3.90 \(\int x^3 (a+b \log (c x^n)) \log (d (e+f x^2)^m) \, dx\)

Optimal. Leaf size=221 \[ \frac{b e^2 m n \text{PolyLog}\left (2,\frac{f x^2}{e}+1\right )}{8 f^2}+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac{e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac{b e^2 m n \log \left (e+f x^2\right )}{16 f^2}+\frac{b e^2 m n \log \left (-\frac{f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac{3 b e m n x^2}{16 f}+\frac{1}{16} b m n x^4 \]

[Out]

(-3*b*e*m*n*x^2)/(16*f) + (b*m*n*x^4)/16 + (e*m*x^2*(a + b*Log[c*x^n]))/(4*f) - (m*x^4*(a + b*Log[c*x^n]))/8 +
(b*e^2*m*n*Log[e + f*x^2])/(16*f^2) + (b*e^2*m*n*Log[-((f*x^2)/e)]*Log[e + f*x^2])/(8*f^2) - (e^2*m*(a + b*Lo
g[c*x^n])*Log[e + f*x^2])/(4*f^2) - (b*n*x^4*Log[d*(e + f*x^2)^m])/16 + (x^4*(a + b*Log[c*x^n])*Log[d*(e + f*x
^2)^m])/4 + (b*e^2*m*n*PolyLog[2, 1 + (f*x^2)/e])/(8*f^2)

________________________________________________________________________________________

Rubi [A]  time = 0.223533, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2454, 2395, 43, 2376, 2394, 2315} \[ \frac{b e^2 m n \text{PolyLog}\left (2,\frac{f x^2}{e}+1\right )}{8 f^2}+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac{e^2 m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{4 f^2}+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac{b e^2 m n \log \left (e+f x^2\right )}{16 f^2}+\frac{b e^2 m n \log \left (-\frac{f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac{3 b e m n x^2}{16 f}+\frac{1}{16} b m n x^4 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*e*m*n*x^2)/(16*f) + (b*m*n*x^4)/16 + (e*m*x^2*(a + b*Log[c*x^n]))/(4*f) - (m*x^4*(a + b*Log[c*x^n]))/8 +
(b*e^2*m*n*Log[e + f*x^2])/(16*f^2) + (b*e^2*m*n*Log[-((f*x^2)/e)]*Log[e + f*x^2])/(8*f^2) - (e^2*m*(a + b*Lo
g[c*x^n])*Log[e + f*x^2])/(4*f^2) - (b*n*x^4*Log[d*(e + f*x^2)^m])/16 + (x^4*(a + b*Log[c*x^n])*Log[d*(e + f*x
^2)^m])/4 + (b*e^2*m*n*PolyLog[2, 1 + (f*x^2)/e])/(8*f^2)

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2376

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

Rule 2394

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

\begin{align*} \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right ) \, dx &=\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-(b n) \int \left (\frac{e m x}{4 f}-\frac{m x^3}{8}-\frac{e^2 m \log \left (e+f x^2\right )}{4 f^2 x}+\frac{1}{4} x^3 \log \left (d \left (e+f x^2\right )^m\right )\right ) \, dx\\ &=-\frac{b e m n x^2}{8 f}+\frac{1}{32} b m n x^4+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac{1}{4} (b n) \int x^3 \log \left (d \left (e+f x^2\right )^m\right ) \, dx+\frac{\left (b e^2 m n\right ) \int \frac{\log \left (e+f x^2\right )}{x} \, dx}{4 f^2}\\ &=-\frac{b e m n x^2}{8 f}+\frac{1}{32} b m n x^4+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac{1}{8} (b n) \operatorname{Subst}\left (\int x \log \left (d (e+f x)^m\right ) \, dx,x,x^2\right )+\frac{\left (b e^2 m n\right ) \operatorname{Subst}\left (\int \frac{\log (e+f x)}{x} \, dx,x,x^2\right )}{8 f^2}\\ &=-\frac{b e m n x^2}{8 f}+\frac{1}{32} b m n x^4+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )+\frac{b e^2 m n \log \left (-\frac{f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}-\frac{1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )-\frac{\left (b e^2 m n\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{f x}{e}\right )}{e+f x} \, dx,x,x^2\right )}{8 f}+\frac{1}{16} (b f m n) \operatorname{Subst}\left (\int \frac{x^2}{e+f x} \, dx,x,x^2\right )\\ &=-\frac{b e m n x^2}{8 f}+\frac{1}{32} b m n x^4+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )+\frac{b e^2 m n \log \left (-\frac{f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}-\frac{1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac{b e^2 m n \text{Li}_2\left (1+\frac{f x^2}{e}\right )}{8 f^2}+\frac{1}{16} (b f m n) \operatorname{Subst}\left (\int \left (-\frac{e}{f^2}+\frac{x}{f}+\frac{e^2}{f^2 (e+f x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{3 b e m n x^2}{16 f}+\frac{1}{16} b m n x^4+\frac{e m x^2 \left (a+b \log \left (c x^n\right )\right )}{4 f}-\frac{1}{8} m x^4 \left (a+b \log \left (c x^n\right )\right )+\frac{b e^2 m n \log \left (e+f x^2\right )}{16 f^2}+\frac{b e^2 m n \log \left (-\frac{f x^2}{e}\right ) \log \left (e+f x^2\right )}{8 f^2}-\frac{e^2 m \left (a+b \log \left (c x^n\right )\right ) \log \left (e+f x^2\right )}{4 f^2}-\frac{1}{16} b n x^4 \log \left (d \left (e+f x^2\right )^m\right )+\frac{1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )+\frac{b e^2 m n \text{Li}_2\left (1+\frac{f x^2}{e}\right )}{8 f^2}\\ \end{align*}

