3.1.79 \(\int x (a+b \log (c x^n))^2 \log (d (e+f x)^m) \, dx\) [79]

Optimal. Leaf size=373 \[ -\frac {3 a b e m n x}{2 f}+\frac {7 b^2 e m n^2 x}{4 f}-\frac {3}{8} b^2 m n^2 x^2-\frac {3 b^2 e m n x \log \left (c x^n\right )}{2 f}+\frac {1}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 f^2}+\frac {b^2 e^2 m n^2 \text {Li}_2\left (-\frac {f x}{e}\right )}{2 f^2}-\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f x}{e}\right )}{f^2}+\frac {b^2 e^2 m n^2 \text {Li}_3\left (-\frac {f x}{e}\right )}{f^2} \]

[Out]

-3/2*a*b*e*m*n*x/f+7/4*b^2*e*m*n^2*x/f-3/8*b^2*m*n^2*x^2-3/2*b^2*e*m*n*x*ln(c*x^n)/f+1/2*b*m*n*x^2*(a+b*ln(c*x
^n))+1/2*e*m*x*(a+b*ln(c*x^n))^2/f-1/4*m*x^2*(a+b*ln(c*x^n))^2-1/4*b^2*e^2*m*n^2*ln(f*x+e)/f^2+1/4*b^2*n^2*x^2
*ln(d*(f*x+e)^m)-1/2*b*n*x^2*(a+b*ln(c*x^n))*ln(d*(f*x+e)^m)+1/2*x^2*(a+b*ln(c*x^n))^2*ln(d*(f*x+e)^m)+1/2*b*e
^2*m*n*(a+b*ln(c*x^n))*ln(1+f*x/e)/f^2-1/2*e^2*m*(a+b*ln(c*x^n))^2*ln(1+f*x/e)/f^2+1/2*b^2*e^2*m*n^2*polylog(2
,-f*x/e)/f^2-b*e^2*m*n*(a+b*ln(c*x^n))*polylog(2,-f*x/e)/f^2+b^2*e^2*m*n^2*polylog(3,-f*x/e)/f^2

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Rubi [A]
time = 0.35, antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 12, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2342, 2341, 2425, 45, 2393, 2332, 2354, 2438, 2395, 2333, 2421, 6724} \begin {gather*} -\frac {b e^2 m n \text {PolyLog}\left (2,-\frac {f x}{e}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+\frac {b^2 e^2 m n^2 \text {PolyLog}\left (2,-\frac {f x}{e}\right )}{2 f^2}+\frac {b^2 e^2 m n^2 \text {PolyLog}\left (3,-\frac {f x}{e}\right )}{f^2}-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^2}-\frac {e^2 m \log \left (\frac {f x}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^2}{2 f^2}+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}+\frac {1}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {3 a b e m n x}{2 f}-\frac {3 b^2 e m n x \log \left (c x^n\right )}{2 f}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {7 b^2 e m n^2 x}{4 f}-\frac {3}{8} b^2 m n^2 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b*e*m*n*x)/(2*f) + (7*b^2*e*m*n^2*x)/(4*f) - (3*b^2*m*n^2*x^2)/8 - (3*b^2*e*m*n*x*Log[c*x^n])/(2*f) + (b
*m*n*x^2*(a + b*Log[c*x^n]))/2 + (e*m*x*(a + b*Log[c*x^n])^2)/(2*f) - (m*x^2*(a + b*Log[c*x^n])^2)/4 - (b^2*e^
2*m*n^2*Log[e + f*x])/(4*f^2) + (b^2*n^2*x^2*Log[d*(e + f*x)^m])/4 - (b*n*x^2*(a + b*Log[c*x^n])*Log[d*(e + f*
x)^m])/2 + (x^2*(a + b*Log[c*x^n])^2*Log[d*(e + f*x)^m])/2 + (b*e^2*m*n*(a + b*Log[c*x^n])*Log[1 + (f*x)/e])/(
2*f^2) - (e^2*m*(a + b*Log[c*x^n])^2*Log[1 + (f*x)/e])/(2*f^2) + (b^2*e^2*m*n^2*PolyLog[2, -((f*x)/e)])/(2*f^2
) - (b*e^2*m*n*(a + b*Log[c*x^n])*PolyLog[2, -((f*x)/e)])/f^2 + (b^2*e^2*m*n^2*PolyLog[3, -((f*x)/e)])/f^2

