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

Optimal. Leaf size=290 \[ \frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}-\frac {b \sqrt {g} n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{3/2}}+\frac {b \sqrt {g} n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 (-f)^{3/2}}+\frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {325, 205, 2416, 2395, 36, 29, 31, 2409, 2394, 2393, 2391} \[ -\frac {b \sqrt {g} n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 (-f)^{3/2}}+\frac {b \sqrt {g} n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 (-f)^{3/2}}+\frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {\sqrt {g} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 (-f)^{3/2}}-\frac {a+b \log \left (c (d+e x)^n\right )}{f x}+\frac {b e n \log (x)}{d f}-\frac {b e n \log (d+e x)}{d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 280, normalized size = 0.97 \[ \frac {f \left (d f \sqrt {g} x \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-d f \sqrt {g} x \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+2 d f \sqrt {-f} \left (a+b \log \left (c (d+e x)^n\right )\right )-b d f \sqrt {g} n x \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b d f \sqrt {g} n x \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )+2 b e (-f)^{3/2} n x (\log (x)-\log (d+e x))\right )}{2 d (-f)^{7/2} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(2*b*e*(-f)^(3/2)*n*x*(Log[x] - Log[d + e*x]) + 2*d*Sqrt[-f]*f*(a + b*Log[c*(d + e*x)^n]) + d*f*Sqrt[g]*x*(
a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])] - d*f*Sqrt[g]*x*(a + b*Log[
c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])] - b*d*f*Sqrt[g]*n*x*PolyLog[2, -((Sqr
t[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))] + b*d*f*Sqrt[g]*n*x*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*
Sqrt[g])]))/(2*d*(-f)^(7/2)*x)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{g x^{4} + f x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.46, size = 722, normalized size = 2.49 \[ \frac {i \pi b g \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \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 )}{2 \sqrt {f g}\, f}-\frac {i \pi b g \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 \sqrt {f g}\, f}-\frac {i \pi b g \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 \sqrt {f g}\, f}+\frac {i \pi b g \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 \sqrt {f g}\, f}+\frac {b g n \arctan \left (\frac {-2 d g +2 \left (e x +d \right ) g}{2 \sqrt {f g}\, e}\right ) \ln \left (e x +d \right )}{\sqrt {f g}\, f}-\frac {b g n \ln \left (\frac {d g +\sqrt {-f g}\, e -\left (e x +d \right ) g}{d g +\sqrt {-f g}\, e}\right ) \ln \left (e x +d \right )}{2 \sqrt {-f g}\, f}+\frac {b g n \ln \left (\frac {-d g +\sqrt {-f g}\, e +\left (e x +d \right ) g}{-d g +\sqrt {-f g}\, e}\right ) \ln \left (e x +d \right )}{2 \sqrt {-f g}\, f}-\frac {b g n \dilog \left (\frac {d g +\sqrt {-f g}\, e -\left (e x +d \right ) g}{d g +\sqrt {-f g}\, e}\right )}{2 \sqrt {-f g}\, f}+\frac {b g n \dilog \left (\frac {-d g +\sqrt {-f g}\, e +\left (e x +d \right ) g}{-d g +\sqrt {-f g}\, e}\right )}{2 \sqrt {-f g}\, f}-\frac {b g \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \ln \relax (c )}{\sqrt {f g}\, f}-\frac {b g \arctan \left (\frac {-2 d g +2 \left (e x +d \right ) g}{2 \sqrt {f g}\, e}\right ) \ln \left (\left (e x +d \right )^{n}\right )}{\sqrt {f g}\, f}-\frac {a g \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{\sqrt {f g}\, f}+\frac {b e n \ln \left (e x \right )}{d f}-\frac {b e n \ln \left (e x +d \right )}{d f}+\frac {i \pi b \,\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 )}{2 f x}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 f x}-\frac {i \pi b \,\mathrm {csgn}\left (i \left (e x +d \right )^{n}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 f x}+\frac {i \pi b \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 f x}-\frac {b \ln \relax (c )}{f x}-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f x}-\frac {a}{f x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {g \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{\sqrt {f g} f} + \frac {1}{f x}\right )} + b \int \frac {\log \left ({\left (e x + d\right )}^{n}\right ) + \log \relax (c)}{g x^{4} + f x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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