3.6.81 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{10}} \, dx\) [581]

Optimal. Leaf size=167 \[ -\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )} \]

[Out]

-1/9*a^3*((b*x^2+a)^2)^(1/2)/x^9/(b*x^2+a)-3/7*a^2*b*((b*x^2+a)^2)^(1/2)/x^7/(b*x^2+a)-3/5*a*b^2*((b*x^2+a)^2)
^(1/2)/x^5/(b*x^2+a)-1/3*b^3*((b*x^2+a)^2)^(1/2)/x^3/(b*x^2+a)

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Rubi [A]
time = 0.03, antiderivative size = 167, 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 = {1126, 276} \begin {gather*} -\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^9 \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^10,x]

[Out]

-1/9*(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x^9*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*x^
7*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*x^5*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2 + b
^2*x^4])/(3*x^3*(a + b*x^2))

Rule 276

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 1126

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^{10}} \, 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^{10}} \, 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^{10}}+\frac {3 a^2 b^4}{x^8}+\frac {3 a b^5}{x^6}+\frac {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}}{9 x^9 \left (a+b x^2\right )}-\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 x^7 \left (a+b x^2\right )}-\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \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 (35 a^3+135 a^2 b x^2+189 a b^2 x^4+105 b^3 x^6\right )}{315 x^9 \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^10,x]

[Out]

-1/315*(Sqrt[(a + b*x^2)^2]*(35*a^3 + 135*a^2*b*x^2 + 189*a*b^2*x^4 + 105*b^3*x^6))/(x^9*(a + b*x^2))

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Maple [A]
time = 0.02, size = 58, normalized size = 0.35

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{3} b^{3} x^{6}-\frac {3}{5} a \,b^{2} x^{4}-\frac {3}{7} a^{2} b \,x^{2}-\frac {1}{9} a^{3}\right )}{\left (b \,x^{2}+a \right ) x^{9}}\)	\(57\)
gosper	\(-\frac {\left (105 b^{3} x^{6}+189 a \,b^{2} x^{4}+135 a^{2} b \,x^{2}+35 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (b \,x^{2}+a \right )^{3} x^{9}}\)	\(58\)
default	\(-\frac {\left (105 b^{3} x^{6}+189 a \,b^{2} x^{4}+135 a^{2} b \,x^{2}+35 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{315 \left (b \,x^{2}+a \right )^{3} x^{9}}\)	\(58\)

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^10,x,method=_RETURNVERBOSE)

[Out]

-1/315*(105*b^3*x^6+189*a*b^2*x^4+135*a^2*b*x^2+35*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3/x^9

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Maxima [A]
time = 0.28, size = 35, normalized size = 0.21 \begin {gather*} -\frac {b^{3}}{3 \, x^{3}} - \frac {3 \, a b^{2}}{5 \, x^{5}} - \frac {3 \, a^{2} b}{7 \, x^{7}} - \frac {a^{3}}{9 \, x^{9}} \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^10,x, algorithm="maxima")

[Out]

-1/3*b^3/x^3 - 3/5*a*b^2/x^5 - 3/7*a^2*b/x^7 - 1/9*a^3/x^9

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Fricas [A]
time = 0.32, size = 37, normalized size = 0.22 \begin {gather*} -\frac {105 \, b^{3} x^{6} + 189 \, a b^{2} x^{4} + 135 \, a^{2} b x^{2} + 35 \, a^{3}}{315 \, x^{9}} \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^10,x, algorithm="fricas")

[Out]

-1/315*(105*b^3*x^6 + 189*a*b^2*x^4 + 135*a^2*b*x^2 + 35*a^3)/x^9

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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^{10}}\, 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**10,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**10, x)

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Giac [A]
time = 6.16, size = 69, normalized size = 0.41 \begin {gather*} -\frac {105 \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 189 \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 135 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 35 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{315 \, x^{9}} \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^10,x, algorithm="giac")

[Out]

-1/315*(105*b^3*x^6*sgn(b*x^2 + a) + 189*a*b^2*x^4*sgn(b*x^2 + a) + 135*a^2*b*x^2*sgn(b*x^2 + a) + 35*a^3*sgn(
b*x^2 + a))/x^9

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Mupad [B]
time = 4.26, size = 151, normalized size = 0.90 \begin {gather*} -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^9\,\left (b\,x^2+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{3\,x^3\,\left (b\,x^2+a\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{5\,x^5\,\left (b\,x^2+a\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{7\,x^7\,\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^10,x)

[Out]

- (a^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(9*x^9*(a + b*x^2)) - (b^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(3*x^3
*(a + b*x^2)) - (3*a*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(5*x^5*(a + b*x^2)) - (3*a^2*b*(a^2 + b^2*x^4 + 2*
a*b*x^2)^(1/2))/(7*x^7*(a + b*x^2))

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