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

Optimal. Leaf size=276 \[ \frac {\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^{3/2}}-\frac {\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^{3/2}}+\frac {a x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac {b \sqrt {-f} n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^{3/2}}+\frac {b \sqrt {-f} n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{2 g^{3/2}}-\frac {b n x}{g} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 205, 2416, 2389, 2295, 2409, 2394, 2393, 2391} \[ -\frac {b \sqrt {-f} n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^{3/2}}+\frac {b \sqrt {-f} n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{2 g^{3/2}}+\frac {\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^{3/2}}-\frac {\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^{3/2}}+\frac {a x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}-\frac {b n x}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, 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 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 {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx &=\int \left (\frac {a+b \log \left (c (d+e x)^n\right )}{g}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \left (f+g x^2\right )}\right ) \, dx\\ &=\frac {\int \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g}-\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{f+g x^2} \, dx}{g}\\ &=\frac {a x}{g}+\frac {b \int \log \left (c (d+e x)^n\right ) \, dx}{g}-\frac {f \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}{g}\\ &=\frac {a x}{g}+\frac {b \operatorname {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x\right )}{e g}-\frac {\sqrt {-f} \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}-\sqrt {g} x} \, dx}{2 g}-\frac {\sqrt {-f} \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}+\sqrt {g} x} \, dx}{2 g}\\ &=\frac {a x}{g}-\frac {b n x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}+\frac {\sqrt {-f} \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 g^{3/2}}-\frac {\sqrt {-f} \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 g^{3/2}}-\frac {\left (b e \sqrt {-f} 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 g^{3/2}}+\frac {\left (b e \sqrt {-f} 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 g^{3/2}}\\ &=\frac {a x}{g}-\frac {b n x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}+\frac {\sqrt {-f} \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 g^{3/2}}-\frac {\sqrt {-f} \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 g^{3/2}}+\frac {\left (b \sqrt {-f} 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 g^{3/2}}-\frac {\left (b \sqrt {-f} 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 g^{3/2}}\\ &=\frac {a x}{g}-\frac {b n x}{g}+\frac {b (d+e x) \log \left (c (d+e x)^n\right )}{e g}+\frac {\sqrt {-f} \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 g^{3/2}}-\frac {\sqrt {-f} \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 g^{3/2}}-\frac {b \sqrt {-f} n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{2 g^{3/2}}+\frac {b \sqrt {-f} n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{2 g^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 263, normalized size = 0.95 \[ \frac {\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\sqrt {-f} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+2 a \sqrt {g} x+\frac {2 b \sqrt {g} (d+e x) \log \left (c (d+e x)^n\right )}{e}-b \sqrt {-f} n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )+b \sqrt {-f} n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )-2 b \sqrt {g} n x}{2 g^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{2}}{g x^{2} + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.30, size = 710, normalized size = 2.57 \[ \frac {i \pi b f \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}\, g}-\frac {i \pi b f \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}\, g}-\frac {i \pi b f \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}\, g}+\frac {i \pi b f \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}\, g}+\frac {b f 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}\, g}-\frac {b f 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}\, g}+\frac {b f 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}\, g}-\frac {i \pi b x \,\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 g}+\frac {i \pi b x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2 g}+\frac {i \pi b x \,\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 g}-\frac {i \pi b x \mathrm {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2 g}-\frac {b f 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}\, g}+\frac {b f 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}\, g}-\frac {b f \arctan \left (\frac {g x}{\sqrt {f g}}\right ) \ln \relax (c )}{\sqrt {f g}\, g}-\frac {b f \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}\, g}-\frac {a f \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{\sqrt {f g}\, g}-\frac {b n x}{g}+\frac {b x \ln \relax (c )}{g}+\frac {b x \ln \left (\left (e x +d \right )^{n}\right )}{g}+\frac {a x}{g}-\frac {b d n}{e g}+\frac {b d \ln \left (\left (e x +d \right )^{n}\right )}{e g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/g*x*ln((e*x+d)^n)+b/e/g*d*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*l
n(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/g*n*x-b/e*n/g*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*(e*x+d)^n)^3*f/g/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+1/2*I*Pi*b/g*x*
csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I*Pi*b/g*x*csgn(I*c*(e*x+d)^n)^3-1/2*I*Pi*b/g*x*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*(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*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*(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*Pi*b/g*x*csgn(I*(e*x+d)
^n)*csgn(I*c*(e*x+d)^n)^2+b/g*x*ln(c)-b*ln(c)*f/g/(f*g)^(1/2)*arctan(1/(f*g)^(1/2)*g*x)+a/g*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 \[ -a {\left (\frac {f \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{\sqrt {f g} g} - \frac {x}{g}\right )} + b \int \frac {x^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{2} \log \relax (c)}{g x^{2} + f}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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