Out]

-1/4*b*d^3*n*x^2-3/16*b*d^2*e*n*x^4-1/12*b*d*e^2*n*x^6-1/64*b*e^3*n*x^8-1/8*b*d^4*n*ln(x)/e+1/8*(e*x^2+d)^4*(a
+b*ln(c*x^n))/e

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Rubi [A]  time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {261, 2334, 12, 266, 43} \[ \frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {3}{16} b d^2 e n x^4-\frac {b d^4 n \log (x)}{8 e}-\frac {1}{4} b d^3 n x^2-\frac {1}{12} b d e^2 n x^6-\frac {1}{64} b e^3 n x^8 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^3*n*x^2)/4 - (3*b*d^2*e*n*x^4)/16 - (b*d*e^2*n*x^6)/12 - (b*e^3*n*x^8)/64 - (b*d^4*n*Log[x])/(8*e) + ((d
+ e*x^2)^4*(a + b*Log[c*x^n]))/(8*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 261

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

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 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 x \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-(b n) \int \frac {\left (d+e x^2\right )^4}{8 e x} \, dx\\ &=\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {(b n) \int \frac {\left (d+e x^2\right )^4}{x} \, dx}{8 e}\\ &=\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {(b n) \operatorname {Subst}\left (\int \frac {(d+e x)^4}{x} \, dx,x,x^2\right )}{16 e}\\ &=\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {(b n) \operatorname {Subst}\left (\int \left (4 d^3 e+\frac {d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx,x,x^2\right )}{16 e}\\ &=-\frac {1}{4} b d^3 n x^2-\frac {3}{16} b d^2 e n x^4-\frac {1}{12} b d e^2 n x^6-\frac {1}{64} b e^3 n x^8-\frac {b d^4 n \log (x)}{8 e}+\frac {\left (d+e x^2\right )^4 \left (a+b \log \left (c x^n\right )\right )}{8 e}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 118, normalized size = 1.30 \[ \frac {1}{192} x^2 \left (24 a \left (4 d^3+6 d^2 e x^2+4 d e^2 x^4+e^3 x^6\right )+24 b \left (4 d^3+6 d^2 e x^2+4 d e^2 x^4+e^3 x^6\right ) \log \left (c x^n\right )-b n \left (48 d^3+36 d^2 e x^2+16 d e^2 x^4+3 e^3 x^6\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(24*a*(4*d^3 + 6*d^2*e*x^2 + 4*d*e^2*x^4 + e^3*x^6) - b*n*(48*d^3 + 36*d^2*e*x^2 + 16*d*e^2*x^4 + 3*e^3*x
^6) + 24*b*(4*d^3 + 6*d^2*e*x^2 + 4*d*e^2*x^4 + e^3*x^6)*Log[c*x^n]))/192

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fricas [B]  time = 0.67, size = 165, normalized size = 1.81 \[ -\frac {1}{64} \, {\left (b e^{3} n - 8 \, a e^{3}\right )} x^{8} - \frac {1}{12} \, {\left (b d e^{2} n - 6 \, a d e^{2}\right )} x^{6} - \frac {3}{16} \, {\left (b d^{2} e n - 4 \, a d^{2} e\right )} x^{4} - \frac {1}{4} \, {\left (b d^{3} n - 2 \, a d^{3}\right )} x^{2} + \frac {1}{8} \, {\left (b e^{3} x^{8} + 4 \, b d e^{2} x^{6} + 6 \, b d^{2} e x^{4} + 4 \, b d^{3} x^{2}\right )} \log \relax (c) + \frac {1}{8} \, {\left (b e^{3} n x^{8} + 4 \, b d e^{2} n x^{6} + 6 \, b d^{2} e n x^{4} + 4 \, b d^{3} n x^{2}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(b*e^3*n - 8*a*e^3)*x^8 - 1/12*(b*d*e^2*n - 6*a*d*e^2)*x^6 - 3/16*(b*d^2*e*n - 4*a*d^2*e)*x^4 - 1/4*(b*d
^3*n - 2*a*d^3)*x^2 + 1/8*(b*e^3*x^8 + 4*b*d*e^2*x^6 + 6*b*d^2*e*x^4 + 4*b*d^3*x^2)*log(c) + 1/8*(b*e^3*n*x^8
+ 4*b*d*e^2*n*x^6 + 6*b*d^2*e*n*x^4 + 4*b*d^3*n*x^2)*log(x)

