3.1945 \(\int \frac{x}{(a+\frac{b}{x^2})^{5/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac{5 b}{2 a^3 \sqrt{a+\frac{b}{x^2}}}+\frac{5 b}{6 a^2 \left (a+\frac{b}{x^2}\right )^{3/2}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{2 a^{7/2}}+\frac{x^2}{2 a \left (a+\frac{b}{x^2}\right )^{3/2}} \]

[Out]

(5*b)/(6*a^2*(a + b/x^2)^(3/2)) + (5*b)/(2*a^3*Sqrt[a + b/x^2]) + x^2/(2*a*(a + b/x^2)^(3/2)) - (5*b*ArcTanh[S
qrt[a + b/x^2]/Sqrt[a]])/(2*a^(7/2))

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Rubi [A]  time = 0.0434574, antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac{5 x^2 \sqrt{a+\frac{b}{x^2}}}{2 a^3}-\frac{5 x^2}{3 a^2 \sqrt{a+\frac{b}{x^2}}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{x^2}{3 a \left (a+\frac{b}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x^2)^(5/2),x]

[Out]

-x^2/(3*a*(a + b/x^2)^(3/2)) - (5*x^2)/(3*a^2*Sqrt[a + b/x^2]) + (5*Sqrt[a + b/x^2]*x^2)/(2*a^3) - (5*b*ArcTan
h[Sqrt[a + b/x^2]/Sqrt[a]])/(2*a^(7/2))

Rule 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.156167, size = 101, normalized size = 1.15 \[ \frac{\sqrt{a} \left (3 a^2 x^4+20 a b x^2+15 b^2\right )-\frac{15 b^{3/2} \left (a x^2+b\right ) \sqrt{\frac{a x^2}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{x}}{6 a^{7/2} \sqrt{a+\frac{b}{x^2}} \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x^2)^(5/2),x]

[Out]

(Sqrt[a]*(15*b^2 + 20*a*b*x^2 + 3*a^2*x^4) - (15*b^(3/2)*(b + a*x^2)*Sqrt[1 + (a*x^2)/b]*ArcSinh[(Sqrt[a]*x)/S
qrt[b]])/x)/(6*a^(7/2)*Sqrt[a + b/x^2]*(b + a*x^2))

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Maple [A]  time = 0.008, size = 85, normalized size = 1. \begin{align*}{\frac{a{x}^{2}+b}{6\,{x}^{5}} \left ( 3\,{x}^{5}{a}^{7/2}+20\,{a}^{5/2}{x}^{3}b+15\,{a}^{3/2}x{b}^{2}-15\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) \left ( a{x}^{2}+b \right ) ^{3/2}ab \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}{a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9624, size = 566, normalized size = 6.43 \begin{align*} \left [\frac{15 \,{\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt{a} \log \left (-2 \, a x^{2} + 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right ) + 2 \,{\left (3 \, a^{3} x^{6} + 20 \, a^{2} b x^{4} + 15 \, a b^{2} x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{12 \,{\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}, \frac{15 \,{\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (3 \, a^{3} x^{6} + 20 \, a^{2} b x^{4} + 15 \, a b^{2} x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{6 \,{\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 4.94806, size = 819, normalized size = 9.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27455, size = 151, normalized size = 1.72 \begin{align*} \frac{1}{6} \, b{\left (\frac{2 \,{\left (a + \frac{6 \,{\left (a x^{2} + b\right )}}{x^{2}}\right )} x^{2}}{{\left (a x^{2} + b\right )} a^{3} \sqrt{\frac{a x^{2} + b}{x^{2}}}} + \frac{15 \, \arctan \left (\frac{\sqrt{\frac{a x^{2} + b}{x^{2}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{3 \, \sqrt{\frac{a x^{2} + b}{x^{2}}}}{{\left (a - \frac{a x^{2} + b}{x^{2}}\right )} a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b*(2*(a + 6*(a*x^2 + b)/x^2)*x^2/((a*x^2 + b)*a^3*sqrt((a*x^2 + b)/x^2)) + 15*arctan(sqrt((a*x^2 + b)/x^2)
/sqrt(-a))/(sqrt(-a)*a^3) - 3*sqrt((a*x^2 + b)/x^2)/((a - (a*x^2 + b)/x^2)*a^3))