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 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 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 2354

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

Rule 2393

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

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
]))

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2425

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((g_.)*(x_))^(q_.),
x_Symbol] :> With[{u = IntHide[(g*x)^q*(a + b*Log[c*x^n])^p, x]}, Dist[Log[d*(e + f*x^m)^r], u, x] - Dist[f*m
*r, Int[Dist[x^(m - 1)/(e + f*x^m), u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && IGtQ[p, 0
] && RationalQ[m] && RationalQ[q]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int x \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right ) \, dx &=\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-(f m) \int \left (\frac {b^2 n^2 x^2}{4 (e+f x)}-\frac {b n x^2 \left (a+b \log \left (c x^n\right )\right )}{2 (e+f x)}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 (e+f x)}\right ) \, dx\\ &=\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} (f m) \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )^2}{e+f x} \, dx+\frac {1}{2} (b f m n) \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e+f x} \, dx-\frac {1}{4} \left (b^2 f m n^2\right ) \int \frac {x^2}{e+f x} \, dx\\ &=\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} (f m) \int \left (-\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{f^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^2}{f}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )^2}{f^2 (e+f x)}\right ) \, dx+\frac {1}{2} (b f m n) \int \left (-\frac {e \left (a+b \log \left (c x^n\right )\right )}{f^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{f}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )}{f^2 (e+f x)}\right ) \, dx-\frac {1}{4} \left (b^2 f m n^2\right ) \int \left (-\frac {e}{f^2}+\frac {x}{f}+\frac {e^2}{f^2 (e+f x)}\right ) \, dx\\ &=\frac {b^2 e m n^2 x}{4 f}-\frac {1}{8} b^2 m n^2 x^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} m \int x \left (a+b \log \left (c x^n\right )\right )^2 \, dx+\frac {(e m) \int \left (a+b \log \left (c x^n\right )\right )^2 \, dx}{2 f}-\frac {\left (e^2 m\right ) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2}{e+f x} \, dx}{2 f}+\frac {1}{2} (b m n) \int x \left (a+b \log \left (c x^n\right )\right ) \, dx-\frac {(b e m n) \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{2 f}+\frac {\left (b e^2 m n\right ) \int \frac {a+b \log \left (c x^n\right )}{e+f x} \, dx}{2 f}\\ &=-\frac {a b e m n x}{2 f}+\frac {b^2 e m n^2 x}{4 f}-\frac {1}{4} b^2 m n^2 x^2+\frac {1}{4} b m n x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 f^2}+\frac {1}{2} (b m n) \int x \left (a+b \log \left (c x^n\right )\right ) \, dx+\frac {\left (b e^2 m n\right ) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{x} \, dx}{f^2}-\frac {(b e m n) \int \left (a+b \log \left (c x^n\right )\right ) \, dx}{f}-\frac {\left (b^2 e m n\right ) \int \log \left (c x^n\right ) \, dx}{2 f}-\frac {\left (b^2 e^2 m n^2\right ) \int \frac {\log \left (1+\frac {f x}{e}\right )}{x} \, dx}{2 f^2}\\ &=-\frac {3 a b e m n x}{2 f}+\frac {3 b^2 e m n^2 x}{4 f}-\frac {3}{8} b^2 m n^2 x^2-\frac {b^2 e m n x \log \left (c x^n\right )}{2 f}+\frac {1}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 f^2}+\frac {b^2 e^2 m n^2 \text {Li}_2\left (-\frac {f x}{e}\right )}{2 f^2}-\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f x}{e}\right )}{f^2}-\frac {\left (b^2 e m n\right ) \int \log \left (c x^n\right ) \, dx}{f}+\frac {\left (b^2 e^2 m n^2\right ) \int \frac {\text {Li}_2\left (-\frac {f x}{e}\right )}{x} \, dx}{f^2}\\ &=-\frac {3 a b e m n x}{2 f}+\frac {7 b^2 e m n^2 x}{4 f}-\frac {3}{8} b^2 m n^2 x^2-\frac {3 