3.1.98 \(\int \frac {(a+b \log (c x^n)) \log (d (e+f x^2)^m)}{x^4} \, dx\) [98]

Optimal. Leaf size=227 \[ -\frac {8 b f m n}{9 e x}-\frac {2 b f^{3/2} m n \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {i b f^{3/2} m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}} \]

[Out]

-8/9*b*f*m*n/e/x-2/9*b*f^(3/2)*m*n*arctan(x*f^(1/2)/e^(1/2))/e^(3/2)-2/3*f*m*(a+b*ln(c*x^n))/e/x-2/3*f^(3/2)*m
*arctan(x*f^(1/2)/e^(1/2))*(a+b*ln(c*x^n))/e^(3/2)-1/9*b*n*ln(d*(f*x^2+e)^m)/x^3-1/3*(a+b*ln(c*x^n))*ln(d*(f*x
^2+e)^m)/x^3+1/3*I*b*f^(3/2)*m*n*polylog(2,-I*x*f^(1/2)/e^(1/2))/e^(3/2)-1/3*I*b*f^(3/2)*m*n*polylog(2,I*x*f^(
1/2)/e^(1/2))/e^(3/2)

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Rubi [A]
time = 0.11, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2505, 331, 211, 2423, 4940, 2438} \begin {gather*} \frac {i b f^{3/2} m n \text {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \text {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {2 f^{3/2} m \text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 b f^{3/2} m n \text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {8 b f m n}{9 e x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*f*m*n)/(9*e*x) - (2*b*f^(3/2)*m*n*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(9*e^(3/2)) - (2*f*m*(a + b*Log[c*x^n]))/
(3*e*x) - (2*f^(3/2)*m*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*x^n]))/(3*e^(3/2)) - (b*n*Log[d*(e + f*x^2)^m]
)/(9*x^3) - ((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/(3*x^3) + ((I/3)*b*f^(3/2)*m*n*PolyLog[2, ((-I)*Sqrt[f]*
x)/Sqrt[e]])/e^(3/2) - ((I/3)*b*f^(3/2)*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/e^(3/2)

Rule 211

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

Rule 331

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 2423

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 2438

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 2505

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 4940

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^4} \, dx &=-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}-(b n) \int \left (-\frac {2 f m}{3 e x^2}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2} x}-\frac {\log \left (d \left (e+f x^2\right )^m\right )}{3 x^4}\right ) \, dx\\ &=-\frac {2 b f m n}{3 e x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {1}{3} (b n) \int \frac {\log \left (d \left (e+f x^2\right )^m\right )}{x^4} \, dx+\frac {\left (2 b f^{3/2} m n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 e^{3/2}}\\ &=-\frac {2 b f m n}{3 e x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {1}{9} (2 b f m n) \int \frac {1}{x^2 \left (e+f x^2\right )} \, dx+\frac {\left (i b f^{3/2} m n\right ) \int \frac {\log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 e^{3/2}}-\frac {\left (i b f^{3/2} m n\right ) \int \frac {\log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{x} \, dx}{3 e^{3/2}}\\ &=-\frac {8 b f m n}{9 e x}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {i b f^{3/2} m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {\left (2 b f^2 m n\right ) \int \frac {1}{e+f x^2} \, dx}{9 e}\\ &=-\frac {8 b f m n}{9 e x}-\frac {2 b f^{3/2} m n \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2}}-\frac {2 f m \left (a+b \log \left (c x^n\right )\right )}{3 e x}-\frac {2 f^{3/2} m \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )}{3 e^{3/2}}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{3 x^3}+\frac {i b f^{3/2} m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {i b f^{3/2} m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{3 e^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.08, size = 362, normalized size = 1.59 \begin {gather*} \frac {-8 b \sqrt {e} f m n x^2-2 b f^{3/2} m n x^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )-6 a \sqrt {e} f m x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {f x^2}{e}\right )+6 b f^{3/2} m n x^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log (x)-6 b \sqrt {e} f m x^2 \log \left (c x^n\right )-6 b f^{3/2} m x^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \log \left (c x^n\right )-3 i b f^{3/2} m n x^3 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+3 i b f^{3/2} m n x^3 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 a e^{3/2} \log \left (d \left (e+f x^2\right )^m\right )-b e^{3/2} n \log \left (d \left (e+f x^2\right )^m\right )-3 b e^{3/2} \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+3 i b f^{3/2} m n x^3 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-3 i b f^{3/2} m n x^3 \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{9 e^{3/2} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*Sqrt[e]*f*m*n*x^2 - 2*b*f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]] - 6*a*Sqrt[e]*f*m*x^2*Hypergeometric
2F1[-1/2, 1, 1/2, -((f*x^2)/e)] + 6*b*f^(3/2)*m*n*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[x] - 6*b*Sqrt[e]*f*m*x^2
*Log[c*x^n] - 6*b*f^(3/2)*m*x^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]]*Log[c*x^n] - (3*I)*b*f^(3/2)*m*n*x^3*Log[x]*Log[1
- (I*Sqrt[f]*x)/Sqrt[e]] + (3*I)*b*f^(3/2)*m*n*x^3*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - 3*a*e^(3/2)*Log[d*(
e + f*x^2)^m] - b*e^(3/2)*n*Log[d*(e + f*x^2)^m] - 3*b*e^(3/2)*Log[c*x^n]*Log[d*(e + f*x^2)^m] + (3*I)*b*f^(3/
2)*m*n*x^3*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] - (3*I)*b*f^(3/2)*m*n*x^3*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(
9*e^(3/2)*x^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.19, size = 2204, normalized size = 9.71

method	result	size
risch	\(\text {Expression too large to display}\)	\(2204\)

