3.430 \(\int \frac{(a+b x^2)^{9/2}}{x^4} \, dx\)

Optimal. Leaf size=128 \[ \frac{105}{16} a^2 b^2 x \sqrt{a+b x^2}+\frac{105}{16} a^3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}+\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x} \]

[Out]

(105*a^2*b^2*x*Sqrt[a + b*x^2])/16 + (35*a*b^2*x*(a + b*x^2)^(3/2))/8 + (7*b^2*x*(a + b*x^2)^(5/2))/2 - (3*b*(
a + b*x^2)^(7/2))/x - (a + b*x^2)^(9/2)/(3*x^3) + (105*a^3*b^(3/2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/16

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Rubi [A]  time = 0.0487097, antiderivative size = 128, normalized size of antiderivative = 1., 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} \[ \frac{105}{16} a^2 b^2 x \sqrt{a+b x^2}+\frac{105}{16} a^3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}+\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^4,x]

[Out]

(105*a^2*b^2*x*Sqrt[a + b*x^2])/16 + (35*a*b^2*x*(a + b*x^2)^(3/2))/8 + (7*b^2*x*(a + b*x^2)^(5/2))/2 - (3*b*(
a + b*x^2)^(7/2))/x - (a + b*x^2)^(9/2)/(3*x^3) + (105*a^3*b^(3/2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/16

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{9/2}}{x^4} \, dx &=-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+(3 b) \int \frac{\left (a+b x^2\right )^{7/2}}{x^2} \, dx\\ &=-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\left (21 b^2\right ) \int \left (a+b x^2\right )^{5/2} \, dx\\ &=\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\frac{1}{2} \left (35 a b^2\right ) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\frac{1}{8} \left (105 a^2 b^2\right ) \int \sqrt{a+b x^2} \, dx\\ &=\frac{105}{16} a^2 b^2 x \sqrt{a+b x^2}+\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\frac{1}{16} \left (105 a^3 b^2\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=\frac{105}{16} a^2 b^2 x \sqrt{a+b x^2}+\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\frac{1}{16} \left (105 a^3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=\frac{105}{16} a^2 b^2 x \sqrt{a+b x^2}+\frac{35}{8} a b^2 x \left (a+b x^2\right )^{3/2}+\frac{7}{2} b^2 x \left (a+b x^2\right )^{5/2}-\frac{3 b \left (a+b x^2\right )^{7/2}}{x}-\frac{\left (a+b x^2\right )^{9/2}}{3 x^3}+\frac{105}{16} a^3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0101454, size = 54, normalized size = 0.42 \[ -\frac{a^4 \sqrt{a+b x^2} \, _2F_1\left (-\frac{9}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x^3 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^4,x]

[Out]

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

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Maple [A]  time = 0.006, size = 146, normalized size = 1.1 \begin{align*} -{\frac{1}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}-{\frac{8\,b}{3\,{a}^{2}x} \left ( b{x}^{2}+a \right ) ^{{\frac{11}{2}}}}+{\frac{8\,{b}^{2}x}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{9}{2}}}}+3\,{\frac{{b}^{2}x \left ( b{x}^{2}+a \right ) ^{7/2}}{a}}+{\frac{7\,{b}^{2}x}{2} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{35\,a{b}^{2}x}{8} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{105\,{b}^{2}{a}^{2}x}{16}\sqrt{b{x}^{2}+a}}+{\frac{105\,{a}^{3}}{16}{b}^{{\frac{3}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(9/2)/x^4,x)

[Out]

-1/3/a/x^3*(b*x^2+a)^(11/2)-8/3*b/a^2/x*(b*x^2+a)^(11/2)+8/3*b^2/a^2*x*(b*x^2+a)^(9/2)+3*b^2/a*x*(b*x^2+a)^(7/
2)+7/2*b^2*x*(b*x^2+a)^(5/2)+35/8*a*b^2*x*(b*x^2+a)^(3/2)+105/16*a^2*b^2*x*(b*x^2+a)^(1/2)+105/16*b^(3/2)*a^3*
ln(x*b^(1/2)+(b*x^2+a)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74687, size = 450, normalized size = 3.52 \begin{align*} \left [\frac{315 \, a^{3} 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 (8 \, b^{4} x^{8} + 50 \, a b^{3} x^{6} + 165 \, a^{2} b^{2} x^{4} - 208 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt{b x^{2} + a}}{96 \, x^{3}}, -\frac{315 \, a^{3} \sqrt{-b} b x^{3} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (8 \, b^{4} x^{8} + 50 \, a b^{3} x^{6} + 165 \, a^{2} b^{2} x^{4} - 208 \, a^{3} b x^{2} - 16 \, a^{4}\right )} \sqrt{b x^{2} + a}}{48 \, x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(315*a^3*b^(3/2)*x^3*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*b^4*x^8 + 50*a*b^3*x^6 + 165
*a^2*b^2*x^4 - 208*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/x^3, -1/48*(315*a^3*sqrt(-b)*b*x^3*arctan(sqrt(-b)*x/s
qrt(b*x^2 + a)) - (8*b^4*x^8 + 50*a*b^3*x^6 + 165*a^2*b^2*x^4 - 208*a^3*b*x^2 - 16*a^4)*sqrt(b*x^2 + a))/x^3]

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Sympy [A]  time = 7.65697, size = 175, normalized size = 1.37 \begin{align*} - \frac{a^{\frac{9}{2}}}{3 x^{3} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{14 a^{\frac{7}{2}} b}{3 x \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{43 a^{\frac{5}{2}} b^{2} x}{48 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{215 a^{\frac{3}{2}} b^{3} x^{3}}{48 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{29 \sqrt{a} b^{4} x^{5}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{105 a^{3} b^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16} + \frac{b^{5} x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(9/2)/x**4,x)

[Out]

-a**(9/2)/(3*x**3*sqrt(1 + b*x**2/a)) - 14*a**(7/2)*b/(3*x*sqrt(1 + b*x**2/a)) - 43*a**(5/2)*b**2*x/(48*sqrt(1
+ b*x**2/a)) + 215*a**(3/2)*b**3*x**3/(48*sqrt(1 + b*x**2/a)) + 29*sqrt(a)*b**4*x**5/(24*sqrt(1 + b*x**2/a))
+ 105*a**3*b**(3/2)*asinh(sqrt(b)*x/sqrt(a))/16 + b**5*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.64402, size = 216, normalized size = 1.69 \begin{align*} -\frac{105}{32} \, a^{3} b^{\frac{3}{2}} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{1}{48} \,{\left (165 \, a^{2} b^{2} + 2 \,{\left (4 \, b^{4} x^{2} + 25 \, a b^{3}\right )} x^{2}\right )} \sqrt{b x^{2} + a} x + \frac{2 \,{\left (15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a^{4} b^{\frac{3}{2}} - 24 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{5} b^{\frac{3}{2}} + 13 \, a^{6} b^{\frac{3}{2}}\right )}}{3 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^4,x, algorithm="giac")

[Out]

-105/32*a^3*b^(3/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 1/48*(165*a^2*b^2 + 2*(4*b^4*x^2 + 25*a*b^3)*x^2)*s
qrt(b*x^2 + a)*x + 2/3*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a^4*b^(3/2) - 24*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^
5*b^(3/2) + 13*a^6*b^(3/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3