Mathematica [C]  time = 0.158681, size = 324, normalized size = 1.47 \[ -\frac{4 b e^2 m n \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )+4 b e^2 m n \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )-4 a f^2 x^4 \log \left (d \left (e+f x^2\right )^m\right )+4 a e^2 m \log \left (e+f x^2\right )-4 a e f m x^2+2 a f^2 m x^4-4 b f^2 x^4 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+4 b e^2 m \log \left (c x^n\right ) \log \left (e+f x^2\right )-4 b e f m x^2 \log \left (c x^n\right )+2 b f^2 m x^4 \log \left (c x^n\right )+b f^2 n x^4 \log \left (d \left (e+f x^2\right )^m\right )-b e^2 m n \log \left (e+f x^2\right )-4 b e^2 m n \log (x) \log \left (e+f x^2\right )+4 b e^2 m n \log (x) \log \left (1-\frac{i \sqrt{f} x}{\sqrt{e}}\right )+4 b e^2 m n \log (x) \log \left (1+\frac{i \sqrt{f} x}{\sqrt{e}}\right )+3 b e f m n x^2-b f^2 m n x^4}{16 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-4*a*e*f*m*x^2 + 3*b*e*f*m*n*x^2 + 2*a*f^2*m*x^4 - b*f^2*m*n*x^4 - 4*b*e*f*m*x^2*Log[c*x^n] + 2*b*f^2*m*x^4*
Log[c*x^n] + 4*b*e^2*m*n*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + 4*b*e^2*m*n*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt
[e]] + 4*a*e^2*m*Log[e + f*x^2] - b*e^2*m*n*Log[e + f*x^2] - 4*b*e^2*m*n*Log[x]*Log[e + f*x^2] + 4*b*e^2*m*Log
[c*x^n]*Log[e + f*x^2] - 4*a*f^2*x^4*Log[d*(e + f*x^2)^m] + b*f^2*n*x^4*Log[d*(e + f*x^2)^m] - 4*b*f^2*x^4*Log
[c*x^n]*Log[d*(e + f*x^2)^m] + 4*b*e^2*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] + 4*b*e^2*m*n*PolyLog[2, (I*Sq
rt[f]*x)/Sqrt[e]])/(16*f^2)

________________________________________________________________________________________

Maple [C]  time = 0.354, size = 2259, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m),x)

[Out]

