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

Optimal. Leaf size=100 \[ \frac{1}{60} \left (45 d^2 e x^4+20 d^3 x^3+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{3}{16} b d^2 e n x^4-\frac{1}{9} b d^3 n x^3-\frac{3}{25} b d e^2 n x^5-\frac{1}{36} b e^3 n x^6 \]

[Out]

-(b*d^3*n*x^3)/9 - (3*b*d^2*e*n*x^4)/16 - (3*b*d*e^2*n*x^5)/25 - (b*e^3*n*x^6)/36 + ((20*d^3*x^3 + 45*d^2*e*x^
4 + 36*d*e^2*x^5 + 10*e^3*x^6)*(a + b*Log[c*x^n]))/60

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Rubi [A]  time = 0.102275, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {43, 2334, 12, 14} \[ \frac{1}{60} \left (45 d^2 e x^4+20 d^3 x^3+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{3}{16} b d^2 e n x^4-\frac{1}{9} b d^3 n x^3-\frac{3}{25} b d e^2 n x^5-\frac{1}{36} b e^3 n x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^3*n*x^3)/9 - (3*b*d^2*e*n*x^4)/16 - (3*b*d*e^2*n*x^5)/25 - (b*e^3*n*x^6)/36 + ((20*d^3*x^3 + 45*d^2*e*x^
4 + 36*d*e^2*x^5 + 10*e^3*x^6)*(a + b*Log[c*x^n]))/60

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 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]]

Rubi steps

\begin{align*} \int x^2 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{1}{60} x^2 \left (20 d^3+45 d^2 e x+36 d e^2 x^2+10 e^3 x^3\right ) \, dx\\ &=\frac{1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{60} (b n) \int x^2 \left (20 d^3+45 d^2 e x+36 d e^2 x^2+10 e^3 x^3\right ) \, dx\\ &=\frac{1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{60} (b n) \int \left (20 d^3 x^2+45 d^2 e x^3+36 d e^2 x^4+10 e^3 x^5\right ) \, dx\\ &=-\frac{1}{9} b d^3 n x^3-\frac{3}{16} b d^2 e n x^4-\frac{3}{25} b d e^2 n x^5-\frac{1}{36} b e^3 n x^6+\frac{1}{60} \left (20 d^3 x^3+45 d^2 e x^4+36 d e^2 x^5+10 e^3 x^6\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0493882, size = 133, normalized size = 1.33 \[ \frac{3}{4} d^2 e x^4 \left (a+b \log \left (c x^n\right )\right )+\frac{1}{3} d^3 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac{3}{5} d e^2 x^5 \left (a+b \log \left (c x^n\right )\right )+\frac{1}{6} e^3 x^6 \left (a+b \log \left (c x^n\right )\right )-\frac{3}{16} b d^2 e n x^4-\frac{1}{9} b d^3 n x^3-\frac{3}{25} b d e^2 n x^5-\frac{1}{36} b e^3 n x^6 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^3*n*x^3)/9 - (3*b*d^2*e*n*x^4)/16 - (3*b*d*e^2*n*x^5)/25 - (b*e^3*n*x^6)/36 + (d^3*x^3*(a + b*Log[c*x^n]
))/3 + (3*d^2*e*x^4*(a + b*Log[c*x^n]))/4 + (3*d*e^2*x^5*(a + b*Log[c*x^n]))/5 + (e^3*x^6*(a + b*Log[c*x^n]))/
6

