3.99 \(\int \frac{(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x^6} \, dx\)

Optimal. Leaf size=267 \[ -\frac{i b f^{5/2} m n \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{16 b f m n}{225 e x^3} \]

[Out]

(-16*b*f*m*n)/(225*e*x^3) + (12*b*f^2*m*n)/(25*e^2*x) + (2*b*f^(5/2)*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(25*e^(5
/2)) - (2*f*m*(a + b*Log[c*x^n]))/(15*e*x^3) + (2*f^2*m*(a + b*Log[c*x^n]))/(5*e^2*x) + (2*f^(5/2)*m*ArcTan[(S
qrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(5*e^(5/2)) - (b*n*Log[d*(e + f*x^2)^m])/(25*x^5) - ((a + b*Log[c*x^n])
*Log[d*(e + f*x^2)^m])/(5*x^5) - ((I/5)*b*f^(5/2)*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/e^(5/2) + ((I/5)*b
*f^(5/2)*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/e^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.189025, antiderivative size = 267, 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 = {2455, 325, 205, 2376, 4848, 2391} \[ -\frac{i b f^{5/2} m n \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{16 b f m n}{225 e x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^6,x]

[Out]

(-16*b*f*m*n)/(225*e*x^3) + (12*b*f^2*m*n)/(25*e^2*x) + (2*b*f^(5/2)*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(25*e^(5
/2)) - (2*f*m*(a + b*Log[c*x^n]))/(15*e*x^3) + (2*f^2*m*(a + b*Log[c*x^n]))/(5*e^2*x) + (2*f^(5/2)*m*ArcTan[(S
qrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(5*e^(5/2)) - (b*n*Log[d*(e + f*x^2)^m])/(25*x^5) - ((a + b*Log[c*x^n])
*Log[d*(e + f*x^2)^m])/(5*x^5) - ((I/5)*b*f^(5/2)*m*n*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]])/e^(5/2) + ((I/5)*b
*f^(5/2)*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/e^(5/2)

Rule 2455

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx &=-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-(b n) \int \left (-\frac{2 f m}{15 e x^4}+\frac{2 f^2 m}{5 e^2 x^2}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2} x}-\frac{\log \left (d \left (e+f x^2\right )^m\right )}{5 x^6}\right ) \, dx\\ &=-\frac{2 b f m n}{45 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{1}{5} (b n) \int \frac{\log \left (d \left (e+f x^2\right )^m\right )}{x^6} \, dx-\frac{\left (2 b f^{5/2} m n\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}\\ &=-\frac{2 b f m n}{45 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac{1}{25} (2 b f m n) \int \frac{1}{x^4 \left (e+f x^2\right )} \, dx-\frac{\left (i b f^{5/2} m n\right ) \int \frac{\log \left (1-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}+\frac{\left (i b f^{5/2} m n\right ) \int \frac{\log \left (1+\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{x} \, dx}{5 e^{5/2}}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{2 b f^2 m n}{5 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}-\frac{\left (2 b f^2 m n\right ) \int \frac{1}{x^2 \left (e+f x^2\right )} \, dx}{25 e}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{12 b f^2 m n}{25 e^2 x}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{\left (2 b f^3 m n\right ) \int \frac{1}{e+f x^2} \, dx}{25 e^2}\\ &=-\frac{16 b f m n}{225 e x^3}+\frac{12 b f^2 m n}{25 e^2 x}+\frac{2 b f^{5/2} m n \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{25 e^{5/2}}-\frac{2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}+\frac{2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}+\frac{2 f^{5/2} m \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}-\frac{b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}-\frac{i b f^{5/2} m n \text{Li}_2\left (-\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}+\frac{i b f^{5/2} m n \text{Li}_2\left (\frac{i \sqrt{f} x}{\sqrt{e}}\right )}{5 e^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.164178, size = 399, normalized size = 1.49 \[ -\frac{45 i b f^{5/2} m n x^5 \text{PolyLog}\left (2,-\frac{i \sqrt{f} x}{\sqrt{e}}\right )-45 i b f^{5/2} m n x^5 \text{PolyLog}\left (2,\frac{i \sqrt{f} x}{\sqrt{e}}\right )+45 a e^{5/2} \log \left (d \left (e+f x^2\right )^m\right )+30 a e^{3/2} f m x^2 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{f x^2}{e}\right )+45 b e^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+30 b e^{3/2} f m x^2 \log \left (c x^n\right )-90 b \sqrt{e} f^2 m x^4 \log \left (c x^n\right )-90 b f^{5/2} m x^5 \log \left (c x^n\right ) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )+9 b e^{5/2} n \log \left (d \left (e+f x^2\right )^m\right )+16 b e^{3/2} f m n x^2-108 b \sqrt{e} f^2 m n x^4-45 i b f^{5/2} m n x^5 \log (x) \log \left (1-\frac{i \sqrt{f} x}{\sqrt{e}}\right )+45 i b f^{5/2} m n x^5 \log (x) \log \left (1+\frac{i \sqrt{f} x}{\sqrt{e}}\right )-18 b f^{5/2} m n x^5 \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )+90 b f^{5/2} m n x^5 \log (x) \tan ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e}}\right )}{225 e^{5/2} x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^6,x]

