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

Optimal. Leaf size=80 \[ -\frac {5}{2} a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}+\frac {5}{2} a b \sqrt {a+b x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {266, 47, 50, 63, 208} \[ -\frac {5}{2} a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}+\frac {5}{2} a b \sqrt {a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^3,x]

[Out]

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

Rule 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {\left (a+b x^2\right )^{5/2}}{x^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {1}{4} (5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{6} b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {1}{4} (5 a b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {5}{2} a b \sqrt {a+b x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {1}{4} \left (5 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {5}{2} a b \sqrt {a+b x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac {1}{2} \left (5 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=\frac {5}{2} a b \sqrt {a+b x^2}+\frac {5}{6} b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{2 x^2}-\frac {5}{2} a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 37, normalized size = 0.46 \[ \frac {b \left (a+b x^2\right )^{7/2} \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {b x^2}{a}+1\right )}{7 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^3,x]

[Out]

(b*(a + b*x^2)^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, 1 + (b*x^2)/a])/(7*a^2)

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 142, normalized size = 1.78 \[ \left [\frac {15 \, a^{\frac {3}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{12 \, x^{2}}, \frac {15 \, \sqrt {-a} a b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{6 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(15*a^(3/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*b^2*x^4 + 14*a*b*x^2 - 3*a^
2)*sqrt(b*x^2 + a))/x^2, 1/6*(15*sqrt(-a)*a*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2*b^2*x^4 + 14*a*b*x^2 -
3*a^2)*sqrt(b*x^2 + a))/x^2]

________________________________________________________________________________________

giac [A]  time = 1.13, size = 82, normalized size = 1.02 \[ \frac {\frac {15 \, a^{2} b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2} + 12 \, \sqrt {b x^{2} + a} a b^{2} - \frac {3 \, \sqrt {b x^{2} + a} a^{2} b}{x^{2}}}{6 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a^2*b^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 2*(b*x^2 + a)^(3/2)*b^2 + 12*sqrt(b*x^2 + a)*a*b^2
- 3*sqrt(b*x^2 + a)*a^2*b/x^2)/b

________________________________________________________________________________________

maple [A]  time = 0.01, size = 88, normalized size = 1.10 \[ -\frac {5 a^{\frac {3}{2}} b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2}+\frac {5 \sqrt {b \,x^{2}+a}\, a b}{2}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b}{6}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b}{2 a}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.38, size = 76, normalized size = 0.95 \[ -\frac {5}{2} \, a^{\frac {3}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {5}{6} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{2 \, a} + \frac {5}{2} \, \sqrt {b x^{2} + a} a b - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{2 \, a x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.95, size = 66, normalized size = 0.82 \[ \frac {b\,{\left (b\,x^2+a\right )}^{3/2}}{3}-\frac {a^2\,\sqrt {b\,x^2+a}}{2\,x^2}+2\,a\,b\,\sqrt {b\,x^2+a}+\frac {a^{3/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^3,x)

[Out]

(b*(a + b*x^2)^(3/2))/3 - (a^2*(a + b*x^2)^(1/2))/(2*x^2) + (a^(3/2)*b*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i
)/2 + 2*a*b*(a + b*x^2)^(1/2)

________________________________________________________________________________________

sympy [A]  time = 3.22, size = 112, normalized size = 1.40 \[ - \frac {a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{2 x^{2}} + \frac {7 a^{\frac {3}{2}} b \sqrt {1 + \frac {b x^{2}}{a}}}{3} + \frac {5 a^{\frac {3}{2}} b \log {\left (\frac {b x^{2}}{a} \right )}}{4} - \frac {5 a^{\frac {3}{2}} b \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2} + \frac {\sqrt {a} b^{2} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)*sqrt(1 + b*x**2/a)/(2*x**2) + 7*a**(3/2)*b*sqrt(1 + b*x**2/a)/3 + 5*a**(3/2)*b*log(b*x**2/a)/4 - 5*a
**(3/2)*b*log(sqrt(1 + b*x**2/a) + 1)/2 + sqrt(a)*b**2*x**2*sqrt(1 + b*x**2/a)/3

________________________________________________________________________________________