\(\int \frac {a+b \log (c (d+e x)^n)}{x^4 (f+g x^2)} \, dx\) [265]

Optimal result

Integrand size = 27, antiderivative size = 388 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=-\frac {b e n}{6 d f x^2}+\frac {b e^2 n}{3 d^2 f x}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e g n \log (x)}{d f^2}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{5/2}}-\frac {g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}-\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}+\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{5/2}} \]

[Out]

-1/6*b*e*n/d/f/x^2+1/3*b*e^2*n/d^2/f/x+1/3*b*e^3*n*ln(x)/d^3/f-b*e*g*n*ln(x)/d/f^2-1/3*b*e^3*n*ln(e*x+d)/d^3/f
+b*e*g*n*ln(e*x+d)/d/f^2+1/3*(-a-b*ln(c*(e*x+d)^n))/f/x^3+g*(a+b*ln(c*(e*x+d)^n))/f^2/x+1/2*g^(3/2)*(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)^(5/2)-1/2*g^(3/2)*(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)^(5/2)-1/2*b*g^(3/2)*n*polylog(2,-(e*x+d)*g^(1/2)/(
e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(5/2)+1/2*b*g^(3/2)*n*polylog(2,(e*x+d)*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(
5/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {331, 211, 2463, 2442, 46, 36, 29, 31, 2456, 2441, 2440, 2438} \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac {g^{3/2} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e^2 n}{3 d^2 f x}-\frac {b e g n \log (x)}{d f^2}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}+\frac {b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 (-f)^{5/2}}-\frac {b e n}{6 d f x^2} \]

[In]

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

[Out]

-1/6*(b*e*n)/(d*f*x^2) + (b*e^2*n)/(3*d^2*f*x) + (b*e^3*n*Log[x])/(3*d^3*f) - (b*e*g*n*Log[x])/(d*f^2) - (b*e^
3*n*Log[d + e*x])/(3*d^3*f) + (b*e*g*n*Log[d + e*x])/(d*f^2) - (a + b*Log[c*(d + e*x)^n])/(3*f*x^3) + (g*(a +
b*Log[c*(d + e*x)^n]))/(f^2*x) + (g^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f
] + d*Sqrt[g])])/(2*(-f)^(5/2)) - (g^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-
f] - d*Sqrt[g])])/(2*(-f)^(5/2)) - (b*g^(3/2)*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(
2*(-f)^(5/2)) + (b*g^(3/2)*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(2*(-f)^(5/2))

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 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 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 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 2442

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 2456

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

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*} \text {integral}& = \int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{f x^4}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x^2}+\frac {g^2 \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^4} \, dx}{f}-\frac {g \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2} \, dx}{f^2}+\frac {g^2 \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x^2} \, dx}{f^2} \\ & = -\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^2 \int \left (\frac {\sqrt {-f} \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {\sqrt {-f} \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 f \left (\sqrt {-f}+\sqrt {g} x\right )}\right ) \, dx}{f^2}+\frac {(b e n) \int \frac {1}{x^3 (d+e x)} \, dx}{3 f}-\frac {(b e g n) \int \frac {1}{x (d+e x)} \, dx}{f^2} \\ & = -\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}-\frac {g^2 \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}-\sqrt {g} x} \, dx}{2 (-f)^{5/2}}-\frac {g^2 \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}+\sqrt {g} x} \, dx}{2 (-f)^{5/2}}+\frac {(b e n) \int \left (\frac {1}{d x^3}-\frac {e}{d^2 x^2}+\frac {e^2}{d^3 x}-\frac {e^3}{d^3 (d+e x)}\right ) \, dx}{3 f}-\frac {(b e g n) \int \frac {1}{x} \, dx}{d f^2}+\frac {\left (b e^2 g n\right ) \int \frac {1}{d+e x} \, dx}{d f^2} \\ & = -\frac {b e n}{6 d f x^2}+\frac {b e^2 n}{3 d^2 f x}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e g n \log (x)}{d f^2}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \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)^{5/2}}-\frac {g^{3/2} \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)^{5/2}}-\frac {\left (b e g^{3/2} n\right ) \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)^{5/2}}+\frac {\left (b e g^{3/2} n\right ) \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)^{5/2}} \\ & = -\frac {b e n}{6 d f x^2}+\frac {b e^2 n}{3 d^2 f x}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e g n \log (x)}{d f^2}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \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)^{5/2}}-\frac {g^{3/2} \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)^{5/2}}+\frac {\left (b g^{3/2} n\right ) \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)^{5/2}}-\frac {\left (b g^{3/2} n\right ) \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)^{5/2}} \\ & = -\frac {b e n}{6 d f x^2}+\frac {b e^2 n}{3 d^2 f x}+\frac {b e^3 n \log (x)}{3 d^3 f}-\frac {b e g n \log (x)}{d f^2}-\frac {b e^3 n \log (d+e x)}{3 d^3 f}+\frac {b e g n \log (d+e x)}{d f^2}-\frac {a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {g^{3/2} \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)^{5/2}}-\frac {g^{3/2} \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)^{5/2}}-\frac {b g^{3/2} n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{5/2}}+\frac {b g^{3/2} n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 (-f)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\frac {1}{6} \left (-\frac {6 b e g n (\log (x)-\log (d+e x))}{d f^2}-\frac {b e n \left (d (d-2 e x)-2 e^2 x^2 \log (x)+2 e^2 x^2 \log (d+e x)\right )}{d^3 f x^2}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f x^3}+\frac {6 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {3 g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 g^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 b g^{3/2} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {3 b g^{3/2} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}\right ) \]

