3.3.30 \(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)^2} \, dx\) [230]

Optimal. Leaf size=224 \[ -\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}} \]

[Out]

-5/18*b*n/d^2/x^3+5/2*b*e*n/d^3/x+1/2*(a+b*ln(c*x^n))/d/x^3/(e*x^2+d)+1/6*(-5*a+b*n-5*b*ln(c*x^n))/d^2/x^3+1/2
*e*(5*a-b*n+5*b*ln(c*x^n))/d^3/x+1/2*e^(3/2)*arctan(x*e^(1/2)/d^(1/2))*(5*a-b*n+5*b*ln(c*x^n))/d^(7/2)-5/4*I*b
*e^(3/2)*n*polylog(2,-I*x*e^(1/2)/d^(1/2))/d^(7/2)+5/4*I*b*e^(3/2)*n*polylog(2,I*x*e^(1/2)/d^(1/2))/d^(7/2)

________________________________________________________________________________________

Rubi [A]
time = 0.25, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2385, 2380, 2341, 211, 2361, 12, 4940, 2438} \begin {gather*} -\frac {5 i b e^{3/2} n \text {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \text {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {e^{3/2} \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a+5 b \log \left (c x^n\right )-b n\right )}{2 d^{7/2}}+\frac {e \left (5 a+5 b \log \left (c x^n\right )-b n\right )}{2 d^3 x}-\frac {5 a+5 b \log \left (c x^n\right )-b n}{6 d^2 x^3}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}+\frac {5 b e n}{2 d^3 x}-\frac {5 b n}{18 d^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*n)/(18*d^2*x^3) + (5*b*e*n)/(2*d^3*x) + (a + b*Log[c*x^n])/(2*d*x^3*(d + e*x^2)) - (5*a - b*n + 5*b*Log[
c*x^n])/(6*d^2*x^3) + (e*(5*a - b*n + 5*b*Log[c*x^n]))/(2*d^3*x) + (e^(3/2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(5*a -
b*n + 5*b*Log[c*x^n]))/(2*d^(7/2)) - (((5*I)/4)*b*e^(3/2)*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/d^(7/2) + (
((5*I)/4)*b*e^(3/2)*n*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]])/d^(7/2)

Rule 12

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2341

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 2361

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 2380

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

Rule 2385

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 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]

Rule 4940

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 {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^2} \, dx &=\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {\int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )} \, dx}{2 d}\\ &=\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {\int \left (\frac {-5 a+b n-5 b \log \left (c x^n\right )}{d x^4}-\frac {e \left (-5 a+b n-5 b \log \left (c x^n\right )\right )}{d^2 x^2}+\frac {e^2 \left (-5 a+b n-5 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^3 \left (d+e x^2\right )}-\frac {\int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^4} \, dx}{2 d^2}+\frac {e \int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{x^2} \, dx}{2 d^3}-\frac {e^2 \int \frac {-5 a+b n-5 b \log \left (c x^n\right )}{d+e x^2} \, dx}{2 d^3}\\ &=-\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 b e^2 n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e} x} \, dx}{2 d^3}\\ &=-\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 b e^{3/2} n\right ) \int \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 d^{7/2}}\\ &=-\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {\left (5 i b e^{3/2} n\right ) \int \frac {\log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{4 d^{7/2}}+\frac {\left (5 i b e^{3/2} n\right ) \int \frac {\log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{4 d^{7/2}}\\ &=-\frac {5 b n}{18 d^2 x^3}+\frac {5 b e n}{2 d^3 x}+\frac {a+b \log \left (c x^n\right )}{2 d x^3 \left (d+e x^2\right )}-\frac {5 a-b n+5 b \log \left (c x^n\right )}{6 d^2 x^3}+\frac {e \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^3 x}+\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (5 a-b n+5 b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 i b e^{3/2} n \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}+\frac {5 i b e^{3/2} n \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )}{4 d^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.46, size = 361, normalized size = 1.61 \begin {gather*} \frac {1}{36} \left (-\frac {4 b n}{d^2 x^3}+\frac {72 b e n}{d^3 x}-\frac {12 \left (a+b \log \left (c x^n\right )\right )}{d^2 x^3}+\frac {72 e \left (a+b \log \left (c x^n\right )\right )}{d^3 x}-\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}-\sqrt {e} x\right )}+\frac {9 e^{3/2} \left (a+b \log \left (c x^n\right )\right )}{d^3 \left (\sqrt {-d}+\sqrt {e} x\right )}-\frac {9 b e^{3/2} n \left (\log (x)-\log \left (\sqrt {-d}-\sqrt {e} x\right )\right )}{(-d)^{7/2}}+\frac {9 b e^{3/2} n \left (\log (x)-\log \left (\sqrt {-d}+\sqrt {e} x\right )\right )}{(-d)^{7/2}}+\frac {45 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}-\frac {45 e^{3/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}-\frac {45 b e^{3/2} n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )}{(-d)^{7/2}}+\frac {45 b e^{3/2} n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{(-d)^{7/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b*n)/(d^2*x^3) + (72*b*e*n)/(d^3*x) - (12*(a + b*Log[c*x^n]))/(d^2*x^3) + (72*e*(a + b*Log[c*x^n]))/(d^3*
x) - (9*e^(3/2)*(a + b*Log[c*x^n]))/(d^3*(Sqrt[-d] - Sqrt[e]*x)) + (9*e^(3/2)*(a + b*Log[c*x^n]))/(d^3*(Sqrt[-
d] + Sqrt[e]*x)) - (9*b*e^(3/2)*n*(Log[x] - Log[Sqrt[-d] - Sqrt[e]*x]))/(-d)^(7/2) + (9*b*e^(3/2)*n*(Log[x] -
Log[Sqrt[-d] + Sqrt[e]*x]))/(-d)^(7/2) + (45*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7
/2) - (45*e^(3/2)*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(7/2) - (45*b*e^(3/2)*n*PolyLog[2
, (Sqrt[e]*x)/Sqrt[-d]])/(-d)^(7/2) + (45*b*e^(3/2)*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(-d)^(7/2))/36

