3.3.32 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x^2)^3} \, dx\) [232]

Optimal. Leaf size=68 \[ -\frac {b n}{8 e^2 \left (d+e x^2\right )}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2373, 272, 45} \begin {gather*} \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n}{8 e^2 \left (d+e x^2\right )}-\frac {b n \log \left (d+e x^2\right )}{8 d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 129, normalized size = 1.90 \begin {gather*} -\frac {2 a d^2+b d^2 n+4 a d e x^2+b d e n x^2-2 b n \left (d+e x^2\right )^2 \log (x)+2 b d \left (d+2 e x^2\right ) \log \left (c x^n\right )+b d^2 n \log \left (d+e x^2\right )+2 b d e n x^2 \log \left (d+e x^2\right )+b e^2 n x^4 \log \left (d+e x^2\right )}{8 d e^2 \left (d+e x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.13, size = 369, normalized size = 5.43

method	result	size
risch	\(-\frac {b \left (2 e \,x^{2}+d \right ) \ln \left (x^{n}\right )}{4 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {-2 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+2 i \pi b d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 \ln \left (x \right ) b \,e^{2} n \,x^{4}+\ln \left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 i \pi b d e \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-4 \ln \left (x \right ) b d e n \,x^{2}+2 \ln \left (e \,x^{2}+d \right ) b d e n \,x^{2}+4 \ln \left (c \right ) b d e \,x^{2}+b d e n \,x^{2}-2 \ln \left (x \right ) b \,d^{2} n +\ln \left (e \,x^{2}+d \right ) b \,d^{2} n +4 x^{2} a d e +2 d^{2} b \ln \left (c \right )+b \,d^{2} n +2 a \,d^{2}}{8 e^{2} d \left (e \,x^{2}+d \right )^{2}}\)	\(369\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 121, normalized size = 1.78 \begin {gather*} -\frac {1}{8} \, b n {\left (\frac {e^{\left (-2\right )} \log \left (x^{2} e + d\right )}{d} - \frac {e^{\left (-2\right )} \log \left (x^{2}\right )}{d} + \frac {1}{x^{2} e^{3} + d e^{2}}\right )} - \frac {{\left (2 \, x^{2} e + d\right )} b \log \left (c x^{n}\right )}{4 \, {\left (x^{4} e^{4} + 2 \, d x^{2} e^{3} + d^{2} e^{2}\right )}} - \frac {{\left (2 \, x^{2} e + d\right )} a}{4 \, {\left (x^{4} e^{4} + 2 \, d x^{2} e^{3} + d^{2} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 124, normalized size = 1.82 \begin {gather*} \frac {2 \, b n x^{4} e^{2} \log \left (x\right ) - b d^{2} n - {\left (b d n + 4 \, a d\right )} x^{2} e - 2 \, a d^{2} - {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \log \left (x^{2} e + d\right ) - 2 \, {\left (2 \, b d x^{2} e + b d^{2}\right )} \log \left (c\right )}{8 \, {\left (d x^{4} e^{4} + 2 \, d^{2} x^{2} e^{3} + d^{3} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (58) = 116\).
time = 201.61, size = 612, normalized size = 9.00 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {a}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 x^{2}}\right ) & \text {for}\: d = 0 \wedge e = 0 \\\frac {\frac {a x^{4}}{4} - \frac {b n x^{4}}{16} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4}}{d^{3}} & \text {for}\: e = 0 \\\frac {- \frac {a}{2 x^{2}} - \frac {b n}{4 x^{2}} - \frac {b \log {\left (c x^{n} \right )}}{2 x^{2}}}{e^{3}} & \text {for}\: d = 0 \\- \frac {2 a d^{2}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {4 a d e x^{2}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b d^{2} n \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b d^{2} n \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b d^{2} n}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {2 b d e n x^{2} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b d e n x^{2}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} - \frac {b e^{2} n x^{4} \log {\left (x + \sqrt {- \frac {d}{e}} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} + \frac {2 b e^{2} x^{4} \log {\left (c x^{n} \right )}}{8 d^{3} e^{2} + 16 d^{2} e^{3} x^{2} + 8 d e^{4} x^{4}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (63) = 126\).
time = 4.44, size = 140, normalized size = 2.06 \begin {gather*} -\frac {b n x^{4} e^{2} \log \left (x^{2} e + d\right ) - 2 \, b n x^{4} e^{2} \log \left (x\right ) + 2 \, b d n x^{2} e \log \left (x^{2} e + d\right ) + b d n x^{2} e + 4 \, b d x^{2} e \log \left (c\right ) + 4 \, a d x^{2} e + b d^{2} n \log \left (x^{2} e + d\right ) + b d^{2} n + 2 \, b d^{2} \log \left (c\right ) + 2 \, a d^{2}}{8 \, {\left (d x^{4} e^{4} + 2 \, d^{2} x^{2} e^{3} + d^{3} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 3.74, size = 129, normalized size = 1.90 \begin {gather*} \frac {b\,n\,\ln \left (x\right )}{4\,d\,e^2}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,x^2}{2\,e}+\frac {b\,d}{4\,e^2}\right )}{d^2+2\,d\,e\,x^2+e^2\,x^4}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d\,e^2}-\frac {\left (2\,a\,e+\frac {b\,e\,n}{2}\right )\,x^2+a\,d+\frac {b\,d\,n}{2}}{4\,d^2\,e^2+8\,d\,e^3\,x^2+4\,e^4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________