3.614 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=249 \[ \frac {b^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {5 a b^4 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 249, 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} \[ -\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {10 a^2 b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {b^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^6,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*x^5*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*x^3*
(a + b*x^2)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(x*(a + b*x^2)) + (10*a^2*b^3*x*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])/(a + b*x^2) + (5*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (b^5*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 )^{5/2}}{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 )^5}{x^6} \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (10 a^2 b^8+\frac {a^5 b^5}{x^6}+\frac {5 a^4 b^6}{x^4}+\frac {10 a^3 b^7}{x^2}+5 a b^9 x^2+b^{10} x^4\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {10 a^2 b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {b^5 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 = 83, normalized size = 0.33 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (-3 a^5-25 a^4 b x^2-150 a^3 b^2 x^4+150 a^2 b^3 x^6+25 a b^4 x^8+3 b^5 x^{10}\right )}{15 x^5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^6,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(-3*a^5 - 25*a^4*b*x^2 - 150*a^3*b^2*x^4 + 150*a^2*b^3*x^6 + 25*a*b^4*x^8 + 3*b^5*x^10))/
(15*x^5*(a + b*x^2))

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fricas [A]  time = 0.81, size = 59, normalized size = 0.24 \[ \frac {3 \, b^{5} x^{10} + 25 \, a b^{4} x^{8} + 150 \, a^{2} b^{3} x^{6} - 150 \, a^{3} b^{2} x^{4} - 25 \, a^{4} b x^{2} - 3 \, a^{5}}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="fricas")

[Out]

1/15*(3*b^5*x^10 + 25*a*b^4*x^8 + 150*a^2*b^3*x^6 - 150*a^3*b^2*x^4 - 25*a^4*b*x^2 - 3*a^5)/x^5

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giac [A]  time = 0.17, size = 106, normalized size = 0.43 \[ \frac {1}{5} \, b^{5} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{3} \, a b^{4} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 10 \, a^{2} b^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {150 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 25 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{15 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="giac")

[Out]

1/5*b^5*x^5*sgn(b*x^2 + a) + 5/3*a*b^4*x^3*sgn(b*x^2 + a) + 10*a^2*b^3*x*sgn(b*x^2 + a) - 1/15*(150*a^3*b^2*x^
4*sgn(b*x^2 + a) + 25*a^4*b*x^2*sgn(b*x^2 + a) + 3*a^5*sgn(b*x^2 + a))/x^5

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maple [A]  time = 0.01, size = 80, normalized size = 0.32 \[ -\frac {\left (-3 b^{5} x^{10}-25 a \,b^{4} x^{8}-150 a^{2} b^{3} x^{6}+150 a^{3} b^{2} x^{4}+25 a^{4} b \,x^{2}+3 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}}}{15 \left (b \,x^{2}+a \right )^{5} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x)

[Out]

-1/15*(-3*b^5*x^10-25*a*b^4*x^8-150*a^2*b^3*x^6+150*a^3*b^2*x^4+25*a^4*b*x^2+3*a^5)*((b*x^2+a)^2)^(5/2)/x^5/(b
*x^2+a)^5

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maxima [A]  time = 1.35, size = 55, normalized size = 0.22 \[ \frac {1}{5} \, b^{5} x^{5} + \frac {5}{3} \, a b^{4} x^{3} + 10 \, a^{2} b^{3} x - \frac {10 \, a^{3} b^{2}}{x} - \frac {5 \, a^{4} b}{3 \, x^{3}} - \frac {a^{5}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="maxima")

[Out]

1/5*b^5*x^5 + 5/3*a*b^4*x^3 + 10*a^2*b^3*x - 10*a^3*b^2/x - 5/3*a^4*b/x^3 - 1/5*a^5/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^6,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**6,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**6, x)

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