3.6.78 \(\int \frac {A+B x^2}{x^3 (a+b x^2)^{3/2}} \, dx\) [578]

Optimal. Leaf size=86 \[ -\frac {3 A b-2 a B}{2 a^2 \sqrt {a+b x^2}}-\frac {A}{2 a x^2 \sqrt {a+b x^2}}+\frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 86, 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 = {457, 79, 53, 65, 214} \begin {gather*} \frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}}-\frac {3 A b-2 a B}{2 a^2 \sqrt {a+b x^2}}-\frac {A}{2 a x^2 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 457

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

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Mathematica [A]
time = 0.14, size = 77, normalized size = 0.90 \begin {gather*} \frac {-a A-3 A b x^2+2 a B x^2}{2 a^2 x^2 \sqrt {a+b x^2}}+\frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 114, normalized size = 1.33

method	result	size
risch	\(-\frac {A \sqrt {b \,x^{2}+a}}{2 a^{2} x^{2}}-\frac {b A}{a^{2} \sqrt {b \,x^{2}+a}}+\frac {B}{a \sqrt {b \,x^{2}+a}}+\frac {3 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A b}{2 a^{\frac {5}{2}}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{a^{\frac {3}{2}}}\)	\(109\)
default	\(A \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )+B \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^3/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 86, normalized size = 1.00 \begin {gather*} -\frac {B \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} + \frac {B}{\sqrt {b x^{2} + a} a} - \frac {3 \, A b}{2 \, \sqrt {b x^{2} + a} a^{2}} - \frac {A}{2 \, \sqrt {b x^{2} + a} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.01, size = 232, normalized size = 2.70 \begin {gather*} \left [-\frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, \frac {{\left ({\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (A a^{2} - {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (73) = 146\).
time = 15.97, size = 262, normalized size = 3.05 \begin {gather*} A \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 ) + B \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{3} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{2} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{2} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(-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))) + B*(2*a**3*sqrt(1 + b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**3*log(b*x**2/a
)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) - 2*a**3*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**
2*b*x**2*log(b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) - 2*a**2*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2
) + 2*a**(7/2)*b*x**2))

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Giac [A]
time = 0.84, size = 99, normalized size = 1.15 \begin {gather*} \frac {{\left (2 \, B a - 3 \, A b\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{2}} + \frac {2 \, {\left (b x^{2} + a\right )} B a - 2 \, B a^{2} - 3 \, {\left (b x^{2} + a\right )} A b + 2 \, A a b}{2 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{2} + a} a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.71, size = 90, normalized size = 1.05 \begin {gather*} \frac {B}{a\,\sqrt {b\,x^2+a}}-\frac {B\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {3\,A\,b}{2\,a^2\,\sqrt {b\,x^2+a}}-\frac {A}{2\,a\,x^2\,\sqrt {b\,x^2+a}}+\frac {3\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^3*(a + b*x^2)^(3/2)),x)

[Out]

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

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