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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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 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 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]

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]]

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*}

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Mathematica [C]  time = 0.02, 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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fricas [A]  time = 0.81, size = 241, normalized size = 3.35 \[ \left [\frac {3 \, {\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt {a} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}, \frac {3 \, {\left (A b^{3} x^{4} + 2 \, A a b^{2} x^{2} + A a^{2} b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, A a b^{2} x^{2} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}\right ] \]

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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giac [A]  time = 0.31, size = 66, normalized size = 0.92 \[ \frac {A \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {B a^{2} - 3 \, {\left (b x^{2} + a\right )} A b - A a b}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} \]

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)

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maple [A]  time = 0.01, size = 75, normalized size = 1.04 \[ \frac {A}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a}-\frac {B}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b}-\frac {A \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {5}{2}}}+\frac {A}{\sqrt {b \,x^{2}+a}\, a^{2}} \]

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*(b*x^2+a)^(1/2)*a^(
1/2))/x)

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maxima [A]  time = 1.04, size = 63, normalized size = 0.88 \[ -\frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {A}{\sqrt {b x^{2} + a} a^{2}} + \frac {A}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \]

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]

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

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mupad [B]  time = 1.11, size = 65, normalized size = 0.90 \[ \frac {\frac {A}{3\,a}+\frac {A\,\left (b\,x^2+a\right )}{a^2}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {B}{3\,b\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 42.17, size = 66, normalized size = 0.92 \[ \frac {A}{a^{2} \sqrt {a + b x^{2}}} + \frac {A \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{a^{2} \sqrt {- a}} - \frac {- A b + B a}{3 a b \left (a + b x^{2}\right )^{\frac {3}{2}}} \]

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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