[Out]

-(16*b*e^(3/2)*f*m*n*x^2 - 108*b*Sqrt[e]*f^2*m*n*x^4 - 18*b*f^(5/2)*m*n*x^5*ArcTan[(Sqrt[f]*x)/Sqrt[e]] + 30*a
*e^(3/2)*f*m*x^2*Hypergeometric2F1[-3/2, 1, -1/2, -((f*x^2)/e)] + 90*b*f^(5/2)*m*n*x^5*ArcTan[(Sqrt[f]*x)/Sqrt
[e]]*Log[x] + 30*b*e^(3/2)*f*m*x^2*Log[c*x^n] - 90*b*Sqrt[e]*f^2*m*x^4*Log[c*x^n] - 90*b*f^(5/2)*m*x^5*ArcTan[
(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (45*I)*b*f^(5/2)*m*n*x^5*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + (45*I)*b*f^
(5/2)*m*n*x^5*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + 45*a*e^(5/2)*Log[d*(e + f*x^2)^m] + 9*b*e^(5/2)*n*Log[d*
(e + f*x^2)^m] + 45*b*e^(5/2)*Log[c*x^n]*Log[d*(e + f*x^2)^m] + (45*I)*b*f^(5/2)*m*n*x^5*PolyLog[2, ((-I)*Sqrt
[f]*x)/Sqrt[e]] - (45*I)*b*f^(5/2)*m*n*x^5*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(225*e^(5/2)*x^5)

________________________________________________________________________________________

Maple [C]  time = 0.194, size = 2385, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^6,x)

[Out]