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 1133, normalized size = 5.06

method	result	size
risch	\(\text {Expression too large to display}\)	\(1133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^2/x^3+5/2*b*ln(c)*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+1/6*I*b*
Pi*csgn(I*c*x^n)^3/d^2/x^3+2*a/d^3*e/x-5/2*b*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))*n*ln(x)+b*n*e^2/d^3/(
-e*d)^(1/2)*ln(x)*ln((-e*x+(-e*d)^(1/2))/(-e*d)^(1/2))-b*n*e^2/d^3/(-e*d)^(1/2)*ln(x)*ln((e*x+(-e*d)^(1/2))/(-
e*d)^(1/2))+1/2*a*e^2/d^3*x/(e*x^2+d)+5/2*a*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+2*b*ln(c)/d^3*e/x-1/3*
a/d^2/x^3-1/3*b/d^2/x^3*ln(x^n)-I*b*Pi*csgn(I*c*x^n)^3/d^3*e/x+1/4*b*n*e^3/d^3*ln(x)/(e*x^2+d)/(-e*d)^(1/2)*ln
((-e*x+(-e*d)^(1/2))/(-e*d)^(1/2))*x^2-1/3*b*ln(c)/d^2/x^3+1/2*b*e^2/d^3*x/(e*x^2+d)*ln(x^n)+5/2*b*e^2/d^3/(e*
d)^(1/2)*arctan(x*e/(e*d)^(1/2))*ln(x^n)-1/4*b*n*e^3/d^3*ln(x)/(e*x^2+d)/(-e*d)^(1/2)*ln((e*x+(-e*d)^(1/2))/(-
e*d)^(1/2))*x^2-1/2*b*n*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+5/4*b*n*e^2/d^3/(-e*d)^(1/2)*dilog((-e*x+(
-e*d)^(1/2))/(-e*d)^(1/2))-5/4*b*n*e^2/d^3/(-e*d)^(1/2)*dilog((e*x+(-e*d)^(1/2))/(-e*d)^(1/2))-1/6*I*b*Pi*csgn
(I*x^n)*csgn(I*c*x^n)^2/d^2/x^3+I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^3*e/x+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2/d^
3*e/x-1/4*b*n*e^2/d^2*ln(x)/(e*x^2+d)/(-e*d)^(1/2)*ln((e*x+(-e*d)^(1/2))/(-e*d)^(1/2))+1/4*b*n*e^2/d^2*ln(x)/(
e*x^2+d)/(-e*d)^(1/2)*ln((-e*x+(-e*d)^(1/2))/(-e*d)^(1/2))+1/2*b*ln(c)*e^2/d^3*x/(e*x^2+d)-1/4*I*b*Pi*csgn(I*c
*x^n)^3*e^2/d^3*x/(e*x^2+d)-5/4*I*b*Pi*csgn(I*c*x^n)^3*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+1/6*I*b*Pi*
csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/d^2/x^3-1/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^2/d^3*x/(e*x^2+d)
-5/4*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+2*b*ln(x^n)/d^3*e/
x+1/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*e^2/d^3*x/(e*x^2+d)-1/9*b*n/d^2/x^3-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I
*c*x^n)/d^3*e/x+5/4*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2*e^2/d^3/(e*d)^(1/2)*arctan(x*e/(e*d)^(1/2))+2*b*e*n/d^3/x
+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^3*x/(e*x^2+d)+5/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/d^3/(e*
d)^(1/2)*arctan(x*e/(e*d)^(1/2))

________________________________________________________________________________________

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

[Out]

1/6*a*((15*x^4*e^2 + 10*d*x^2*e - 2*d^2)/(d^3*x^5*e + d^4*x^3) + 15*arctan(x*e^(1/2)/sqrt(d))*e^(3/2)/d^(7/2))
+ b*integrate((log(c) + log(x^n))/(x^8*e^2 + 2*d*x^6*e + d^2*x^4), x)

________________________________________________________________________________________

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

[Out]

integral((b*log(c*x^n) + a)/(x^8*e^2 + 2*d*x^6*e + d^2*x^4), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \left (d + e x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________