3.3.31 \(\int \frac {x^5 (a+b \log (c x^n))}{(d+e x^2)^3} \, dx\) [231]

Optimal. Leaf size=152 \[ \frac {b d n}{8 e^3 \left (d+e x^2\right )}+\frac {b n \log (x)}{4 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{4 e^3 \left (d+e x^2\right )^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {3 b n \log \left (d+e x^2\right )}{8 e^3}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}+\frac {b n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e^3} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {272, 45, 2393, 2376, 46, 2373, 266, 2375, 2438} \begin {gather*} \frac {b n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{4 e^3 \left (d+e x^2\right )^2}+\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {b d n}{8 e^3 \left (d+e x^2\right )}+\frac {3 b n \log \left (d+e x^2\right )}{8 e^3}+\frac {b n \log (x)}{4 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

(b*d*n)/(8*e^3*(d + e*x^2)) + (b*n*Log[x])/(4*e^3) - (d^2*(a + b*Log[c*x^n]))/(4*e^3*(d + e*x^2)^2) - (x^2*(a
+ b*Log[c*x^n]))/(e^2*(d + e*x^2)) + (3*b*n*Log[d + e*x^2])/(8*e^3) + ((a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/
(2*e^3) + (b*n*PolyLog[2, -((e*x^2)/d)])/(4*e^3)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2373

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])/(d*f*(m + 1))), x] - Dist[b*(n/(d*(m + 1))), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rule 2375

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^p/(e*r)), x] - Dist[b*f^m*n*(p/(e*r)), Int[Log[1 + e*(x^r/d)]*((
a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2376

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :
> Simp[f^m*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^p/(e*r*(q + 1))), x] - Dist[b*f^m*n*(p/(e*r*(q + 1))), Int[
(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[
m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rule 2393

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 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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.36, size = 498, normalized size = 3.28 \begin {gather*} \frac {-2 d^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+8 d \left (d+e x^2\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )+4 \left (d+e x^2\right )^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )+b n \left (d^2+d e x^2-4 d e x^2 \log (x)-6 e^2 x^4 \log (x)+3 d^2 \log \left (i \sqrt {d}-\sqrt {e} x\right )+6 d e x^2 \log \left (i \sqrt {d}-\sqrt {e} x\right )+3 e^2 x^4 \log \left (i \sqrt {d}-\sqrt {e} x\right )+3 d^2 \log \left (i \sqrt {d}+\sqrt {e} x\right )+6 d e x^2 \log \left (i \sqrt {d}+\sqrt {e} x\right )+3 e^2 x^4 \log \left (i \sqrt {d}+\sqrt {e} x\right )+4 d^2 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+8 d e x^2 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 e^2 x^4 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 d^2 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+8 d e x^2 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 e^2 x^4 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 \left (d+e x^2\right )^2 \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+4 \left (d+e x^2\right )^2 \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )}{8 e^3 \left (d+e x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

(-2*d^2*(a - b*n*Log[x] + b*Log[c*x^n]) + 8*d*(d + e*x^2)*(a - b*n*Log[x] + b*Log[c*x^n]) + 4*(d + e*x^2)^2*(a
- b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] + b*n*(d^2 + d*e*x^2 - 4*d*e*x^2*Log[x] - 6*e^2*x^4*Log[x] + 3*d^
2*Log[I*Sqrt[d] - Sqrt[e]*x] + 6*d*e*x^2*Log[I*Sqrt[d] - Sqrt[e]*x] + 3*e^2*x^4*Log[I*Sqrt[d] - Sqrt[e]*x] + 3
*d^2*Log[I*Sqrt[d] + Sqrt[e]*x] + 6*d*e*x^2*Log[I*Sqrt[d] + Sqrt[e]*x] + 3*e^2*x^4*Log[I*Sqrt[d] + Sqrt[e]*x]
+ 4*d^2*Log[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 8*d*e*x^2*Log[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 4*e^2*x^4*Lo
g[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 4*d^2*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 8*d*e*x^2*Log[x]*Log[1 + (
I*Sqrt[e]*x)/Sqrt[d]] + 4*e^2*x^4*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 4*(d + e*x^2)^2*PolyLog[2, ((-I)*Sqr
t[e]*x)/Sqrt[d]] + 4*(d + e*x^2)^2*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))/(8*e^3*(d + e*x^2)^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 727, normalized size = 4.78

