3.5.8 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=163 \[ -\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 163, 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 {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}-\frac {b^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^8,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^7*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*x^5*
(a + b*x^2)) - (a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^3*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(x*(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^8} \, 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^8} \, 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 (\frac {a^3 b^3}{x^8}+\frac {3 a^2 b^4}{x^6}+\frac {3 a b^5}{x^4}+\frac {b^6}{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}}{7 x^7 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 61, normalized size = 0.37 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (5 a^3+21 a^2 b x^2+35 a b^2 x^4+35 b^3 x^6\right )}{35 x^7 \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^8,x]

[Out]

-1/35*(Sqrt[(a + b*x^2)^2]*(5*a^3 + 21*a^2*b*x^2 + 35*a*b^2*x^4 + 35*b^3*x^6))/(x^7*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 17.90, size = 61, normalized size = 0.37 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (-5 a^3-21 a^2 b x^2-35 a b^2 x^4-35 b^3 x^6\right )}{35 x^7 \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^8,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(-5*a^3 - 21*a^2*b*x^2 - 35*a*b^2*x^4 - 35*b^3*x^6))/(35*x^7*(a + b*x^2))

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fricas [A]  time = 0.77, size = 37, normalized size = 0.23 \begin {gather*} -\frac {35 \, b^{3} x^{6} + 35 \, a b^{2} x^{4} + 21 \, a^{2} b x^{2} + 5 \, a^{3}}{35 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

-1/35*(35*b^3*x^6 + 35*a*b^2*x^4 + 21*a^2*b*x^2 + 5*a^3)/x^7

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giac [A]  time = 0.22, size = 69, normalized size = 0.42 \begin {gather*} -\frac {35 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 21 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{35 \, x^{7}} \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^8,x, algorithm="giac")

[Out]

-1/35*(35*b^3*x^6*sgn(b*x^2 + a) + 35*a*b^2*x^4*sgn(b*x^2 + a) + 21*a^2*b*x^2*sgn(b*x^2 + a) + 5*a^3*sgn(b*x^2
+ a))/x^7

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maple [A]  time = 0.01, size = 58, normalized size = 0.36 \begin {gather*} -\frac {\left (35 b^{3} x^{6}+35 a \,b^{2} x^{4}+21 a^{2} b \,x^{2}+5 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (b \,x^{2}+a \right )^{3} x^{7}} \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^8,x)

[Out]

-1/35*(35*b^3*x^6+35*a*b^2*x^4+21*a^2*b*x^2+5*a^3)*((b*x^2+a)^2)^(3/2)/x^7/(b*x^2+a)^3

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maxima [A]  time = 1.27, size = 35, normalized size = 0.21 \begin {gather*} -\frac {b^{3}}{x} - \frac {a b^{2}}{x^{3}} - \frac {3 \, a^{2} b}{5 \, x^{5}} - \frac {a^{3}}{7 \, x^{7}} \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^8,x, algorithm="maxima")

[Out]

-b^3/x - a*b^2/x^3 - 3/5*a^2*b/x^5 - 1/7*a^3/x^7

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mupad [B]  time = 4.25, size = 151, normalized size = 0.93 \begin {gather*} -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x\,\left (b\,x^2+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{x^3\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{5\,x^5\,\left (b\,x^2+a\right )} \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^8,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(x*(a
+ b*x^2)) - (a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(x^3*(a + b*x^2)) - (3*a^2*b*(a^2 + b^2*x^4 + 2*a*b*x^2)
^(1/2))/(5*x^5*(a + b*x^2))

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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^{8}}\, 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**8,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**8, x)

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