3.2.86 \(\int \frac {(d+e x^2)^2 (a+b \log (c x^n))}{x} \, dx\) [186]

Optimal. Leaf size=89 \[ -\frac {1}{2} b d e n x^2-\frac {1}{16} b e^2 n x^4-\frac {1}{2} b d^2 n \log ^2(x)+d e x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+d^2 \log (x) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {272, 45, 2372, 2338} \begin {gather*} d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+d e x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^2 x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b d^2 n \log ^2(x)-\frac {1}{2} b d e n x^2-\frac {1}{16} b e^2 n x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*d*e*n*x^2) - (b*e^2*n*x^4)/16 - (b*d^2*n*Log[x]^2)/2 + d*e*x^2*(a + b*Log[c*x^n]) + (e^2*x^4*(a + b*Lo
g[c*x^n]))/4 + d^2*Log[x]*(a + b*Log[c*x^n])

Rule 45

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 272

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 2338

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]

Rule 2372

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

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Mathematica [A]
time = 0.04, size = 82, normalized size = 0.92 \begin {gather*} \frac {1}{16} \left (-8 b d e n x^2-b e^2 n x^4+16 d e x^2 \left (a+b \log \left (c x^n\right )\right )+4 e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {8 d^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.25, size = 3072, normalized size = 34.52

