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

Optimal. Leaf size=72 \[ a^{5/2} \left (-\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+a^2 \sqrt {a+b x^2}+\frac {1}{3} a \left (a+b x^2\right )^{3/2}+\frac {1}{5} \left (a+b x^2\right )^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, 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, 50, 63, 208} \[ a^2 \sqrt {a+b x^2}+a^{5/2} \left (-\tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\right )+\frac {1}{3} a \left (a+b x^2\right )^{3/2}+\frac {1}{5} \left (a+b x^2\right )^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x,x]

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 62, normalized size = 0.86 \[ \frac {1}{15} \sqrt {a+b x^2} \left (23 a^2+11 a b x^2+3 b^2 x^4\right )-a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x,x]

[Out]

(Sqrt[a + b*x^2]*(23*a^2 + 11*a*b*x^2 + 3*b^2*x^4))/15 - a^(5/2)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

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fricas [A]  time = 0.96, size = 126, normalized size = 1.75 \[ \left [\frac {1}{2} \, a^{\frac {5}{2}} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + \frac {1}{15} \, {\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\right )} \sqrt {b x^{2} + a}, \sqrt {-a} a^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + \frac {1}{15} \, {\left (3 \, b^{2} x^{4} + 11 \, a b x^{2} + 23 \, a^{2}\right )} \sqrt {b x^{2} + a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*a^(5/2)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 1/15*(3*b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt
(b*x^2 + a), sqrt(-a)*a^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + 1/15*(3*b^2*x^4 + 11*a*b*x^2 + 23*a^2)*sqrt(b*x^2
+ a)]

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giac [A]  time = 1.19, size = 62, normalized size = 0.86 \[ \frac {a^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a + \sqrt {b x^{2} + a} a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 66, normalized size = 0.92 \[ -a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\sqrt {b \,x^{2}+a}\, a^{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} a}{3}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 54, normalized size = 0.75 \[ -a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a + \sqrt {b x^{2} + a} a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.68, size = 59, normalized size = 0.82 \[ \frac {a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+\frac {{\left (b\,x^2+a\right )}^{5/2}}{5}+a^2\,\sqrt {b\,x^2+a}+a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x,x)

[Out]

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

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sympy [A]  time = 3.56, size = 105, normalized size = 1.46 \[ \frac {23 a^{\frac {5}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{15} + \frac {a^{\frac {5}{2}} \log {\left (\frac {b x^{2}}{a} \right )}}{2} - a^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )} + \frac {11 a^{\frac {3}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{15} + \frac {\sqrt {a} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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