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

Optimal. Leaf size=90 \[ \frac{\sqrt{a+b x^2} (3 A b-4 a B)}{8 a^2 x^2}-\frac{b (3 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{A \sqrt{a+b x^2}}{4 a x^4} \]

[Out]

-(A*Sqrt[a + b*x^2])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2])/(8*a^2*x^2) - (b*(3*A*b - 4*a*B)*ArcTanh[Sq
rt[a + b*x^2]/Sqrt[a]])/(8*a^(5/2))

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Rubi [A]  time = 0.0678825, antiderivative size = 90, 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} \[ \frac{\sqrt{a+b x^2} (3 A b-4 a B)}{8 a^2 x^2}-\frac{b (3 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{A \sqrt{a+b x^2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[a + b*x^2])/(4*a*x^4) + ((3*A*b - 4*a*B)*Sqrt[a + b*x^2])/(8*a^2*x^2) - (b*(3*A*b - 4*a*B)*ArcTanh[Sq
rt[a + b*x^2]/Sqrt[a]])/(8*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^5 \sqrt{a+b x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{A \sqrt{a+b x^2}}{4 a x^4}+\frac{\left (-\frac{3 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{A \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2}}{8 a^2 x^2}+\frac{(b (3 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{16 a^2}\\ &=-\frac{A \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2}}{8 a^2 x^2}+\frac{(3 A b-4 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{8 a^2}\\ &=-\frac{A \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 A b-4 a B) \sqrt{a+b x^2}}{8 a^2 x^2}-\frac{b (3 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.156196, size = 83, normalized size = 0.92 \[ \frac{\sqrt{a+b x^2} \left (-\frac{2 a^2 \left (A+2 B x^2\right )}{x^4}+\frac{b (4 a B-3 A b) \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{\sqrt{\frac{b x^2}{a}+1}}+\frac{3 a A b}{x^2}\right )}{8 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*((3*a*A*b)/x^2 - (2*a^2*(A + 2*B*x^2))/x^4 + (b*(-3*A*b + 4*a*B)*ArcTanh[Sqrt[1 + (b*x^2)/a]]
)/Sqrt[1 + (b*x^2)/a]))/(8*a^3)

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Maple [A]  time = 0.007, size = 119, normalized size = 1.3 \begin{align*} -{\frac{A}{4\,a{x}^{4}}\sqrt{b{x}^{2}+a}}+{\frac{3\,Ab}{8\,{a}^{2}{x}^{2}}\sqrt{b{x}^{2}+a}}-{\frac{3\,A{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{B}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{Bb}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^5/(b*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.70152, size = 406, normalized size = 4.51 \begin{align*} \left [-\frac{{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt{a} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{16 \, a^{3} x^{4}}, -\frac{{\left (4 \, B a b - 3 \, A b^{2}\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, A a^{2} +{\left (4 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{8 \, a^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*((4*B*a*b - 3*A*b^2)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*A*a^2 + (4*
B*a^2 - 3*A*a*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^4), -1/8*((4*B*a*b - 3*A*b^2)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(
b*x^2 + a)) + (2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x^2)*sqrt(b*x^2 + a))/(a^3*x^4)]

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Sympy [A]  time = 28.9176, size = 150, normalized size = 1.67 \begin{align*} - \frac{A}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A \sqrt{b}}{8 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 A b^{\frac{3}{2}}}{8 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 A b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 a^{\frac{5}{2}}} - \frac{B \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 a x} + \frac{B b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**5/(b*x**2+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.11397, size = 163, normalized size = 1.81 \begin{align*} -\frac{\frac{{\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a b^{2} - 4 \, \sqrt{b x^{2} + a} B a^{2} b^{2} - 3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b^{3} + 5 \, \sqrt{b x^{2} + a} A a b^{3}}{a^{2} b^{2} x^{4}}}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^5/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/8*((4*B*a*b^2 - 3*A*b^3)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (4*(b*x^2 + a)^(3/2)*B*a*b^2 - 4
*sqrt(b*x^2 + a)*B*a^2*b^2 - 3*(b*x^2 + a)^(3/2)*A*b^3 + 5*sqrt(b*x^2 + a)*A*a*b^3)/(a^2*b^2*x^4))/b