3.5.18 \(\int \frac {(a+b x^2)^{9/2}}{x^3} \, dx\) [418]

Optimal. Leaf size=118 \[ \frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}-\frac {9}{2} a^{7/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

3/2*a^2*b*(b*x^2+a)^(3/2)+9/10*a*b*(b*x^2+a)^(5/2)+9/14*b*(b*x^2+a)^(7/2)-1/2*(b*x^2+a)^(9/2)/x^2-9/2*a^(7/2)*
b*arctanh((b*x^2+a)^(1/2)/a^(1/2))+9/2*a^3*b*(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 52, 65, 214} \begin {gather*} -\frac {9}{2} a^{7/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(9/2)/x^3,x]

[Out]

(9*a^3*b*Sqrt[a + b*x^2])/2 + (3*a^2*b*(a + b*x^2)^(3/2))/2 + (9*a*b*(a + b*x^2)^(5/2))/10 + (9*b*(a + b*x^2)^
(7/2))/14 - (a + b*x^2)^(9/2)/(2*x^2) - (9*a^(7/2)*b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{9/2}}{x^3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{9/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{4} (9 b) \text {Subst}\left (\int \frac {(a+b x)^{7/2}}{x} \, dx,x,x^2\right )\\ &=\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{4} (9 a b) \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{4} \left (9 a^2 b\right ) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{4} \left (9 a^3 b\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{4} \left (9 a^4 b\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}+\frac {1}{2} \left (9 a^4\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=\frac {9}{2} a^3 b \sqrt {a+b x^2}+\frac {3}{2} a^2 b \left (a+b x^2\right )^{3/2}+\frac {9}{10} a b \left (a+b x^2\right )^{5/2}+\frac {9}{14} b \left (a+b x^2\right )^{7/2}-\frac {\left (a+b x^2\right )^{9/2}}{2 x^2}-\frac {9}{2} a^{7/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 90, normalized size = 0.76 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-35 a^4+388 a^3 b x^2+156 a^2 b^2 x^4+58 a b^3 x^6+10 b^4 x^8\right )}{70 x^2}-\frac {9}{2} a^{7/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(9/2)/x^3,x]

[Out]

(Sqrt[a + b*x^2]*(-35*a^4 + 388*a^3*b*x^2 + 156*a^2*b^2*x^4 + 58*a*b^3*x^6 + 10*b^4*x^8))/(70*x^2) - (9*a^(7/2
)*b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

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Maple [A]
time = 0.09, size = 119, normalized size = 1.01

method	result	size
risch	\(-\frac {a^{4} \sqrt {b \,x^{2}+a}}{2 x^{2}}+\frac {b^{4} x^{6} \sqrt {b \,x^{2}+a}}{7}+\frac {29 b^{3} a \,x^{4} \sqrt {b \,x^{2}+a}}{35}+\frac {78 b^{2} a^{2} x^{2} \sqrt {b \,x^{2}+a}}{35}+\frac {194 a^{3} b \sqrt {b \,x^{2}+a}}{35}-\frac {9 b \,a^{\frac {7}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2}\)	\(118\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}}}{2 a \,x^{2}}+\frac {9 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {9}{2}}}{9}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\right )\right )}{2 a}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(9/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, size = 106, normalized size = 0.90 \begin {gather*} -\frac {9}{2} \, a^{\frac {7}{2}} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {9}{14} \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b + \frac {{\left (b x^{2} + a\right )}^{\frac {9}{2}} b}{2 \, a} + \frac {9}{10} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b + \frac {3}{2} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b + \frac {9}{2} \, \sqrt {b x^{2} + a} a^{3} b - \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}}}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^3,x, algorithm="maxima")

[Out]

-9/2*a^(7/2)*b*arcsinh(a/(sqrt(a*b)*abs(x))) + 9/14*(b*x^2 + a)^(7/2)*b + 1/2*(b*x^2 + a)^(9/2)*b/a + 9/10*(b*
x^2 + a)^(5/2)*a*b + 3/2*(b*x^2 + a)^(3/2)*a^2*b + 9/2*sqrt(b*x^2 + a)*a^3*b - 1/2*(b*x^2 + a)^(11/2)/(a*x^2)

