3.5.29 \(\int \frac {(a+b x^2)^{9/2}}{x^2} \, dx\) [429]

Optimal. Leaf size=123 \[ \frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {315}{128} a^4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[Out]

105/64*a^2*b*x*(b*x^2+a)^(3/2)+21/16*a*b*x*(b*x^2+a)^(5/2)+9/8*b*x*(b*x^2+a)^(7/2)-(b*x^2+a)^(9/2)/x+315/128*a
^4*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))*b^(1/2)+315/128*a^3*b*x*(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 123, 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 = {283, 201, 223, 212} \begin {gather*} \frac {315}{128} a^4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^2,x]

[Out]

(315*a^3*b*x*Sqrt[a + b*x^2])/128 + (105*a^2*b*x*(a + b*x^2)^(3/2))/64 + (21*a*b*x*(a + b*x^2)^(5/2))/16 + (9*
b*x*(a + b*x^2)^(7/2))/8 - (a + b*x^2)^(9/2)/x + (315*a^4*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/128

Rule 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 283

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^2} \, dx &=-\frac {\left (a+b x^2\right )^{9/2}}{x}+(9 b) \int \left (a+b x^2\right )^{7/2} \, dx\\ &=\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {1}{8} (63 a b) \int \left (a+b x^2\right )^{5/2} \, dx\\ &=\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {1}{16} \left (105 a^2 b\right ) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {1}{64} \left (315 a^3 b\right ) \int \sqrt {a+b x^2} \, dx\\ &=\frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {1}{128} \left (315 a^4 b\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {1}{128} \left (315 a^4 b\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {315}{128} a^3 b x \sqrt {a+b x^2}+\frac {105}{64} a^2 b x \left (a+b x^2\right )^{3/2}+\frac {21}{16} a b x \left (a+b x^2\right )^{5/2}+\frac {9}{8} b x \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{x}+\frac {315}{128} a^4 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 95, normalized size = 0.77 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-128 a^4+325 a^3 b x^2+210 a^2 b^2 x^4+88 a b^3 x^6+16 b^4 x^8\right )}{128 x}-\frac {315}{128} a^4 \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^2,x]

[Out]

(Sqrt[a + b*x^2]*(-128*a^4 + 325*a^3*b*x^2 + 210*a^2*b^2*x^4 + 88*a*b^3*x^6 + 16*b^4*x^8))/(128*x) - (315*a^4*
Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/128

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Maple [A]
time = 0.06, size = 124, normalized size = 1.01

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-16 b^{4} x^{8}-88 a \,b^{3} x^{6}-210 a^{2} b^{2} x^{4}-325 a^{3} b \,x^{2}+128 a^{4}\right )}{128 x}+\frac {315 a^{4} \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128}\)	\(83\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{a x}+\frac {10 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {9}{2}}}{10}+\frac {9 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8}+\frac {7 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8}\right )}{10}\right )}{a}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(9/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/a/x*(b*x^2+a)^(11/2)+10*b/a*(1/10*x*(b*x^2+a)^(9/2)+9/10*a*(1/8*x*(b*x^2+a)^(7/2)+7/8*a*(1/6*x*(b*x^2+a)^(5
/2)+5/6*a*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2)))))))

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Maxima [A]
time = 0.28, size = 91, normalized size = 0.74 \begin {gather*} \frac {9}{8} \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b x + \frac {21}{16} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b x + \frac {105}{64} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b x + \frac {315}{128} \, \sqrt {b x^{2} + a} a^{3} b x + \frac {315}{128} \, a^{4} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {9}{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/8*(b*x^2 + a)^(7/2)*b*x + 21/16*(b*x^2 + a)^(5/2)*a*b*x + 105/64*(b*x^2 + a)^(3/2)*a^2*b*x + 315/128*sqrt(b*
x^2 + a)*a^3*b*x + 315/128*a^4*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - (b*x^2 + a)^(9/2)/x

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Fricas [A]
time = 0.91, size = 184, normalized size = 1.50 \begin {gather*} \left [\frac {315 \, a^{4} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (16 \, b^{4} x^{8} + 88 \, a b^{3} x^{6} + 210 \, a^{2} b^{2} x^{4} + 325 \, a^{3} b x^{2} - 128 \, a^{4}\right )} \sqrt {b x^{2} + a}}{256 \, x}, -\frac {315 \, a^{4} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (16 \, b^{4} x^{8} + 88 \, a b^{3} x^{6} + 210 \, a^{2} b^{2} x^{4} + 325 \, a^{3} b x^{2} - 128 \, a^{4}\right )} \sqrt {b x^{2} + a}}{128 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/256*(315*a^4*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(16*b^4*x^8 + 88*a*b^3*x^6 + 210
*a^2*b^2*x^4 + 325*a^3*b*x^2 - 128*a^4)*sqrt(b*x^2 + a))/x, -1/128*(315*a^4*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(
b*x^2 + a)) - (16*b^4*x^8 + 88*a*b^3*x^6 + 210*a^2*b^2*x^4 + 325*a^3*b*x^2 - 128*a^4)*sqrt(b*x^2 + a))/x]

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Sympy [A]
time = 9.83, size = 173, normalized size = 1.41 \begin {gather*} - \frac {a^{\frac {9}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {197 a^{\frac {7}{2}} b x}{128 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {535 a^{\frac {5}{2}} b^{2} x^{3}}{128 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {149 a^{\frac {3}{2}} b^{3} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 \sqrt {a} b^{4} x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {315 a^{4} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128} + \frac {b^{5} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(9/2)/(x*sqrt(1 + b*x**2/a)) + 197*a**(7/2)*b*x/(128*sqrt(1 + b*x**2/a)) + 535*a**(5/2)*b**2*x**3/(128*sqr
t(1 + b*x**2/a)) + 149*a**(3/2)*b**3*x**5/(64*sqrt(1 + b*x**2/a)) + 13*sqrt(a)*b**4*x**7/(16*sqrt(1 + b*x**2/a
)) + 315*a**4*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/128 + b**5*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]
time = 0.90, size = 115, normalized size = 0.93 \begin {gather*} -\frac {315}{256} \, a^{4} \sqrt {b} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, a^{5} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{128} \, {\left (325 \, a^{3} b + 2 \, {\left (105 \, a^{2} b^{2} + 4 \, {\left (2 \, b^{4} x^{2} + 11 \, a b^{3}\right )} x^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-315/256*a^4*sqrt(b)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2*a^5*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a
) + 1/128*(325*a^3*b + 2*(105*a^2*b^2 + 4*(2*b^4*x^2 + 11*a*b^3)*x^2)*x^2)*sqrt(b*x^2 + a)*x

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Mupad [B]
time = 6.03, size = 40, normalized size = 0.33 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{9/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {9}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(9/2)/x^2,x)

[Out]

-((a + b*x^2)^(9/2)*hypergeom([-9/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(9/2))

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