b^2 e m n x \log \left (c x^n\right )}{2 f}+\frac {1}{2} b m n x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {e m x \left (a+b \log \left (c x^n\right )\right )^2}{2 f}-\frac {1}{4} m x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {b^2 e^2 m n^2 \log (e+f x)}{4 f^2}+\frac {1}{4} b^2 n^2 x^2 \log \left (d (e+f x)^m\right )-\frac {1}{2} b n x^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log \left (d (e+f x)^m\right )+\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {f x}{e}\right )}{2 f^2}-\frac {e^2 m \left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x}{e}\right )}{2 f^2}+\frac {b^2 e^2 m n^2 \text {Li}_2\left (-\frac {f x}{e}\right )}{2 f^2}-\frac {b e^2 m n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f x}{e}\right )}{f^2}+\frac {b^2 e^2 m n^2 \text {Li}_3\left (-\frac {f x}{e}\right )}{f^2}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 674, normalized size = 1.81 \begin {gather*} \frac {4 a^2 e f m x-12 a b e f m n x+14 b^2 e f m n^2 x-2 a^2 f^2 m x^2+4 a b f^2 m n x^2-3 b^2 f^2 m n^2 x^2+8 a b e f m x \log \left (c x^n\right )-12 b^2 e f m n x \log \left (c x^n\right )-4 a b f^2 m x^2 \log \left (c x^n\right )+4 b^2 f^2 m n x^2 \log \left (c x^n\right )+4 b^2 e f m x \log ^2\left (c x^n\right )-2 b^2 f^2 m x^2 \log ^2\left (c x^n\right )-4 a^2 e^2 m \log (e+f x)+4 a b e^2 m n \log (e+f x)-2 b^2 e^2 m n^2 \log (e+f x)+8 a b e^2 m n \log (x) \log (e+f x)-4 b^2 e^2 m n^2 \log (x) \log (e+f x)-4 b^2 e^2 m n^2 \log ^2(x) \log (e+f x)-8 a b e^2 m \log \left (c x^n\right ) \log (e+f x)+4 b^2 e^2 m n \log \left (c x^n\right ) \log (e+f x)+8 b^2 e^2 m n \log (x) \log \left (c x^n\right ) \log (e+f x)-4 b^2 e^2 m \log ^2\left (c x^n\right ) \log (e+f x)+4 a^2 f^2 x^2 \log \left (d (e+f x)^m\right )-4 a b f^2 n x^2 \log \left (d (e+f x)^m\right )+2 b^2 f^2 n^2 x^2 \log \left (d (e+f x)^m\right )+8 a b f^2 x^2 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )-4 b^2 f^2 n x^2 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )+4 b^2 f^2 x^2 \log ^2\left (c x^n\right ) \log \left (d (e+f x)^m\right )-8 a b e^2 m n \log (x) \log \left (1+\frac {f x}{e}\right )+4 b^2 e^2 m n^2 \log (x) \log \left (1+\frac {f x}{e}\right )+4 b^2 e^2 m n^2 \log ^2(x) \log \left (1+\frac {f x}{e}\right )-8 b^2 e^2 m n \log (x) \log \left (c x^n\right ) \log \left (1+\frac {f x}{e}\right )+4 b e^2 m n \left (-2 a+b n-2 b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f x}{e}\right )+8 b^2 e^2 m n^2 \text {Li}_3\left (-\frac {f x}{e}\right )}{8 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*e*f*m*x - 12*a*b*e*f*m*n*x + 14*b^2*e*f*m*n^2*x - 2*a^2*f^2*m*x^2 + 4*a*b*f^2*m*n*x^2 - 3*b^2*f^2*m*n^2
*x^2 + 8*a*b*e*f*m*x*Log[c*x^n] - 12*b^2*e*f*m*n*x*Log[c*x^n] - 4*a*b*f^2*m*x^2*Log[c*x^n] + 4*b^2*f^2*m*n*x^2
*Log[c*x^n] + 4*b^2*e*f*m*x*Log[c*x^n]^2 - 2*b^2*f^2*m*x^2*Log[c*x^n]^2 - 4*a^2*e^2*m*Log[e + f*x] + 4*a*b*e^2
*m*n*Log[e + f*x] - 2*b^2*e^2*m*n^2*Log[e + f*x] + 8*a*b*e^2*m*n*Log[x]*Log[e + f*x] - 4*b^2*e^2*m*n^2*Log[x]*
Log[e + f*x] - 4*b^2*e^2*m*n^2*Log[x]^2*Log[e + f*x] - 8*a*b*e^2*m*Log[c*x^n]*Log[e + f*x] + 4*b^2*e^2*m*n*Log
[c*x^n]*Log[e + f*x] + 8*b^2*e^2*m*n*Log[x]*Log[c*x^n]*Log[e + f*x] - 4*b^2*e^2*m*Log[c*x^n]^2*Log[e + f*x] +
4*a^2*f^2*x^2*Log[d*(e + f*x)^m] - 4*a*b*f^2*n*x^2*Log[d*(e + f*x)^m] + 2*b^2*f^2*n^2*x^2*Log[d*(e + f*x)^m] +
8*a*b*f^2*x^2*Log[c*x^n]*Log[d*(e + f*x)^m] - 4*b^2*f^2*n*x^2*Log[c*x^n]*Log[d*(e + f*x)^m] + 4*b^2*f^2*x^2*L
og[c*x^n]^2*Log[d*(e + f*x)^m] - 8*a*b*e^2*m*n*Log[x]*Log[1 + (f*x)/e] + 4*b^2*e^2*m*n^2*Log[x]*Log[1 + (f*x)/
e] + 4*b^2*e^2*m*n^2*Log[x]^2*Log[1 + (f*x)/e] - 8*b^2*e^2*m*n*Log[x]*Log[c*x^n]*Log[1 + (f*x)/e] + 4*b*e^2*m*
n*(-2*a + b*n - 2*b*Log[c*x^n])*PolyLog[2, -((f*x)/e)] + 8*b^2*e^2*m*n^2*PolyLog[3, -((f*x)/e)])/(8*f^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.54, size = 11828, normalized size = 31.71

