3.4.93 \(\int \frac {(a+b x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=82 \[ b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {b^2 \sqrt {a+b x^2}}{x}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3} \]

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Rubi [A]  time = 0.03, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {277, 217, 206} \begin {gather*} -\frac {b^2 \sqrt {a+b x^2}}{x}+b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^6,x]

[Out]

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

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^6} \, dx &=-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}+b \int \frac {\left (a+b x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}+b^2 \int \frac {\sqrt {a+b x^2}}{x^2} \, dx\\ &=-\frac {b^2 \sqrt {a+b x^2}}{x}-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}+b^3 \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=-\frac {b^2 \sqrt {a+b x^2}}{x}-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}+b^3 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=-\frac {b^2 \sqrt {a+b x^2}}{x}-\frac {b \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {\left (a+b x^2\right )^{5/2}}{5 x^5}+b^{5/2} \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 = 0.66 \begin {gather*} -\frac {a^2 \sqrt {a+b x^2} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \sqrt {\frac {b x^2}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^6,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 68, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-3 a^2-11 a b x^2-23 b^2 x^4\right )}{15 x^5}-b^{5/2} \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^(5/2)/x^6,x]

[Out]

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

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fricas [A]  time = 0.85, size = 140, normalized size = 1.71 \begin {gather*} \left [\frac {15 \, b^{\frac {5}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (23 \, b^{2} x^{4} + 11 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{30 \, x^{5}}, -\frac {15 \, \sqrt {-b} b^{2} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (23 \, b^{2} x^{4} + 11 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{15 \, x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/30*(15*b^(5/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(23*b^2*x^4 + 11*a*b*x^2 + 3*a^2)*sq
rt(b*x^2 + a))/x^5, -1/15*(15*sqrt(-b)*b^2*x^5*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (23*b^2*x^4 + 11*a*b*x^2 +
3*a^2)*sqrt(b*x^2 + a))/x^5]

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giac [B]  time = 1.14, size = 168, normalized size = 2.05 \begin {gather*} -\frac {1}{2} \, b^{\frac {5}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a b^{\frac {5}{2}} - 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{2} b^{\frac {5}{2}} + 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{3} b^{\frac {5}{2}} - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{4} b^{\frac {5}{2}} + 23 \, a^{5} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="giac")

[Out]

-1/2*b^(5/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a*b^(5/2) - 90*(s
qrt(b)*x - sqrt(b*x^2 + a))^6*a^2*b^(5/2) + 140*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^3*b^(5/2) - 70*(sqrt(b)*x -
sqrt(b*x^2 + a))^2*a^4*b^(5/2) + 23*a^5*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5

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maple [A]  time = 0.01, size = 130, normalized size = 1.59 \begin {gather*} b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {\sqrt {b \,x^{2}+a}\, b^{3} x}{a}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3} x}{3 a^{2}}+\frac {8 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3} x}{15 a^{3}}-\frac {8 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}{15 a^{3} x}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}{15 a^{2} x^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^6,x)

[Out]

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

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maxima [A]  time = 1.40, size = 104, normalized size = 1.27 \begin {gather*} \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} b^{3} x}{a} + b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{15 \, a^{2} x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^6,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^6,x)

[Out]

int((a + b*x^2)^(5/2)/x^6, x)

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sympy [A]  time = 3.66, size = 105, normalized size = 1.28 \begin {gather*} - \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {23 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} - \frac {b^{\frac {5}{2}} \log {\left (\frac {a}{b x^{2}} \right )}}{2} + b^{\frac {5}{2}} \log {\left (\sqrt {\frac {a}{b x^{2}} + 1} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**6,x)

[Out]

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

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