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

Optimal. Leaf size=88 \[ \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\sqrt {a+b x^2} (A b-4 a B)}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, 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, 47, 63, 208} \[ \frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\sqrt {a+b x^2} (A b-4 a B)}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x} (A+B x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {\left (-\frac {A b}{2}+2 a B\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,x^2\right )}{4 a}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}-\frac {(b (A b-4 a B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{16 a}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}-\frac {(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}\\ &=\frac {(A b-4 a B) \sqrt {a+b x^2}}{8 a x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^4}+\frac {b (A b-4 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 93, normalized size = 1.06 \[ \frac {-\left (a+b x^2\right ) \left (2 a \left (A+2 B x^2\right )+A b x^2\right )-b x^4 \sqrt {\frac {b x^2}{a}+1} (4 a B-A b) \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{8 a x^4 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.84, size = 170, normalized size = 1.93 \[ \left [-\frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {a} x^{4} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, a^{2} x^{4}}, \frac {{\left (4 \, B a b - A b^{2}\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, A a^{2} + {\left (4 \, B a^{2} + A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, a^{2} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 120, normalized size = 1.36 \[ \frac {\frac {{\left (4 \, B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{2} - 4 \, \sqrt {b x^{2} + a} B a^{2} b^{2} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3} + \sqrt {b x^{2} + a} A a b^{3}}{a b^{2} x^{4}}}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 153, normalized size = 1.74 \[ \frac {A \,b^{2} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}}}-\frac {B b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}}-\frac {\sqrt {b \,x^{2}+a}\, A \,b^{2}}{8 a^{2}}+\frac {\sqrt {b \,x^{2}+a}\, B b}{2 a}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A b}{8 a^{2} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B}{2 a \,x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A}{4 a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 130, normalized size = 1.48 \[ -\frac {B b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, \sqrt {a}} + \frac {A b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {3}{2}}} + \frac {\sqrt {b x^{2} + a} B b}{2 \, a} - \frac {\sqrt {b x^{2} + a} A b^{2}}{8 \, a^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{2 \, a x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{4 \, a x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.70, size = 93, normalized size = 1.06 \[ \frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{3/2}}-\frac {B\,\sqrt {b\,x^2+a}}{2\,x^2}-\frac {A\,\sqrt {b\,x^2+a}}{8\,x^4}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a\,x^4}-\frac {B\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 143.85, size = 144, normalized size = 1.64 \[ - \frac {A a}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 A \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {A b^{\frac {3}{2}}}{8 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {3}{2}}} - \frac {B \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 x} - \frac {B b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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