3.431 \(\int \frac {(a+b x^2)^{9/2}}{x^6} \, dx\)

Optimal. Leaf size=129 \[ \frac {63}{8} a^2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}+\frac {63}{8} a b^3 x \sqrt {a+b x^2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3} \]

[Out]

21/4*b^3*x*(b*x^2+a)^(3/2)-21/5*b^2*(b*x^2+a)^(5/2)/x-3/5*b*(b*x^2+a)^(7/2)/x^3-1/5*(b*x^2+a)^(9/2)/x^5+63/8*a
^2*b^(5/2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))+63/8*a*b^3*x*(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {277, 195, 217, 206} \[ \frac {63}{8} a^2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}+\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}+\frac {63}{8} a b^3 x \sqrt {a+b x^2}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^6,x]

[Out]

(63*a*b^3*x*Sqrt[a + b*x^2])/8 + (21*b^3*x*(a + b*x^2)^(3/2))/4 - (21*b^2*(a + b*x^2)^(5/2))/(5*x) - (3*b*(a +
b*x^2)^(7/2))/(5*x^3) - (a + b*x^2)^(9/2)/(5*x^5) + (63*a^2*b^(5/2)*ArcTanh[(Sqrt[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 )^{9/2}}{x^6} \, dx &=-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {1}{5} (9 b) \int \frac {\left (a+b x^2\right )^{7/2}}{x^4} \, dx\\ &=-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {1}{5} \left (21 b^2\right ) \int \frac {\left (a+b x^2\right )^{5/2}}{x^2} \, dx\\ &=-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\left (21 b^3\right ) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {1}{4} \left (63 a b^3\right ) \int \sqrt {a+b x^2} \, dx\\ &=\frac {63}{8} a b^3 x \sqrt {a+b x^2}+\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {1}{8} \left (63 a^2 b^3\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {63}{8} a b^3 x \sqrt {a+b x^2}+\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {1}{8} \left (63 a^2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {63}{8} a b^3 x \sqrt {a+b x^2}+\frac {21}{4} b^3 x \left (a+b x^2\right )^{3/2}-\frac {21 b^2 \left (a+b x^2\right )^{5/2}}{5 x}-\frac {3 b \left (a+b x^2\right )^{7/2}}{5 x^3}-\frac {\left (a+b x^2\right )^{9/2}}{5 x^5}+\frac {63}{8} a^2 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 = 0.42 \[ -\frac {a^4 \sqrt {a+b x^2} \, _2F_1\left (-\frac {9}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^6,x]

[Out]

-1/5*(a^4*Sqrt[a + b*x^2]*Hypergeometric2F1[-9/2, -5/2, -3/2, -((b*x^2)/a)])/(x^5*Sqrt[1 + (b*x^2)/a])

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fricas [A]  time = 1.04, size = 191, normalized size = 1.48 \[ \left [\frac {315 \, a^{2} b^{\frac {5}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (10 \, b^{4} x^{8} + 85 \, a b^{3} x^{6} - 288 \, a^{2} b^{2} x^{4} - 56 \, a^{3} b x^{2} - 8 \, a^{4}\right )} \sqrt {b x^{2} + a}}{80 \, x^{5}}, -\frac {315 \, a^{2} \sqrt {-b} b^{2} x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (10 \, b^{4} x^{8} + 85 \, a b^{3} x^{6} - 288 \, a^{2} b^{2} x^{4} - 56 \, a^{3} b x^{2} - 8 \, a^{4}\right )} \sqrt {b x^{2} + a}}{40 \, x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^6,x, algorithm="fricas")

[Out]

[1/80*(315*a^2*b^(5/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(10*b^4*x^8 + 85*a*b^3*x^6 - 28
8*a^2*b^2*x^4 - 56*a^3*b*x^2 - 8*a^4)*sqrt(b*x^2 + a))/x^5, -1/40*(315*a^2*sqrt(-b)*b^2*x^5*arctan(sqrt(-b)*x/
sqrt(b*x^2 + a)) - (10*b^4*x^8 + 85*a*b^3*x^6 - 288*a^2*b^2*x^4 - 56*a^3*b*x^2 - 8*a^4)*sqrt(b*x^2 + a))/x^5]