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Maple [C]  time = 0.223, size = 600, normalized size = 6. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/5*a*d*e^2*x^5+3/4*a*d^2*e*x^4-1/6*I*Pi*b*d^3*x^3*csgn(I*c*x^n)^3-1/12*I*Pi*b*e^3*x^6*csgn(I*c*x^n)^3+3/4*ln(
c)*b*d^2*e*x^4+3/5*ln(c)*b*d*e^2*x^5+3/10*I*Pi*b*d*e^2*x^5*csgn(I*x^n)*csgn(I*c*x^n)^2+3/8*I*Pi*b*d^2*e*x^4*cs
gn(I*x^n)*csgn(I*c*x^n)^2-1/6*I*Pi*b*d^3*x^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3/8*I*Pi*b*d^2*e*x^4*csgn(I*c
*x^n)^2*csgn(I*c)+3/10*I*Pi*b*d*e^2*x^5*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
)*csgn(I*c)+1/3*ln(c)*b*d^3*x^3+1/6*ln(c)*b*e^3*x^6-3/10*I*Pi*b*d*e^2*x^5*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-
3/8*I*Pi*b*d^2*e*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/60*b*x^3*(10*e^3*x^3+36*d*e^2*x^2+45*d^2*e*x+20*d^3
)*ln(x^n)+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)+1/6*a*
e^3*x^6+1/3*a*d^3*x^3-3/8*I*Pi*b*d^2*e*x^4*csgn(I*c*x^n)^3-3/10*I*Pi*b*d*e^2*x^5*csgn(I*c*x^n)^3+1/6*I*Pi*b*d^
3*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6*I*Pi*b*d^3*x^3*csgn(I*c*x^n)^2*csgn(I*c)-1/9*b*d^3*n*x^3-1/36*b*e^3*n*x^
6-3/16*b*d^2*e*n*x^4-3/25*b*d*e^2*n*x^5

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Maxima [A]  time = 1.17195, size = 193, normalized size = 1.93 \begin{align*} -\frac{1}{36} \, b e^{3} n x^{6} + \frac{1}{6} \, b e^{3} x^{6} \log \left (c x^{n}\right ) - \frac{3}{25} \, b d e^{2} n x^{5} + \frac{1}{6} \, a e^{3} x^{6} + \frac{3}{5} \, b d e^{2} x^{5} \log \left (c x^{n}\right ) - \frac{3}{16} \, b d^{2} e n x^{4} + \frac{3}{5} \, a d e^{2} x^{5} + \frac{3}{4} \, b d^{2} e x^{4} \log \left (c x^{n}\right ) - \frac{1}{9} \, b d^{3} n x^{3} + \frac{3}{4} \, a d^{2} e x^{4} + \frac{1}{3} \, b d^{3} x^{3} \log \left (c x^{n}\right ) + \frac{1}{3} \, a d^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*b*e^3*n*x^6 + 1/6*b*e^3*x^6*log(c*x^n) - 3/25*b*d*e^2*n*x^5 + 1/6*a*e^3*x^6 + 3/5*b*d*e^2*x^5*log(c*x^n)
- 3/16*b*d^2*e*n*x^4 + 3/5*a*d*e^2*x^5 + 3/4*b*d^2*e*x^4*log(c*x^n) - 1/9*b*d^3*n*x^3 + 3/4*a*d^2*e*x^4 + 1/3
*b*d^3*x^3*log(c*x^n) + 1/3*a*d^3*x^3

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Fricas [A]  time = 1.01714, size = 398, normalized size = 3.98 \begin{align*} -\frac{1}{36} \,{\left (b e^{3} n - 6 \, a e^{3}\right )} x^{6} - \frac{3}{25} \,{\left (b d e^{2} n - 5 \, a d e^{2}\right )} x^{5} - \frac{3}{16} \,{\left (b d^{2} e n - 4 \, a d^{2} e\right )} x^{4} - \frac{1}{9} \,{\left (b d^{3} n - 3 \, a d^{3}\right )} x^{3} + \frac{1}{60} \,{\left (10 \, b e^{3} x^{6} + 36 \, b d e^{2} x^{5} + 45 \, b d^{2} e x^{4} + 20 \, b d^{3} x^{3}\right )} \log \left (c\right ) + \frac{1}{60} \,{\left (10 \, b e^{3} n x^{6} + 36 \, b d e^{2} n x^{5} + 45 \, b d^{2} e n x^{4} + 20 \, b d^{3} n x^{3}\right )} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(b*e^3*n - 6*a*e^3)*x^6 - 3/25*(b*d*e^2*n - 5*a*d*e^2)*x^5 - 3/16*(b*d^2*e*n - 4*a*d^2*e)*x^4 - 1/9*(b*d
^3*n - 3*a*d^3)*x^3 + 1/60*(10*b*e^3*x^6 + 36*b*d*e^2*x^5 + 45*b*d^2*e*x^4 + 20*b*d^3*x^3)*log(c) + 1/60*(10*b
*e^3*n*x^6 + 36*b*d*e^2*n*x^5 + 45*b*d^2*e*n*x^4 + 20*b*d^3*n*x^3)*log(x)

