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

Optimal. Leaf size=118 \[ \frac{3 b (5 A b-4 a B)}{8 a^3 \sqrt{a+b x^2}}+\frac{5 A b-4 a B}{8 a^2 x^2 \sqrt{a+b x^2}}-\frac{3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{7/2}}-\frac{A}{4 a x^4 \sqrt{a+b x^2}} \]

[Out]

(3*b*(5*A*b - 4*a*B))/(8*a^3*Sqrt[a + b*x^2]) - A/(4*a*x^4*Sqrt[a + b*x^2]) + (5*A*b - 4*a*B)/(8*a^2*x^2*Sqrt[
a + b*x^2]) - (3*b*(5*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*a^(7/2))

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Rubi [A]  time = 0.0882975, antiderivative size = 120, normalized size of antiderivative = 1.02, number of steps used = 6, 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} \[ \frac{3 \sqrt{a+b x^2} (5 A b-4 a B)}{8 a^3 x^2}-\frac{5 A b-4 a B}{4 a^2 x^2 \sqrt{a+b x^2}}-\frac{3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{7/2}}-\frac{A}{4 a x^4 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]

[Out]

-A/(4*a*x^4*Sqrt[a + b*x^2]) - (5*A*b - 4*a*B)/(4*a^2*x^2*Sqrt[a + b*x^2]) + (3*(5*A*b - 4*a*B)*Sqrt[a + b*x^2
])/(8*a^3*x^2) - (3*b*(5*A*b - 4*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(8*a^(7/2))

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 \left (a+b x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^3 (a+b x)^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{A}{4 a x^4 \sqrt{a+b x^2}}+\frac{\left (-\frac{5 A b}{2}+2 a B\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{A}{4 a x^4 \sqrt{a+b x^2}}-\frac{5 A b-4 a B}{4 a^2 x^2 \sqrt{a+b x^2}}-\frac{(3 (5 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{8 a^2}\\ &=-\frac{A}{4 a x^4 \sqrt{a+b x^2}}-\frac{5 A b-4 a B}{4 a^2 x^2 \sqrt{a+b x^2}}+\frac{3 (5 A b-4 a B) \sqrt{a+b x^2}}{8 a^3 x^2}+\frac{(3 b (5 A b-4 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{16 a^3}\\ &=-\frac{A}{4 a x^4 \sqrt{a+b x^2}}-\frac{5 A b-4 a B}{4 a^2 x^2 \sqrt{a+b x^2}}+\frac{3 (5 A b-4 a B) \sqrt{a+b x^2}}{8 a^3 x^2}+\frac{(3 (5 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^3}\\ &=-\frac{A}{4 a x^4 \sqrt{a+b x^2}}-\frac{5 A b-4 a B}{4 a^2 x^2 \sqrt{a+b x^2}}+\frac{3 (5 A b-4 a B) \sqrt{a+b x^2}}{8 a^3 x^2}-\frac{3 b (5 A b-4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0188725, size = 60, normalized size = 0.51 \[ \frac{b x^4 (5 A b-4 a B) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^2}{a}+1\right )-a^2 A}{4 a^3 x^4 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^5*(a + b*x^2)^(3/2)),x]

[Out]

(-(a^2*A) + b*(5*A*b - 4*a*B)*x^4*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x^2)/a])/(4*a^3*x^4*Sqrt[a + b*x^2])

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Maple [A]  time = 0.01, size = 153, normalized size = 1.3 \begin{align*} -{\frac{A}{4\,a{x}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{5\,Ab}{8\,{a}^{2}{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{15\,A{b}^{2}}{8\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{15\,A{b}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}}-{\frac{B}{2\,a{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,Bb}{2\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,Bb}{2}\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^5/(b*x^2+a)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64314, size = 628, normalized size = 5.32 \begin{align*} \left [-\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\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 \,{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{16 \,{\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}, -\frac{3 \,{\left ({\left (4 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} +{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \,{\left (4 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} + 2 \, A a^{3} +{\left (4 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{8 \,{\left (a^{4} b x^{6} + a^{5} x^{4}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A*a*b^2)*x^4)*sqrt(a)*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sq
rt(a) + 2*a)/x^2) + 2*(3*(4*B*a^2*b - 5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(
a^4*b*x^6 + a^5*x^4), -1/8*(3*((4*B*a*b^2 - 5*A*b^3)*x^6 + (4*B*a^2*b - 5*A*a*b^2)*x^4)*sqrt(-a)*arctan(sqrt(-
a)/sqrt(b*x^2 + a)) + (3*(4*B*a^2*b - 5*A*a*b^2)*x^4 + 2*A*a^3 + (4*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(
a^4*b*x^6 + a^5*x^4)]

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Sympy [A]  time = 41.5856, size = 180, normalized size = 1.53 \begin{align*} A \left (- \frac{1}{4 a \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{5 \sqrt{b}}{8 a^{2} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{15 b^{\frac{3}{2}}}{8 a^{3} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 a^{\frac{7}{2}}}\right ) + B \left (- \frac{1}{2 a \sqrt{b} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 \sqrt{b}}{2 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{5}{2}}}\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)**(3/2),x)

[Out]

A*(-1/(4*a*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + 5*sqrt(b)/(8*a**2*x**3*sqrt(a/(b*x**2) + 1)) + 15*b**(3/2)/(8*
a**3*x*sqrt(a/(b*x**2) + 1)) - 15*b**2*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(7/2))) + B*(-1/(2*a*sqrt(b)*x**3*sqrt
(a/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(5/2)))

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Giac [A]  time = 1.13359, size = 185, normalized size = 1.57 \begin{align*} -\frac{3 \,{\left (4 \, B a b - 5 \, A b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{3}} - \frac{B a b - A b^{2}}{\sqrt{b x^{2} + a} a^{3}} - \frac{4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a b - 4 \, \sqrt{b x^{2} + a} B a^{2} b - 7 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b^{2} + 9 \, \sqrt{b x^{2} + a} A a b^{2}}{8 \, a^{3} b^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*(4*B*a*b - 5*A*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) - (B*a*b - A*b^2)/(sqrt(b*x^2 + a)*a^
3) - 1/8*(4*(b*x^2 + a)^(3/2)*B*a*b - 4*sqrt(b*x^2 + a)*B*a^2*b - 7*(b*x^2 + a)^(3/2)*A*b^2 + 9*sqrt(b*x^2 + a
)*A*a*b^2)/(a^3*b^2*x^4)