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

Optimal. Leaf size=267 \[ -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{5 x^5}+\frac {2 f^{5/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^{5/2}}+\frac {2 f^2 m \left (a+b \log \left (c x^n\right )\right )}{5 e^2 x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{15 e x^3}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{25 x^5}-\frac {i b f^{5/2} m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}+\frac {i b f^{5/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{5 e^{5/2}}+\frac {2 b f^{5/2} m n \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{25 e^{5/2}}+\frac {12 b f^2 m n}{25 e^2 x}-\frac {16 b f m n}{225 e x^3} \]

[Out]

-16/225*b*f*m*n/e/x^3+12/25*b*f^2*m*n/e^2/x+2/25*b*f^(5/2)*m*n*arctan(x*f^(1/2)/e^(1/2))/e^(5/2)-2/15*f*m*(a+b
*ln(c*x^n))/e/x^3+2/5*f^2*m*(a+b*ln(c*x^n))/e^2/x+2/5*f^(5/2)*m*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x^n))/e^(5
/2)-1/25*b*n*ln(d*(f*x^2+e)^m)/x^5-1/5*(a+b*ln(c*x^n))*ln(d*(f*x^2+e)^m)/x^5-1/5*I*b*f^(5/2)*m*n*polylog(2,-I*
x*f^(1/2)/e^(1/2))/e^(5/2)+1/5*I*b*f^(5/2)*m*n*polylog(2,I*x*f^(1/2)/e^(1/2))/e^(5/2)

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Rubi [A]  time = 0.19, antiderivative size = 267, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.18, size = 399, normalized size = 1.49 \[ -\frac {45 a e^{5/2} \log \left (d \left (e+f x^2\right )^m\right )+30 a e^{3/2} f m x^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {f x^2}{e}\right )+45 b e^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+30 b e^{3/2} f m x^2 \log \left (c x^n\right )-90 b f^{5/2} m x^5 \log \left (c x^n\right ) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-90 b \sqrt {e} f^2 m x^4 \log \left (c x^n\right )+9 b e^{5/2} n \log \left (d \left (e+f x^2\right )^m\right )+16 b e^{3/2} f m n x^2+45 i b f^{5/2} m n x^5 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-45 i b f^{5/2} m n x^5 \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )-45 i b f^{5/2} m n x^5 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+45 i b f^{5/2} m n x^5 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-18 b f^{5/2} m n x^5 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )+90 b f^{5/2} m n x^5 \log (x) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-108 b \sqrt {e} f^2 m n x^4}{225 e^{5/2} x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/225*(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*A
rcTan[(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]])/(e^(5/2)*x^5)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{6}}\,{d x} \]

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)

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maple [C]  time = 0.59, size = 2385, normalized size = 8.93 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (5 \, b m \log \left (x^{n}\right ) + {\left (m n + 5 \, m \log \relax (c)\right )} b + 5 \, a m\right )} \log \left (f x^{2} + e\right )}{25 \, x^{5}} + \int \frac {25 \, b e \log \relax (c) \log \relax (d) + {\left (5 \, {\left (2 \, f m + 5 \, f \log \relax (d)\right )} a + {\left (2 \, f m n + 5 \, {\left (2 \, f m + 5 \, f \log \relax (d)\right )} \log \relax (c)\right )} b\right )} x^{2} + 25 \, a e \log \relax (d) + 5 \, {\left ({\left (2 \, f m + 5 \, f \log \relax (d)\right )} b x^{2} + 5 \, b e \log \relax (d)\right )} \log \left (x^{n}\right )}{25 \, {\left (f x^{8} + e x^{6}\right )}}\,{d x} \]

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]

-1/25*(5*b*m*log(x^n) + (m*n + 5*m*log(c))*b + 5*a*m)*log(f*x^2 + e)/x^5 + integrate(1/25*(25*b*e*log(c)*log(d
) + (5*(2*f*m + 5*f*log(d))*a + (2*f*m*n + 5*(2*f*m + 5*f*log(d))*log(c))*b)*x^2 + 25*a*e*log(d) + 5*((2*f*m +
5*f*log(d))*b*x^2 + 5*b*e*log(d))*log(x^n))/(f*x^8 + e*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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