3.459 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^3}{x^7} \, dx\)

Optimal. Leaf size=79 \[ -\frac {a^6}{6 x^6}-\frac {3 a^5 b}{2 x^4}-\frac {15 a^4 b^2}{2 x^2}+20 a^3 b^3 \log (x)+\frac {15}{2} a^2 b^4 x^2+\frac {3}{2} a b^5 x^4+\frac {b^6 x^6}{6} \]

[Out]

-1/6*a^6/x^6-3/2*a^5*b/x^4-15/2*a^4*b^2/x^2+15/2*a^2*b^4*x^2+3/2*a*b^5*x^4+1/6*b^6*x^6+20*a^3*b^3*ln(x)

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Rubi [A]  time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 266, 43} \[ -\frac {15 a^4 b^2}{2 x^2}+\frac {15}{2} a^2 b^4 x^2+20 a^3 b^3 \log (x)-\frac {3 a^5 b}{2 x^4}-\frac {a^6}{6 x^6}+\frac {3}{2} a b^5 x^4+\frac {b^6 x^6}{6} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^7,x]

[Out]

-a^6/(6*x^6) - (3*a^5*b)/(2*x^4) - (15*a^4*b^2)/(2*x^2) + (15*a^2*b^4*x^2)/2 + (3*a*b^5*x^4)/2 + (b^6*x^6)/6 +
20*a^3*b^3*Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

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 266

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]]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^3}{x^7} \, dx &=\frac {\int \frac {\left (a b+b^2 x^2\right )^6}{x^7} \, dx}{b^6}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a b+b^2 x\right )^6}{x^4} \, dx,x,x^2\right )}{2 b^6}\\ &=\frac {\operatorname {Subst}\left (\int \left (15 a^2 b^{10}+\frac {a^6 b^6}{x^4}+\frac {6 a^5 b^7}{x^3}+\frac {15 a^4 b^8}{x^2}+\frac {20 a^3 b^9}{x}+6 a b^{11} x+b^{12} x^2\right ) \, dx,x,x^2\right )}{2 b^6}\\ &=-\frac {a^6}{6 x^6}-\frac {3 a^5 b}{2 x^4}-\frac {15 a^4 b^2}{2 x^2}+\frac {15}{2} a^2 b^4 x^2+\frac {3}{2} a b^5 x^4+\frac {b^6 x^6}{6}+20 a^3 b^3 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 79, normalized size = 1.00 \[ -\frac {a^6}{6 x^6}-\frac {3 a^5 b}{2 x^4}-\frac {15 a^4 b^2}{2 x^2}+20 a^3 b^3 \log (x)+\frac {15}{2} a^2 b^4 x^2+\frac {3}{2} a b^5 x^4+\frac {b^6 x^6}{6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^3/x^7,x]

[Out]

-1/6*a^6/x^6 - (3*a^5*b)/(2*x^4) - (15*a^4*b^2)/(2*x^2) + (15*a^2*b^4*x^2)/2 + (3*a*b^5*x^4)/2 + (b^6*x^6)/6 +
20*a^3*b^3*Log[x]

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fricas [A]  time = 0.96, size = 71, normalized size = 0.90 \[ \frac {b^{6} x^{12} + 9 \, a b^{5} x^{10} + 45 \, a^{2} b^{4} x^{8} + 120 \, a^{3} b^{3} x^{6} \log \relax (x) - 45 \, a^{4} b^{2} x^{4} - 9 \, a^{5} b x^{2} - a^{6}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^7,x, algorithm="fricas")

[Out]

1/6*(b^6*x^12 + 9*a*b^5*x^10 + 45*a^2*b^4*x^8 + 120*a^3*b^3*x^6*log(x) - 45*a^4*b^2*x^4 - 9*a^5*b*x^2 - a^6)/x
^6

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giac [A]  time = 0.15, size = 81, normalized size = 1.03 \[ \frac {1}{6} \, b^{6} x^{6} + \frac {3}{2} \, a b^{5} x^{4} + \frac {15}{2} \, a^{2} b^{4} x^{2} + 10 \, a^{3} b^{3} \log \left (x^{2}\right ) - \frac {110 \, a^{3} b^{3} x^{6} + 45 \, a^{4} b^{2} x^{4} + 9 \, a^{5} b x^{2} + a^{6}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^7,x, algorithm="giac")

[Out]

1/6*b^6*x^6 + 3/2*a*b^5*x^4 + 15/2*a^2*b^4*x^2 + 10*a^3*b^3*log(x^2) - 1/6*(110*a^3*b^3*x^6 + 45*a^4*b^2*x^4 +
9*a^5*b*x^2 + a^6)/x^6

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maple [A]  time = 0.01, size = 68, normalized size = 0.86 \[ \frac {b^{6} x^{6}}{6}+\frac {3 a \,b^{5} x^{4}}{2}+\frac {15 a^{2} b^{4} x^{2}}{2}+20 a^{3} b^{3} \ln \relax (x )-\frac {15 a^{4} b^{2}}{2 x^{2}}-\frac {3 a^{5} b}{2 x^{4}}-\frac {a^{6}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^3/x^7,x)

[Out]

-1/6*a^6/x^6-3/2*a^5*b/x^4-15/2*a^4*b^2/x^2+15/2*a^2*b^4*x^2+3/2*a*b^5*x^4+1/6*b^6*x^6+20*a^3*b^3*ln(x)

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maxima [A]  time = 1.39, size = 70, normalized size = 0.89 \[ \frac {1}{6} \, b^{6} x^{6} + \frac {3}{2} \, a b^{5} x^{4} + \frac {15}{2} \, a^{2} b^{4} x^{2} + 10 \, a^{3} b^{3} \log \left (x^{2}\right ) - \frac {45 \, a^{4} b^{2} x^{4} + 9 \, a^{5} b x^{2} + a^{6}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^3/x^7,x, algorithm="maxima")

[Out]

1/6*b^6*x^6 + 3/2*a*b^5*x^4 + 15/2*a^2*b^4*x^2 + 10*a^3*b^3*log(x^2) - 1/6*(45*a^4*b^2*x^4 + 9*a^5*b*x^2 + a^6
)/x^6

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mupad [B]  time = 4.34, size = 70, normalized size = 0.89 \[ \frac {b^6\,x^6}{6}-\frac {\frac {a^6}{6}+\frac {3\,a^5\,b\,x^2}{2}+\frac {15\,a^4\,b^2\,x^4}{2}}{x^6}+\frac {3\,a\,b^5\,x^4}{2}+\frac {15\,a^2\,b^4\,x^2}{2}+20\,a^3\,b^3\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^3/x^7,x)

[Out]

(b^6*x^6)/6 - (a^6/6 + (3*a^5*b*x^2)/2 + (15*a^4*b^2*x^4)/2)/x^6 + (3*a*b^5*x^4)/2 + (15*a^2*b^4*x^2)/2 + 20*a
^3*b^3*log(x)

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sympy [A]  time = 0.32, size = 76, normalized size = 0.96 \[ 20 a^{3} b^{3} \log {\relax (x )} + \frac {15 a^{2} b^{4} x^{2}}{2} + \frac {3 a b^{5} x^{4}}{2} + \frac {b^{6} x^{6}}{6} + \frac {- a^{6} - 9 a^{5} b x^{2} - 45 a^{4} b^{2} x^{4}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**3/x**7,x)

[Out]

20*a**3*b**3*log(x) + 15*a**2*b**4*x**2/2 + 3*a*b**5*x**4/2 + b**6*x**6/6 + (-a**6 - 9*a**5*b*x**2 - 45*a**4*b
**2*x**4)/(6*x**6)

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