method	result	size
risch	\(\text {Expression too large to display}\)	\(3072\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/4*x^4*b*e^2+b*d*e*x^2+b*d^2+b*d^2*ln(x))*ln(x^n)+1/16*(16*x^4*a^2*e^2+32*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*x^
n)*csgn(I*c*x^n)+24*I*Pi*b^2*d*e*n*x^2*csgn(I*c*x^n)^3-64*I*Pi*a*b*d*e*x^2*csgn(I*c*x^n)^3-64*I*ln(c)*Pi*b^2*d
*e*x^2*csgn(I*c*x^n)^3-64*I*ln(c)*Pi*b^2*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-16*I*Pi*ln(c)*b^2*e^2*x^4*csg
n(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+4*I*Pi*b^2*e^2*n*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-16*I*Pi*a*b*e^2*x^4*
csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-64*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+64*I*ln(x)*ln(c)*Pi*b^
2*d^2*csgn(I*c)*csgn(I*c*x^n)^2-40*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*c*x^n)^2+64*I*ln(x)*Pi*a*b*d^2*csgn(I
*c)*csgn(I*c*x^n)^2+64*I*ln(x)*ln(c)*Pi*b^2*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+16*I*Pi*a*b*e^2*x^4*csgn(I*x^n)*cs
gn(I*c*x^n)^2+16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-40*I*ln(x)*Pi*b^2*d^2*n*csgn(I*x^n)*csgn(I
*c*x^n)^2+64*I*ln(x)*Pi*a*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-4*I*Pi*b^2*e^2*n*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+6
4*a^2*d*e*x^2+64*a^2*d^2*ln(x)-64*b^2*d^2*ln(c)*n+16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^2-16*Pi^2*
b^2*d*e*x^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+32*Pi^2*b^2*d*e*x^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n
)^3-16*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*c*x^n)^4+32*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^5+32*Pi^2*b^2*d*e*x^2*
csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-64*Pi^2*b^2*d*e*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-4*I*Pi*b^2*e
^2*n*x^4*csgn(I*c)*csgn(I*c*x^n)^2-16*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*c*x^n)^2*ln(x)^2-16*I*Pi*b^2*d^2*n*csgn(
I*x^n)*csgn(I*c*x^n)^2*ln(x)^2+64*d^2*b^2*ln(c)^2+64*a^2*d^2+16*I*Pi*a*b*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^2-64*
I*ln(x)*ln(c)*Pi*b^2*d^2*csgn(I*c*x^n)^3+40*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c*x^n)^3+16*ln(x)*b^2*d^2*n^2+128*d^2*
a*b*ln(c)-16*Pi^2*b^2*d^2*csgn(I*c*x^n)^6-64*I*ln(c)*Pi*b^2*d^2*csgn(I*c*x^n)^3+8*b^2*d^2*n^2*ln(x)^2+16*b^2*d
^2*n^2+64*ln(c)^2*b^2*d*e*x^2-80*ln(x)*a*b*d^2*n-64*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4+4*I*Pi*
b^2*e^2*n*x^4*csgn(I*c*x^n)^3+b^2*e^2*n^2*x^4-64*b*d^2*n*a+32*Pi^2*b^2*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^5+64*
ln(x)*ln(c)^2*b^2*d^2-80*ln(x)*ln(c)*b^2*d^2*n+16*ln(c)^2*b^2*e^2*x^4+64*I*Pi*a*b*d*e*x^2*csgn(I*c)*csgn(I*c*x
^n)^2+64*I*Pi*a*b*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+8*Pi^2*b^2*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^5+128*ln(x)*l
n(c)*a*b*d^2-32*I*Pi*b^2*d^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2-48*b*n*a*d*e*x^2-4*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csg
n(I*x^n)^2*csgn(I*c*x^n)^2+8*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+8*Pi^2*b^2*e^2*x^4*csgn(
I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-16*Pi^2*b^2*e^2*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-16*ln(x)*Pi^2*b^2
*d^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2-16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*c*x^n)^3-32*ln(x)^2*ln(c)*b^2*
d^2*n-32*ln(x)^2*b*d^2*n*a-8*a*b*e^2*n*x^4+64*I*ln(c)*Pi*b^2*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+64*I*Pi*ln(c)
*b^2*d*e*x^2*csgn(I*c)*csgn(I*c*x^n)^2-24*I*Pi*b^2*d*e*n*x^2*csgn(I*c)*csgn(I*c*x^n)^2+32*I*Pi*b^2*d^2*n*csgn(
I*c*x^n)^3-64*I*Pi*a*b*d^2*csgn(I*c*x^n)^3-16*ln(x)*Pi^2*b^2*d^2*csgn(I*c*x^n)^6-64*I*ln(x)*Pi*a*b*d^2*csgn(I*
c)*csgn(I*x^n)*csgn(I*c*x^n)+16*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*ln(x)^2-4*Pi^2*b^2*e^2*x^4*
csgn(I*x^n)^2*csgn(I*c*x^n)^4-16*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+32*Pi^2*b^2*d^2*csgn(I
*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+32*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-24*I*Pi*b^2*d*e*n*x^
2*csgn(I*x^n)*csgn(I*c*x^n)^2-4*Pi^2*b^2*e^2*x^4*csgn(I*c*x^n)^6+8*b^2*d*e*n^2*x^2-64*I*ln(c)*Pi*b^2*d*e*x^2*c
sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+64*I*ln(c)*Pi*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^2+8*Pi^2*b^2*e^2*x^4*csgn(I*x
^n)*csgn(I*c*x^n)^5-4*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csgn(I*c*x^n)^4+64*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2-1
6*Pi^2*b^2*d*e*x^2*csgn(I*c*x^n)^6-16*ln(x)*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*c*x^n)^4+32*ln(x)*Pi^2*b^2*d^2*csg
n(I*c)*csgn(I*c*x^n)^5-16*ln(x)*Pi^2*b^2*d^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+32*ln(x)*Pi^2*b^2*d^2*csgn(I*x^n)*c
sgn(I*c*x^n)^5-16*I*Pi*a*b*e^2*x^4*csgn(I*c*x^n)^3-16*Pi^2*b^2*d*e*x^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4-16*Pi^2*b
^2*d^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+32*Pi^2*b^2*d^2*csgn(I*x^n)*csgn(I*c*x^n)^5-64*I*ln(x)*ln(c)*Pi*b^2*d^2*c
sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-48*ln(c)*b^2*d*e*n*x^2+128*ln(c)*a*b*d*e*x^2-16*Pi^2*b^2*d*e*x^2*csgn(I*c)^
2*csgn(I*c*x^n)^4+32*Pi^2*b^2*d*e*x^2*csgn(I*c)*csgn(I*c*x^n)^5+40*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*x^n)*
csgn(I*c*x^n)+32*ln(x)*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+32*ln(x)*Pi^2*b^2*d^2*csgn(I*c)*cs
gn(I*x^n)^2*csgn(I*c*x^n)^3-64*ln(x)*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4+16*I*Pi*b^2*d^2*n*csgn
(I*c*x^n)^3*ln(x)^2+64*I*Pi*a*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+24*I*Pi*b^2*d*e*n*x^2*csgn(I*c)*csgn(I*x^n)*cs
gn(I*c*x^n)-64*I*Pi*a*b*d*e*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-32*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*c*x^n)^
2+64*I*ln(c)*Pi*b^2*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-8*ln(c)*b^2*e^2*n*x^4+32*ln(c)*a*b*e^2*x^4-64*I*ln(x)*Pi*a
*b*d^2*csgn(I*c*x^n)^3)/(-2*I*b*Pi*csgn(I*c)*cs...