1/20*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^5*b*csgn(I*c*x^n)^3+1/10*I*Pi*csgn(I*d*(f*x^2+e)^m)^3/x^5*a-1/10*I*Pi*csgn
(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b/x^5*ln(x^n)-1/10*I*ln(d)/x^5*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/10*I*ln(d)/x^5*b
*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+2/5*m*f^3*b/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^n)-16/225*b*f*m*n/e/x
^3+12/25*b*f^2*m*n/e^2/x-1/50*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b/x^5*n+1/20*Pi^2*csgn(I*d)*csgn(I*d*(f*x
^2+e)^m)^2/x^5*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/20*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2/x^5*b*csgn(
I*c)*csgn(I*c*x^n)^2+1/20*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2/x^5*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1
/20*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^5*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/20*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e
)^m)*csgn(I*d*(f*x^2+e)^m)/x^5*b*csgn(I*c*x^n)^3-2/15*m*f/e/x^3*a-1/15*I*m*f/e/x^3*b*Pi*csgn(I*c)*csgn(I*c*x^n
)^2-1/15*I*m*f/e/x^3*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/5*I
*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3+1/20*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^
2/x^5*b*csgn(I*c)*csgn(I*c*x^n)^2-1/5*ln(d)/x^5*a+1/10*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m
)*b/x^5*ln(x^n)+1/50*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b/x^5*n-1/20*Pi^2*csgn(I*d)*csgn
(I*d*(f*x^2+e)^m)^2/x^5*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/20*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)
^m)^2/x^5*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/20*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/
x^5*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/15*I*m*f/e/x^3*b*
Pi*csgn(I*c*x^n)^3+1/10*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^5*b*ln(c)-1/20*Pi^2*csgn(I*
d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^5*b*csgn(I*c)*csgn(I*c*x^n)^2-1/20*Pi^2*csgn(I*d)*csgn(I*(f*x^2
+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^5*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/10*I*ln(d)/x^5*b*Pi*csgn(I*c)*csgn(I*x^n)*csg
n(I*c*x^n)-1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*c*x^n)^3-2/5*m*f^3*b/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*ln(x)+
1/5*m*f^3*b*n/e^2/(-e*f)^(1/2)*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/5*m*f^3*b*n/e^2/(-e*f)^(1/2)*ln(x)
*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/5*I*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I
*c*x^n)^2+1/5*I*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/15*I*m*f/e/x^
3*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/5*I*m*f^2/e^2/x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+(-1/5*b/
x^5*ln(x^n)-1/50*(-5*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+5*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+5*I*b*Pi*cs
gn(I*x^n)*csgn(I*c*x^n)^2-5*I*b*Pi*csgn(I*c*x^n)^3+10*b*ln(c)+2*b*n+10*a)/x^5)*ln((f*x^2+e)^m)-1/5*ln(d)*b/x^5
*ln(x^n)-1/5*ln(d)/x^5*b*ln(c)-1/25*ln(d)*b/x^5*n-1/50*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b/x^5*
n+1/10*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^5*a-1/10*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m
)^2/x^5*b*ln(c)+2/5/e^2*f^2*m/x*b*ln(c)+2/5*m*f^2*b*ln(x^n)/e^2/x-1/10*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^
2+e)^m)^2/x^5*b*ln(c)-1/10*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b/x^5*ln(x^n)+2/5/e^2*f^2*m/x*a-1/
20*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^5*b*csgn(I*c*x^n)^3-1/20*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^5*b*csgn(I
*c)*csgn(I*c*x^n)^2-1/20*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^5*b*csgn(I*x^n)*csgn(I*c*x^n)^2-2/15*m*f*b/e/x^3*ln(x^
n)+1/10*I*Pi*csgn(I*d*(f*x^2+e)^m)^3/x^5*b*ln(c)+1/10*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*b/x^5*ln(x^n)+1/50*I*Pi*csg
n(I*d*(f*x^2+e)^m)^3*b/x^5*n+1/10*I*ln(d)/x^5*b*Pi*csgn(I*c*x^n)^3-1/20*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x
^2+e)^m)^2/x^5*b*csgn(I*c*x^n)^3-1/10*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^5*a-1/10*I*Pi*csgn(I*(f*x^2+e)^
m)*csgn(I*d*(f*x^2+e)^m)^2/x^5*a+2/25*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*n+1/5*m*f^3*b*n/e^2/(-e*
f)^(1/2)*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))-1/5*m*f^3*b*n/e^2/(-e*f)^(1/2)*dilog((f*x+(-e*f)^(1/2))/(-e*f
)^(1/2))+2/5*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*ln(c)+2/5*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^
(1/2))*a-2/15*m*f/e/x^3*b*ln(c)-1/5*I*m*f^3/e^2/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*x^n)
*csgn(I*c*x^n)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^6, 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((a+b*ln(c*x**n))*ln(d*(f*x**2+e)**m)/x**6,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(f*x^2+e)^m)/x^6,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^6, x)