3.1.88 \(\int \frac {(d+e x)^2 (a+b \log (c x^n))^2}{x^2} \, dx\) [88]

Optimal. Leaf size=133 \[ -\frac {2 b^2 d^2 n^2}{x}-2 a b e^2 n x+2 b^2 e^2 n^2 x-2 b^2 e^2 n x \log \left (c x^n\right )-\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{x}+e^2 x \left (a+b \log \left (c x^n\right )\right )^2+\frac {2 d e \left (a+b \log \left (c x^n\right )\right )^3}{3 b n} \]

[Out]

-2*b^2*d^2*n^2/x-2*a*b*e^2*n*x+2*b^2*e^2*n^2*x-2*b^2*e^2*n*x*ln(c*x^n)-2*b*d^2*n*(a+b*ln(c*x^n))/x-d^2*(a+b*ln
(c*x^n))^2/x+e^2*x*(a+b*ln(c*x^n))^2+2/3*d*e*(a+b*ln(c*x^n))^3/b/n

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Rubi [A]
time = 0.12, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2395, 2333, 2332, 2342, 2341, 2339, 30} \begin {gather*} -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{x}-\frac {2 b d^2 n \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {2 d e \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}+e^2 x \left (a+b \log \left (c x^n\right )\right )^2-2 a b e^2 n x-2 b^2 e^2 n x \log \left (c x^n\right )-\frac {2 b^2 d^2 n^2}{x}+2 b^2 e^2 n^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