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Maxima [A]
time = 0.29, size = 88, normalized size = 0.99 \begin {gather*} -\frac {1}{16} \, b n x^{4} e^{2} + \frac {1}{4} \, b x^{4} e^{2} \log \left (c x^{n}\right ) + \frac {1}{4} \, a x^{4} e^{2} - \frac {1}{2} \, b d n x^{2} e + b d x^{2} e \log \left (c x^{n}\right ) + a d x^{2} e + \frac {b d^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + a d^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 100, normalized size = 1.12 \begin {gather*} -\frac {1}{16} \, {\left (b n - 4 \, a\right )} x^{4} e^{2} + \frac {1}{2} \, b d^{2} n \log \left (x\right )^{2} - \frac {1}{2} \, {\left (b d n - 2 \, a d\right )} x^{2} e + \frac {1}{4} \, {\left (b x^{4} e^{2} + 4 \, b d x^{2} e\right )} \log \left (c\right ) + \frac {1}{4} \, {\left (b n x^{4} e^{2} + 4 \, b d n x^{2} e + 4 \, b d^{2} \log \left (c\right ) + 4 \, a d^{2}\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.70, size = 133, normalized size = 1.49 \begin {gather*} \begin {cases} \frac {a d^{2} \log {\left (c x^{n} \right )}}{n} + a d e x^{2} + \frac {a e^{2} x^{4}}{4} + \frac {b d^{2} \log {\left (c x^{n} \right )}^{2}}{2 n} - \frac {b d e n x^{2}}{2} + b d e x^{2} \log {\left (c x^{n} \right )} - \frac {b e^{2} n x^{4}}{16} + \frac {b e^{2} x^{4} \log {\left (c x^{n} \right )}}{4} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (d^{2} \log {\left (x \right )} + d e x^{2} + \frac {e^{2} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**2*log(c*x**n)/n + a*d*e*x**2 + a*e**2*x**4/4 + b*d**2*log(c*x**n)**2/(2*n) - b*d*e*n*x**2/2 +
b*d*e*x**2*log(c*x**n) - b*e**2*n*x**4/16 + b*e**2*x**4*log(c*x**n)/4, Ne(n, 0)), ((a + b*log(c))*(d**2*log(x)
+ d*e*x**2 + e**2*x**4/4), True))

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Giac [A]
time = 4.01, size = 105, normalized size = 1.18 \begin {gather*} \frac {1}{4} \, b n x^{4} e^{2} \log \left (x\right ) - \frac {1}{16} \, b n x^{4} e^{2} + \frac {1}{4} \, b x^{4} e^{2} \log \left (c\right ) + b d n x^{2} e \log \left (x\right ) + \frac {1}{4} \, a x^{4} e^{2} - \frac {1}{2} \, b d n x^{2} e + b d x^{2} e \log \left (c\right ) + \frac {1}{2} \, b d^{2} n \log \left (x\right )^{2} + a d x^{2} e + b d^{2} \log \left (c\right ) \log \left (x\right ) + a d^{2} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.66, size = 80, normalized size = 0.90 \begin {gather*} \ln \left (c\,x^n\right )\,\left (\frac {b\,e^2\,x^4}{4}+b\,d\,e\,x^2\right )+\frac {e^2\,x^4\,\left (4\,a-b\,n\right )}{16}+a\,d^2\,\ln \left (x\right )+\frac {b\,d^2\,{\ln \left (c\,x^n\right )}^2}{2\,n}+\frac {d\,e\,x^2\,\left (2\,a-b\,n\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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