3.515 \(\int \frac{1}{x^3 (a+b x^2)^{5/2}} \, dx\)

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

[Out]

(-5*b)/(6*a^2*(a + b*x^2)^(3/2)) - 1/(2*a*x^2*(a + b*x^2)^(3/2)) - (5*b)/(2*a^3*Sqrt[a + b*x^2]) + (5*b*ArcTan
h[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0495339, antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{5 \sqrt{a+b x^2}}{2 a^3 x^2}+\frac{5}{3 a^2 x^2 \sqrt{a+b x^2}}+\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{7/2}}+\frac{1}{3 a x^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)^(5/2)),x]

[Out]

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

Mathematica [C]  time = 0.0071367, size = 37, normalized size = 0.42 \[ -\frac{b \, _2F_1\left (-\frac{3}{2},2;-\frac{1}{2};\frac{b x^2}{a}+1\right )}{3 a^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 78, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,b}{6\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,b}{2\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{5\,b}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^2+a)^(5/2),x)

[Out]

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

________________________________________________________________________________________

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/x^3/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.42624, size = 527, normalized size = 5.99 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \sqrt{a} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (15 \, a b^{2} x^{4} + 20 \, a^{2} b x^{2} + 3 \, a^{3}\right )} \sqrt{b x^{2} + a}}{12 \,{\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, -\frac{15 \,{\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (15 \, a b^{2} x^{4} + 20 \, a^{2} b x^{2} + 3 \, a^{3}\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 4.96205, size = 864, normalized size = 9.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**2+a)**(5/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 2.13894, size = 100, normalized size = 1.14 \begin{align*} -\frac{1}{6} \, b{\left (\frac{15 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{2 \,{\left (6 \, b x^{2} + 7 \, a\right )}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}} + \frac{3 \, \sqrt{b x^{2} + a}}{a^{3} b x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-1/6*b*(15*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*(6*b*x^2 + 7*a)/((b*x^2 + a)^(3/2)*a^3) + 3*sqr
t(b*x^2 + a)/(a^3*b*x^2))