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

Optimal. Leaf size=388 \[ -\frac{b g^{3/2} n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 (-f)^{5/2}}+\frac{b g^{3/2} n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 (-f)^{5/2}}+\frac{g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac{g^{3/2} \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac{g^{3/2} \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac{a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac{b e^2 n}{3 d^2 f x}+\frac{b e^3 n \log (x)}{3 d^3 f}-\frac{b e^3 n \log (d+e x)}{3 d^3 f}-\frac{b e g n \log (x)}{d f^2}+\frac{b e g n \log (d+e x)}{d f^2}-\frac{b e n}{6 d f x^2} \]

[Out]

-(b*e*n)/(6*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))

________________________________________________________________________________________

Rubi [A]  time = 0.37595, antiderivative size = 388, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 12, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {325, 205, 2416, 2395, 44, 36, 29, 31, 2409, 2394, 2393, 2391} \[ -\frac{b g^{3/2} n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{2 (-f)^{5/2}}+\frac{b g^{3/2} n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{2 (-f)^{5/2}}+\frac{g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac{g^{3/2} \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac{g^{3/2} \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{5/2}}-\frac{a+b \log \left (c (d+e x)^n\right )}{3 f x^3}+\frac{b e^2 n}{3 d^2 f x}+\frac{b e^3 n \log (x)}{3 d^3 f}-\frac{b e^3 n \log (d+e x)}{3 d^3 f}-\frac{b e g n \log (x)}{d f^2}+\frac{b e g n \log (d+e x)}{d f^2}-\frac{b e n}{6 d f x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*n)/(6*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 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 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 2416

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]

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 44

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] && L
tQ[m + n + 2, 0])

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 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 2409

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 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 2393

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 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]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c (d+e x)^n\right )}{x^4 \left (f+g x^2\right )} \, dx &=\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 ) \operatorname{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 ) \operatorname{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]  time = 0.348928, size = 350, normalized size = 0.9 \[ \frac{1}{6} \left (-\frac{3 b g^{3/2} n \text{PolyLog}\left (2,-\frac{\sqrt{g} (d+e x)}{e \sqrt{-f}-d \sqrt{g}}\right )}{(-f)^{5/2}}+\frac{3 b g^{3/2} n \text{PolyLog}\left (2,\frac{\sqrt{g} (d+e x)}{d \sqrt{g}+e \sqrt{-f}}\right )}{(-f)^{5/2}}+\frac{6 g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac{3 g^{3/2} \log \left (\frac{e \left (\sqrt{-f}-\sqrt{g} x\right )}{d \sqrt{g}+e \sqrt{-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(-f)^{5/2}}-\frac{3 g^{3/2} \log \left (\frac{e \left (\sqrt{-f}+\sqrt{g} x\right )}{e \sqrt{-f}-d \sqrt{g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{(-f)^{5/2}}-\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f x^3}-\frac{b e n \left (2 e^2 x^2 \log (d+e x)+d (d-2 e x)-2 e^2 x^2 \log (x)\right )}{d^3 f x^2}-\frac{6 b e g n (\log (x)-\log (d+e x))}{d f^2}\right ) \]

Antiderivative was successfully verified.

[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

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Maple [C]  time = 0.602, size = 983, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g x^{6} + f x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)