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

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 107, normalized size = 0.80 \begin {gather*} -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )^2}{x}+e^2 x \left (a+b \log \left (c x^n\right )\right )^2+\frac {2 d e \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-2 b e^2 n x \left (a-b n+b \log \left (c x^n\right )\right )-\frac {2 b d^2 n \left (a+b n+b \log \left (c x^n\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.28, size = 2521, normalized size = 18.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(2521\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*(-2*d*e*x*ln(x)-e^2*x^2+d^2)/x*ln(x^n)^2-b*(I*Pi*b*e^2*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2*I*ln(x)*
Pi*b*d*e*csgn(I*c*x^n)^3*x-I*Pi*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*Pi*b*e^2*x^2*csgn(I*x^n)*csgn(I*c*
x^n)^2+I*Pi*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2-I*Pi*b*d^2*csgn(I*c*x^n)^3-I*Pi*b*e^2*x^2*csgn(I*c)*csgn(I*c*x^n)^
2+2*I*ln(x)*Pi*b*d*e*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x-2*I*ln(x)*Pi*b*d*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x+I*
Pi*b*e^2*x^2*csgn(I*c*x^n)^3-2*I*ln(x)*Pi*b*d*e*csgn(I*c)*csgn(I*c*x^n)^2*x+I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^
n)^2+2*b*d*e*n*ln(x)^2*x-4*ln(x)*ln(c)*b*d*e*x-2*ln(c)*b*e^2*x^2+2*b*e^2*n*x^2-4*ln(x)*a*d*e*x-2*a*e^2*x^2+2*d
^2*b*ln(c)+2*b*d^2*n+2*a*d^2)/x*ln(x^n)+1/12*(12*a^2*e^2*x^2-24*I*ln(x)*Pi*a*b*d*e*csgn(I*c*x^n)^3*x+24*ln(x)*
ln(c)^2*b^2*d*e*x+8*e*d*b^2*n^2*ln(x)^3*x-24*b^2*d^2*ln(c)*n+12*I*Pi*b^2*d*e*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*
x^n)*ln(x)^2*x-24*I*ln(x)*Pi*a*b*d*e*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x-24*I*ln(x)*Pi*ln(c)*b^2*d*e*csgn(I*
c)*csgn(I*x^n)*csgn(I*c*x^n)*x-12*I*Pi*b^2*d^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2+12*I*n*Pi*b^2*e^2*x^2*csgn(I*c*x^
n)^3-12*d^2*b^2*ln(c)^2-12*a^2*d^2+3*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2-6*Pi^2*b^2*d^2*csg
n(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3-6*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3+12*Pi^2*b^2*d^2*cs
gn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-24*d^2*a*b*ln(c)-12*I*Pi*b^2*d*e*n*csgn(I*c)*csgn(I*c*x^n)^2*ln(x)^2*x-12*
I*Pi*b^2*d*e*n*csgn(I*x^n)*csgn(I*c*x^n)^2*ln(x)^2*x-24*b^2*d^2*n^2+3*Pi^2*b^2*d^2*csgn(I*c*x^n)^6+24*b^2*e^2*
n^2*x^2-24*b*d^2*n*a+12*ln(c)^2*b^2*e^2*x^2-6*Pi^2*b^2*d^2*csgn(I*x^n)*csgn(I*c*x^n)^5+3*Pi^2*b^2*d^2*csgn(I*c
)^2*csgn(I*c*x^n)^4-6*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^5+3*Pi^2*b^2*d^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4-12*I
*Pi*a*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-6*ln(x)*Pi^2*b^2*d*e*csgn(I*c*x^n)^6*x-12*I*ln(c)*Pi*b^2*e^2*x^2*csgn(
I*c)*csgn(I*x^n)*csgn(I*c*x^n)+12*I*Pi*b^2*d*e*n*csgn(I*c*x^n)^3*ln(x)^2*x-24*I*ln(x)*Pi*ln(c)*b^2*d*e*csgn(I*
c*x^n)^3*x+24*ln(x)*e*d*a^2*x-6*ln(x)*Pi^2*b^2*d*e*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2*x+12*ln(x)*Pi^2*b
^2*d*e*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3*x+12*ln(x)*Pi^2*b^2*d*e*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3
*x-24*ln(x)*Pi^2*b^2*d*e*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4*x-12*I*Pi*a*b*e^2*x^2*csgn(I*c)*csgn(I*x^n)*csg
n(I*c*x^n)-24*n*ln(c)*b^2*e^2*x^2+24*ln(c)*a*b*e^2*x^2-12*I*ln(c)*Pi*b^2*e^2*x^2*csgn(I*c*x^n)^3-12*I*Pi*a*b*e
^2*x^2*csgn(I*c*x^n)^3-12*I*Pi*ln(c)*b^2*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-12*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*c*x^
n)^2-3*Pi^2*b^2*e^2*x^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+6*Pi^2*b^2*e^2*x^2*csgn(I*c)^2*csgn(I*x^n)*c
sgn(I*c*x^n)^3-3*Pi^2*b^2*e^2*x^2*csgn(I*c)^2*csgn(I*c*x^n)^4+6*Pi^2*b^2*e^2*x^2*csgn(I*c)*csgn(I*c*x^n)^5-6*l
n(x)*Pi^2*b^2*d*e*csgn(I*c)^2*csgn(I*c*x^n)^4*x+12*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+12*I*Pi*
ln(c)*b^2*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-12*I*n*Pi*b^2*e^2*x^2*csgn(I*c)*csgn(I*c*x^n)^2-12*I*Pi*ln(c
)*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^2-12*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*c*x^n)^2+12*I*Pi*ln(c)*b^2*d^2*csgn(I*c
*x^n)^3+12*I*Pi*a*b*d^2*csgn(I*c*x^n)^3+12*I*Pi*b^2*d^2*n*csgn(I*c*x^n)^3+24*I*ln(x)*Pi*ln(c)*b^2*d*e*csgn(I*c
)*csgn(I*c*x^n)^2*x+12*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+12*I*ln(c)*Pi*b^2*e^2*x^2*csgn(I*c)*cs
gn(I*c*x^n)^2+24*I*ln(x)*Pi*a*b*d*e*csgn(I*c)*csgn(I*c*x^n)^2*x+24*I*ln(x)*Pi*a*b*d*e*csgn(I*x^n)*csgn(I*c*x^n
)^2*x+12*I*n*Pi*b^2*e^2*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-3*Pi^2*b^2*e^2*x^2*csgn(I*x^n)^2*csgn(I*c*x^n)
^4+6*Pi^2*b^2*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^5+24*I*ln(x)*Pi*ln(c)*b^2*d*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x+6*
Pi^2*b^2*e^2*x^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-24*b*n*x^2*a*e^2-3*Pi^2*b^2*e^2*x^2*csgn(I*c*x^n)^6+1
2*I*Pi*a*b*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+12*ln(x)*Pi^2*b^2*d*e*csgn(I*c)*csgn(I*c*x^n)^5*x-6*ln(x)*Pi^2*
b^2*d*e*csgn(I*x^n)^2*csgn(I*c*x^n)^4*x+12*ln(x)*Pi^2*b^2*d*e*csgn(I*x^n)*csgn(I*c*x^n)^5*x-12*I*n*Pi*b^2*e^2*
x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+12*I*ln(c)*Pi*b^2*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+12*I*Pi*a*b*e^2*x^2*csgn
(I*c)*csgn(I*c*x^n)^2-12*Pi^2*b^2*e^2*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4+48*ln(x)*ln(c)*a*b*d*e*x-24*a*
b*d*e*n*ln(x)^2*x-24*ln(c)*b^2*d*e*n*ln(x)^2*x)/x

