3.392 \(\int \frac{(a+b x^2)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=80 \[ -\frac{5}{2} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{5}{6} b \left (a+b x^2\right )^{3/2}+\frac{5}{2} a b \sqrt{a+b x^2} \]

[Out]

(5*a*b*Sqrt[a + b*x^2])/2 + (5*b*(a + b*x^2)^(3/2))/6 - (a + b*x^2)^(5/2)/(2*x^2) - (5*a^(3/2)*b*ArcTanh[Sqrt[
a + b*x^2]/Sqrt[a]])/2

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Rubi [A]  time = 0.0449643, antiderivative size = 80, 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} \[ -\frac{5}{2} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{5}{6} b \left (a+b x^2\right )^{3/2}+\frac{5}{2} a b \sqrt{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^3,x]

[Out]

(5*a*b*Sqrt[a + b*x^2])/2 + (5*b*(a + b*x^2)^(3/2))/6 - (a + b*x^2)^(5/2)/(2*x^2) - (5*a^(3/2)*b*ArcTanh[Sqrt[
a + b*x^2]/Sqrt[a]])/2

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^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{1}{4} (5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{6} b \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{1}{4} (5 a b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{5}{2} a b \sqrt{a+b x^2}+\frac{5}{6} b \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{1}{4} \left (5 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{5}{2} a b \sqrt{a+b x^2}+\frac{5}{6} b \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}+\frac{1}{2} \left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=\frac{5}{2} a b \sqrt{a+b x^2}+\frac{5}{6} b \left (a+b x^2\right )^{3/2}-\frac{\left (a+b x^2\right )^{5/2}}{2 x^2}-\frac{5}{2} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0094557, size = 37, normalized size = 0.46 \[ \frac{b \left (a+b x^2\right )^{7/2} \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{b x^2}{a}+1\right )}{7 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^3,x]

[Out]

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

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Maple [A]  time = 0.006, size = 88, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{b}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,b}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,b}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{5\,ab}{2}\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^3,x)

[Out]

-1/2/a/x^2*(b*x^2+a)^(7/2)+1/2*b/a*(b*x^2+a)^(5/2)+5/6*b*(b*x^2+a)^(3/2)-5/2*b*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^
2+a)^(1/2))/x)+5/2*a*b*(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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62712, size = 342, normalized size = 4.28 \begin{align*} \left [\frac{15 \, a^{\frac{3}{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 (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt{b x^{2} + a}}{12 \, x^{2}}, \frac{15 \, \sqrt{-a} a b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, b^{2} x^{4} + 14 \, a b x^{2} - 3 \, a^{2}\right )} \sqrt{b x^{2} + a}}{6 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/12*(15*a^(3/2)*b*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*b^2*x^4 + 14*a*b*x^2 - 3*a^
2)*sqrt(b*x^2 + a))/x^2, 1/6*(15*sqrt(-a)*a*b*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (2*b^2*x^4 + 14*a*b*x^2 -
3*a^2)*sqrt(b*x^2 + a))/x^2]

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Sympy [A]  time = 3.23765, size = 112, normalized size = 1.4 \begin{align*} - \frac{a^{\frac{5}{2}} \sqrt{1 + \frac{b x^{2}}{a}}}{2 x^{2}} + \frac{7 a^{\frac{3}{2}} b \sqrt{1 + \frac{b x^{2}}{a}}}{3} + \frac{5 a^{\frac{3}{2}} b \log{\left (\frac{b x^{2}}{a} \right )}}{4} - \frac{5 a^{\frac{3}{2}} b \log{\left (\sqrt{1 + \frac{b x^{2}}{a}} + 1 \right )}}{2} + \frac{\sqrt{a} b^{2} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**3,x)

[Out]

-a**(5/2)*sqrt(1 + b*x**2/a)/(2*x**2) + 7*a**(3/2)*b*sqrt(1 + b*x**2/a)/3 + 5*a**(3/2)*b*log(b*x**2/a)/4 - 5*a
**(3/2)*b*log(sqrt(1 + b*x**2/a) + 1)/2 + sqrt(a)*b**2*x**2*sqrt(1 + b*x**2/a)/3

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Giac [A]  time = 2.0765, size = 99, normalized size = 1.24 \begin{align*} \frac{1}{6} \,{\left (\frac{15 \, a^{2} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 2 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} + 12 \, \sqrt{b x^{2} + a} a - \frac{3 \, \sqrt{b x^{2} + a} a^{2}}{b x^{2}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/6*(15*a^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 2*(b*x^2 + a)^(3/2) + 12*sqrt(b*x^2 + a)*a - 3*sqrt(b*
x^2 + a)*a^2/(b*x^2))*b