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

Optimal. Leaf size=20 \[ \frac {1}{3} \left (a x+b \log ^2\left (c x^n\right )\right )^3 \]

[Out]

1/3*(a*x+b*ln(c*x^n)^2)^3

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Rubi [A]  time = 0.15, 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} \[ \frac {1}{3} \left (a x+b \log ^2\left (c x^n\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x + b*Log[c*x^n]^2)^3/3

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \[ \frac {1}{3} \left (a x+b \log ^2\left (c x^n\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x + b*Log[c*x^n]^2)^3/3

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fricas [B]  time = 0.46, size = 195, normalized size = 9.75 \[ \frac {1}{3} \, b^{3} n^{6} \log \relax (x)^{6} + 2 \, b^{3} n^{5} \log \relax (c) \log \relax (x)^{5} + a b^{2} x \log \relax (c)^{4} + a^{2} b x^{2} \log \relax (c)^{2} + \frac {1}{3} \, a^{3} x^{3} + {\left (5 \, b^{3} n^{4} \log \relax (c)^{2} + a b^{2} n^{4} x\right )} \log \relax (x)^{4} + \frac {4}{3} \, {\left (5 \, b^{3} n^{3} \log \relax (c)^{3} + 3 \, a b^{2} n^{3} x \log \relax (c)\right )} \log \relax (x)^{3} + {\left (5 \, b^{3} n^{2} \log \relax (c)^{4} + 6 \, a b^{2} n^{2} x \log \relax (c)^{2} + a^{2} b n^{2} x^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (b^{3} n \log \relax (c)^{5} + 2 \, a b^{2} n x \log \relax (c)^{3} + a^{2} b n x^{2} \log \relax (c)\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 198, normalized size = 9.90 \[ \frac {1}{3} \, b^{3} n^{6} \log \relax (x)^{6} + 2 \, b^{3} n^{5} \log \relax (c) \log \relax (x)^{5} + 2 \, b^{3} n \log \relax (c)^{5} \log \relax (x) + a b^{2} x \log \relax (c)^{4} + a^{2} b x^{2} \log \relax (c)^{2} + \frac {1}{3} \, a^{3} x^{3} + {\left (5 \, b^{3} n^{4} \log \relax (c)^{2} + a b^{2} n^{4} x\right )} \log \relax (x)^{4} + \frac {4}{3} \, {\left (5 \, b^{3} n^{3} \log \relax (c)^{3} + 3 \, a b^{2} n^{3} x \log \relax (c)\right )} \log \relax (x)^{3} + {\left (5 \, b^{3} n^{2} \log \relax (c)^{4} + 6 \, a b^{2} n^{2} x \log \relax (c)^{2} + a^{2} b n^{2} x^{2}\right )} \log \relax (x)^{2} + 2 \, {\left (2 \, a b^{2} n x \log \relax (c)^{3} + a^{2} b n x^{2} \log \relax (c)\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.89, size = 16321, normalized size = 816.05 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [B]  time = 0.49, size = 211, normalized size = 10.55 \[ \frac {1}{3} \, b^{3} \log \left (c x^{n}\right )^{6} + 4 \, a b^{2} n x \log \left (c x^{n}\right )^{3} + a b^{2} x \log \left (c x^{n}\right )^{4} - \frac {1}{2} \, a^{2} b n^{2} x^{2} + a^{2} b n x^{2} \log \left (c x^{n}\right ) + a^{2} b x^{2} \log \left (c x^{n}\right )^{2} + \frac {1}{3} \, a^{3} x^{3} - 12 \, {\left (n x \log \left (c x^{n}\right )^{2} + 2 \, {\left (n^{2} x - n x \log \left (c x^{n}\right )\right )} n\right )} a b^{2} n + \frac {1}{2} \, {\left (n^{2} x^{2} - 2 \, n x^{2} \log \left (c x^{n}\right )\right )} a^{2} b - 4 \, {\left (n x \log \left (c x^{n}\right )^{3} - 3 \, {\left (n x \log \left (c x^{n}\right )^{2} + 2 \, {\left (n^{2} x - n x \log \left (c x^{n}\right )\right )} n\right )} n\right )} a b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^3*log(c*x^n)^6 + 4*a*b^2*n*x*log(c*x^n)^3 + a*b^2*x*log(c*x^n)^4 - 1/2*a^2*b*n^2*x^2 + a^2*b*n*x^2*log(c
*x^n) + a^2*b*x^2*log(c*x^n)^2 + 1/3*a^3*x^3 - 12*(n*x*log(c*x^n)^2 + 2*(n^2*x - n*x*log(c*x^n))*n)*a*b^2*n +
1/2*(n^2*x^2 - 2*n*x^2*log(c*x^n))*a^2*b - 4*(n*x*log(c*x^n)^3 - 3*(n*x*log(c*x^n)^2 + 2*(n^2*x - n*x*log(c*x^
n))*n)*n)*a*b^2

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mupad [B]  time = 0.33, size = 52, normalized size = 2.60 \[ \frac {a^3\,x^3}{3}+a^2\,b\,x^2\,{\ln \left (c\,x^n\right )}^2+a\,b^2\,x\,{\ln \left (c\,x^n\right )}^4+\frac {b^3\,{\ln \left (c\,x^n\right )}^6}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^3*log(c*x^n)^6)/3 + (a^3*x^3)/3 + a^2*b*x^2*log(c*x^n)^2 + a*b^2*x*log(c*x^n)^4

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sympy [A]  time = 12.32, size = 221, normalized size = 11.05 \[ \frac {a^{3} x^{3}}{3} + a^{2} b n^{2} x^{2} \log {\relax (x )}^{2} - a^{2} b n^{2} x^{2} \log {\relax (x )} + 2 a^{2} b n x^{2} \log {\relax (c )} \log {\relax (x )} - a^{2} b n x^{2} \log {\relax (c )} + a^{2} b n x^{2} \log {\left (c x^{n} \right )} + a^{2} b x^{2} \log {\relax (c )}^{2} + a b^{2} n^{4} x \log {\relax (x )}^{4} + 4 a b^{2} n^{3} x \log {\relax (c )} \log {\relax (x )}^{3} + 6 a b^{2} n^{2} x \log {\relax (c )}^{2} \log {\relax (x )}^{2} + 4 a b^{2} n x \log {\relax (c )}^{3} \log {\relax (x )} + a b^{2} x \log {\relax (c )}^{4} - 2 b^{3} n \left (\begin {cases} - \log {\relax (c )}^{5} \log {\relax (x )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{6}}{6 n} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x**3/3 + a**2*b*n**2*x**2*log(x)**2 - a**2*b*n**2*x**2*log(x) + 2*a**2*b*n*x**2*log(c)*log(x) - a**2*b*n*
x**2*log(c) + a**2*b*n*x**2*log(c*x**n) + a**2*b*x**2*log(c)**2 + a*b**2*n**4*x*log(x)**4 + 4*a*b**2*n**3*x*lo
g(c)*log(x)**3 + 6*a*b**2*n**2*x*log(c)**2*log(x)**2 + 4*a*b**2*n*x*log(c)**3*log(x) + a*b**2*x*log(c)**4 - 2*
b**3*n*Piecewise((-log(c)**5*log(x), Eq(n, 0)), (-log(c*x**n)**6/(6*n), True))

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