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Sympy [B]  time = 7.322, size = 230, normalized size = 2.3 \begin{align*} \frac{a d^{3} x^{3}}{3} + \frac{3 a d^{2} e x^{4}}{4} + \frac{3 a d e^{2} x^{5}}{5} + \frac{a e^{3} x^{6}}{6} + \frac{b d^{3} n x^{3} \log{\left (x \right )}}{3} - \frac{b d^{3} n x^{3}}{9} + \frac{b d^{3} x^{3} \log{\left (c \right )}}{3} + \frac{3 b d^{2} e n x^{4} \log{\left (x \right )}}{4} - \frac{3 b d^{2} e n x^{4}}{16} + \frac{3 b d^{2} e x^{4} \log{\left (c \right )}}{4} + \frac{3 b d e^{2} n x^{5} \log{\left (x \right )}}{5} - \frac{3 b d e^{2} n x^{5}}{25} + \frac{3 b d e^{2} x^{5} \log{\left (c \right )}}{5} + \frac{b e^{3} n x^{6} \log{\left (x \right )}}{6} - \frac{b e^{3} n x^{6}}{36} + \frac{b e^{3} x^{6} \log{\left (c \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*x**3/3 + 3*a*d**2*e*x**4/4 + 3*a*d*e**2*x**5/5 + a*e**3*x**6/6 + b*d**3*n*x**3*log(x)/3 - b*d**3*n*x**3
/9 + b*d**3*x**3*log(c)/3 + 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 + 3*b
*d*e**2*n*x**5*log(x)/5 - 3*b*d*e**2*n*x**5/25 + 3*b*d*e**2*x**5*log(c)/5 + 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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Giac [A]  time = 1.29318, size = 234, normalized size = 2.34 \begin{align*} \frac{1}{6} \, b n x^{6} e^{3} \log \left (x\right ) + \frac{3}{5} \, b d n x^{5} e^{2} \log \left (x\right ) + \frac{3}{4} \, b d^{2} n x^{4} e \log \left (x\right ) - \frac{1}{36} \, b n x^{6} e^{3} - \frac{3}{25} \, b d n x^{5} e^{2} - \frac{3}{16} \, b d^{2} n x^{4} e + \frac{1}{6} \, b x^{6} e^{3} \log \left (c\right ) + \frac{3}{5} \, b d x^{5} e^{2} \log \left (c\right ) + \frac{3}{4} \, b d^{2} x^{4} e \log \left (c\right ) + \frac{1}{3} \, b d^{3} n x^{3} \log \left (x\right ) - \frac{1}{9} \, b d^{3} n x^{3} + \frac{1}{6} \, a x^{6} e^{3} + \frac{3}{5} \, a d x^{5} e^{2} + \frac{3}{4} \, a d^{2} x^{4} e + \frac{1}{3} \, b d^{3} x^{3} \log \left (c\right ) + \frac{1}{3} \, a d^{3} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*n*x^6*e^3*log(x) + 3/5*b*d*n*x^5*e^2*log(x) + 3/4*b*d^2*n*x^4*e*log(x) - 1/36*b*n*x^6*e^3 - 3/25*b*d*n*x
^5*e^2 - 3/16*b*d^2*n*x^4*e + 1/6*b*x^6*e^3*log(c) + 3/5*b*d*x^5*e^2*log(c) + 3/4*b*d^2*x^4*e*log(c) + 1/3*b*d
^3*n*x^3*log(x) - 1/9*b*d^3*n*x^3 + 1/6*a*x^6*e^3 + 3/5*a*d*x^5*e^2 + 3/4*a*d^2*x^4*e + 1/3*b*d^3*x^3*log(c) +
1/3*a*d^3*x^3