3.236 \(\int \frac {x^4 (a+b \log (c x^n))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=211 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {3 i b n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}+\frac {3 i b n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}-\frac {b n x}{8 e^2 \left (d+e x^2\right )} \]

[Out]

-1/8*b*n*x/e^2/(e*x^2+d)+1/4*d*x*(a+b*ln(c*x^n))/e^2/(e*x^2+d)^2-5/8*x*(a+b*ln(c*x^n))/e^2/(e*x^2+d)+1/2*b*n*a
rctan(x*e^(1/2)/d^(1/2))/e^(5/2)/d^(1/2)+3/8*arctan(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/e^(5/2)/d^(1/2)-3/16*I*
b*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/e^(5/2)/d^(1/2)+3/16*I*b*n*polylog(2,I*x*e^(1/2)/d^(1/2))/e^(5/2)/d^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {288, 205, 2351, 2323, 2324, 12, 4848, 2391, 199} \[ -\frac {3 i b n \text {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}+\frac {3 i b n \text {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

-(b*n*x)/(8*e^2*(d + e*x^2)) + (b*n*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*Sqrt[d]*e^(5/2)) + (d*x*(a + b*Log[c*x^n])
)/(4*e^2*(d + e*x^2)^2) - (5*x*(a + b*Log[c*x^n]))/(8*e^2*(d + e*x^2)) + (3*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b
*Log[c*x^n]))/(8*Sqrt[d]*e^(5/2)) - (((3*I)/16)*b*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*e^(5/2)) +
(((3*I)/16)*b*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*e^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 2323

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(q + 1
)*(a + b*Log[c*x^n]))/(2*d*(q + 1)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[c*
x^n]), x], x] + Dist[(b*n)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &&
LtQ[q, -1]

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx &=\int \left (\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^3}-\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^2}+\frac {a+b \log \left (c x^n\right )}{e^2 \left (d+e x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{e^2}-\frac {(2 d) \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^2} \, dx}{e^2}+\frac {d^2 \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^3} \, dx}{e^2}\\ &=\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d} e^{5/2}}-\frac {\int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{e^2}+\frac {(3 d) \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^2} \, dx}{4 e^2}+\frac {(b n) \int \frac {1}{d+e x^2} \, dx}{e^2}-\frac {(b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{e^2}-\frac {(b d n) \int \frac {1}{\left (d+e x^2\right )^2} \, dx}{4 e^2}\\ &=-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {3 \int \frac {a+b \log \left (c x^n\right )}{d+e x^2} \, dx}{8 e^2}-\frac {(b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{\sqrt {d} e^{5/2}}-\frac {(b n) \int \frac {1}{d+e x^2} \, dx}{8 e^2}-\frac {(3 b n) \int \frac {1}{d+e x^2} \, dx}{8 e^2}+\frac {(b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{e^2}\\ &=-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {(i b n) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 \sqrt {d} e^{5/2}}+\frac {(i b n) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 \sqrt {d} e^{5/2}}+\frac {(b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{\sqrt {d} e^{5/2}}-\frac {(3 b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{8 e^2}\\ &=-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {i b n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {i b n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {(i b n) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 \sqrt {d} e^{5/2}}-\frac {(i b n) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 \sqrt {d} e^{5/2}}-\frac {(3 b n) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{8 \sqrt {d} e^{5/2}}\\ &=-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {(3 i b n) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{16 \sqrt {d} e^{5/2}}+\frac {(3 i b n) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{16 \sqrt {d} e^{5/2}}\\ &=-\frac {b n x}{8 e^2 \left (d+e x^2\right )}+\frac {b n \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 \sqrt {d} e^{5/2}}+\frac {d x \left (a+b \log \left (c x^n\right )\right )}{4 e^2 \left (d+e x^2\right )^2}-\frac {5 x \left (a+b \log \left (c x^n\right )\right )}{8 e^2 \left (d+e x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{8 \sqrt {d} e^{5/2}}-\frac {3 i b n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}+\frac {3 i b n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{16 \sqrt {d} e^{5/2}}\\ \end {align*}

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Mathematica [B]  time = 1.27, size = 495, normalized size = 2.35 \[ \frac {-\frac {3 \log \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}}+\frac {3 \log \left (\frac {d \sqrt {e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}}+\frac {5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}-\sqrt {e} x}-\frac {5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {-d}+\sqrt {e} x}-\frac {\sqrt {-d} \left (a+b \log \left (c x^n\right )\right )}{\left (\sqrt {-d}-\sqrt {e} x\right )^2}+\frac {\sqrt {-d} \left (a+b \log \left (c x^n\right )\right )}{\left (\sqrt {-d}+\sqrt {e} x\right )^2}+\frac {3 b n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{\sqrt {-d}}-\frac {3 b n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{\sqrt {-d}}-\frac {5 b n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{\sqrt {-d}}+\frac {5 b n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{\sqrt {-d}}-\frac {b n \left (\log (x) \left (d-\sqrt {-d} \sqrt {e} x\right )+\left (\sqrt {-d} \sqrt {e} x-d\right ) \log \left (\sqrt {-d}+\sqrt {e} x\right )+d\right )}{d \left (\sqrt {-d}+\sqrt {e} x\right )}+\frac {b n \left (\log (x) \left (\sqrt {-d} \sqrt {e} x+d\right )-\left (\sqrt {-d} \sqrt {e} x+d\right ) \log \left (d \sqrt {e} x+(-d)^{3/2}\right )+d\right )}{d \left (\sqrt {-d}-\sqrt {e} x\right )}}{16 e^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

