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

Optimal. Leaf size=130 \[ -\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{b d^5 n \log (x)}{40 e^2}+\frac{1}{60} b d^2 e n x^6+\frac{b d^4 n x^2}{20 e}+\frac{3}{80} b d^3 n x^4+\frac{1}{320} b d e^2 n x^8-\frac{b n \left (d+e x^2\right )^5}{100 e^2} \]

[Out]

(b*d^4*n*x^2)/(20*e) + (3*b*d^3*n*x^4)/80 + (b*d^2*e*n*x^6)/60 + (b*d*e^2*n*x^8)/320 - (b*n*(d + e*x^2)^5)/(10
0*e^2) + (b*d^5*n*Log[x])/(40*e^2) - (((5*d*(d + e*x^2)^4)/e^2 - (4*(d + e*x^2)^5)/e^2)*(a + b*Log[c*x^n]))/40

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Rubi [A]  time = 0.15226, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {266, 43, 2334, 12, 446, 80} \[ -\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{b d^5 n \log (x)}{40 e^2}+\frac{1}{60} b d^2 e n x^6+\frac{b d^4 n x^2}{20 e}+\frac{3}{80} b d^3 n x^4+\frac{1}{320} b d e^2 n x^8-\frac{b n \left (d+e x^2\right )^5}{100 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^4*n*x^2)/(20*e) + (3*b*d^3*n*x^4)/80 + (b*d^2*e*n*x^6)/60 + (b*d*e^2*n*x^8)/320 - (b*n*(d + e*x^2)^5)/(10
0*e^2) + (b*d^5*n*Log[x])/(40*e^2) - (((5*d*(d + e*x^2)^4)/e^2 - (4*(d + e*x^2)^5)/e^2)*(a + b*Log[c*x^n]))/40

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 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 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])

Rule 12

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rubi steps

\begin{align*} \int x^3 \left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{\left (d+e x^2\right )^4 \left (-d+4 e x^2\right )}{40 e^2 x} \, dx\\ &=-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(b n) \int \frac{\left (d+e x^2\right )^4 \left (-d+4 e x^2\right )}{x} \, dx}{40 e^2}\\ &=-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{(b n) \operatorname{Subst}\left (\int \frac{(d+e x)^4 (-d+4 e x)}{x} \, dx,x,x^2\right )}{80 e^2}\\ &=-\frac{b n \left (d+e x^2\right )^5}{100 e^2}-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{(b d n) \operatorname{Subst}\left (\int \frac{(d+e x)^4}{x} \, dx,x,x^2\right )}{80 e^2}\\ &=-\frac{b n \left (d+e x^2\right )^5}{100 e^2}-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{(b d 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 )}{80 e^2}\\ &=\frac{b d^4 n x^2}{20 e}+\frac{3}{80} b d^3 n x^4+\frac{1}{60} b d^2 e n x^6+\frac{1}{320} b d e^2 n x^8-\frac{b n \left (d+e x^2\right )^5}{100 e^2}+\frac{b d^5 n \log (x)}{40 e^2}-\frac{1}{40} \left (\frac{5 d \left (d+e x^2\right )^4}{e^2}-\frac{4 \left (d+e x^2\right )^5}{e^2}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0523048, size = 120, normalized size = 0.92 \[ \frac{x^4 \left (120 a \left (20 d^2 e x^2+10 d^3+15 d e^2 x^4+4 e^3 x^6\right )+120 b \left (20 d^2 e x^2+10 d^3+15 d e^2 x^4+4 e^3 x^6\right ) \log \left (c x^n\right )-b n \left (400 d^2 e x^2+300 d^3+225 d e^2 x^4+48 e^3 x^6\right )\right )}{4800} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(120*a*(10*d^3 + 20*d^2*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6) - b*n*(300*d^3 + 400*d^2*e*x^2 + 225*d*e^2*x^4
+ 48*e^3*x^6) + 120*b*(10*d^3 + 20*d^2*e*x^2 + 15*d*e^2*x^4 + 4*e^3*x^6)*Log[c*x^n]))/4800

