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

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

[Out]

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

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Rubi [A]  time = 0.05, 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 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]

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]]

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*}

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Mathematica [C]  time = 0.01, 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]

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

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fricas [A]  time = 1.01, size = 241, normalized size = 2.74 \[ \left [\frac {15 \, {\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{2} x^{4} + 20 \, a^{2} b x^{2} + 3 \, a^{3}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, -\frac {15 \, {\left (b^{3} x^{6} + 2 \, a b^{2} x^{4} + a^{2} b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, a b^{2} x^{4} + 20 \, a^{2} b x^{2} + 3 \, a^{3}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \]

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)]

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giac [A]  time = 1.08, size = 73, normalized size = 0.83 \[ -\frac {5 \, b \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{3}} - \frac {6 \, {\left (b x^{2} + a\right )} b + a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {\sqrt {b x^{2} + a}}{2 \, a^{3} x^{2}} \]

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]

-5/2*b*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) - 1/3*(6*(b*x^2 + a)*b + a*b)/((b*x^2 + a)^(3/2)*a^3) -
1/2*sqrt(b*x^2 + a)/(a^3*x^2)

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maple [A]  time = 0.01, size = 78, normalized size = 0.89 \[ -\frac {5 b}{6 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}+\frac {5 b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {7}{2}}}-\frac {5 b}{2 \sqrt {b \,x^{2}+a}\, a^{3}}-\frac {1}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,x^{2}} \]

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/a^(7/2)*b*ln((2*a+2*(b*x^2+
a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 1.35, size = 66, normalized size = 0.75 \[ \frac {5 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} - \frac {5 \, b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {1}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} \]

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]

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

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mupad [B]  time = 5.25, size = 73, normalized size = 0.83 \[ \frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}-\frac {1}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {10\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {5\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^2)^(5/2)),x)

[Out]

(5*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(7/2)) - 1/(2*a*x^2*(a + b*x^2)^(3/2)) - (10*b)/(3*a^2*(a + b*x^2)
^(3/2)) - (5*b^2*x^2)/(2*a^3*(a + b*x^2)^(3/2))

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sympy [B]  time = 5.17, size = 864, normalized size = 9.82 \[ - \frac {6 a^{17} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {46 a^{16} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {15 a^{16} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {30 a^{16} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {70 a^{15} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {45 a^{15} b^{2} x^{4} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {90 a^{15} b^{2} x^{4} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {30 a^{14} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {45 a^{14} b^{3} x^{6} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {90 a^{14} b^{3} x^{6} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {15 a^{13} b^{4} x^{8} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {30 a^{13} b^{4} x^{8} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} \]

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)

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