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giac [B]  time = 0.28, size = 173, normalized size = 1.90 \[ \frac {1}{8} \, b n x^{8} e^{3} \log \relax (x) - \frac {1}{64} \, b n x^{8} e^{3} + \frac {1}{8} \, b x^{8} e^{3} \log \relax (c) + \frac {1}{2} \, b d n x^{6} e^{2} \log \relax (x) + \frac {1}{8} \, a x^{8} e^{3} - \frac {1}{12} \, b d n x^{6} e^{2} + \frac {1}{2} \, b d x^{6} e^{2} \log \relax (c) + \frac {3}{4} \, b d^{2} n x^{4} e \log \relax (x) + \frac {1}{2} \, a d x^{6} e^{2} - \frac {3}{16} \, b d^{2} n x^{4} e + \frac {3}{4} \, b d^{2} x^{4} e \log \relax (c) + \frac {3}{4} \, a d^{2} x^{4} e + \frac {1}{2} \, b d^{3} n x^{2} \log \relax (x) - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{2} \, b d^{3} x^{2} \log \relax (c) + \frac {1}{2} \, a d^{3} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b*n*x^8*e^3*log(x) - 1/64*b*n*x^8*e^3 + 1/8*b*x^8*e^3*log(c) + 1/2*b*d*n*x^6*e^2*log(x) + 1/8*a*x^8*e^3 -
1/12*b*d*n*x^6*e^2 + 1/2*b*d*x^6*e^2*log(c) + 3/4*b*d^2*n*x^4*e*log(x) + 1/2*a*d*x^6*e^2 - 3/16*b*d^2*n*x^4*e
+ 3/4*b*d^2*x^4*e*log(c) + 3/4*a*d^2*x^4*e + 1/2*b*d^3*n*x^2*log(x) - 1/4*b*d^3*n*x^2 + 1/2*b*d^3*x^2*log(c) +
1/2*a*d^3*x^2

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maple [C]  time = 0.24, size = 601, normalized size = 6.60 \[ \frac {a \,d^{3} x^{2}}{2}+\frac {3 a \,d^{2} e \,x^{4}}{4}+\frac {b d \,e^{2} x^{6} \ln \relax (c )}{2}+\frac {3 b \,d^{2} e \,x^{4} \ln \relax (c )}{4}+\frac {b \,e^{3} x^{8} \ln \relax (c )}{8}+\frac {b \,d^{3} x^{2} \ln \relax (c )}{2}+\frac {\left (e^{3} x^{6}+4 d \,e^{2} x^{4}+6 d^{2} e \,x^{2}+4 d^{3}\right ) b \,x^{2} \ln \left (x^{n}\right )}{8}+\frac {a d \,e^{2} x^{6}}{2}+\frac {a \,e^{3} x^{8}}{8}-\frac {i \pi b \,e^{3} x^{8} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{16}+\frac {i \pi b d \,e^{2} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,e^{3} x^{8} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{16}+\frac {i \pi b \,e^{3} x^{8} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{16}-\frac {i \pi b d \,e^{2} x^{6} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}-\frac {3 i \pi b \,d^{2} e \,x^{4} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {b \,d^{3} n \,x^{2}}{4}+\frac {i \pi b d \,e^{2} x^{6} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {3 i \pi b \,d^{2} e \,x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {3 i \pi b \,d^{2} e \,x^{4} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b \,d^{3} 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 {b \,e^{3} n \,x^{8}}{64}-\frac {i \pi b d \,e^{2} x^{6} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4}-\frac {3 i \pi b \,d^{2} e \,x^{4} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}-\frac {i \pi b \,e^{3} x^{8} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{16}-\frac {i \pi b \,d^{3} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}-\frac {b d \,e^{2} n \,x^{6}}{12}-\frac {3 b \,d^{2} e n \,x^{4}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*d^3*x^2+3/4*a*d^2*e*x^4+1/2*b*d*e^2*x^6*ln(c)+3/4*b*d^2*e*x^4*ln(c)+1/8*ln(c)*b*e^3*x^8+1/2*b*d^3*x^2*ln
(c)+1/8*b*x^2*(e^3*x^6+4*d*e^2*x^4+6*d^2*e*x^2+4*d^3)*ln(x^n)+1/2*a*d*e^2*x^6+1/8*a*e^3*x^8-3/8*I*Pi*b*d^2*e*x
^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*I*Pi*b*d*e^2*x^6*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*b*d^3*n*x^
2+3/8*I*Pi*b*d^2*e*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*d^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*
I*Pi*b*d*e^2*x^6*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*b*d*e^2*x^6*csgn(I*c*x^n)^2*csgn(I*c)+3/8*I*Pi*b*d^2*e*x
^4*csgn(I*c*x^n)^2*csgn(I*c)-1/16*I*Pi*b*e^3*x^8*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/64*b*e^3*n*x^8-1/4*I*Pi
*b*d^3*x^2*csgn(I*c*x^n)^3-1/16*I*Pi*b*e^3*x^8*csgn(I*c*x^n)^3-3/8*I*Pi*b*d^2*e*x^4*csgn(I*c*x^n)^3+1/4*I*Pi*b
*d^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*d*e^2*x^6*csgn(I*c*x^n)^3+1/16*I*Pi*b*e^3*x^8*csgn(I*c*x^n)^2*
csgn(I*c)+1/16*I*Pi*b*e^3*x^8*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*b*d^3*x^2*csgn(I*c*x^n)^2*csgn(I*c)-1/12*b*
d*e^2*n*x^6-3/16*b*d^2*e*n*x^4

