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

Optimal. Leaf size=74 \[ \frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{9} b d^2 n x^3-\frac{1}{8} b d e n x^4-\frac{1}{25} b e^2 n x^5 \]

[Out]

-(b*d^2*n*x^3)/9 - (b*d*e*n*x^4)/8 - (b*e^2*n*x^5)/25 + ((10*d^2*x^3 + 15*d*e*x^4 + 6*e^2*x^5)*(a + b*Log[c*x^
n]))/30

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Rubi [A]  time = 0.079526, antiderivative size = 74, 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} \[ \frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{9} b d^2 n x^3-\frac{1}{8} b d e n x^4-\frac{1}{25} b e^2 n x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d^2*n*x^3)/9 - (b*d*e*n*x^4)/8 - (b*e^2*n*x^5)/25 + ((10*d^2*x^3 + 15*d*e*x^4 + 6*e^2*x^5)*(a + b*Log[c*x^
n]))/30

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)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{1}{30} x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right ) \, dx\\ &=\frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{30} (b n) \int x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right ) \, dx\\ &=\frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{30} (b n) \int \left (10 d^2 x^2+15 d e x^3+6 e^2 x^4\right ) \, dx\\ &=-\frac{1}{9} b d^2 n x^3-\frac{1}{8} b d e n x^4-\frac{1}{25} b e^2 n x^5+\frac{1}{30} \left (10 d^2 x^3+15 d e x^4+6 e^2 x^5\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.039512, size = 81, normalized size = 1.09 \[ \frac{x^3 \left (60 a \left (10 d^2+15 d e x+6 e^2 x^2\right )+60 b \left (10 d^2+15 d e x+6 e^2 x^2\right ) \log \left (c x^n\right )-b n \left (200 d^2+225 d e x+72 e^2 x^2\right )\right )}{1800} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(60*a*(10*d^2 + 15*d*e*x + 6*e^2*x^2) - b*n*(200*d^2 + 225*d*e*x + 72*e^2*x^2) + 60*b*(10*d^2 + 15*d*e*x
+ 6*e^2*x^2)*Log[c*x^n]))/1800

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Maple [C]  time = 0.21, size = 432, normalized size = 5.8 \begin{align*}{\frac{b{x}^{3} \left ( 6\,{e}^{2}{x}^{2}+15\,dex+10\,{d}^{2} \right ) \ln \left ({x}^{n} \right ) }{30}}-{\frac{i}{6}}\pi \,b{d}^{2}{x}^{3}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{6}}\pi \,b{d}^{2}{x}^{3}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{6}}\pi \,b{d}^{2}{x}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{10}}\pi \,b{e}^{2}{x}^{5}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}+{\frac{\ln \left ( c \right ) b{e}^{2}{x}^{5}}{5}}-{\frac{b{e}^{2}n{x}^{5}}{25}}+{\frac{a{e}^{2}{x}^{5}}{5}}+{\frac{i}{10}}\pi \,b{e}^{2}{x}^{5} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{4}}\pi \,bde{x}^{4} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{6}}\pi \,b{d}^{2}{x}^{3} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -{\frac{i}{10}}\pi \,b{e}^{2}{x}^{5} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{\ln \left ( c \right ) bde{x}^{4}}{2}}-{\frac{bden{x}^{4}}{8}}+{\frac{ade{x}^{4}}{2}}-{\frac{i}{4}}\pi \,bde{x}^{4} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-{\frac{i}{4}}\pi \,bde{x}^{4}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{10}}\pi \,b{e}^{2}{x}^{5}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{4}}\pi \,bde{x}^{4}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}+{\frac{\ln \left ( c \right ) b{d}^{2}{x}^{3}}{3}}-{\frac{b{d}^{2}n{x}^{3}}{9}}+{\frac{a{d}^{2}{x}^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*b*x^3*(6*e^2*x^2+15*d*e*x+10*d^2)*ln(x^n)-1/6*I*Pi*b*d^2*x^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/6*I*Pi
*b*d^2*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2-1/6*I*Pi*b*d^2*x^3*csgn(I*c*x^n)^3+1/10*I*Pi*b*e^2*x^5*csgn(I*x^n)*csgn
(I*c*x^n)^2+1/5*ln(c)*b*e^2*x^5-1/25*b*e^2*n*x^5+1/5*a*e^2*x^5+1/10*I*Pi*b*e^2*x^5*csgn(I*c*x^n)^2*csgn(I*c)+1
/4*I*Pi*b*d*e*x^4*csgn(I*c*x^n)^2*csgn(I*c)+1/6*I*Pi*b*d^2*x^3*csgn(I*c*x^n)^2*csgn(I*c)-1/10*I*Pi*b*e^2*x^5*c
sgn(I*c*x^n)^3+1/2*ln(c)*b*d*e*x^4-1/8*b*d*e*n*x^4+1/2*a*d*e*x^4-1/4*I*Pi*b*d*e*x^4*csgn(I*c*x^n)^3-1/4*I*Pi*b
*d*e*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/10*I*Pi*b*e^2*x^5*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*I*Pi*
b*d*e*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*ln(c)*b*d^2*x^3-1/9*b*d^2*n*x^3+1/3*a*d^2*x^3