method	result	size
risch	\(-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e \,x^{2}+d \right )}{4 e^{3}}+\frac {b n \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{3}}+\frac {b n \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{3}}+\frac {b \ln \left (c \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {a \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e^{3}}+\frac {b \ln \left (x^{n}\right ) d}{e^{3} \left (e \,x^{2}+d \right )}+\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e^{3}}+\frac {b n \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{3}}+\frac {b n \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{3}}-\frac {b n \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}-\frac {a \,d^{2}}{4 e^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {a d}{e^{3} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (c \right ) d^{2}}{4 e^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {b \ln \left (c \right ) d}{e^{3} \left (e \,x^{2}+d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{4 e^{3}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{8 e^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d}{2 e^{3} \left (e \,x^{2}+d \right )}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d^{2}}{8 e^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d}{2 e^{3} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) d^{2}}{4 e^{3} \left (e \,x^{2}+d \right )^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d}{2 e^{3} \left (e \,x^{2}+d \right )}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d^{2}}{8 e^{3} \left (e \,x^{2}+d \right )^{2}}-\frac {3 b n \ln \left (x \right )}{4 e^{3}}+\frac {3 b n \ln \left (e \,x^{2}+d \right )}{8 e^{3}}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d^{2}}{8 e^{3} \left (e \,x^{2}+d \right )^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d}{2 e^{3} \left (e \,x^{2}+d \right )}+\frac {b d n}{8 e^{3} \left (e \,x^{2}+d \right )}\)	\(727\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*ln(c*x^n))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

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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(x^5*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/4*(2*e^(-3)*log(x^2*e + d) + (4*d*x^2*e + 3*d^2)/(x^4*e^5 + 2*d*x^2*e^4 + d^2*e^3))*a + b*integrate((x^5*log
(c) + x^5*log(x^n))/(x^6*e^3 + 3*d*x^4*e^2 + 3*d^2*x^2*e + d^3), 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(x^5*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^5*log(c*x^n) + a*x^5)/(x^6*e^3 + 3*d*x^4*e^2 + 3*d^2*x^2*e + d^3), x)

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Sympy [A]
time = 72.70, size = 403, normalized size = 2.65 \begin {gather*} \frac {a d^{2} \left (\begin {cases} \frac {x^{2}}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x^{2}\right )^{2}} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {a d \left (\begin {cases} \frac {x^{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x^{2}} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {a \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {b d^{2} n \left (\begin {cases} \frac {x^{2}}{2 d^{3}} & \text {for}\: e = 0 \\- \frac {1}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac {\log {\left (x \right )}}{2 d^{2} e} + \frac {\log {\left (\frac {d}{e} + x^{2} \right )}}{4 d^{2} e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x^{2}}{d^{3}} & \text {for}\: e = 0 \\- \frac {1}{2 e \left (d + e x^{2}\right )^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e^{2}} + \frac {b d n \left (\begin {cases} \frac {x^{2}}{2 d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x^{2} \right )}}{2 d e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {b d \left (\begin {cases} \frac {x^{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*ln(c*x**n))/(e*x**2+d)**3,x)

[Out]

a*d**2*Piecewise((x**2/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x**2)**2), True))/(2*e**2) - a*d*Piecewise((x**2/d**2,
Eq(e, 0)), (-1/(d*e + e**2*x**2), True))/e**2 + a*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2
*e**2) - b*d**2*n*Piecewise((x**2/(2*d**3), Eq(e, 0)), (-1/(4*d**2*e + 4*d*e**2*x**2) - log(x)/(2*d**2*e) + lo
g(d/e + x**2)/(4*d**2*e), True))/(2*e**2) + b*d**2*Piecewise((x**2/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x**2)**2),
True))*log(c*x**n)/(2*e**2) + b*d*n*Piecewise((x**2/(2*d**2), Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x**2)/(2*
d*e), True))/e**2 - b*d*Piecewise((x**2/d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))*log(c*x**n)/e**2 - b*n*
Piecewise((x**2/(2*d), Eq(e, 0)), (Piecewise((-polylog(2, e*x**2*exp_polar(I*pi)/d)/2, (Abs(x) < 1) & (1/Abs(x
) < 1)), (log(d)*log(x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1/x) - polylog(2,
e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1),
()), ((), (0, 0)), x)*log(d) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, True))/e, True))/(2*e**2) + b*Piecewis
e((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/(2*e**2)

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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(x^5*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^5/(x^2*e + d)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*log(c*x^n)))/(d + e*x^2)^3,x)

[Out]

int((x^5*(a + b*log(c*x^n)))/(d + e*x^2)^3, x)

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