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

Optimal. Leaf size=148 \[ \frac {\left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 f}-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \left (e+f x^2\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}-\frac {b e n \log \left (-\frac {f x^2}{e}\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}-\frac {b e m n \text {Li}_2\left (\frac {f x^2}{e}+1\right )}{4 f}+\frac {1}{2} b m n x^2 \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2454, 2389, 2295, 2376, 2475, 2411, 43, 2351, 2317, 2391} \[ -\frac {b e m n \text {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{4 f}+\frac {\left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 f}-\frac {1}{2} m x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \left (e+f x^2\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}-\frac {b e n \log \left (-\frac {f x^2}{e}\right ) \log \left (d \left (e+f x^2\right )^m\right )}{4 f}+\frac {1}{2} b m n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2295

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

Rule 2317

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 2351

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 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 2389

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

Rule 2391

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 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 266, normalized size = 1.80 \[ \frac {2 a f x^2 \log \left (d \left (e+f x^2\right )^m\right )+2 a e \log \left (d \left (e+f x^2\right )^m\right )-2 a f m x^2+2 b f x^2 \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b e m \log \left (c x^n\right ) \log \left (e+f x^2\right )-2 b f m x^2 \log \left (c x^n\right )-b f n x^2 \log \left (d \left (e+f x^2\right )^m\right )+2 b e m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b e m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b e m n \log \left (e+f x^2\right )-2 b e m n \log (x) \log \left (e+f x^2\right )+2 b e m n \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b e m n \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b f m n x^2}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x \log \left (c x^{n}\right ) + a x\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} x \log \left ({\left (f x^{2} + e\right )}^{m} d\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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*log((f*x^2 + e)^m*d), x)

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maple [C]  time = 0.72, size = 2068, normalized size = 13.97 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*e*m/f*ln(f*x^2+e)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*a*m*x^2+(1/2*b*x^2*ln(x^n)+1/4*x^2*(I*Pi
*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I
*c)*csgn(I*c*x^n)^2+2*b*ln(c)-b*n+2*a))*ln((f*x^2+e)^m)+1/2*b*e*n*m/f*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+
1/2*b*e*n*m/f*dilog((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/4*I*Pi*a*x^2*csgn(I*d*(f*x^2+e)^m)^3+1/2*a*x^2*ln(d)+1/
2*m/f*b*ln(x^n)*e*ln(f*x^2+e)-1/8*I*Pi*b*n*x^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2+1/4*I*Pi*csgn(I*(f*
x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b*x^2*ln(x^n)+1/4*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b*x^2*ln(x^n)-1/4*I
*x^2*Pi*ln(d)*b*csgn(I*c*x^n)^3-1/4*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b*x^2*ln(x^n)+1/4
*I*Pi*b*m*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/8*I*Pi*b*n*x^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x
^2+e)^m)-1/2*b*m*x^2*ln(x^n)+1/2*b*x^2*ln(d)*ln(x^n)-1/4*b*n*x^2*ln(d)+1/2*b*x^2*ln(c)*ln(d)-1/2*b*m*x^2*ln(c)
-1/8*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^2*b*csgn(I*c*x^n)^3+1/4*I*x^2*Pi*ln(d)*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*P
i^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn(I*c*x^n)^2*csgn(I*c)-1/8*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f
*x^2+e)^m)^2*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn
(I*c*x^n)^2*csgn(I*c)-1/8*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/8*Pi^2*csgn
(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^2*b*csgn(I*c*x^n)^3-1/4*I*Pi*a*x^2*csgn(I*d)*csgn(I*(f*x^2+e
)^m)*csgn(I*d*(f*x^2+e)^m)+1/4*I*ln(c)*Pi*b*x^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2-1/8*I*Pi*b*n*x^2*c
sgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2-1/4*I*Pi*b*m*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*m*x^2*csgn(I*c*x^n)^2
*csgn(I*c)+1/4*I*ln(c)*Pi*b*x^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2-1/8*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(
I*d*(f*x^2+e)^m)*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/4*I*e*m/f*ln(f*x^2+e)*Pi*b*csgn(I*x^n)*csgn(I*c*x
^n)^2+1/4*I*e*m/f*ln(f*x^2+e)*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)-1/8*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*
csgn(I*x^n)*csgn(I*c*x^n)^2+1/8*I*Pi*b*n*x^2*csgn(I*d*(f*x^2+e)^m)^3+1/4*I*Pi*a*x^2*csgn(I*(f*x^2+e)^m)*csgn(I
*d*(f*x^2+e)^m)^2-1/4*I*ln(c)*Pi*b*x^2*csgn(I*d*(f*x^2+e)^m)^3+1/2*e*m/f*ln(f*x^2+e)*a-1/4*b*e/f*m*n*ln(f*x^2+
e)+1/2*e*m/f*ln(f*x^2+e)*b*ln(c)+1/4*I*Pi*b*m*x^2*csgn(I*c*x^n)^3+1/4*I*Pi*a*x^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^
m)^2-1/4*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*b*x^2*ln(x^n)+1/8*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*x^2*b*csgn(I*x^n)*csgn(I*
c*x^n)^2+1/2*b*m*n*x^2+1/4*I*x^2*Pi*ln(d)*b*csgn(I*c*x^n)^2*csgn(I*c)-1/4*I*x^2*Pi*ln(d)*b*csgn(I*x^n)*csgn(I*
c*x^n)*csgn(I*c)+1/8*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)^
2+1/8*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn(I*c*x^n)^3+1/8*Pi^2*csgn(I*d*(f*x^2+e)^m)^3*
x^2*b*csgn(I*c*x^n)^2*csgn(I*c)+1/8*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn(I*c*x^n)^3-1/2*b*e/f*m*n
*ln(x)*ln(f*x^2+e)+1/2*b*e*n*m/f*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/2*b*e*n*m/f*ln(x)*ln((f*x+(-e*f)
^(1/2))/(-e*f)^(1/2))+1/8*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*x^2*b*csgn(I*c*x^n)^2*csgn(
I*c)+1/8*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/8*Pi^2*csgn(I*(f*x
^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*x^2*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*I*e*m/f*ln(f*x^2+e)*Pi*b*csgn(I
*c*x^n)^3-1/4*I*ln(c)*Pi*b*x^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left (2 \, b m x^{2} \log \left (x^{n}\right ) - {\left ({\left (m n - 2 \, m \log \relax (c)\right )} b - 2 \, a m\right )} x^{2}\right )} \log \left (f x^{2} + e\right ) + \int -\frac {{\left (2 \, {\left (f m - f \log \relax (d)\right )} a - {\left (f m n - 2 \, {\left (f m - f \log \relax (d)\right )} \log \relax (c)\right )} b\right )} x^{3} - 2 \, {\left (b e \log \relax (c) \log \relax (d) + a e \log \relax (d)\right )} x + 2 \, {\left ({\left (f m - f \log \relax (d)\right )} b x^{3} - b e x \log \relax (d)\right )} \log \left (x^{n}\right )}{2 \, {\left (f x^{2} + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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