3.584 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x^{16}} \, dx\)

Optimal. Leaf size=167 \[ -\frac{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 x^{15} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )} \]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*x^15*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*x
^13*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*x^11*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(9*x^9*(a + b*x^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0411472, antiderivative size = 167, normalized size of antiderivative = 1., 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} \[ -\frac{a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{15 x^{15} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^16,x]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*x^15*(a + b*x^2)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*x
^13*(a + b*x^2)) - (3*a*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*x^11*(a + b*x^2)) - (b^3*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(9*x^9*(a + b*x^2))

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]

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]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{x^{16}} \, 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^{16}} \, 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^{16}}+\frac{3 a^2 b^4}{x^{14}}+\frac{3 a b^5}{x^{12}}+\frac{b^6}{x^{10}}\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}}{15 x^{15} \left (a+b x^2\right )}-\frac{3 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 x^{13} \left (a+b x^2\right )}-\frac{3 a b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}{11 x^{11} \left (a+b x^2\right )}-\frac{b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}{9 x^9 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0141802, size = 61, normalized size = 0.37 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (1485 a^2 b x^2+429 a^3+1755 a b^2 x^4+715 b^3 x^6\right )}{6435 x^{15} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x^16,x]

[Out]

-(Sqrt[(a + b*x^2)^2]*(429*a^3 + 1485*a^2*b*x^2 + 1755*a*b^2*x^4 + 715*b^3*x^6))/(6435*x^15*(a + b*x^2))

________________________________________________________________________________________

Maple [A]  time = 0.163, size = 58, normalized size = 0.4 \begin{align*} -{\frac{715\,{b}^{3}{x}^{6}+1755\,a{x}^{4}{b}^{2}+1485\,{a}^{2}b{x}^{2}+429\,{a}^{3}}{6435\,{x}^{15} \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

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^16,x)

[Out]

-1/6435*(715*b^3*x^6+1755*a*b^2*x^4+1485*a^2*b*x^2+429*a^3)*((b*x^2+a)^2)^(3/2)/x^15/(b*x^2+a)^3

________________________________________________________________________________________

Maxima [A]  time = 0.997737, size = 50, normalized size = 0.3 \begin{align*} -\frac{715 \, b^{3} x^{6} + 1755 \, a b^{2} x^{4} + 1485 \, a^{2} b x^{2} + 429 \, a^{3}}{6435 \, x^{15}} \end{align*}

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^16,x, algorithm="maxima")

[Out]

-1/6435*(715*b^3*x^6 + 1755*a*b^2*x^4 + 1485*a^2*b*x^2 + 429*a^3)/x^15

________________________________________________________________________________________

Fricas [A]  time = 1.49762, size = 97, normalized size = 0.58 \begin{align*} -\frac{715 \, b^{3} x^{6} + 1755 \, a b^{2} x^{4} + 1485 \, a^{2} b x^{2} + 429 \, a^{3}}{6435 \, x^{15}} \end{align*}

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^16,x, algorithm="fricas")

[Out]

-1/6435*(715*b^3*x^6 + 1755*a*b^2*x^4 + 1485*a^2*b*x^2 + 429*a^3)/x^15

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}}{x^{16}}\, dx \end{align*}

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**16,x)

[Out]

Integral(((a + b*x**2)**2)**(3/2)/x**16, x)

________________________________________________________________________________________

Giac [A]  time = 1.14869, size = 93, normalized size = 0.56 \begin{align*} -\frac{715 \, b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 1755 \, a b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 1485 \, a^{2} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 429 \, a^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{6435 \, x^{15}} \end{align*}

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^16,x, algorithm="giac")

[Out]

-1/6435*(715*b^3*x^6*sgn(b*x^2 + a) + 1755*a*b^2*x^4*sgn(b*x^2 + a) + 1485*a^2*b*x^2*sgn(b*x^2 + a) + 429*a^3*
sgn(b*x^2 + a))/x^15