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Maple [C]  time = 0.208, size = 602, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*d^2*e*x^6+3/8*a*d*e^2*x^8-1/100*b*e^3*n*x^10-1/8*I*Pi*b*d^3*x^4*csgn(I*c*x^n)^3-1/8*I*Pi*b*d^3*x^4*csgn(
I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*I*Pi*b*d^2*e*x^6*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*b*d^2*e*x^6*csgn(I*c*
x^n)^2*csgn(I*c)+3/16*I*Pi*b*d*e^2*x^8*csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*ln(c)*b*d^2*e*x^6+3/8*ln(c)*b*d*e^2*x^8
+1/4*ln(c)*b*d^3*x^4+3/16*I*Pi*b*d*e^2*x^8*csgn(I*c*x^n)^2*csgn(I*c)+1/4*a*d^3*x^4-1/20*I*Pi*b*e^3*x^10*csgn(I
*c*x^n)^3+1/40*b*x^4*(4*e^3*x^6+15*d*e^2*x^4+20*d^2*e*x^2+10*d^3)*ln(x^n)+1/10*ln(c)*b*e^3*x^10-3/16*I*Pi*b*d*
e^2*x^8*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/20*I*Pi*b*e^3*x^10*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3/16*I*Pi
*b*d*e^2*x^8*csgn(I*c*x^n)^3+1/10*a*e^3*x^10+1/8*I*Pi*b*d^3*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-1/12*b*d^2*e*n*x^6
-3/64*b*d*e^2*n*x^8+1/20*I*Pi*b*e^3*x^10*csgn(I*x^n)*csgn(I*c*x^n)^2+1/20*I*Pi*b*e^3*x^10*csgn(I*c*x^n)^2*csgn
(I*c)+1/8*I*Pi*b*d^3*x^4*csgn(I*c*x^n)^2*csgn(I*c)-1/16*b*d^3*n*x^4-1/4*I*Pi*b*d^2*e*x^6*csgn(I*c*x^n)^3-1/4*I
*Pi*b*d^2*e*x^6*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)

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Maxima [A]  time = 1.13128, size = 193, normalized size = 1.48 \begin{align*} -\frac{1}{100} \, b e^{3} n x^{10} + \frac{1}{10} \, b e^{3} x^{10} \log \left (c x^{n}\right ) + \frac{1}{10} \, a e^{3} x^{10} - \frac{3}{64} \, b d e^{2} n x^{8} + \frac{3}{8} \, b d e^{2} x^{8} \log \left (c x^{n}\right ) + \frac{3}{8} \, a d e^{2} x^{8} - \frac{1}{12} \, b d^{2} e n x^{6} + \frac{1}{2} \, b d^{2} e x^{6} \log \left (c x^{n}\right ) + \frac{1}{2} \, a d^{2} e x^{6} - \frac{1}{16} \, b d^{3} n x^{4} + \frac{1}{4} \, b d^{3} x^{4} \log \left (c x^{n}\right ) + \frac{1}{4} \, a d^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/100*b*e^3*n*x^10 + 1/10*b*e^3*x^10*log(c*x^n) + 1/10*a*e^3*x^10 - 3/64*b*d*e^2*n*x^8 + 3/8*b*d*e^2*x^8*log(
c*x^n) + 3/8*a*d*e^2*x^8 - 1/12*b*d^2*e*n*x^6 + 1/2*b*d^2*e*x^6*log(c*x^n) + 1/2*a*d^2*e*x^6 - 1/16*b*d^3*n*x^
4 + 1/4*b*d^3*x^4*log(c*x^n) + 1/4*a*d^3*x^4

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Fricas [A]  time = 1.3549, size = 404, normalized size = 3.11 \begin{align*} -\frac{1}{100} \,{\left (b e^{3} n - 10 \, a e^{3}\right )} x^{10} - \frac{3}{64} \,{\left (b d e^{2} n - 8 \, a d e^{2}\right )} x^{8} - \frac{1}{12} \,{\left (b d^{2} e n - 6 \, a d^{2} e\right )} x^{6} - \frac{1}{16} \,{\left (b d^{3} n - 4 \, a d^{3}\right )} x^{4} + \frac{1}{40} \,{\left (4 \, b e^{3} x^{10} + 15 \, b d e^{2} x^{8} + 20 \, b d^{2} e x^{6} + 10 \, b d^{3} x^{4}\right )} \log \left (c\right ) + \frac{1}{40} \,{\left (4 \, b e^{3} n x^{10} + 15 \, b d e^{2} n x^{8} + 20 \, b d^{2} e n x^{6} + 10 \, b d^{3} n x^{4}\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/100*(b*e^3*n - 10*a*e^3)*x^10 - 3/64*(b*d*e^2*n - 8*a*d*e^2)*x^8 - 1/12*(b*d^2*e*n - 6*a*d^2*e)*x^6 - 1/16*
(b*d^3*n - 4*a*d^3)*x^4 + 1/40*(4*b*e^3*x^10 + 15*b*d*e^2*x^8 + 20*b*d^2*e*x^6 + 10*b*d^3*x^4)*log(c) + 1/40*(
4*b*e^3*n*x^10 + 15*b*d*e^2*n*x^8 + 20*b*d^2*e*n*x^6 + 10*b*d^3*n*x^4)*log(x)

