3.215 \(\int \frac {a+b \log (c x^n)}{x^5 (d+e x^2)} \, dx\)

Optimal. Leaf size=121 \[ -\frac {e^2 \log \left (\frac {d}{e x^2}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {b e^2 n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{4 d^3}+\frac {b e n}{4 d^2 x^2}-\frac {b n}{16 d x^4} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 149, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {266, 44, 2351, 2304, 2301, 2337, 2391} \[ -\frac {b e^2 n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 d^3}+\frac {e^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^3 n}-\frac {e^2 \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^3}+\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}-\frac {a+b \log \left (c x^n\right )}{4 d x^4}+\frac {b e n}{4 d^2 x^2}-\frac {b n}{16 d x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n)/(16*d*x^4) + (b*e*n)/(4*d^2*x^2) - (a + b*Log[c*x^n])/(4*d*x^4) + (e*(a + b*Log[c*x^n]))/(2*d^2*x^2) +
(e^2*(a + b*Log[c*x^n])^2)/(2*b*d^3*n) - (e^2*(a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2*d^3) - (b*e^2*n*PolyLo
g[2, -((e*x^2)/d)])/(4*d^3)

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 266

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2337

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 196, normalized size = 1.62 \[ -\frac {\frac {4 d^2 \left (a+b \log \left (c x^n\right )\right )}{x^4}+8 e^2 \log \left (\frac {\sqrt {e} x}{\sqrt {-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )+8 e^2 \log \left (\frac {d \sqrt {e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {8 d e \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac {8 e^2 \left (a+b \log \left (c x^n\right )\right )^2}{b n}+\frac {b d^2 n}{x^4}+8 b e^2 n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b e^2 n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )-\frac {4 b d e n}{x^2}}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((b*d^2*n)/x^4 - (4*b*d*e*n)/x^2 + (4*d^2*(a + b*Log[c*x^n]))/x^4 - (8*d*e*(a + b*Log[c*x^n]))/x^2 - (8*
e^2*(a + b*Log[c*x^n])^2)/(b*n) + 8*e^2*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 8*e^2*(a + b*Log[c*
x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)] + 8*b*e^2*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]] + 8*b*e^2*n*PolyLog[2, (d
*Sqrt[e]*x)/(-d)^(3/2)])/d^3

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e x^{7} + d x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e*x^7 + d*x^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.21, size = 805, normalized size = 6.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a {\left (\frac {2 \, e^{2} \log \left (e x^{2} + d\right )}{d^{3}} - \frac {4 \, e^{2} \log \relax (x)}{d^{3}} - \frac {2 \, e x^{2} - d}{d^{2} x^{4}}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{e x^{7} + d x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a*(2*e^2*log(e*x^2 + d)/d^3 - 4*e^2*log(x)/d^3 - (2*e*x^2 - d)/(d^2*x^4)) + b*integrate((log(c) + log(x^n
))/(e*x^7 + d*x^5), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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