[In]

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

[Out]

((-6*b*e*g*n*(Log[x] - Log[d + e*x]))/(d*f^2) - (b*e*n*(d*(d - 2*e*x) - 2*e^2*x^2*Log[x] + 2*e^2*x^2*Log[d + e
*x]))/(d^3*f*x^2) - (2*(a + b*Log[c*(d + e*x)^n]))/(f*x^3) + (6*g*(a + b*Log[c*(d + e*x)^n]))/(f^2*x) + (3*g^(
3/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2) - (3*g^(3
/2)*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(-f)^(5/2) - (3*b*g^(
3/2)*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(-f)^(5/2) + (3*b*g^(3/2)*n*PolyLog[2, (Sq
rt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2))/6

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.75 (sec) , antiderivative size = 598, normalized size of antiderivative = 1.54

method	result	size
risch	\(-\frac {b \,g^{2} \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) n \ln \left (e x +d \right )}{f^{2} \sqrt {f g}}+\frac {b \,g^{2} \arctan \left (\frac {2 g \left (e x +d \right )-2 d g}{2 e \sqrt {f g}}\right ) \ln \left (\left (e x +d \right )^{n}\right )}{f^{2} \sqrt {f g}}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{3 f \,x^{3}}+\frac {b \ln \left (\left (e x +d \right )^{n}\right ) g}{f^{2} x}-\frac {e b n g \ln \left (e x \right )}{f^{2} d}+\frac {b e g n \ln \left (e x +d \right )}{d \,f^{2}}-\frac {b e n}{6 d f \,x^{2}}+\frac {b \,e^{2} n}{3 d^{2} f x}+\frac {e^{3} b n \ln \left (e x \right )}{3 f \,d^{3}}-\frac {b \,e^{3} n \ln \left (e x +d \right )}{3 d^{3} f}+\frac {b n \,g^{2} \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} \sqrt {-f g}}-\frac {b n \,g^{2} \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} \sqrt {-f g}}+\frac {b n \,g^{2} \operatorname {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} \sqrt {-f g}}-\frac {b n \,g^{2} \operatorname {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} \sqrt {-f g}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} b}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} b}{2}+b \ln \left (c \right )+a \right ) \left (-\frac {1}{3 f \,x^{3}}+\frac {g}{f^{2} x}+\frac {g^{2} \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{f^{2} \sqrt {f g}}\right )\)	\(598\)

[In]

int((a+b*ln(c*(e*x+d)^n))/x^4/(g*x^2+f),x,method=_RETURNVERBOSE)

[Out]

-b*g^2/f^2/(f*g)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*n*ln(e*x+d)+b*g^2/f^2/(f*g)^(1/2)*arctan(
1/2*(2*g*(e*x+d)-2*d*g)/e/(f*g)^(1/2))*ln((e*x+d)^n)-1/3*b*ln((e*x+d)^n)/f/x^3+b*ln((e*x+d)^n)/f^2*g/x-e*b*n/f
^2*g/d*ln(e*x)+b*e*g*n*ln(e*x+d)/d/f^2-1/6*b*e*n/d/f/x^2+1/3*b*e^2*n/d^2/f/x+1/3*e^3*b*n/f/d^3*ln(e*x)-1/3*b*e
^3*n*ln(e*x+d)/d^3/f+1/2*b*n*g^2/f^2*ln(e*x+d)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+
d*g))-1/2*b*n*g^2/f^2*ln(e*x+d)/(-f*g)^(1/2)*ln((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+1/2*b*n*g
^2/f^2/(-f*g)^(1/2)*dilog((e*(-f*g)^(1/2)-g*(e*x+d)+d*g)/(e*(-f*g)^(1/2)+d*g))-1/2*b*n*g^2/f^2/(-f*g)^(1/2)*di
log((e*(-f*g)^(1/2)+g*(e*x+d)-d*g)/(e*(-f*g)^(1/2)-d*g))+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*
x+d)^n)+1/2*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*b*
Pi*csgn(I*c*(e*x+d)^n)^3+b*ln(c)+a)*(-1/3/f/x^3+1/f^2*g/x+g^2/f^2/(f*g)^(1/2)*arctan(g*x/(f*g)^(1/2)))

Fricas [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="fricas")

[Out]

integral((b*log((e*x + d)^n*c) + a)/(g*x^6 + f*x^4), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate((a+b*ln(c*(e*x+d)**n))/x**4/(g*x**2+f),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="maxima")

[Out]

1/3*a*(3*g^2*arctan(g*x/sqrt(f*g))/(sqrt(f*g)*f^2) + (3*g*x^2 - f)/(f^2*x^3)) + b*integrate((log((e*x + d)^n)
+ log(c))/(g*x^6 + f*x^4), x)

Giac [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/x^4/(g*x^2+f),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^4\,\left (g\,x^2+f\right )} \,d x \]

[In]

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

[Out]

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