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Maxima [A]
time = 0.32, size = 198, normalized size = 1.49 \begin {gather*} b^{2} x e^{2} \log \left (c x^{n}\right )^{2} + \frac {2 \, b^{2} d e \log \left (c x^{n}\right )^{3}}{3 \, n} - 2 \, b^{2} d^{2} {\left (\frac {n^{2}}{x} + \frac {n \log \left (c x^{n}\right )}{x}\right )} - 2 \, a b n x e^{2} + 2 \, a b x e^{2} \log \left (c x^{n}\right ) - \frac {b^{2} d^{2} \log \left (c x^{n}\right )^{2}}{x} + \frac {2 \, a b d e \log \left (c x^{n}\right )^{2}}{n} + 2 \, a^{2} d e \log \left (x\right ) - \frac {2 \, a b d^{2} n}{x} + 2 \, {\left (n^{2} x - n x \log \left (c x^{n}\right )\right )} b^{2} e^{2} + a^{2} x e^{2} - \frac {2 \, a b d^{2} \log \left (c x^{n}\right )}{x} - \frac {a^{2} d^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (128) = 256\).
time = 0.36, size = 278, normalized size = 2.09 \begin {gather*} \frac {2 \, b^{2} d n^{2} x e \log \left (x\right )^{3} - 6 \, b^{2} d^{2} n^{2} - 6 \, a b d^{2} n - 3 \, a^{2} d^{2} + 3 \, {\left (2 \, b^{2} n^{2} - 2 \, a b n + a^{2}\right )} x^{2} e^{2} + 3 \, {\left (b^{2} x^{2} e^{2} - b^{2} d^{2}\right )} \log \left (c\right )^{2} + 3 \, {\left (b^{2} n^{2} x^{2} e^{2} + 2 \, b^{2} d n x e \log \left (c\right ) - b^{2} d^{2} n^{2} + 2 \, a b d n x e\right )} \log \left (x\right )^{2} - 6 \, {\left (b^{2} d^{2} n + a b d^{2} + {\left (b^{2} n - a b\right )} x^{2} e^{2}\right )} \log \left (c\right ) + 6 \, {\left (b^{2} d x e \log \left (c\right )^{2} - b^{2} d^{2} n^{2} - a b d^{2} n + a^{2} d x e - {\left (b^{2} n^{2} - a b n\right )} x^{2} e^{2} + {\left (b^{2} n x^{2} e^{2} - b^{2} d^{2} n + 2 \, a b d x e\right )} \log \left (c\right )\right )} \log \left (x\right )}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b^2*d*n^2*x*e*log(x)^3 - 6*b^2*d^2*n^2 - 6*a*b*d^2*n - 3*a^2*d^2 + 3*(2*b^2*n^2 - 2*a*b*n + a^2)*x^2*e^
2 + 3*(b^2*x^2*e^2 - b^2*d^2)*log(c)^2 + 3*(b^2*n^2*x^2*e^2 + 2*b^2*d*n*x*e*log(c) - b^2*d^2*n^2 + 2*a*b*d*n*x
*e)*log(x)^2 - 6*(b^2*d^2*n + a*b*d^2 + (b^2*n - a*b)*x^2*e^2)*log(c) + 6*(b^2*d*x*e*log(c)^2 - b^2*d^2*n^2 -
a*b*d^2*n + a^2*d*x*e - (b^2*n^2 - a*b*n)*x^2*e^2 + (b^2*n*x^2*e^2 - b^2*d^2*n + 2*a*b*d*x*e)*log(c))*log(x))/
x

