3.6.65 \(\int \frac {A+B x^2}{x^5 \sqrt {a+b x^2}} \, dx\) [565]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 90, 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, 44, 65, 214} \begin {gather*} -\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 {\sqrt {a+b x^2} (3 A b-4 a B)}{8 a^2 x^2}-\frac {A \sqrt {a+b x^2}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(A*Sqrt[a + b*x^2])/(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[
Sqrt[a + b*x^2]/Sqrt[a]])/(8*a^(5/2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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^5 \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \text {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 ) \text {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)) \text {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) \text {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*}

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Mathematica [A]
time = 0.13, size = 78, normalized size = 0.87 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2 a A+3 A b x^2-4 a B x^2\right )}{8 a^2 x^4}+\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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 124, normalized size = 1.38

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 A b \,x^{2}+4 B a \,x^{2}+2 A a \right )}{8 a^{2} x^{4}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A}{8 a^{\frac {5}{2}}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) B}{2 a^{\frac {3}{2}}}\)	\(100\)
default	\(A \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+B \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(124\)

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,method=_RETURNVERBOSE)

[Out]

A*(-1/4/a/x^4*(b*x^2+a)^(1/2)-3/4*b/a*(-1/2*(b*x^2+a)^(1/2)/a/x^2+1/2*b/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1
/2))/x)))+B*(-1/2*(b*x^2+a)^(1/2)/a/x^2+1/2*b/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))

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Maxima [A]
time = 0.30, size = 96, normalized size = 1.07 \begin {gather*} \frac {B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {3 \, A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a} B}{2 \, a x^{2}} + \frac {3 \, \sqrt {b x^{2} + a} A b}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} A}{4 \, a x^{4}} \end {gather*}

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]

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

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Fricas [A]
time = 1.97, size = 171, normalized size = 1.90 \begin {gather*} \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 {gather*}

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 = 25.88, size = 150, normalized size = 1.67 \begin {gather*} - \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 {gather*}

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 = 1.41, size = 121, normalized size = 1.34 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.70, size = 99, normalized size = 1.10 \begin {gather*} \frac {3\,A\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {5\,A\,\sqrt {b\,x^2+a}}{8\,a\,x^4}-\frac {3\,A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {B\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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