\(\int \frac {x (a+b \log (c x^n))}{(d+e x^2)^3} \, dx\) [233]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2376, 272, 46} \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e}+\frac {b n \log (x)}{4 d^2 e}+\frac {b n}{8 d e \left (d+e x^2\right )} \]

[In]

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

[Out]

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

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 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 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]

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {b n \log (x)}{4 e \left (d+e x^2\right )^2}+\frac {-a-b \left (-n \log (x)+\log \left (c x^n\right )\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.74

method	result	size
parallelrisch	\(\frac {-\ln \left (e \,x^{2}+d \right ) x^{4} b \,e^{3} n^{2}+2 x^{4} \ln \left (c \,x^{n}\right ) b \,e^{3} n -2 \ln \left (e \,x^{2}+d \right ) x^{2} b d \,e^{2} n^{2}+4 x^{2} \ln \left (c \,x^{n}\right ) b d \,e^{2} n +x^{2} b d \,e^{2} n^{2}-\ln \left (e \,x^{2}+d \right ) b \,d^{2} e \,n^{2}+b \,d^{2} e \,n^{2}-2 a \,d^{2} e n}{8 n \,d^{2} e^{2} \left (e \,x^{2}+d \right )^{2}}\)	\(143\)
risch	\(-\frac {b \ln \left (x^{n}\right )}{4 e \left (e \,x^{2}+d \right )^{2}}-\frac {\ln \left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}-2 \ln \left (x \right ) b \,e^{2} n \,x^{4}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 \ln \left (e \,x^{2}+d \right ) b d e n \,x^{2}-4 \ln \left (x \right ) b d e n \,x^{2}-b d e n \,x^{2}+\ln \left (e \,x^{2}+d \right ) b \,d^{2} n -2 \ln \left (x \right ) b \,d^{2} n +2 d^{2} b \ln \left (c \right )-b \,d^{2} n +2 a \,d^{2}}{8 d^{2} e \left (e \,x^{2}+d \right )^{2}}\)	\(243\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b d e n x^{2} + b d^{2} n - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2} - {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (e x^{2} + d\right ) + 2 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2}\right )} \log \left (x\right )}{8 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 619 vs. \(2 (71) = 142\).

Time = 169.10 (sec) , antiderivative size = 619, normalized size of antiderivative = 7.55 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {a}{4 x^{4}} - \frac {b n}{16 x^{4}} - \frac {b \log {\left (c x^{n} \right )}}{4 x^{4}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {\frac {a x^{2}}{2} - \frac {b n x^{2}}{4} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2}}{d^{3}} & \text {for}\: e = 0 \\\frac {- \frac {a}{4 x^{4}} - \frac {b n}{16 x^{4}} - \frac {b \log {\left (c x^{n} \right )}}{4 x^{4}}}{e^{3}} & \text {for}\: d = 0 \\- \frac {2 a d^{2}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b d^{2} n \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b d^{2} n \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {b d^{2} n}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {b d e n x^{2}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {4 b d e x^{2} \log {\left (c x^{n} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} + \frac {2 b e^{2} x^{4} \log {\left (c x^{n} \right )}}{8 d^{4} e + 16 d^{3} e^{2} x^{2} + 8 d^{2} e^{3} x^{4}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*(-a/(4*x**4) - b*n/(16*x**4) - b*log(c*x**n)/(4*x**4)), Eq(d, 0) & Eq(e, 0)), ((a*x**2/2 - b*n*
x**2/4 + b*x**2*log(c*x**n)/2)/d**3, Eq(e, 0)), ((-a/(4*x**4) - b*n/(16*x**4) - b*log(c*x**n)/(4*x**4))/e**3,
Eq(d, 0)), (-2*a*d**2/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - b*d**2*n*log(x - sqrt(-d/e))/(8*d**4
*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - b*d**2*n*log(x + sqrt(-d/e))/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d*
*2*e**3*x**4) + b*d**2*n/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - 2*b*d*e*n*x**2*log(x - sqrt(-d/e)
)/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - 2*b*d*e*n*x**2*log(x + sqrt(-d/e))/(8*d**4*e + 16*d**3*e
**2*x**2 + 8*d**2*e**3*x**4) + b*d*e*n*x**2/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) + 4*b*d*e*x**2*l
og(c*x**n)/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - b*e**2*n*x**4*log(x - sqrt(-d/e))/(8*d**4*e + 1
6*d**3*e**2*x**2 + 8*d**2*e**3*x**4) - b*e**2*n*x**4*log(x + sqrt(-d/e))/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**
2*e**3*x**4) + 2*b*e**2*x**4*log(c*x**n)/(8*d**4*e + 16*d**3*e**2*x**2 + 8*d**2*e**3*x**4), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {1}{8} \, b n {\left (\frac {1}{d e^{2} x^{2} + d^{2} e} - \frac {\log \left (e x^{2} + d\right )}{d^{2} e} + \frac {\log \left (x^{2}\right )}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac {a}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.41 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b d n \log \left (x\right )}{4 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} + \frac {b e n x^{2} + b d n - 2 \, b d \log \left (c\right ) - 2 \, a d}{8 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} - \frac {b n \log \left (e x^{2} + d\right )}{8 \, d^{2} e} + \frac {b n \log \left (x\right )}{4 \, d^{2} e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {\frac {b\,n}{2}-a+\frac {b\,e\,n\,x^2}{2\,d}}{4\,d^2\,e+8\,d\,e^2\,x^2+4\,e^3\,x^4}-\frac {b\,\ln \left (c\,x^n\right )}{4\,e\,\left (d^2+2\,d\,e\,x^2+e^2\,x^4\right )}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d^2\,e}+\frac {b\,n\,\ln \left (x\right )}{4\,d^2\,e} \]

[In]

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

[Out]

((b*n)/2 - a + (b*e*n*x^2)/(2*d))/(4*d^2*e + 4*e^3*x^4 + 8*d*e^2*x^2) - (b*log(c*x^n))/(4*e*(d^2 + e^2*x^4 + 2
*d*e*x^2)) - (b*n*log(d + e*x^2))/(8*d^2*e) + (b*n*log(x))/(4*d^2*e)