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

Optimal. Leaf size=72 \[ \frac{A}{a^2 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0492278, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {446, 78, 51, 63, 208} \[ \frac{A}{a^2 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x*(a + b*x^2)^(5/2)),x]

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{A+B x^2}{x \left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x (a+b x)^{5/2}} \, dx,x,x^2\right )\\ &=\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{A \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,x^2\right )}{2 a}\\ &=\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{A}{a^2 \sqrt{a+b x^2}}+\frac{A \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{2 a^2}\\ &=\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{A}{a^2 \sqrt{a+b x^2}}+\frac{A \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{a^2 b}\\ &=\frac{A b-a B}{3 a b \left (a+b x^2\right )^{3/2}}+\frac{A}{a^2 \sqrt{a+b x^2}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0202502, size = 61, normalized size = 0.85 \[ \frac{a (A b-a B)+3 A b \left (a+b x^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b x^2}{a}+1\right )}{3 a^2 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x*(a + b*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.007, size = 75, normalized size = 1. \begin{align*} -{\frac{B}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{A}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{A}{{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*B/b/(b*x^2+a)^(3/2)+1/3*A/a/(b*x^2+a)^(3/2)+A/a^2/(b*x^2+a)^(1/2)-A/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(
1/2))/x)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98621, size = 521, normalized size = 7.24 \begin{align*} \left [\frac{3 \,{\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt{a} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}, \frac{3 \,{\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(A*b^3*x^4 + 2*A*a*b^2*x^2 + A*a^2*b)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*
(3*A*a*b^2*x^2 - B*a^3 + 4*A*a^2*b)*sqrt(b*x^2 + a))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b), 1/3*(3*(A*b^3*x^4
+ 2*A*a*b^2*x^2 + A*a^2*b)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*A*a*b^2*x^2 - B*a^3 + 4*A*a^2*b)*sqr
t(b*x^2 + a))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)]

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Sympy [A]  time = 17.7912, size = 66, normalized size = 0.92 \begin{align*} \frac{A}{a^{2} \sqrt{a + b x^{2}}} + \frac{A \operatorname{atan}{\left (\frac{\sqrt{a + b x^{2}}}{\sqrt{- a}} \right )}}{a^{2} \sqrt{- a}} - \frac{- A b + B a}{3 a b \left (a + b x^{2}\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A/(a**2*sqrt(a + b*x**2)) + A*atan(sqrt(a + b*x**2)/sqrt(-a))/(a**2*sqrt(-a)) - (-A*b + B*a)/(3*a*b*(a + b*x**
2)**(3/2))

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Giac [A]  time = 1.12456, size = 89, normalized size = 1.24 \begin{align*} \frac{A \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} - \frac{B a^{2} - 3 \,{\left (b x^{2} + a\right )} A b - A a b}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) - 1/3*(B*a^2 - 3*(b*x^2 + a)*A*b - A*a*b)/((b*x^2 + a)^(3/2)
*a^2*b)