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

Optimal. Leaf size=195 \[ -\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 x^2}+\frac {f m \log (x) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {f m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}+\frac {b f m n \text {Li}_2\left (\frac {f x^2}{e}+1\right )}{4 e}-\frac {b f m n \log \left (e+f x^2\right )}{4 e}+\frac {b f m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{4 e}-\frac {b f m n \log ^2(x)}{2 e}+\frac {b f m n \log (x)}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {2454, 2395, 36, 29, 31, 2376, 2301, 2394, 2315} \[ \frac {b f m n \text {PolyLog}\left (2,\frac {f x^2}{e}+1\right )}{4 e}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^2\right )^m\right )}{2 x^2}+\frac {f m \log (x) \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {f m \log \left (e+f x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \log \left (d \left (e+f x^2\right )^m\right )}{4 x^2}-\frac {b f m n \log \left (e+f x^2\right )}{4 e}+\frac {b f m n \log \left (-\frac {f x^2}{e}\right ) \log \left (e+f x^2\right )}{4 e}-\frac {b f m n \log ^2(x)}{2 e}+\frac {b f m n \log (x)}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 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 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

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

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Mathematica [C]  time = 0.13, size = 298, normalized size = 1.53 \[ -\frac {2 a e \log \left (d \left (e+f x^2\right )^m\right )+2 a f m x^2 \log \left (e+f x^2\right )-4 a f m x^2 \log (x)+2 b e \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b f m x^2 \log \left (c x^n\right ) \log \left (e+f x^2\right )-4 b f m x^2 \log (x) \log \left (c x^n\right )+b e n \log \left (d \left (e+f x^2\right )^m\right )+2 b f m n x^2 \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b f m n x^2 \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b f m n x^2 \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b f m n x^2 \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+b f m n x^2 \log \left (e+f x^2\right )-2 b f m n x^2 \log (x) \log \left (e+f x^2\right )+2 b f m n x^2 \log ^2(x)-2 b f m n x^2 \log (x)}{4 e x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[d*(e + f*x^2)^m])/x^3,x]

[Out]

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

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x^{3}}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

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

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

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^3,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.68, size = 2101, normalized size = 10.77 \[ \text {Expression too large to display} \]

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^3,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (2 \, b m \log \left (x^{n}\right ) + {\left (m n + 2 \, m \log \relax (c)\right )} b + 2 \, a m\right )} \log \left (f x^{2} + e\right )}{4 \, x^{2}} + \int \frac {2 \, b e \log \relax (c) \log \relax (d) + {\left (2 \, {\left (f m + f \log \relax (d)\right )} a + {\left (f m n + 2 \, {\left (f m + f \log \relax (d)\right )} \log \relax (c)\right )} b\right )} x^{2} + 2 \, a e \log \relax (d) + 2 \, {\left ({\left (f m + f \log \relax (d)\right )} b x^{2} + b e \log \relax (d)\right )} \log \left (x^{n}\right )}{2 \, {\left (f x^{5} + e x^{3}\right )}}\,{d x} \]

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^3,x, algorithm="maxima")

[Out]

-1/4*(2*b*m*log(x^n) + (m*n + 2*m*log(c))*b + 2*a*m)*log(f*x^2 + e)/x^2 + integrate(1/2*(2*b*e*log(c)*log(d) +
(2*(f*m + f*log(d))*a + (f*m*n + 2*(f*m + f*log(d))*log(c))*b)*x^2 + 2*a*e*log(d) + 2*((f*m + f*log(d))*b*x^2
+ b*e*log(d))*log(x^n))/(f*x^5 + e*x^3), x)

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

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^3,x)

[Out]

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

[Out]

Timed out

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