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

Optimal. Leaf size=86 \[ \frac{15}{8} b^2 \sqrt{a+b x^2}-\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2} \]

[Out]

(15*b^2*Sqrt[a + b*x^2])/8 - (5*b*(a + b*x^2)^(3/2))/(8*x^2) - (a + b*x^2)^(5/2)/(4*x^4) - (15*Sqrt[a]*b^2*Arc
Tanh[Sqrt[a + b*x^2]/Sqrt[a]])/8

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Rubi [A]  time = 0.0481995, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ \frac{15}{8} b^2 \sqrt{a+b x^2}-\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^5,x]

[Out]

(15*b^2*Sqrt[a + b*x^2])/8 - (5*b*(a + b*x^2)^(3/2))/(8*x^2) - (a + b*x^2)^(5/2)/(4*x^4) - (15*Sqrt[a]*b^2*Arc
Tanh[Sqrt[a + b*x^2]/Sqrt[a]])/8

Rule 266

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]]

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 63

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 208

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{5/2}}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac{1}{8} (5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac{1}{16} \left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{15}{8} b^2 \sqrt{a+b x^2}-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac{1}{16} \left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{15}{8} b^2 \sqrt{a+b x^2}-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac{1}{8} (15 a b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{15}{8} b^2 \sqrt{a+b x^2}-\frac{5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac{\left (a+b x^2\right )^{5/2}}{4 x^4}-\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0097769, size = 39, normalized size = 0.45 \[ -\frac{b^2 \left (a+b x^2\right )^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^5,x]

[Out]

-(b^2*(a + b*x^2)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 + (b*x^2)/a])/(7*a^3)

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Maple [A]  time = 0.006, size = 116, normalized size = 1.4 \begin{align*} -{\frac{1}{4\,a{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,b}{8\,{a}^{2}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{3\,{b}^{2}}{8\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{2}}{8\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{15\,{b}^{2}}{8}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{15\,{b}^{2}}{8}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a/x^4*(b*x^2+a)^(7/2)-3/8*b/a^2/x^2*(b*x^2+a)^(7/2)+3/8*b^2/a^2*(b*x^2+a)^(5/2)+5/8*b^2/a*(b*x^2+a)^(3/2)
-15/8*b^2*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+15/8*b^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)^(5/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60684, size = 342, normalized size = 3.98 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{2} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (8 \, b^{2} x^{4} - 9 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt{b x^{2} + a}}{16 \, x^{4}}, \frac{15 \, \sqrt{-a} b^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, b^{2} x^{4} - 9 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt{b x^{2} + a}}{8 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*sqrt(a)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*b^2*x^4 - 9*a*b*x^2 - 2*a
^2)*sqrt(b*x^2 + a))/x^4, 1/8*(15*sqrt(-a)*b^2*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*b^2*x^4 - 9*a*b*x^2 -
2*a^2)*sqrt(b*x^2 + a))/x^4]

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Sympy [A]  time = 3.84351, size = 117, normalized size = 1.36 \begin{align*} - \frac{15 \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8} - \frac{a^{3}}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{11 a^{2} \sqrt{b}}{8 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{a b^{\frac{3}{2}}}{8 x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{b^{\frac{5}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 - a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 11*a**2*sqrt(b)/(
8*x**3*sqrt(a/(b*x**2) + 1)) - a*b**(3/2)/(8*x*sqrt(a/(b*x**2) + 1)) + b**(5/2)*x/sqrt(a/(b*x**2) + 1)

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Giac [A]  time = 3.23042, size = 103, normalized size = 1.2 \begin{align*} \frac{1}{8} \,{\left (\frac{15 \, a \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 8 \, \sqrt{b x^{2} + a} - \frac{9 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a - 7 \, \sqrt{b x^{2} + a} a^{2}}{b^{2} x^{4}}\right )} b^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(15*a*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(b*x^2 + a) - (9*(b*x^2 + a)^(3/2)*a - 7*sqrt(b*x^
2 + a)*a^2)/(b^2*x^4))*b^2