3.1956 \(\int \frac{1}{(a+\frac{b}{x^2})^{5/2} x^8} \, dx\)

Optimal. Leaf size=95 \[ -\frac{5 \sqrt{a+\frac{b}{x^2}}}{2 b^3 x}+\frac{5}{3 b^2 x^3 \sqrt{a+\frac{b}{x^2}}}+\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{2 b^{7/2}}+\frac{1}{3 b x^5 \left (a+\frac{b}{x^2}\right )^{3/2}} \]

[Out]

1/(3*b*(a + b/x^2)^(3/2)*x^5) + 5/(3*b^2*Sqrt[a + b/x^2]*x^3) - (5*Sqrt[a + b/x^2])/(2*b^3*x) + (5*a*ArcTanh[S
qrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0486969, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {335, 288, 321, 217, 206} \[ -\frac{5 \sqrt{a+\frac{b}{x^2}}}{2 b^3 x}+\frac{5}{3 b^2 x^3 \sqrt{a+\frac{b}{x^2}}}+\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{2 b^{7/2}}+\frac{1}{3 b x^5 \left (a+\frac{b}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x^8),x]

[Out]

1/(3*b*(a + b/x^2)^(3/2)*x^5) + 5/(3*b^2*Sqrt[a + b/x^2]*x^3) - (5*Sqrt[a + b/x^2])/(2*b^3*x) + (5*a*ArcTanh[S
qrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(7/2))

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^{5/2} x^8} \, dx &=-\operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^2\right )^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{3 b \left (a+\frac{b}{x^2}\right )^{3/2} x^5}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^2\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{3 b}\\ &=\frac{1}{3 b \left (a+\frac{b}{x^2}\right )^{3/2} x^5}+\frac{5}{3 b^2 \sqrt{a+\frac{b}{x^2}} x^3}-\frac{5 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{b^2}\\ &=\frac{1}{3 b \left (a+\frac{b}{x^2}\right )^{3/2} x^5}+\frac{5}{3 b^2 \sqrt{a+\frac{b}{x^2}} x^3}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{2 b^3 x}+\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{2 b^3}\\ &=\frac{1}{3 b \left (a+\frac{b}{x^2}\right )^{3/2} x^5}+\frac{5}{3 b^2 \sqrt{a+\frac{b}{x^2}} x^3}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{2 b^3 x}+\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )}{2 b^3}\\ &=\frac{1}{3 b \left (a+\frac{b}{x^2}\right )^{3/2} x^5}+\frac{5}{3 b^2 \sqrt{a+\frac{b}{x^2}} x^3}-\frac{5 \sqrt{a+\frac{b}{x^2}}}{2 b^3 x}+\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{2 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0113773, size = 47, normalized size = 0.49 \[ -\frac{a \left (a x^2+b\right ) \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{a x^2}{b}+1\right )}{3 b^2 x^5 \left (a+\frac{b}{x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x^8),x]

[Out]

-(a*(b + a*x^2)*Hypergeometric2F1[-3/2, 2, -1/2, 1 + (a*x^2)/b])/(3*b^2*(a + b/x^2)^(5/2)*x^5)

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 92, normalized size = 1. \begin{align*} -{\frac{a{x}^{2}+b}{6\,{x}^{7}} \left ( 15\,{b}^{3/2}{x}^{4}{a}^{2}+20\,{b}^{5/2}{x}^{2}a-15\,\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ) \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{2}ab+3\,{b}^{7/2} \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^(5/2)/x^8,x)

[Out]

-1/6*(a*x^2+b)*(15*b^(3/2)*x^4*a^2+20*b^(5/2)*x^2*a-15*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*(a*x^2+b)^(3/2)*x^2
*a*b+3*b^(7/2))/((a*x^2+b)/x^2)^(5/2)/x^7/b^(9/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.8945, size = 570, normalized size = 6. \begin{align*} \left [\frac{15 \,{\left (a^{3} x^{5} + 2 \, a^{2} b x^{3} + a b^{2} x\right )} \sqrt{b} \log \left (-\frac{a x^{2} + 2 \, \sqrt{b} x \sqrt{\frac{a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \,{\left (15 \, a^{2} b x^{4} + 20 \, a b^{2} x^{2} + 3 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{12 \,{\left (a^{2} b^{4} x^{5} + 2 \, a b^{5} x^{3} + b^{6} x\right )}}, -\frac{15 \,{\left (a^{3} x^{5} + 2 \, a^{2} b x^{3} + a b^{2} x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (15 \, a^{2} b x^{4} + 20 \, a b^{2} x^{2} + 3 \, b^{3}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{6 \,{\left (a^{2} b^{4} x^{5} + 2 \, a b^{5} x^{3} + b^{6} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^8,x, algorithm="fricas")

[Out]

[1/12*(15*(a^3*x^5 + 2*a^2*b*x^3 + a*b^2*x)*sqrt(b)*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2
) - 2*(15*a^2*b*x^4 + 20*a*b^2*x^2 + 3*b^3)*sqrt((a*x^2 + b)/x^2))/(a^2*b^4*x^5 + 2*a*b^5*x^3 + b^6*x), -1/6*(
15*(a^3*x^5 + 2*a^2*b*x^3 + a*b^2*x)*sqrt(-b)*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (15*a^2*b
*x^4 + 20*a*b^2*x^2 + 3*b^3)*sqrt((a*x^2 + b)/x^2))/(a^2*b^4*x^5 + 2*a*b^5*x^3 + b^6*x)]

________________________________________________________________________________________

Sympy [B]  time = 7.32211, size = 864, normalized size = 9.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(5/2)/x**8,x)

[Out]

-15*a**4*b**13*x**8*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*
b**(39/2)*x**2) + 30*a**4*b**13*x**8*log(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x
**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 30*a**3*b**14*x**6*sqrt(a*x**2/b + 1)/(12*a**3*b**(33/2)*x**8
+ 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 45*a**3*b**14*x**6*log(a*x**2/b)/(12*a*
*3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) + 90*a**3*b**14*x**6*log
(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)
*x**2) - 70*a**2*b**15*x**4*sqrt(a*x**2/b + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/
2)*x**4 + 12*b**(39/2)*x**2) - 45*a**2*b**15*x**4*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x*
*6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) + 90*a**2*b**15*x**4*log(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33
/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 46*a*b**16*x**2*sqrt(a*x**2/b +
1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2) - 15*a*b**16*x
**2*log(a*x**2/b)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2)
+ 30*a*b**16*x**2*log(sqrt(a*x**2/b + 1) + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36*a*b**(37/2
)*x**4 + 12*b**(39/2)*x**2) - 6*b**17*sqrt(a*x**2/b + 1)/(12*a**3*b**(33/2)*x**8 + 36*a**2*b**(35/2)*x**6 + 36
*a*b**(37/2)*x**4 + 12*b**(39/2)*x**2)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{2}}\right )}^{\frac{5}{2}} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^8,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(5/2)*x^8), x)