method	result	size
risch	\(\text {Expression too large to display}\)	\(11828\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*b^2*f*m*x*e - (f^2*m - 2*f^2*log(d))*b^2*x^2 - 2*b^2*m*e^2*log(f*x + e))*log(x^n)^2 + (2*b^2*f^2*x^2*l
og(x^n)^2 + 2*(2*a*b*f^2 - (f^2*n - 2*f^2*log(c))*b^2)*x^2*log(x^n) + (2*a^2*f^2 - 2*(f^2*n - 2*f^2*log(c))*a*
b + (f^2*n^2 - 2*f^2*n*log(c) + 2*f^2*log(c)^2)*b^2)*x^2)*log((f*x + e)^m))/f^2 + integrate(-1/4*((2*(f^3*m -
2*f^3*log(d))*a^2 - 2*(f^3*m*n - 2*(f^3*m - 2*f^3*log(d))*log(c))*a*b + (f^3*m*n^2 - 2*f^3*m*n*log(c) + 2*(f^3
*m - 2*f^3*log(d))*log(c)^2)*b^2)*x^3 - 4*(b^2*f^2*log(c)^2*log(d) + 2*a*b*f^2*log(c)*log(d) + a^2*f^2*log(d))
*x^2*e + 2*(2*b^2*f*m*n*x*e^2 + 2*((f^3*m - 2*f^3*log(d))*a*b - (f^3*m*n - f^3*n*log(d) - (f^3*m - 2*f^3*log(d
))*log(c))*b^2)*x^3 - (4*a*b*f^2*log(d) - (f^2*m*n + 2*f^2*n*log(d) - 4*f^2*log(c)*log(d))*b^2)*x^2*e - 2*(b^2
*f*m*n*x*e^2 + b^2*m*n*e^3)*log(f*x + e))*log(x^n))/(f^3*x^2 + f^2*x*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x*log(c*x^n)^2 + 2*a*b*x*log(c*x^n) + a^2*x)*log((f*x + e)^m*d), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*x^n) + a)^2*x*log((f*x + e)^m*d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x\,\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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