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

Optimal. Leaf size=77 \[ -\frac {a^6}{2 x^2}+6 a^5 b \log (x)+\frac {15}{2} a^4 b^2 x^2+5 a^3 b^3 x^4+\frac {5}{2} a^2 b^4 x^6+\frac {3}{4} a b^5 x^8+\frac {b^6 x^{10}}{10} \]

[Out]

-1/2*a^6/x^2+15/2*a^4*b^2*x^2+5*a^3*b^3*x^4+5/2*a^2*b^4*x^6+3/4*a*b^5*x^8+1/10*b^6*x^10+6*a^5*b*ln(x)

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Rubi [A]  time = 0.06, antiderivative size = 77, 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} \[ \frac {5}{2} a^2 b^4 x^6+5 a^3 b^3 x^4+\frac {15}{2} a^4 b^2 x^2+6 a^5 b \log (x)-\frac {a^6}{2 x^2}+\frac {3}{4} a b^5 x^8+\frac {b^6 x^{10}}{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 77, normalized size = 1.00 \[ -\frac {a^6}{2 x^2}+6 a^5 b \log (x)+\frac {15}{2} a^4 b^2 x^2+5 a^3 b^3 x^4+\frac {5}{2} a^2 b^4 x^6+\frac {3}{4} a b^5 x^8+\frac {b^6 x^{10}}{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a^6/x^2 + (15*a^4*b^2*x^2)/2 + 5*a^3*b^3*x^4 + (5*a^2*b^4*x^6)/2 + (3*a*b^5*x^8)/4 + (b^6*x^10)/10 + 6*a^
5*b*Log[x]

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fricas [A]  time = 0.98, size = 72, normalized size = 0.94 \[ \frac {2 \, b^{6} x^{12} + 15 \, a b^{5} x^{10} + 50 \, a^{2} b^{4} x^{8} + 100 \, a^{3} b^{3} x^{6} + 150 \, a^{4} b^{2} x^{4} + 120 \, a^{5} b x^{2} \log \relax (x) - 10 \, a^{6}}{20 \, x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

1/20*(2*b^6*x^12 + 15*a*b^5*x^10 + 50*a^2*b^4*x^8 + 100*a^3*b^3*x^6 + 150*a^4*b^2*x^4 + 120*a^5*b*x^2*log(x) -
10*a^6)/x^2

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giac [A]  time = 0.15, size = 79, normalized size = 1.03 \[ \frac {1}{10} \, b^{6} x^{10} + \frac {3}{4} \, a b^{5} x^{8} + \frac {5}{2} \, a^{2} b^{4} x^{6} + 5 \, a^{3} b^{3} x^{4} + \frac {15}{2} \, a^{4} b^{2} x^{2} + 3 \, a^{5} b \log \left (x^{2}\right ) - \frac {6 \, a^{5} b x^{2} + a^{6}}{2 \, x^{2}} \]

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^3,x, algorithm="giac")

[Out]

1/10*b^6*x^10 + 3/4*a*b^5*x^8 + 5/2*a^2*b^4*x^6 + 5*a^3*b^3*x^4 + 15/2*a^4*b^2*x^2 + 3*a^5*b*log(x^2) - 1/2*(6
*a^5*b*x^2 + a^6)/x^2

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maple [A]  time = 0.01, size = 68, normalized size = 0.88 \[ \frac {b^{6} x^{10}}{10}+\frac {3 a \,b^{5} x^{8}}{4}+\frac {5 a^{2} b^{4} x^{6}}{2}+5 a^{3} b^{3} x^{4}+\frac {15 a^{4} b^{2} x^{2}}{2}+6 a^{5} b \ln \relax (x )-\frac {a^{6}}{2 x^{2}} \]

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^3,x)

[Out]

-1/2*a^6/x^2+15/2*a^4*b^2*x^2+5*a^3*b^3*x^4+5/2*a^2*b^4*x^6+3/4*a*b^5*x^8+1/10*b^6*x^10+6*a^5*b*ln(x)

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maxima [A]  time = 1.43, size = 69, normalized size = 0.90 \[ \frac {1}{10} \, b^{6} x^{10} + \frac {3}{4} \, a b^{5} x^{8} + \frac {5}{2} \, a^{2} b^{4} x^{6} + 5 \, a^{3} b^{3} x^{4} + \frac {15}{2} \, a^{4} b^{2} x^{2} + 3 \, a^{5} b \log \left (x^{2}\right ) - \frac {a^{6}}{2 \, x^{2}} \]

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^3,x, algorithm="maxima")

[Out]

1/10*b^6*x^10 + 3/4*a*b^5*x^8 + 5/2*a^2*b^4*x^6 + 5*a^3*b^3*x^4 + 15/2*a^4*b^2*x^2 + 3*a^5*b*log(x^2) - 1/2*a^
6/x^2

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mupad [B]  time = 0.04, size = 67, normalized size = 0.87 \[ \frac {b^6\,x^{10}}{10}-\frac {a^6}{2\,x^2}+\frac {3\,a\,b^5\,x^8}{4}+6\,a^5\,b\,\ln \relax (x)+\frac {15\,a^4\,b^2\,x^2}{2}+5\,a^3\,b^3\,x^4+\frac {5\,a^2\,b^4\,x^6}{2} \]

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^3,x)

[Out]

(b^6*x^10)/10 - a^6/(2*x^2) + (3*a*b^5*x^8)/4 + 6*a^5*b*log(x) + (15*a^4*b^2*x^2)/2 + 5*a^3*b^3*x^4 + (5*a^2*b
^4*x^6)/2

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sympy [A]  time = 0.20, size = 76, normalized size = 0.99 \[ - \frac {a^{6}}{2 x^{2}} + 6 a^{5} b \log {\relax (x )} + \frac {15 a^{4} b^{2} x^{2}}{2} + 5 a^{3} b^{3} x^{4} + \frac {5 a^{2} b^{4} x^{6}}{2} + \frac {3 a b^{5} x^{8}}{4} + \frac {b^{6} x^{10}}{10} \]

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**3,x)

[Out]

-a**6/(2*x**2) + 6*a**5*b*log(x) + 15*a**4*b**2*x**2/2 + 5*a**3*b**3*x**4 + 5*a**2*b**4*x**6/2 + 3*a*b**5*x**8
/4 + b**6*x**10/10

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