(-((Sqrt[-d]*(a + b*Log[c*x^n]))/(Sqrt[-d] - Sqrt[e]*x)^2) + (5*(a + b*Log[c*x^n]))/(Sqrt[-d] - Sqrt[e]*x) + (
Sqrt[-d]*(a + b*Log[c*x^n]))/(Sqrt[-d] + Sqrt[e]*x)^2 - (5*(a + b*Log[c*x^n]))/(Sqrt[-d] + Sqrt[e]*x) - (5*b*n
*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]))/Sqrt[-d] + (5*b*n*(Log[x] - Log[Sqrt[-d] + Sqrt[e]*x]))/Sqrt[-d] - (b*n
*(d + (d - Sqrt[-d]*Sqrt[e]*x)*Log[x] + (-d + Sqrt[-d]*Sqrt[e]*x)*Log[Sqrt[-d] + Sqrt[e]*x]))/(d*(Sqrt[-d] + S
qrt[e]*x)) - (3*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/Sqrt[-d] + (b*n*(d + (d + Sqrt[-d]*Sqrt[e]*x
)*Log[x] - (d + Sqrt[-d]*Sqrt[e]*x)*Log[(-d)^(3/2) + d*Sqrt[e]*x]))/(d*(Sqrt[-d] - Sqrt[e]*x)) + (3*(a + b*Log
[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])/Sqrt[-d] + (3*b*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]])/Sqrt[-d] - (3*
b*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/Sqrt[-d])/(16*e^(5/2))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{4} \log \left (c x^{n}\right ) + a x^{4}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^4*log(c*x^n) + a*x^4)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}}{{\left (e x^{2} + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^4/(e*x^2 + d)^3, x)

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maple [C]  time = 0.36, size = 1311, normalized size = 6.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*ln(c*x^n)+a)/(e*x^2+d)^3,x)

[Out]

3/16*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/(e*x^2+d)^2*d/e^2*x-3/8*b*ln(c)/(e*x^2+d)^2*d/e^2*x-3/8*b/e^2/
(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*n*ln(x)+1/2*b*n/e^2/(-d*e)^(1/2)*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/
2))-1/2*b*n/e^2/(-d*e)^(1/2)*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-b*n/e^2*ln(x)*x/(e*x^2+d)+b*n/e*ln(x)/(
e*x^2+d)^2*x^3-5/8*a/(e*x^2+d)^2/e*x^3+3/8*a/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)+3/16*I*b*Pi*csgn(I*x^n)
*csgn(I*c*x^n)^2/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)-5/16*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/(e*x^2+d)^2
/e*x^3+3/16*b*n*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4-3/16*b*n*ln(x)/(e*x^2+
d)^2/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^4+b*n*d/e^2*ln(x)/(e*x^2+d)^2*x+3/8*b*n*d/e*ln(x)/(e*x
^2+d)^2/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-3/8*b*n*d/e*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((e
*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-3/16*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^2/(d*e)^(1/2)*arctan(1/(d
*e)^(1/2)*e*x)+5/16*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/(e*x^2+d)^2/e*x^3-3/8*b/(e*x^2+d)^2*d/e^2*x*ln(
x^n)-5/16*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/(e*x^2+d)^2/e*x^3-3/16*I*b*Pi*csgn(I*c*x^n)^3/e^2/(d*e)^(1/2)*arcta
n(1/(d*e)^(1/2)*e*x)-3/16*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/(e*x^2+d)^2*d/e^2*x+5/16*I*b*Pi*csgn(I*c*x^n)^3/(
e*x^2+d)^2/e*x^3-3/16*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/(e*x^2+d)^2*d/e^2*x+1/2*b*n*d/e^2*ln(x)/(e*x^2+d)/(-d*e
)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+3/16*b*n*d^2/e^2*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1
/2))/(-d*e)^(1/2))-3/16*b*n*d^2/e^2*ln(x)/(e*x^2+d)^2/(-d*e)^(1/2)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n
/e*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2+1/2*b*n/e*ln(x)/(e*x^2+d)/(-d*e)^(1/2
)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))*x^2-1/2*b*n*d/e^2*ln(x)/(e*x^2+d)/(-d*e)^(1/2)*ln((-e*x+(-d*e)^(1/2))/(-
d*e)^(1/2))+3/16*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)-5/8*b/(e*x^2+d)^2/
e*x^3*ln(x^n)+3/8*b/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*ln(x^n)-3/8*a/(e*x^2+d)^2*d/e^2*x+3/8*b*ln(c)/e^
2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)-5/8*b*ln(c)/(e*x^2+d)^2/e*x^3+1/2*b*n/e^2/(d*e)^(1/2)*arctan(1/(d*e)^(
1/2)*e*x)+3/16*b*n/e^2/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-3/16*b*n/e^2/(-d*e)^(1/2)*dilog((e
*x+(-d*e)^(1/2))/(-d*e)^(1/2))+3/16*I*b*Pi*csgn(I*c*x^n)^3/(e*x^2+d)^2*d/e^2*x-1/8*b*n*x/e^2/(e*x^2+d)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*log(c*x^n)))/(d + e*x^2)^3,x)

[Out]

int((x^4*(a + b*log(c*x^n)))/(d + e*x^2)^3, 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**4*(a+b*ln(c*x**n))/(e*x**2+d)**3,x)

[Out]

Timed out

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