3.3.69 \(\int \frac {a+b \log (c (d+e x)^n)}{x (f+g x^2)^2} \, dx\) [269]

Optimal. Leaf size=383 \[ -\frac {b d e \sqrt {g} n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{3/2} \left (e^2 f+d^2 g\right )}-\frac {b e^2 n \log (d+e x)}{2 f \left (e^2 f+d^2 g\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}+\frac {b e^2 n \log \left (f+g x^2\right )}{4 f \left (e^2 f+d^2 g\right )}-\frac {b n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}-\frac {b n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 f^2}+\frac {b n \text {Li}_2\left (1+\frac {e x}{d}\right )}{f^2} \]

[Out]

-1/2*b*e^2*n*ln(e*x+d)/f/(d^2*g+e^2*f)+1/2*(a+b*ln(c*(e*x+d)^n))/f/(g*x^2+f)+ln(-e*x/d)*(a+b*ln(c*(e*x+d)^n))/
f^2+1/4*b*e^2*n*ln(g*x^2+f)/f/(d^2*g+e^2*f)-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2
)+d*g^(1/2)))/f^2-1/2*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)+x*g^(1/2))/(e*(-f)^(1/2)-d*g^(1/2)))/f^2+b*n*poly
log(2,1+e*x/d)/f^2-1/2*b*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/f^2-1/2*b*n*polylog(2,(e*x+d)*
g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/f^2-1/2*b*d*e*n*arctan(x*g^(1/2)/f^(1/2))*g^(1/2)/f^(3/2)/(d^2*g+e^2*f)

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Rubi [A]
time = 0.33, antiderivative size = 383, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.482, Rules used = {272, 46, 2463, 2441, 2352, 2460, 720, 31, 649, 211, 266, 2440, 2438} \begin {gather*} -\frac {b n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 f^2}-\frac {b n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 f^2}+\frac {b n \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{f^2}-\frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}-\frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f^2}+\frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 f \left (f+g x^2\right )}-\frac {b d e \sqrt {g} n \text {ArcTan}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 f^{3/2} \left (d^2 g+e^2 f\right )}+\frac {b e^2 n \log \left (f+g x^2\right )}{4 f \left (d^2 g+e^2 f\right )}-\frac {b e^2 n \log (d+e x)}{2 f \left (d^2 g+e^2 f\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

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 266

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 720

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

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 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

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 2460

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_)^(r_.))^(q_.), x_
Symbol] :> Simp[(f + g*x^r)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^p/(g*r*(q + 1))), x] - Dist[b*e*n*(p/(g*r*(q +
1))), Int[(f + g*x^r)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, n, q, r}, x] && EqQ[m, r - 1] && NeQ[q, -1] && IGtQ[p, 0]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.90, size = 521, normalized size = 1.36 \begin {gather*} \frac {a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )}{2 f^2+2 f g x^2}+\frac {\log (x) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{f^2}-\frac {\left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right ) \log \left (f+g x^2\right )}{2 f^2}+\frac {b n \left (\frac {\sqrt {f} \left (-i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (i \sqrt {f}-\sqrt {g} x\right )\right )}{\left (e \sqrt {f}-i d \sqrt {g}\right ) \left (\sqrt {f}+i \sqrt {g} x\right )}+\frac {\sqrt {f} \left (i \sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {f}-i \sqrt {g} x\right ) \log \left (i \sqrt {f}+\sqrt {g} x\right )\right )}{\left (e \sqrt {f}+i d \sqrt {g}\right ) \left (\sqrt {f}-i \sqrt {g} x\right )}-2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )+\text {Li}_2\left (-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )\right )-2 \left (\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )+\text {Li}_2\left (\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}+i d \sqrt {g}}\right )\right )+4 \left (\log \left (-\frac {e x}{d}\right ) \log (d+e x)+\text {Li}_2\left (1+\frac {e x}{d}\right )\right )\right )}{4 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.39, size = 910, normalized size = 2.38

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{2 f \left (g \,x^{2}+f \right )}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (g \,x^{2}+f \right )}{4 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )}{4 f \left (g \,x^{2}+f \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right ) \ln \left (x \right )}{2 f^{2}}-\frac {a \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b \ln \left (c \right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}+\frac {b \ln \left (c \right )}{2 f \left (g \,x^{2}+f \right )}+\frac {a}{2 f \left (g \,x^{2}+f \right )}+\frac {a \ln \left (x \right )}{f^{2}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) \ln \left (x \right )}{f^{2}}+\frac {b n \ln \left (e x +d \right ) \ln \left (g \,x^{2}+f \right )}{2 f^{2}}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f^{2}}-\frac {b n \ln \left (e x +d \right ) \ln \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f^{2}}+\frac {b \ln \left (c \right ) \ln \left (x \right )}{f^{2}}-\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +d}{d}\right )}{f^{2}}-\frac {b n \dilog \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f^{2}}-\frac {b n \dilog \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f^{2}}-\frac {b n \dilog \left (\frac {e x +d}{d}\right )}{f^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (g \,x^{2}+f \right )}{4 f^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} \ln \left (x \right )}{2 f^{2}}-\frac {b e n g d \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{2 f \left (d^{2} g +f \,e^{2}\right ) \sqrt {f g}}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 f^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (x \right )}{2 f^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} \ln \left (g \,x^{2}+f \right )}{4 f^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4 f \left (g \,x^{2}+f \right )}+\frac {i b \pi \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4 f \left (g \,x^{2}+f \right )}-\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{4 f \left (g \,x^{2}+f \right )}-\frac {b \,e^{2} n \ln \left (e x +d \right )}{2 f \left (d^{2} g +f \,e^{2}\right )}+\frac {b \,e^{2} n \ln \left (g \,x^{2}+f \right )}{4 f \left (d^{2} g +f \,e^{2}\right )}\)	\(910\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/x/(g*x^2+f)^2,x,method=_RETURNVERBOSE)

[Out]

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

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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*(e*x+d)^n))/x/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

1/2*a*(1/(f*g*x^2 + f^2) - log(g*x^2 + f)/f^2 + 2*log(x)/f^2) + b*integrate((log((x*e + d)^n) + log(c))/(g^2*x
^5 + 2*f*g*x^3 + f^2*x), 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*(e*x+d)^n))/x/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral((b*log((x*e + d)^n*c) + a)/(g^2*x^5 + 2*f*g*x^3 + f^2*x), 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*(e*x+d)**n))/x/(g*x**2+f)**2,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*(e*x+d)^n))/x/(g*x^2+f)^2,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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