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giac [A]  time = 1.08, size = 200, normalized size = 1.55 \[ -\frac {63}{16} \, a^{2} b^{\frac {5}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {1}{8} \, {\left (2 \, b^{4} x^{2} + 17 \, a b^{3}\right )} \sqrt {b x^{2} + a} x + \frac {4 \, {\left (25 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} a^{3} b^{\frac {5}{2}} - 75 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a^{4} b^{\frac {5}{2}} + 105 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{5} b^{\frac {5}{2}} - 65 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{6} b^{\frac {5}{2}} + 18 \, a^{7} b^{\frac {5}{2}}\right )}}{5 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^6,x, algorithm="giac")

[Out]

-63/16*a^2*b^(5/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 1/8*(2*b^4*x^2 + 17*a*b^3)*sqrt(b*x^2 + a)*x + 4/5*(
25*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b^(5/2) - 75*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a^4*b^(5/2) + 105*(sqrt(b)
*x - sqrt(b*x^2 + a))^4*a^5*b^(5/2) - 65*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^6*b^(5/2) + 18*a^7*b^(5/2))/((sqrt(
b)*x - sqrt(b*x^2 + a))^2 - a)^5

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maple [A]  time = 0.01, size = 166, normalized size = 1.29 \[ \frac {63 a^{2} b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8}+\frac {63 \sqrt {b \,x^{2}+a}\, a \,b^{3} x}{8}+\frac {21 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3} x}{4}+\frac {21 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3} x}{5 a}+\frac {18 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{3} x}{5 a^{2}}+\frac {16 \left (b \,x^{2}+a \right )^{\frac {9}{2}} b^{3} x}{5 a^{3}}-\frac {16 \left (b \,x^{2}+a \right )^{\frac {11}{2}} b^{2}}{5 a^{3} x}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {11}{2}} b}{5 a^{2} x^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{5 a \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(9/2)/x^6,x)

[Out]

-1/5/a/x^5*(b*x^2+a)^(11/2)-2/5/a^2*b/x^3*(b*x^2+a)^(11/2)-16/5/a^3*b^2/x*(b*x^2+a)^(11/2)+16/5/a^3*b^3*x*(b*x
^2+a)^(9/2)+18/5/a^2*b^3*x*(b*x^2+a)^(7/2)+21/5/a*b^3*x*(b*x^2+a)^(5/2)+21/4*b^3*x*(b*x^2+a)^(3/2)+63/8*a*b^3*
x*(b*x^2+a)^(1/2)+63/8*a^2*b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.45, size = 140, normalized size = 1.09 \[ \frac {21}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3} x + \frac {18 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3} x}{5 \, a^{2}} + \frac {21 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3} x}{5 \, a} + \frac {63}{8} \, \sqrt {b x^{2} + a} a b^{3} x + \frac {63}{8} \, a^{2} b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {16 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} b^{2}}{5 \, a^{2} x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} b}{5 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}}}{5 \, a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^6,x, algorithm="maxima")

[Out]

21/4*(b*x^2 + a)^(3/2)*b^3*x + 18/5*(b*x^2 + a)^(7/2)*b^3*x/a^2 + 21/5*(b*x^2 + a)^(5/2)*b^3*x/a + 63/8*sqrt(b
*x^2 + a)*a*b^3*x + 63/8*a^2*b^(5/2)*arcsinh(b*x/sqrt(a*b)) - 16/5*(b*x^2 + a)^(9/2)*b^2/(a^2*x) - 2/5*(b*x^2
+ a)^(11/2)*b/(a^2*x^3) - 1/5*(b*x^2 + a)^(11/2)/(a*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{9/2}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(9/2)/x^6,x)

[Out]

int((a + b*x^2)^(9/2)/x^6, x)

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sympy [A]  time = 7.82, size = 175, normalized size = 1.36 \[ - \frac {a^{\frac {9}{2}}}{5 x^{5} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {8 a^{\frac {7}{2}} b}{5 x^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {43 a^{\frac {5}{2}} b^{2}}{5 x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {203 a^{\frac {3}{2}} b^{3} x}{40 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {19 \sqrt {a} b^{4} x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {63 a^{2} b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8} + \frac {b^{5} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(9/2)/x**6,x)

[Out]

-a**(9/2)/(5*x**5*sqrt(1 + b*x**2/a)) - 8*a**(7/2)*b/(5*x**3*sqrt(1 + b*x**2/a)) - 43*a**(5/2)*b**2/(5*x*sqrt(
1 + b*x**2/a)) - 203*a**(3/2)*b**3*x/(40*sqrt(1 + b*x**2/a)) + 19*sqrt(a)*b**4*x**3/(8*sqrt(1 + b*x**2/a)) + 6
3*a**2*b**(5/2)*asinh(sqrt(b)*x/sqrt(a))/8 + b**5*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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