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^4,x,method=_RETURNVERBOSE)

[Out]

(-1/3*b/x^3*ln(x^n)-1/18*(-3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+3*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+3*I
*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-3*I*b*Pi*csgn(I*c*x^n)^3+6*b*ln(c)+2*b*n+6*a)/x^3)*ln((f*x^2+e)^m)+1/3*I*m*f
^2/e/(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)-1/3*m*f^2*b*n/e/(-e*f)^(1/2)
*ln(x)*ln((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/3*m*f^2*b*n/e/(-e*f)^(1/2)*ln(x)*ln((f*x+(-e*f)^(1/2))/(-e*f)^(1
/2))+1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^3*b*csgn(I*c*x^n)^3+1/6*I*Pi*csgn(I*d*(f*x^2+e)^m)^3/x^3*a-1/3/x^3*ln
(d)*a-1/6*I/x^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*c*x^n)^2-2/3*m*f^2*b/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*ln(x^n)
+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/12*Pi^2*csgn(I*(f*x^2+e)^m)*c
sgn(I*d*(f*x^2+e)^m)^2/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^3*b*csgn(
I*c*x^n)^3-2/3*m*f*b*ln(x^n)/e/x-1/3*ln(d)*b/x^3*ln(x^n)+2/3*m*f^2*b/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*n*l
n(x)-1/3*I*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/3*I*m*f^2/e/(e*f)^(1
/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c*x^n)^3-1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^3*b*csgn(I*c)*csgn(I*c*x^
n)^2-1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*
d*(f*x^2+e)^m)^2/x^3*b*csgn(I*c*x^n)^3-1/6*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2/x^3*a+1/6*I*Pi*csg
n(I*d*(f*x^2+e)^m)^3/x^3*b*ln(c)+1/6*I*Pi*csgn(I*d*(f*x^2+e)^m)^3*b/x^3*ln(x^n)+1/18*I*Pi*csgn(I*d*(f*x^2+e)^m
)^3*b*n/x^3+1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^3*a-2/3*m*f^2/e/(e*f)^(1/2)*arctan(
x*f/(e*f)^(1/2))*b*ln(c)-1/6*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b/x^3*ln(x^n)+1/12*Pi^2*csgn(I*d
)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^3*b*csgn(I*c*x^n)^3+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/
x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-2/3/e*f*m/x*b*ln(c)-1/3*I*m*f/e/x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/3*I*m*f/e/x
*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-8/9*b*f*m*n/e/x+1/12*Pi^2*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m
)/x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2/3/e*f*m/x*a+1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^
2/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/12*Pi^2*csgn(I*d*(f*x^2+e)^m)^3/x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n
)-1/18*I*Pi*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2*b*n/x^3-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^
3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/12*Pi^2*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)^2/x^3*b*csgn(I*c)*
csgn(I*x^n)*csgn(I*c*x^n)+1/3*I*m*f/e/x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/18*I*Pi*csgn(I*d)*csgn(I*d*
(f*x^2+e)^m)^2*b*n/x^3-1/6*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2/x^3*b*ln(c)-1/6*I*Pi*csgn(I*(f*x^2+e)^m)*csg
n(I*d*(f*x^2+e)^m)^2/x^3*b*ln(c)-1/6*I*Pi*csgn(I*d)*csgn(I*d*(f*x^2+e)^m)^2*b/x^3*ln(x^n)-1/6*I*Pi*csgn(I*d)*c
sgn(I*d*(f*x^2+e)^m)^2/x^3*a-1/9/x^3*ln(d)*b*n-1/3/x^3*ln(d)*ln(c)*b-1/6*I/x^3*Pi*ln(d)*b*csgn(I*x^n)*csgn(I*c
*x^n)^2+1/3*I*m*f/e/x*b*Pi*csgn(I*c*x^n)^3+1/6*I/x^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/12*Pi^2*
csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*d)*csgn(I
*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*c
sgn(I*d*(f*x^2+e)^m)/x^3*b*ln(c)+1/6*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b/x^3*ln(x^n)+1/
18*I*Pi*csgn(I*d)*csgn(I*(f*x^2+e)^m)*csgn(I*d*(f*x^2+e)^m)*b*n/x^3-2/9*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(
1/2))*b*n-1/3*m*f^2*b*n/e/(-e*f)^(1/2)*dilog((-f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/3*m*f^2*b*n/e/(-e*f)^(1/2)*di
log((f*x+(-e*f)^(1/2))/(-e*f)^(1/2))+1/6*I/x^3*Pi*ln(d)*b*csgn(I*c*x^n)^3-2/3*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(
e*f)^(1/2))*a-1/3*I*m*f^2/e/(e*f)^(1/2)*arctan(x*f/(e*f)^(1/2))*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^4,x, algorithm="maxima")

[Out]

-1/9*(3*b*m*log(x^n) + (m*n + 3*m*log(c))*b + 3*a*m)*log(f*x^2 + e)/x^3 + integrate(1/9*((3*(2*f*m + 3*f*log(d
))*a + (2*f*m*n + 3*(2*f*m + 3*f*log(d))*log(c))*b)*x^2 + 9*(b*log(c)*log(d) + a*log(d))*e + 3*((2*f*m + 3*f*l
og(d))*b*x^2 + 3*b*e*log(d))*log(x^n))/(f*x^6 + x^4*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)*log((f*x^2 + e)^m*d)/x^4, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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**4,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (d\,{\left (f\,x^2+e\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \end {gather*}

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^4,x)

[Out]

int((log(d*(e + f*x^2)^m)*(a + b*log(c*x^n)))/x^4, x)

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