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Sympy [A]
time = 0.56, size = 255, normalized size = 1.92 \begin {gather*} \begin {cases} - \frac {a^{2} d^{2}}{x} + \frac {2 a^{2} d e \log {\left (c x^{n} \right )}}{n} + a^{2} e^{2} x - \frac {2 a b d^{2} n}{x} - \frac {2 a b d^{2} \log {\left (c x^{n} \right )}}{x} + \frac {2 a b d e \log {\left (c x^{n} \right )}^{2}}{n} - 2 a b e^{2} n x + 2 a b e^{2} x \log {\left (c x^{n} \right )} - \frac {2 b^{2} d^{2} n^{2}}{x} - \frac {2 b^{2} d^{2} n \log {\left (c x^{n} \right )}}{x} - \frac {b^{2} d^{2} \log {\left (c x^{n} \right )}^{2}}{x} + \frac {2 b^{2} d e \log {\left (c x^{n} \right )}^{3}}{3 n} + 2 b^{2} e^{2} n^{2} x - 2 b^{2} e^{2} n x \log {\left (c x^{n} \right )} + b^{2} e^{2} x \log {\left (c x^{n} \right )}^{2} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right )^{2} \left (- \frac {d^{2}}{x} + 2 d e \log {\left (x \right )} + e^{2} x\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*d**2/x + 2*a**2*d*e*log(c*x**n)/n + a**2*e**2*x - 2*a*b*d**2*n/x - 2*a*b*d**2*log(c*x**n)/x +
2*a*b*d*e*log(c*x**n)**2/n - 2*a*b*e**2*n*x + 2*a*b*e**2*x*log(c*x**n) - 2*b**2*d**2*n**2/x - 2*b**2*d**2*n*l
og(c*x**n)/x - b**2*d**2*log(c*x**n)**2/x + 2*b**2*d*e*log(c*x**n)**3/(3*n) + 2*b**2*e**2*n**2*x - 2*b**2*e**2
*n*x*log(c*x**n) + b**2*e**2*x*log(c*x**n)**2, Ne(n, 0)), ((a + b*log(c))**2*(-d**2/x + 2*d*e*log(x) + e**2*x)
, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (128) = 256\).
time = 2.99, size = 329, normalized size = 2.47 \begin {gather*} \frac {2 \, b^{2} d n^{2} x e \log \left (x\right )^{3} + 3 \, b^{2} n^{2} x^{2} e^{2} \log \left (x\right )^{2} + 6 \, b^{2} d n x e \log \left (c\right ) \log \left (x\right )^{2} - 6 \, b^{2} n^{2} x^{2} e^{2} \log \left (x\right ) + 6 \, b^{2} n x^{2} e^{2} \log \left (c\right ) \log \left (x\right ) + 6 \, b^{2} d x e \log \left (c\right )^{2} \log \left (x\right ) - 3 \, b^{2} d^{2} n^{2} \log \left (x\right )^{2} + 6 \, a b d n x e \log \left (x\right )^{2} + 6 \, b^{2} n^{2} x^{2} e^{2} - 6 \, b^{2} n x^{2} e^{2} \log \left (c\right ) + 3 \, b^{2} x^{2} e^{2} \log \left (c\right )^{2} - 6 \, b^{2} d^{2} n^{2} \log \left (x\right ) + 6 \, a b n x^{2} e^{2} \log \left (x\right ) - 6 \, b^{2} d^{2} n \log \left (c\right ) \log \left (x\right ) + 12 \, a b d x e \log \left (c\right ) \log \left (x\right ) - 6 \, b^{2} d^{2} n^{2} - 6 \, a b n x^{2} e^{2} - 6 \, b^{2} d^{2} n \log \left (c\right ) + 6 \, a b x^{2} e^{2} \log \left (c\right ) - 3 \, b^{2} d^{2} \log \left (c\right )^{2} - 6 \, a b d^{2} n \log \left (x\right ) + 6 \, a^{2} d x e \log \left (x\right ) - 6 \, a b d^{2} n + 3 \, a^{2} x^{2} e^{2} - 6 \, a b d^{2} \log \left (c\right ) - 3 \, a^{2} d^{2}}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.74, size = 228, normalized size = 1.71 \begin {gather*} \ln \left (x\right )\,\left (2\,d\,e\,a^2+4\,d\,e\,a\,b\,n+4\,d\,e\,b^2\,n^2\right )-\frac {a^2\,d^2+2\,a\,b\,d^2\,n+2\,b^2\,d^2\,n^2}{x}-\ln \left (c\,x^n\right )\,\left (\frac {2\,b\,\left (a+b\,n\right )\,d^2+4\,b\,\left (a+b\,n\right )\,d\,e\,x+2\,b\,\left (a-b\,n\right )\,e^2\,x^2}{x}-4\,b\,e^2\,x\,\left (a-b\,n\right )\right )+{\ln \left (c\,x^n\right )}^2\,\left (2\,b^2\,e^2\,x-\frac {b^2\,d^2+2\,b^2\,d\,e\,x+b^2\,e^2\,x^2}{x}+\frac {2\,b\,d\,e\,\left (a+b\,n\right )}{n}\right )+e^2\,x\,\left (a^2-2\,a\,b\,n+2\,b^2\,n^2\right )+\frac {2\,b^2\,d\,e\,{\ln \left (c\,x^n\right )}^3}{3\,n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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