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

Optimal. Leaf size=83 \[ \frac {15}{8} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {5}{4} b x \left (a+b x^2\right )^{3/2}+\frac {15}{8} a b x \sqrt {a+b x^2} \]

[Out]

5/4*b*x*(b*x^2+a)^(3/2)-(b*x^2+a)^(5/2)/x+15/8*a^2*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)+15/8*a*b*x*(b*x^
2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {277, 195, 217, 206} \[ \frac {15}{8} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {5}{4} b x \left (a+b x^2\right )^{3/2}+\frac {15}{8} a b x \sqrt {a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^2,x]

[Out]

(15*a*b*x*Sqrt[a + b*x^2])/8 + (5*b*x*(a + b*x^2)^(3/2))/4 - (a + b*x^2)^(5/2)/x + (15*a^2*Sqrt[b]*ArcTanh[(Sq
rt[b]*x)/Sqrt[a + b*x^2]])/8

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

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 277

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{5/2}}{x^2} \, dx &=-\frac {\left (a+b x^2\right )^{5/2}}{x}+(5 b) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {5}{4} b x \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {1}{4} (15 a b) \int \sqrt {a+b x^2} \, dx\\ &=\frac {15}{8} a b x \sqrt {a+b x^2}+\frac {5}{4} b x \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {1}{8} \left (15 a^2 b\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {15}{8} a b x \sqrt {a+b x^2}+\frac {5}{4} b x \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {1}{8} \left (15 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {15}{8} a b x \sqrt {a+b x^2}+\frac {5}{4} b x \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{5/2}}{x}+\frac {15}{8} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 52, normalized size = 0.63 \[ -\frac {a^2 \sqrt {a+b x^2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};-\frac {b x^2}{a}\right )}{x \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^2,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 140, normalized size = 1.69 \[ \left [\frac {15 \, a^{2} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} x^{4} + 9 \, a b x^{2} - 8 \, a^{2}\right )} \sqrt {b x^{2} + a}}{16 \, x}, -\frac {15 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} x^{4} + 9 \, a b x^{2} - 8 \, a^{2}\right )} \sqrt {b x^{2} + a}}{8 \, x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*a^2*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*x^4 + 9*a*b*x^2 - 8*a^2)*sq
rt(b*x^2 + a))/x, -1/8*(15*a^2*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*x^4 + 9*a*b*x^2 - 8*a^2)
*sqrt(b*x^2 + a))/x]

________________________________________________________________________________________

giac [A]  time = 1.11, size = 87, normalized size = 1.05 \[ -\frac {15}{16} \, a^{2} \sqrt {b} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, a^{3} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{8} \, {\left (2 \, b^{2} x^{2} + 9 \, a b\right )} \sqrt {b x^{2} + a} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/16*a^2*sqrt(b)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2*a^3*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)
+ 1/8*(2*b^2*x^2 + 9*a*b)*sqrt(b*x^2 + a)*x

________________________________________________________________________________________

maple [A]  time = 0.00, size = 85, normalized size = 1.02 \[ \frac {15 a^{2} \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}+\frac {15 \sqrt {b \,x^{2}+a}\, a b x}{8}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b x}{4}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b x}{a}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/x*(b*x^2+a)^(7/2)+1/a*b*x*(b*x^2+a)^(5/2)+5/4*b*x*(b*x^2+a)^(3/2)+15/8*a*b*x*(b*x^2+a)^(1/2)+15/8*a^2*b^(
1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.32, size = 59, normalized size = 0.71 \[ \frac {5}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b x + \frac {15}{8} \, \sqrt {b x^{2} + a} a b x + \frac {15}{8} \, a^{2} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*(b*x^2 + a)^(3/2)*b*x + 15/8*sqrt(b*x^2 + a)*a*b*x + 15/8*a^2*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - (b*x^2 + a)
^(5/2)/x

________________________________________________________________________________________

mupad [B]  time = 5.05, size = 40, normalized size = 0.48 \[ -\frac {{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^2,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 3.63, size = 117, normalized size = 1.41 \[ - \frac {a^{\frac {5}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{\frac {3}{2}} b x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 \sqrt {a} b^{2} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {b^{3} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)/(x*sqrt(1 + b*x**2/a)) + a**(3/2)*b*x/(8*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b**2*x**3/(8*sqrt(1 + b*x*
*2/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/8 + b**3*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

________________________________________________________________________________________