3.24 \(\int x^3 (a+b \log (c x^n)) \log (d (\frac {1}{d}+f x^2)) \, dx\)

Optimal. Leaf size=180 \[ -\frac {\log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^2 f^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{4 d f}+\frac {1}{4} x^4 \log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2\left (-d f x^2\right )}{8 d^2 f^2}+\frac {b n \log \left (d f x^2+1\right )}{16 d^2 f^2}-\frac {3 b n x^2}{16 d f}-\frac {1}{16} b n x^4 \log \left (d f x^2+1\right )+\frac {1}{16} b n x^4 \]

[Out]

-3/16*b*n*x^2/d/f+1/16*b*n*x^4+1/4*x^2*(a+b*ln(c*x^n))/d/f-1/8*x^4*(a+b*ln(c*x^n))+1/16*b*n*ln(d*f*x^2+1)/d^2/
f^2-1/16*b*n*x^4*ln(d*f*x^2+1)-1/4*(a+b*ln(c*x^n))*ln(d*f*x^2+1)/d^2/f^2+1/4*x^4*(a+b*ln(c*x^n))*ln(d*f*x^2+1)
-1/8*b*n*polylog(2,-d*f*x^2)/d^2/f^2

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2454, 2395, 43, 2376, 2391} \[ -\frac {b n \text {PolyLog}\left (2,-d f x^2\right )}{8 d^2 f^2}-\frac {\log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^2 f^2}+\frac {1}{4} x^4 \log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{4 d f}-\frac {1}{8} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log \left (d f x^2+1\right )}{16 d^2 f^2}-\frac {3 b n x^2}{16 d f}-\frac {1}{16} b n x^4 \log \left (d f x^2+1\right )+\frac {1}{16} b n x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Log[c*x^n])*Log[d*(d^(-1) + f*x^2)],x]

[Out]

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

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 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 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 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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.11, size = 348, normalized size = 1.93 \[ -\frac {a \log \left (d f x^2+1\right )}{4 d^2 f^2}+\frac {a x^2}{4 d f}+\frac {1}{4} a x^4 \log \left (d f x^2+1\right )-\frac {a x^4}{8}+\frac {b \left (n-4 \left (\log \left (c x^n\right )-n \log (x)\right )\right ) \log \left (d f x^2+1\right )}{16 d^2 f^2}+\frac {b x^2 \left (4 \left (\log \left (c x^n\right )-n \log (x)\right )-n\right )}{16 d f}+\frac {1}{16} b x^4 \left (4 \left (\log \left (c x^n\right )-n \log (x)\right )+4 n \log (x)-n\right ) \log \left (d f x^2+1\right )+\frac {1}{32} b x^4 \left (n-4 \left (\log \left (c x^n\right )-n \log (x)\right )\right )-\frac {1}{2} b d f n \left (\frac {\text {Li}_2\left (-i \sqrt {d} \sqrt {f} x\right )+\log (x) \log \left (1+i \sqrt {d} \sqrt {f} x\right )}{2 d^3 f^3}+\frac {\text {Li}_2\left (i \sqrt {d} \sqrt {f} x\right )+\log (x) \log \left (1-i \sqrt {d} \sqrt {f} x\right )}{2 d^3 f^3}-\frac {\frac {1}{2} x^2 \log (x)-\frac {x^2}{4}}{d^2 f^2}+\frac {\frac {1}{4} x^4 \log (x)-\frac {x^4}{16}}{d f}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Log[c*x^n])*Log[d*(d^(-1) + f*x^2)],x]

[Out]

(a*x^2)/(4*d*f) - (a*x^4)/8 + (b*x^4*(n - 4*(-(n*Log[x]) + Log[c*x^n])))/32 + (b*x^2*(-n + 4*(-(n*Log[x]) + Lo
g[c*x^n])))/(16*d*f) - (a*Log[1 + d*f*x^2])/(4*d^2*f^2) + (a*x^4*Log[1 + d*f*x^2])/4 + (b*(n - 4*(-(n*Log[x])
+ Log[c*x^n]))*Log[1 + d*f*x^2])/(16*d^2*f^2) + (b*x^4*(-n + 4*n*Log[x] + 4*(-(n*Log[x]) + Log[c*x^n]))*Log[1
+ d*f*x^2])/16 - (b*d*f*n*(-((-1/4*x^2 + (x^2*Log[x])/2)/(d^2*f^2)) + (-1/16*x^4 + (x^4*Log[x])/4)/(d*f) + (Lo
g[x]*Log[1 + I*Sqrt[d]*Sqrt[f]*x] + PolyLog[2, (-I)*Sqrt[d]*Sqrt[f]*x])/(2*d^3*f^3) + (Log[x]*Log[1 - I*Sqrt[d
]*Sqrt[f]*x] + PolyLog[2, I*Sqrt[d]*Sqrt[f]*x])/(2*d^3*f^3)))/2

