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

Optimal. Leaf size=126 \[ \frac {e \log \left (\frac {d}{e x^2}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-b n\right )}{4 d^3}-\frac {4 a+4 b \log \left (c x^n\right )-b n}{4 d^2 x^2}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}-\frac {b e n \text {Li}_2\left (-\frac {d}{e x^2}\right )}{2 d^3}-\frac {b n}{2 d^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 159, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2340, 266, 44, 2351, 2304, 2301, 2337, 2391} \[ \frac {b e n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{2 d^3}-\frac {e \left (4 a+4 b \log \left (c x^n\right )-b n\right )^2}{16 b d^3 n}+\frac {e \log \left (\frac {e x^2}{d}+1\right ) \left (4 a+4 b \log \left (c x^n\right )-b n\right )}{4 d^3}-\frac {4 a+4 b \log \left (c x^n\right )-b n}{4 d^2 x^2}+\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )}-\frac {b n}{2 d^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n)/(2*d^2*x^2) + (a + b*Log[c*x^n])/(2*d*x^2*(d + e*x^2)) - (4*a - b*n + 4*b*Log[c*x^n])/(4*d^2*x^2) - (e*
(4*a - b*n + 4*b*Log[c*x^n])^2)/(16*b*d^3*n) + (e*(4*a - b*n + 4*b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(4*d^3) + (
b*e*n*PolyLog[2, -((e*x^2)/d)])/(2*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 2340

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> -Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*Log[c*x^n]))/(2*d*f*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(f*x)^m*
(d + e*x^2)^(q + 1)*(a*(m + 2*q + 3) + b*n + b*(m + 2*q + 3)*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d, e, f, m
, n}, x] && ILtQ[q, -1] && ILtQ[m, 0]

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

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Mathematica [C]  time = 0.57, size = 334, normalized size = 2.65 \[ \frac {4 e \log \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-\frac {2 d e \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d+e x^2}-\frac {2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{x^2}-8 e \log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )+b n \left (\frac {e^{3/2} x \log (x)}{\sqrt {e} x-i \sqrt {d}}+\frac {e \left (-\sqrt {d}+i \sqrt {e} x\right ) \log \left (\sqrt {e} x+i \sqrt {d}\right )-i e^{3/2} x \log (x)}{\sqrt {d}-i \sqrt {e} x}+4 e \left (\text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )+4 e \left (\text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-e \log \left (-\sqrt {e} x+i \sqrt {d}\right )-\frac {2 d \log (x)+d}{x^2}-4 e \log ^2(x)\right )}{4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (c x^{n}\right ) + a}{e^{2} x^{7} + 2 \, d e x^{5} + d^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e^2*x^7 + 2*d*e*x^5 + d^2*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 817, normalized size = 6.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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