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

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

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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} \begin {gather*} \frac {b^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {a b^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {3 a^2 b 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}}{x \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x*(a + b*x^2))) + (3*a^2*b*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*
x^2) + (a*b^2*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*x^2) + (b^3*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*
(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^2} \, 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^2} \, 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 (3 a^2 b^4+\frac {a^3 b^3}{x^2}+3 a b^5 x^2+b^6 x^4\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}}{x \left (a+b x^2\right )}+\frac {3 a^2 b x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {a b^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {b^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 60, normalized size = 0.38 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (-5 a^3+15 a^2 b x^2+5 a b^2 x^4+b^3 x^6\right )}{5 x \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 9.71, size = 60, normalized size = 0.38 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (-5 a^3+15 a^2 b x^2+5 a b^2 x^4+b^3 x^6\right )}{5 x \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 36, normalized size = 0.23 \begin {gather*} \frac {b^{3} x^{6} + 5 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} - 5 \, a^{3}}{5 \, x} \end {gather*}

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^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 64, normalized size = 0.41 \begin {gather*} \frac {1}{5} \, b^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + a b^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{2} b x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{x} \end {gather*}

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^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 58, normalized size = 0.37 \begin {gather*} -\frac {\left (-b^{3} x^{6}-5 a \,b^{2} x^{4}-15 a^{2} b \,x^{2}+5 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (b \,x^{2}+a \right )^{3} x} \end {gather*}

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^2,x)

[Out]

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

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maxima [A]  time = 1.34, size = 32, normalized size = 0.20 \begin {gather*} \frac {1}{5} \, b^{3} x^{5} + a b^{2} x^{3} + 3 \, a^{2} b x - \frac {a^{3}}{x} \end {gather*}

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^2,x, algorithm="maxima")

[Out]

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

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

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^2,x)

[Out]

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

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

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**2,x)

[Out]

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

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