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Maxima [A]  time = 1.22446, size = 135, normalized size = 1.82 \begin{align*} -\frac{1}{25} \, b e^{2} n x^{5} + \frac{1}{5} \, b e^{2} x^{5} \log \left (c x^{n}\right ) - \frac{1}{8} \, b d e n x^{4} + \frac{1}{5} \, a e^{2} x^{5} + \frac{1}{2} \, b d e x^{4} \log \left (c x^{n}\right ) - \frac{1}{9} \, b d^{2} n x^{3} + \frac{1}{2} \, a d e x^{4} + \frac{1}{3} \, b d^{2} x^{3} \log \left (c x^{n}\right ) + \frac{1}{3} \, a d^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/25*b*e^2*n*x^5 + 1/5*b*e^2*x^5*log(c*x^n) - 1/8*b*d*e*n*x^4 + 1/5*a*e^2*x^5 + 1/2*b*d*e*x^4*log(c*x^n) - 1/
9*b*d^2*n*x^3 + 1/2*a*d*e*x^4 + 1/3*b*d^2*x^3*log(c*x^n) + 1/3*a*d^2*x^3

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Fricas [A]  time = 0.993726, size = 288, normalized size = 3.89 \begin{align*} -\frac{1}{25} \,{\left (b e^{2} n - 5 \, a e^{2}\right )} x^{5} - \frac{1}{8} \,{\left (b d e n - 4 \, a d e\right )} x^{4} - \frac{1}{9} \,{\left (b d^{2} n - 3 \, a d^{2}\right )} x^{3} + \frac{1}{30} \,{\left (6 \, b e^{2} x^{5} + 15 \, b d e x^{4} + 10 \, b d^{2} x^{3}\right )} \log \left (c\right ) + \frac{1}{30} \,{\left (6 \, b e^{2} n x^{5} + 15 \, b d e n x^{4} + 10 \, b d^{2} 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)^2*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

-1/25*(b*e^2*n - 5*a*e^2)*x^5 - 1/8*(b*d*e*n - 4*a*d*e)*x^4 - 1/9*(b*d^2*n - 3*a*d^2)*x^3 + 1/30*(6*b*e^2*x^5
+ 15*b*d*e*x^4 + 10*b*d^2*x^3)*log(c) + 1/30*(6*b*e^2*n*x^5 + 15*b*d*e*n*x^4 + 10*b*d^2*n*x^3)*log(x)

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Sympy [B]  time = 21.0525, size = 151, normalized size = 2.04 \begin{align*} \frac{a d^{2} x^{3}}{3} + \frac{a d e x^{4}}{2} + \frac{a e^{2} x^{5}}{5} + \frac{b d^{2} n x^{3} \log{\left (x \right )}}{3} - \frac{b d^{2} n x^{3}}{9} + \frac{b d^{2} x^{3} \log{\left (c \right )}}{3} + \frac{b d e n x^{4} \log{\left (x \right )}}{2} - \frac{b d e n x^{4}}{8} + \frac{b d e x^{4} \log{\left (c \right )}}{2} + \frac{b e^{2} n x^{5} \log{\left (x \right )}}{5} - \frac{b e^{2} n x^{5}}{25} + \frac{b e^{2} x^{5} \log{\left (c \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**2*x**3/3 + a*d*e*x**4/2 + a*e**2*x**5/5 + b*d**2*n*x**3*log(x)/3 - b*d**2*n*x**3/9 + b*d**2*x**3*log(c)/3
+ b*d*e*n*x**4*log(x)/2 - b*d*e*n*x**4/8 + b*d*e*x**4*log(c)/2 + b*e**2*n*x**5*log(x)/5 - b*e**2*n*x**5/25 +
b*e**2*x**5*log(c)/5

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Giac [A]  time = 1.33896, size = 166, normalized size = 2.24 \begin{align*} \frac{1}{5} \, b n x^{5} e^{2} \log \left (x\right ) + \frac{1}{2} \, b d n x^{4} e \log \left (x\right ) - \frac{1}{25} \, b n x^{5} e^{2} - \frac{1}{8} \, b d n x^{4} e + \frac{1}{5} \, b x^{5} e^{2} \log \left (c\right ) + \frac{1}{2} \, b d x^{4} e \log \left (c\right ) + \frac{1}{3} \, b d^{2} n x^{3} \log \left (x\right ) - \frac{1}{9} \, b d^{2} n x^{3} + \frac{1}{5} \, a x^{5} e^{2} + \frac{1}{2} \, a d x^{4} e + \frac{1}{3} \, b d^{2} x^{3} \log \left (c\right ) + \frac{1}{3} \, a d^{2} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b*n*x^5*e^2*log(x) + 1/2*b*d*n*x^4*e*log(x) - 1/25*b*n*x^5*e^2 - 1/8*b*d*n*x^4*e + 1/5*b*x^5*e^2*log(c) +
1/2*b*d*x^4*e*log(c) + 1/3*b*d^2*n*x^3*log(x) - 1/9*b*d^2*n*x^3 + 1/5*a*x^5*e^2 + 1/2*a*d*x^4*e + 1/3*b*d^2*x^
3*log(c) + 1/3*a*d^2*x^3