1/8*I/f^2*e^2*m*ln(f*x^2+e)*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/4/f^2*e^2*m*ln(f*x^2+e)*b*ln(c)-1/4/f^2
*e^2*m*ln(f*x^2+e)*a+1/4/f*x^2*a*e*m-1/4*b*e^2*m*n/f^2*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*b*e^2*m*n/f
^2*dilog((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*m/f^2*b*ln(x^n)*e^2*ln(f*x^2+e)+1/16*Pi^2*csgn(I*d)*csgn(I*(f*x^
2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^4*b*csgn(I*c)*csgn(I*c*x^n)^2+1/16*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d
*(f*x^2+e)^m)*x^4*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/16*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*c)*cs
gn(I*x^n)*csgn(I*c*x^n)+1/16*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)-1/8*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b*x^4*ln(x^n)+1/32*I*Pi*b*n*x^4*csgn(I*
d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)+(1/4*x^4*b*ln(x^n)+1/16*x^4*(-2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn
(I*c*x^n)+2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*b*Pi*csgn(I*c*x^n)^3+4*b
*ln(c)-b*n+4*a))*ln((f*x^2+e)^m)+1/16*b*e^2*m*n*ln(f*x^2+e)/f^2-3/16*b*e*m*n*x^2/f+1/16*b*m*n*x^4-1/8*I*ln(c)*
Pi*b*x^4*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)+1/16*I*Pi*b*m*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*
x^n)-1/8*I/f*Pi*b*e*m*csgn(I*c*x^n)^3*x^2+1/8*I/f^2*e^2*m*ln(f*x^2+e)*Pi*b*csgn(I*c*x^n)^3+1/16*Pi^2*csgn(I*(f
*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*c*x^n)^3+1/32*I*Pi*b*n*x^4*csgn(I*d*(f*x^2+e)^m)^3-1/8*I*ln(c)
*Pi*b*x^4*csgn(I*d*(f*x^2+e)^m)^3+1/16*I*Pi*b*m*x^4*csgn(I*c*x^n)^3+1/8*I*Pi*a*x^4*csgn(I*d)*csgn(I*d*(f*x^2+e
)^m)^2+1/8*I*Pi*a*x^4*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2-1/8*I/f^2*e^2*m*ln(f*x^2+e)*Pi*b*csgn(I*x^n)
*csgn(I*c*x^n)^2+1/8*I/f*Pi*b*e*m*csgn(I*c)*csgn(I*c*x^n)^2*x^2+1/8*I/f*Pi*b*e*m*csgn(I*x^n)*csgn(I*c*x^n)^2*x
^2-1/8*m*b*ln(x^n)*x^4+1/4*ln(d)*b*x^4*ln(x^n)-1/8*I/f*Pi*b*e*m*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x^2+1/16*P
i^2*csgn(I*d*(f*x^2+e)^m)^3*x^4*b*csgn(I*c)*csgn(I*c*x^n)^2+1/16*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^4*b*csgn(I*x^n
)*csgn(I*c*x^n)^2+1/16*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*c*x^n)^3-1/8*I*Pi*csgn(I*d*(f*x^2+e
)^m)^3*b*x^4*ln(x^n)-1/16*I*Pi*b*m*x^4*csgn(I*c)*csgn(I*c*x^n)^2-1/16*I*Pi*b*m*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2
-1/8*x^4*ln(c)*b*m-1/16*ln(d)*b*n*x^4+1/4*x^4*ln(c)*ln(d)*b-1/4*b*e^2*m*n/f^2*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e
*f)^(1/2))-1/4*b*e^2*m*n/f^2*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/4*b*e^2*m*n/f^2*ln(x)*ln(f*x^2+e)-1/1
6*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^4*b*csgn(I*c*x^n)^3-1/8*I*Pi*a*x^4*csgn(I*d*(f*x^2+e)^m)^3-1/8*x^4*a*m+1/8*I*
Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b*x^4*ln(x^n)-1/16*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b
*csgn(I*c)*csgn(I*c*x^n)^2-1/16*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*x^n)*csgn(I*c*x^
n)^2+1/4*x^4*ln(d)*a+1/8*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b*x^4*ln(x^n)-1/32*I*Pi*b*n*x^4*csgn
(I*d)*csgn(I*d*(f*x^2+e)^m)^2-1/32*I*Pi*b*n*x^4*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2-1/16*Pi^2*csgn(I*d
)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^4*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*I/f^2*e^2*m*ln(f*x^2
+e)*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2+1/8*I*x^4*Pi*ln(d)*b*csgn(I*c)*csgn(I*c*x^n)^2+1/8*I*x^4*Pi*ln(d)*b*csgn(I*
x^n)*csgn(I*c*x^n)^2-1/16*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^4*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/16*Pi^2*csg
n(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^4*b*csgn(I*c*x^n)^3-1/16*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^
m)^2*x^4*b*csgn(I*c)*csgn(I*c*x^n)^2-1/16*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^4*b*csgn(I*x^n)*csgn(I*c*x^
n)^2-1/8*I*x^4*Pi*ln(d)*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/8*I*x^4*Pi*ln(d)*b*csgn(I*c*x^n)^3+1/4*m/f*b*l
n(x^n)*e*x^2+1/4/f*ln(c)*x^2*b*e*m-1/8*I*Pi*a*x^4*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)+1/8*I*ln
(c)*Pi*b*x^4*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2+1/8*I*ln(c)*Pi*b*x^4*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^
2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{16} \,{\left (4 \, b x^{4} \log \left (x^{n}\right ) -{\left (b{\left (n - 4 \, \log \left (c\right )\right )} - 4 \, a\right )} x^{4}\right )} \log \left ({\left (f x^{2} + e\right )}^{m}\right ) + \int -\frac{{\left (4 \,{\left (f m - 2 \, f \log \left (d\right )\right )} a -{\left (f m n - 4 \,{\left (f m - 2 \, f \log \left (d\right )\right )} \log \left (c\right )\right )} b\right )} x^{5} - 8 \,{\left (b e \log \left (c\right ) \log \left (d\right ) + a e \log \left (d\right )\right )} x^{3} + 4 \,{\left ({\left (f m - 2 \, f \log \left (d\right )\right )} b x^{5} - 2 \, b e x^{3} \log \left (d\right )\right )} \log \left (x^{n}\right )}{8 \,{\left (f x^{2} + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*b*x^4*log(x^n) - (b*(n - 4*log(c)) - 4*a)*x^4)*log((f*x^2 + e)^m) + integrate(-1/8*((4*(f*m - 2*f*log(
d))*a - (f*m*n - 4*(f*m - 2*f*log(d))*log(c))*b)*x^5 - 8*(b*e*log(c)*log(d) + a*e*log(d))*x^3 + 4*((f*m - 2*f*
log(d))*b*x^5 - 2*b*e*x^3*log(d))*log(x^n))/(f*x^2 + e), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{3} \log \left (c x^{n}\right ) + a x^{3}\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3*log(c*x^n) + a*x^3)*log((f*x^2 + e)^m*d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*x^n) + a)*x^3*log((f*x^2 + e)^m*d), x)