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maxima [A]  time = 0.49, size = 143, normalized size = 1.57 \[ -\frac {1}{64} \, b e^{3} n x^{8} + \frac {1}{8} \, b e^{3} x^{8} \log \left (c x^{n}\right ) + \frac {1}{8} \, a e^{3} x^{8} - \frac {1}{12} \, b d e^{2} n x^{6} + \frac {1}{2} \, b d e^{2} x^{6} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d e^{2} x^{6} - \frac {3}{16} \, b d^{2} e n x^{4} + \frac {3}{4} \, b d^{2} e x^{4} \log \left (c x^{n}\right ) + \frac {3}{4} \, a d^{2} e x^{4} - \frac {1}{4} \, b d^{3} n x^{2} + \frac {1}{2} \, b d^{3} x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d^{3} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*b*e^3*n*x^8 + 1/8*b*e^3*x^8*log(c*x^n) + 1/8*a*e^3*x^8 - 1/12*b*d*e^2*n*x^6 + 1/2*b*d*e^2*x^6*log(c*x^n)
+ 1/2*a*d*e^2*x^6 - 3/16*b*d^2*e*n*x^4 + 3/4*b*d^2*e*x^4*log(c*x^n) + 3/4*a*d^2*e*x^4 - 1/4*b*d^3*n*x^2 + 1/2
*b*d^3*x^2*log(c*x^n) + 1/2*a*d^3*x^2

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mupad [B]  time = 3.49, size = 113, normalized size = 1.24 \[ \ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^2}{2}+\frac {3\,b\,d^2\,e\,x^4}{4}+\frac {b\,d\,e^2\,x^6}{2}+\frac {b\,e^3\,x^8}{8}\right )+\frac {d^3\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e^3\,x^8\,\left (8\,a-b\,n\right )}{64}+\frac {3\,d^2\,e\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {d\,e^2\,x^6\,\left (6\,a-b\,n\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^n)*((b*d^3*x^2)/2 + (b*e^3*x^8)/8 + (3*b*d^2*e*x^4)/4 + (b*d*e^2*x^6)/2) + (d^3*x^2*(2*a - b*n))/4 + (
e^3*x^8*(8*a - b*n))/64 + (3*d^2*e*x^4*(4*a - b*n))/16 + (d*e^2*x^6*(6*a - b*n))/12

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sympy [B]  time = 9.99, size = 223, normalized size = 2.45 \[ \frac {a d^{3} x^{2}}{2} + \frac {3 a d^{2} e x^{4}}{4} + \frac {a d e^{2} x^{6}}{2} + \frac {a e^{3} x^{8}}{8} + \frac {b d^{3} n x^{2} \log {\relax (x )}}{2} - \frac {b d^{3} n x^{2}}{4} + \frac {b d^{3} x^{2} \log {\relax (c )}}{2} + \frac {3 b d^{2} e n x^{4} \log {\relax (x )}}{4} - \frac {3 b d^{2} e n x^{4}}{16} + \frac {3 b d^{2} e x^{4} \log {\relax (c )}}{4} + \frac {b d e^{2} n x^{6} \log {\relax (x )}}{2} - \frac {b d e^{2} n x^{6}}{12} + \frac {b d e^{2} x^{6} \log {\relax (c )}}{2} + \frac {b e^{3} n x^{8} \log {\relax (x )}}{8} - \frac {b e^{3} n x^{8}}{64} + \frac {b e^{3} x^{8} \log {\relax (c )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*x**2/2 + 3*a*d**2*e*x**4/4 + a*d*e**2*x**6/2 + a*e**3*x**8/8 + b*d**3*n*x**2*log(x)/2 - b*d**3*n*x**2/4
+ b*d**3*x**2*log(c)/2 + 3*b*d**2*e*n*x**4*log(x)/4 - 3*b*d**2*e*n*x**4/16 + 3*b*d**2*e*x**4*log(c)/4 + b*d*e
**2*n*x**6*log(x)/2 - b*d*e**2*n*x**6/12 + b*d*e**2*x**6*log(c)/2 + b*e**3*n*x**8*log(x)/8 - b*e**3*n*x**8/64
+ b*e**3*x**8*log(c)/8

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