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Sympy [A]  time = 31.2569, size = 223, normalized size = 1.72 \begin{align*} \frac{a d^{3} x^{4}}{4} + \frac{a d^{2} e x^{6}}{2} + \frac{3 a d e^{2} x^{8}}{8} + \frac{a e^{3} x^{10}}{10} + \frac{b d^{3} n x^{4} \log{\left (x \right )}}{4} - \frac{b d^{3} n x^{4}}{16} + \frac{b d^{3} x^{4} \log{\left (c \right )}}{4} + \frac{b d^{2} e n x^{6} \log{\left (x \right )}}{2} - \frac{b d^{2} e n x^{6}}{12} + \frac{b d^{2} e x^{6} \log{\left (c \right )}}{2} + \frac{3 b d e^{2} n x^{8} \log{\left (x \right )}}{8} - \frac{3 b d e^{2} n x^{8}}{64} + \frac{3 b d e^{2} x^{8} \log{\left (c \right )}}{8} + \frac{b e^{3} n x^{10} \log{\left (x \right )}}{10} - \frac{b e^{3} n x^{10}}{100} + \frac{b e^{3} x^{10} \log{\left (c \right )}}{10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*x**4/4 + a*d**2*e*x**6/2 + 3*a*d*e**2*x**8/8 + a*e**3*x**10/10 + b*d**3*n*x**4*log(x)/4 - b*d**3*n*x**4
/16 + b*d**3*x**4*log(c)/4 + b*d**2*e*n*x**6*log(x)/2 - b*d**2*e*n*x**6/12 + b*d**2*e*x**6*log(c)/2 + 3*b*d*e*
*2*n*x**8*log(x)/8 - 3*b*d*e**2*n*x**8/64 + 3*b*d*e**2*x**8*log(c)/8 + b*e**3*n*x**10*log(x)/10 - b*e**3*n*x**
10/100 + b*e**3*x**10*log(c)/10

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Giac [A]  time = 1.32529, size = 234, normalized size = 1.8 \begin{align*} \frac{1}{10} \, b n x^{10} e^{3} \log \left (x\right ) - \frac{1}{100} \, b n x^{10} e^{3} + \frac{1}{10} \, b x^{10} e^{3} \log \left (c\right ) + \frac{3}{8} \, b d n x^{8} e^{2} \log \left (x\right ) + \frac{1}{10} \, a x^{10} e^{3} - \frac{3}{64} \, b d n x^{8} e^{2} + \frac{3}{8} \, b d x^{8} e^{2} \log \left (c\right ) + \frac{1}{2} \, b d^{2} n x^{6} e \log \left (x\right ) + \frac{3}{8} \, a d x^{8} e^{2} - \frac{1}{12} \, b d^{2} n x^{6} e + \frac{1}{2} \, b d^{2} x^{6} e \log \left (c\right ) + \frac{1}{2} \, a d^{2} x^{6} e + \frac{1}{4} \, b d^{3} n x^{4} \log \left (x\right ) - \frac{1}{16} \, b d^{3} n x^{4} + \frac{1}{4} \, b d^{3} x^{4} \log \left (c\right ) + \frac{1}{4} \, a d^{3} x^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b*n*x^10*e^3*log(x) - 1/100*b*n*x^10*e^3 + 1/10*b*x^10*e^3*log(c) + 3/8*b*d*n*x^8*e^2*log(x) + 1/10*a*x^1
0*e^3 - 3/64*b*d*n*x^8*e^2 + 3/8*b*d*x^8*e^2*log(c) + 1/2*b*d^2*n*x^6*e*log(x) + 3/8*a*d*x^8*e^2 - 1/12*b*d^2*
n*x^6*e + 1/2*b*d^2*x^6*e*log(c) + 1/2*a*d^2*x^6*e + 1/4*b*d^3*n*x^4*log(x) - 1/16*b*d^3*n*x^4 + 1/4*b*d^3*x^4
*log(c) + 1/4*a*d^3*x^4