3.401 \(\int \frac {(a+b x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=86 \[ \frac {5}{2} a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {5}{2} b^2 x \sqrt {a+b x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{3 x}-\frac {\left (a+b x^2\right )^{5/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 86, 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} \[ \frac {5}{2} b^2 x \sqrt {a+b x^2}+\frac {5}{2} a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{5/2}}{3 x^3}-\frac {5 b \left (a+b x^2\right )^{3/2}}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^4,x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.63 \[ -\frac {a^2 \sqrt {a+b x^2} \, _2F_1\left (-\frac {5}{2},-\frac {3}{2};-\frac {1}{2};-\frac {b x^2}{a}\right )}{3 x^3 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^4,x]

[Out]

-1/3*(a^2*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((b*x^2)/a)])/(x^3*Sqrt[1 + (b*x^2)/a])

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fricas [A]  time = 0.93, size = 141, normalized size = 1.64 \[ \left [\frac {15 \, a b^{\frac {3}{2}} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (3 \, b^{2} x^{4} - 14 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{12 \, x^{3}}, -\frac {15 \, a \sqrt {-b} b x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, b^{2} x^{4} - 14 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{6 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.12, size = 132, normalized size = 1.53 \[ \frac {1}{2} \, \sqrt {b x^{2} + a} b^{2} x - \frac {5}{4} \, a b^{\frac {3}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} b^{\frac {3}{2}} - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} b^{\frac {3}{2}} + 7 \, a^{4} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*b^2*x - 5/4*a*b^(3/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/3*(9*(sqrt(b)*x - sqrt(b*x^
2 + a))^4*a^2*b^(3/2) - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*b^(3/2) + 7*a^4*b^(3/2))/((sqrt(b)*x - sqrt(b*x
^2 + a))^2 - a)^3

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maple [A]  time = 0.01, size = 110, normalized size = 1.28 \[ \frac {5 a \,b^{\frac {3}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {5 \sqrt {b \,x^{2}+a}\, b^{2} x}{2}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2} x}{3 a}+\frac {4 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2} x}{3 a^{2}}-\frac {4 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}{3 a^{2} x}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^4,x)

[Out]

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

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maxima [A]  time = 1.32, size = 84, normalized size = 0.98 \[ \frac {5}{2} \, \sqrt {b x^{2} + a} b^{2} x + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2} x}{3 \, a} + \frac {5}{2} \, a b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{3 \, a x} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{3 \, a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^4,x)

[Out]

int((a + b*x^2)^(5/2)/x^4, x)

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sympy [A]  time = 3.16, size = 112, normalized size = 1.30 \[ - \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {7 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} - \frac {5 a b^{\frac {3}{2}} \log {\left (\frac {a}{b x^{2}} \right )}}{4} + \frac {5 a b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a}{b x^{2}} + 1} + 1 \right )}}{2} + \frac {b^{\frac {5}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**4,x)

[Out]

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

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