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{3} \log \left (d f x^{2} + 1\right ) \log \left (c x^{n}\right ) + a x^{3} \log \left (d f x^{2} + 1\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*log(d*(1/d+f*x^2)),x, algorithm="fricas")

[Out]

integral(b*x^3*log(d*f*x^2 + 1)*log(c*x^n) + a*x^3*log(d*f*x^2 + 1), x)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*log(d*(1/d+f*x^2)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [C]  time = 0.31, size = 827, normalized size = 4.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*ln(c*x^n)+a)*ln(d*(1/d+f*x^2)),x)

[Out]

(1/4*b*x^4*ln(d*(1/d+f*x^2))-1/8*b*(d^2*f^2*x^4-2*d*f*x^2+2*ln(d*(1/d+f*x^2)))/d^2/f^2)*ln(x^n)-1/8*a*x^4+1/4*
b*ln(c)*x^4*ln(d*f*x^2+1)+1/4*a/d/f*x^2-1/4*a/d^2/f^2*ln(d*f*x^2+1)-1/8*b*x^4*ln(c)+1/4*a*x^4*ln(d*f*x^2+1)+1/
16*b*n*x^4+1/16*b*n*ln(d*f*x^2+1)/d^2/f^2-1/8*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^4*ln(d*f*x^2+1)+1/4
*n*b/d^2/f^2*ln(x)*ln(d*f*x^2+1)-1/4/d^2/f^2*b*n*dilog(1-x*(-d*f)^(1/2))-1/4/d^2/f^2*b*n*dilog(1+x*(-d*f)^(1/2
))-1/4/d^2/f^2*ln(d*f*x^2+1)*b*ln(c)+1/4/d/f*x^2*b*ln(c)-1/8*I/d^2/f^2*ln(d*f*x^2+1)*Pi*b*csgn(I*c*x^n)^2*csgn
(I*c)+1/8*I/d/f*x^2*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)-1/8*I/d^2/f^2*ln(d*f*x^2+1)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^
2+1/8*I/d/f*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/16*I*Pi*b*x^4*csgn(I*c*x^n)^3-1/4/d^2/f^2*b*n*ln(x)*ln(1+x*
(-d*f)^(1/2))-1/4/d^2/f^2*b*n*ln(x)*ln(1-x*(-d*f)^(1/2))-1/8*I*Pi*b*csgn(I*c*x^n)^3*x^4*ln(d*f*x^2+1)-1/16*I*P
i*b*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-1/16*I*Pi*b*x^4*csgn(I*c*x^n)^2*csgn(I*c)-1/8*I/d/f*x^2*Pi*b*csgn(I*c*x^n)
^3+1/16*I*Pi*b*x^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/8*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)*x^4*ln(d*f*x^2+1)+
1/8*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*x^4*ln(d*f*x^2+1)-1/8*I/d/f*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c
)+1/8*I/d^2/f^2*ln(d*f*x^2+1)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/8*I/d^2/f^2*ln(d*f*x^2+1)*Pi*b*csgn(I
*c*x^n)^3-1/16*b*n*x^4*ln(d*f*x^2+1)-3/16*b*n*x^2/d/f

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, {\left (4 \, b x^{4} \log \left (x^{n}\right ) - {\left (b {\left (n - 4 \, \log \relax (c)\right )} - 4 \, a\right )} x^{4}\right )} \log \left (d f x^{2} + 1\right ) - \int \frac {4 \, b d f x^{5} \log \left (x^{n}\right ) + {\left (4 \, a d f - {\left (d f n - 4 \, d f \log \relax (c)\right )} b\right )} x^{5}}{8 \, {\left (d f x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))*log(d*(1/d+f*x^2)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\ln \left (d\,\left (f\,x^2+\frac {1}{d}\right )\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n)),x)

[Out]

int(x^3*log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n)), x)

________________________________________________________________________________________

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**3*(a+b*ln(c*x**n))*ln(d*(1/d+f*x**2)),x)

[Out]

Timed out

________________________________________________________________________________________