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

Optimal. Leaf size=20 \[ -\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2} \]

[Out]

-1/2/(a*x+b*ln(c*x^n)^2)^2

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Rubi [A]  time = 0.16, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2561, 2544} \[ -\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*(a*x + b*Log[c*x^n]^2)^2)

Rule 2544

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

Rule 2561

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \[ -\frac {1}{2 \left (a x+b \log ^2\left (c x^n\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*1/(a*x + b*Log[c*x^n]^2)^2

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fricas [B]  time = 0.45, size = 101, normalized size = 5.05 \[ -\frac {1}{2 \, {\left (b^{2} n^{4} \log \relax (x)^{4} + 4 \, b^{2} n^{3} \log \relax (c) \log \relax (x)^{3} + b^{2} \log \relax (c)^{4} + 2 \, a b x \log \relax (c)^{2} + a^{2} x^{2} + 2 \, {\left (3 \, b^{2} n^{2} \log \relax (c)^{2} + a b n^{2} x\right )} \log \relax (x)^{2} + 4 \, {\left (b^{2} n \log \relax (c)^{3} + a b n x \log \relax (c)\right )} \log \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 306, normalized size = 15.30 \[ -\frac {4 \, a b n^{2} x + a^{2} x^{2}}{2 \, {\left (4 \, a b^{3} n^{6} x \log \relax (x)^{4} + 16 \, a b^{3} n^{5} x \log \relax (c) \log \relax (x)^{3} + a^{2} b^{2} n^{4} x^{2} \log \relax (x)^{4} + 24 \, a b^{3} n^{4} x \log \relax (c)^{2} \log \relax (x)^{2} + 4 \, a^{2} b^{2} n^{3} x^{2} \log \relax (c) \log \relax (x)^{3} + 16 \, a b^{3} n^{3} x \log \relax (c)^{3} \log \relax (x) + 8 \, a^{2} b^{2} n^{4} x^{2} \log \relax (x)^{2} + 6 \, a^{2} b^{2} n^{2} x^{2} \log \relax (c)^{2} \log \relax (x)^{2} + 4 \, a b^{3} n^{2} x \log \relax (c)^{4} + 16 \, a^{2} b^{2} n^{3} x^{2} \log \relax (c) \log \relax (x) + 4 \, a^{2} b^{2} n x^{2} \log \relax (c)^{3} \log \relax (x) + 2 \, a^{3} b n^{2} x^{3} \log \relax (x)^{2} + 8 \, a^{2} b^{2} n^{2} x^{2} \log \relax (c)^{2} + a^{2} b^{2} x^{2} \log \relax (c)^{4} + 4 \, a^{3} b n x^{3} \log \relax (c) \log \relax (x) + 4 \, a^{3} b n^{2} x^{3} + 2 \, a^{3} b x^{3} \log \relax (c)^{2} + a^{4} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a*b*n^2*x + a^2*x^2)/(4*a*b^3*n^6*x*log(x)^4 + 16*a*b^3*n^5*x*log(c)*log(x)^3 + a^2*b^2*n^4*x^2*log(x)
^4 + 24*a*b^3*n^4*x*log(c)^2*log(x)^2 + 4*a^2*b^2*n^3*x^2*log(c)*log(x)^3 + 16*a*b^3*n^3*x*log(c)^3*log(x) + 8
*a^2*b^2*n^4*x^2*log(x)^2 + 6*a^2*b^2*n^2*x^2*log(c)^2*log(x)^2 + 4*a*b^3*n^2*x*log(c)^4 + 16*a^2*b^2*n^3*x^2*
log(c)*log(x) + 4*a^2*b^2*n*x^2*log(c)^3*log(x) + 2*a^3*b*n^2*x^3*log(x)^2 + 8*a^2*b^2*n^2*x^2*log(c)^2 + a^2*
b^2*x^2*log(c)^4 + 4*a^3*b*n*x^3*log(c)*log(x) + 4*a^3*b*n^2*x^3 + 2*a^3*b*x^3*log(c)^2 + a^4*x^4)

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maple [C]  time = 0.32, size = 447, normalized size = 22.35 \[ -\frac {8}{\left (\pi ^{2} b \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-2 \pi ^{2} b \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+\pi ^{2} b \mathrm {csgn}\left (i c \right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{4}-2 \pi ^{2} b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+4 \pi ^{2} b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{4}-2 \pi ^{2} b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{5}+\pi ^{2} b \mathrm {csgn}\left (i x^{n}\right )^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{4}-2 \pi ^{2} b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{5}+\pi ^{2} b \mathrm {csgn}\left (i c \,x^{n}\right )^{6}+4 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (c )+4 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (x^{n}\right )-4 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (c )-4 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (x^{n}\right )-4 i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (c )-4 i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (x^{n}\right )+4 i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (c )+4 i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (x^{n}\right )-4 b \ln \relax (c )^{2}-8 b \ln \relax (c ) \ln \left (x^{n}\right )-4 b \ln \left (x^{n}\right )^{2}-4 a x \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/(Pi^2*b*csgn(I*x^n)^2*csgn(I*c*x^n)^4-2*Pi^2*b*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3+Pi^2*b*csgn(I*c)^2*c
sgn(I*x^n)^2*csgn(I*c*x^n)^2-2*Pi^2*b*csgn(I*x^n)*csgn(I*c*x^n)^5+4*Pi^2*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)
^4-2*Pi^2*b*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+Pi^2*b*csgn(I*c*x^n)^6-2*Pi^2*b*csgn(I*c)*csgn(I*c*x^n)^5+
Pi^2*b*csgn(I*c)^2*csgn(I*c*x^n)^4+4*I*b*ln(x^n)*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+4*I*b*ln(x^n)*Pi*csgn(
I*c*x^n)^3+4*I*b*ln(c)*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-4*I*b*ln(x^n)*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-4*I
*b*ln(c)*Pi*csgn(I*c*x^n)^2*csgn(I*c)-4*I*b*ln(c)*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-4*I*b*ln(x^n)*Pi*csgn(I*c*x^n
)^2*csgn(I*c)+4*I*b*ln(c)*Pi*csgn(I*c*x^n)^3-4*b*ln(c)^2-8*b*ln(c)*ln(x^n)-4*b*ln(x^n)^2-4*a*x)^2

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maxima [B]  time = 0.85, size = 95, normalized size = 4.75 \[ -\frac {1}{2 \, {\left (b^{2} \log \relax (c)^{4} + 4 \, b^{2} \log \relax (c) \log \left (x^{n}\right )^{3} + b^{2} \log \left (x^{n}\right )^{4} + 2 \, a b x \log \relax (c)^{2} + a^{2} x^{2} + 2 \, {\left (3 \, b^{2} \log \relax (c)^{2} + a b x\right )} \log \left (x^{n}\right )^{2} + 4 \, {\left (b^{2} \log \relax (c)^{3} + a b x \log \relax (c)\right )} \log \left (x^{n}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 39, normalized size = 1.95 \[ -\frac {1}{2\,a^2\,x^2+4\,a\,b\,x\,{\ln \left (c\,x^n\right )}^2+2\,b^2\,{\ln \left (c\,x^n\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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