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Fricas [A]
time = 1.26, size = 188, normalized size = 1.59 \begin {gather*} \left [\frac {315 \, a^{\frac {7}{2}} b x^{2} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (10 \, b^{4} x^{8} + 58 \, a b^{3} x^{6} + 156 \, a^{2} b^{2} x^{4} + 388 \, a^{3} b x^{2} - 35 \, a^{4}\right )} \sqrt {b x^{2} + a}}{140 \, x^{2}}, \frac {315 \, \sqrt {-a} a^{3} b x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (10 \, b^{4} x^{8} + 58 \, a b^{3} x^{6} + 156 \, a^{2} b^{2} x^{4} + 388 \, a^{3} b x^{2} - 35 \, a^{4}\right )} \sqrt {b x^{2} + a}}{70 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^3,x, algorithm="fricas")

[Out]

[1/140*(315*a^(7/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(10*b^4*x^8 + 58*a*b^3*x^6 +
156*a^2*b^2*x^4 + 388*a^3*b*x^2 - 35*a^4)*sqrt(b*x^2 + a))/x^2, 1/70*(315*sqrt(-a)*a^3*b*x^2*arctan(sqrt(-a)/
sqrt(b*x^2 + a)) + (10*b^4*x^8 + 58*a*b^3*x^6 + 156*a^2*b^2*x^4 + 388*a^3*b*x^2 - 35*a^4)*sqrt(b*x^2 + a))/x^2
]

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Sympy [A]
time = 10.31, size = 167, normalized size = 1.42 \begin {gather*} - \frac {a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}}}{2 x^{2}} + \frac {194 a^{\frac {7}{2}} b \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {9 a^{\frac {7}{2}} b \log {\left (\frac {b x^{2}}{a} \right )}}{4} - \frac {9 a^{\frac {7}{2}} b \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2} + \frac {78 a^{\frac {5}{2}} b^{2} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {29 a^{\frac {3}{2}} b^{3} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{35} + \frac {\sqrt {a} b^{4} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(9/2)/x**3,x)

[Out]

-a**(9/2)*sqrt(1 + b*x**2/a)/(2*x**2) + 194*a**(7/2)*b*sqrt(1 + b*x**2/a)/35 + 9*a**(7/2)*b*log(b*x**2/a)/4 -
9*a**(7/2)*b*log(sqrt(1 + b*x**2/a) + 1)/2 + 78*a**(5/2)*b**2*x**2*sqrt(1 + b*x**2/a)/35 + 29*a**(3/2)*b**3*x*
*4*sqrt(1 + b*x**2/a)/35 + sqrt(a)*b**4*x**6*sqrt(1 + b*x**2/a)/7

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Giac [A]
time = 1.05, size = 116, normalized size = 0.98 \begin {gather*} \frac {\frac {315 \, a^{4} b^{2} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 10 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2} + 28 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b^{2} + 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 280 \, \sqrt {b x^{2} + a} a^{3} b^{2} - \frac {35 \, \sqrt {b x^{2} + a} a^{4} b}{x^{2}}}{70 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(9/2)/x^3,x, algorithm="giac")

[Out]

1/70*(315*a^4*b^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 10*(b*x^2 + a)^(7/2)*b^2 + 28*(b*x^2 + a)^(5/2)*
a*b^2 + 70*(b*x^2 + a)^(3/2)*a^2*b^2 + 280*sqrt(b*x^2 + a)*a^3*b^2 - 35*sqrt(b*x^2 + a)*a^4*b/x^2)/b

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Mupad [B]
time = 5.45, size = 95, normalized size = 0.81 \begin {gather*} \frac {b\,{\left (b\,x^2+a\right )}^{7/2}}{7}+4\,a^3\,b\,\sqrt {b\,x^2+a}+a^2\,b\,{\left (b\,x^2+a\right )}^{3/2}-\frac {a^4\,\sqrt {b\,x^2+a}}{2\,x^2}+\frac {2\,a\,b\,{\left (b\,x^2+a\right )}^{5/2}}{5}+\frac {a^{7/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,9{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(9/2)/x^3,x)

[Out]

(b*(a + b*x^2)^(7/2))/7 + 4*a^3*b*(a + b*x^2)^(1/2) + a^2*b*(a + b*x^2)^(3/2) - (a^4*(a + b*x^2)^(1/2))/(2*x^2
) + (a^(7/2)*b*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*9i)/2 + (2*a*b*(a + b*x^2)^(5/2))/5

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