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

Optimal. Leaf size=131 \[ 3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac{3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{3}{2} b d^2 e n \log ^2(x)-\frac{b d^3 n}{4 x^2}-\frac{3}{4} b d e^2 n x^2-\frac{1}{16} b e^3 n x^4 \]

[Out]

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

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Rubi [A]  time = 0.124834, antiderivative size = 100, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {266, 43, 2334, 12, 14, 2301} \[ -\frac{1}{4} \left (-12 d^2 e \log (x)+\frac{2 d^3}{x^2}-6 d e^2 x^2-e^3 x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{3}{2} b d^2 e n \log ^2(x)-\frac{b d^3 n}{4 x^2}-\frac{3}{4} b d e^2 n x^2-\frac{1}{16} b e^3 n x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.081901, size = 115, normalized size = 0.88 \[ \frac{1}{16} \left (\frac{24 d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac{8 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^2}+24 d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+4 e^3 x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{4 b d^3 n}{x^2}-12 b d e^2 n x^2-b e^3 n x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*d^3*n)/x^2 - 12*b*d*e^2*n*x^2 - b*e^3*n*x^4 - (8*d^3*(a + b*Log[c*x^n]))/x^2 + 24*d*e^2*x^2*(a + b*Log[
c*x^n]) + 4*e^3*x^4*(a + b*Log[c*x^n]) + (24*d^2*e*(a + b*Log[c*x^n])^2)/(b*n))/16

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Maple [C]  time = 0.242, size = 604, 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((e*x^2+d)^3*(a+b*ln(c*x^n))/x^3,x)

[Out]

-1/4*b*(-e^3*x^6-6*d*e^2*x^4-12*d^2*e*ln(x)*x^2+2*d^3)/x^2*ln(x^n)-1/16*(-24*ln(c)*b*d*e^2*x^4+8*a*d^3+24*I*ln
(x)*Pi*b*d^2*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^2-24*I*ln(x)*Pi*b*d^2*e*csgn(I*c*x^n)^2*csgn(I*c)*x^2+24*
I*ln(x)*Pi*b*d^2*e*csgn(I*c*x^n)^3*x^2-12*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^2*csgn(I*c)+2*I*Pi*b*e^3*x^6*csgn(I*x
^n)*csgn(I*c*x^n)*csgn(I*c)-24*a*d*e^2*x^4+8*ln(c)*b*d^3-4*ln(c)*b*e^3*x^6-48*ln(x)*a*d^2*e*x^2+12*I*Pi*b*d*e^
2*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-4*I*Pi*b*d^3*csgn(I*c*x^n)^3-4*a*e^3*x^6-24*I*ln(x)*Pi*b*d^2*e*csgn(
I*x^n)*csgn(I*c*x^n)^2*x^2+4*b*d^3*n+4*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2-12*I*Pi*b*d*e^2*x^4*csgn(I*x^n)*
csgn(I*c*x^n)^2-48*ln(x)*ln(c)*b*d^2*e*x^2+24*b*d^2*e*n*ln(x)^2*x^2+4*I*Pi*b*d^3*csgn(I*c*x^n)^2*csgn(I*c)+2*I
*Pi*b*e^3*x^6*csgn(I*c*x^n)^3+12*I*Pi*b*d*e^2*x^4*csgn(I*c*x^n)^3-2*I*Pi*b*e^3*x^6*csgn(I*x^n)*csgn(I*c*x^n)^2
+b*e^3*n*x^6+12*b*d*e^2*n*x^4-4*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-2*I*Pi*b*e^3*x^6*csgn(I*c*x^n)^
2*csgn(I*c))/x^2

