3.3.64 \(\int \frac {a+b \log (c (d+e x)^n)}{x^2 (f+g x^2)} \, dx\) [264]

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

[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 = 0.24, 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 = {331, 211, 2463, 2442, 36, 29, 31, 2456, 2441, 2440, 2438} \begin {gather*} -\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} \end {gather*}

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 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*} \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 ) \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)^{3/2}}-\frac {\left (b \sqrt {g} 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)^{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 = 0.12, size = 280, normalized size = 0.97 \begin {gather*} \frac {f \left (2 b e (-f)^{3/2} n x (\log (x)-\log (d+e x))+2 d \sqrt {-f} f \left (a+b \log \left (c (d+e x)^n\right )\right )+d f \sqrt {g} x \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 )-d f \sqrt {g} x \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 )-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)}{e \sqrt {-f}+d \sqrt {g}}\right )\right )}{2 d (-f)^{7/2} x} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.42, size = 722, normalized size = 2.49

method	result	size
risch	\(-\frac {b \ln \left (\left (e x +d \right )^{n}\right )}{f x}+\frac {b g \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 \sqrt {f g}}-\frac {b g \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 \sqrt {f g}}+\frac {b e n \ln \left (e x \right )}{f d}-\frac {b e n \ln \left (e x +d \right )}{d f}-\frac {b n g \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 \sqrt {-f g}}+\frac {b n g \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 \sqrt {-f g}}-\frac {b n g \dilog \left (\frac {e \sqrt {-f g}-g \left (e x +d \right )+d g}{e \sqrt {-f g}+d g}\right )}{2 f \sqrt {-f g}}+\frac {b n g \dilog \left (\frac {e \sqrt {-f g}+g \left (e x +d \right )-d g}{e \sqrt {-f g}-d g}\right )}{2 f \sqrt {-f g}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2} g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{2 f \sqrt {f g}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 f x}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3} g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{2 f \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}}{2 f x}+\frac {i b \pi \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 f x}-\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} g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{2 f \sqrt {f g}}+\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 ) g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{2 f \sqrt {f g}}+\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 )}{2 f x}-\frac {b \ln \left (c \right ) g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{f \sqrt {f g}}-\frac {b \ln \left (c \right )}{f x}-\frac {a g \arctan \left (\frac {x g}{\sqrt {f g}}\right )}{f \sqrt {f g}}-\frac {a}{f x}\)	\(722\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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