3.579 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=158 \[ -\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}+\frac {b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )} \]

[Out]

-1/5*a^3*((b*x^2+a)^2)^(1/2)/x^5/(b*x^2+a)-a^2*b*((b*x^2+a)^2)^(1/2)/x^3/(b*x^2+a)-3*a*b^2*((b*x^2+a)^2)^(1/2)
/x/(b*x^2+a)+b^3*x*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.04, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1112, 270} \[ -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^6,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*x^5*(a + b*x^2)) - (a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^3*(a +
b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x*(a + b*x^2)) + (b^3*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
/(a + b*x^2)

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^6} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^3}{x^6} \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^6+\frac {a^3 b^3}{x^6}+\frac {3 a^2 b^4}{x^4}+\frac {3 a b^5}{x^2}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 59, normalized size = 0.37 \[ -\frac {\sqrt {\left (a+b x^2\right )^2} \left (a^3+5 a^2 b x^2+15 a b^2 x^4-5 b^3 x^6\right )}{5 x^5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^6,x]

[Out]

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

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fricas [A]  time = 0.97, size = 37, normalized size = 0.23 \[ \frac {5 \, b^{3} x^{6} - 15 \, a b^{2} x^{4} - 5 \, a^{2} b x^{2} - a^{3}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 66, normalized size = 0.42 \[ b^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {15 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 56, normalized size = 0.35 \[ -\frac {\left (-5 b^{3} x^{6}+15 a \,b^{2} x^{4}+5 a^{2} b \,x^{2}+a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (b \,x^{2}+a \right )^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^6,x)

[Out]

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

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maxima [A]  time = 1.32, size = 32, normalized size = 0.20 \[ b^{3} x - \frac {3 \, a b^{2}}{x} - \frac {a^{2} b}{x^{3}} - \frac {a^{3}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

b^3*x - 3*a*b^2/x - a^2*b/x^3 - 1/5*a^3/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^6,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x**6,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**6, x)

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