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Maxima [A]  time = 1.05643, size = 180, normalized size = 1.37 \begin{align*} -\frac{1}{16} \, b e^{3} n x^{4} + \frac{1}{4} \, b e^{3} x^{4} \log \left (c x^{n}\right ) + \frac{1}{4} \, a e^{3} x^{4} - \frac{3}{4} \, b d e^{2} n x^{2} + \frac{3}{2} \, b d e^{2} x^{2} \log \left (c x^{n}\right ) + \frac{3}{2} \, a d e^{2} x^{2} + \frac{3 \, b d^{2} e \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d^{2} e \log \left (x\right ) - \frac{b d^{3} n}{4 \, x^{2}} - \frac{b d^{3} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac{a d^{3}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.55742, size = 356, normalized size = 2.72 \begin{align*} \frac{24 \, b d^{2} e n x^{2} \log \left (x\right )^{2} -{\left (b e^{3} n - 4 \, a e^{3}\right )} x^{6} - 4 \, b d^{3} n - 12 \,{\left (b d e^{2} n - 2 \, a d e^{2}\right )} x^{4} - 8 \, a d^{3} + 4 \,{\left (b e^{3} x^{6} + 6 \, b d e^{2} x^{4} - 2 \, b d^{3}\right )} \log \left (c\right ) + 4 \,{\left (b e^{3} n x^{6} + 6 \, b d e^{2} n x^{4} + 12 \, b d^{2} e x^{2} \log \left (c\right ) + 12 \, a d^{2} e x^{2} - 2 \, b d^{3} n\right )} \log \left (x\right )}{16 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(24*b*d^2*e*n*x^2*log(x)^2 - (b*e^3*n - 4*a*e^3)*x^6 - 4*b*d^3*n - 12*(b*d*e^2*n - 2*a*d*e^2)*x^4 - 8*a*d
^3 + 4*(b*e^3*x^6 + 6*b*d*e^2*x^4 - 2*b*d^3)*log(c) + 4*(b*e^3*n*x^6 + 6*b*d*e^2*n*x^4 + 12*b*d^2*e*x^2*log(c)
+ 12*a*d^2*e*x^2 - 2*b*d^3*n)*log(x))/x^2

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Sympy [A]  time = 9.21979, size = 209, normalized size = 1.6 \begin{align*} - \frac{a d^{3}}{2 x^{2}} + 3 a d^{2} e \log{\left (x \right )} + \frac{3 a d e^{2} x^{2}}{2} + \frac{a e^{3} x^{4}}{4} - \frac{b d^{3} n \log{\left (x \right )}}{2 x^{2}} - \frac{b d^{3} n}{4 x^{2}} - \frac{b d^{3} \log{\left (c \right )}}{2 x^{2}} + \frac{3 b d^{2} e n \log{\left (x \right )}^{2}}{2} + 3 b d^{2} e \log{\left (c \right )} \log{\left (x \right )} + \frac{3 b d e^{2} n x^{2} \log{\left (x \right )}}{2} - \frac{3 b d e^{2} n x^{2}}{4} + \frac{3 b d e^{2} x^{2} \log{\left (c \right )}}{2} + \frac{b e^{3} n x^{4} \log{\left (x \right )}}{4} - \frac{b e^{3} n x^{4}}{16} + \frac{b e^{3} x^{4} \log{\left (c \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28888, size = 216, normalized size = 1.65 \begin{align*} \frac{4 \, b n x^{6} e^{3} \log \left (x\right ) - b n x^{6} e^{3} + 4 \, b x^{6} e^{3} \log \left (c\right ) + 24 \, b d n x^{4} e^{2} \log \left (x\right ) + 24 \, b d^{2} n x^{2} e \log \left (x\right )^{2} + 4 \, a x^{6} e^{3} - 12 \, b d n x^{4} e^{2} + 24 \, b d x^{4} e^{2} \log \left (c\right ) + 48 \, b d^{2} x^{2} e \log \left (c\right ) \log \left (x\right ) + 24 \, a d x^{4} e^{2} + 48 \, a d^{2} x^{2} e \log \left (x\right ) - 8 \, b d^{3} n \log \left (x\right ) - 4 \, b d^{3} n - 8 \, b d^{3} \log \left (c\right ) - 8 \, a d^{3}}{16 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*b*n*x^6*e^3*log(x) - b*n*x^6*e^3 + 4*b*x^6*e^3*log(c) + 24*b*d*n*x^4*e^2*log(x) + 24*b*d^2*n*x^2*e*log
(x)^2 + 4*a*x^6*e^3 - 12*b*d*n*x^4*e^2 + 24*b*d*x^4*e^2*log(c) + 48*b*d^2*x^2*e*log(c)*log(x) + 24*a*d*x^4*e^2
+ 48*a*d^2*x^2*e*log(x) - 8*b*d^3*n*log(x) - 4*b*d^3*n - 8*b*d^3*log(c) - 8*a*d^3)/x^2