\(\int \frac {A+B x^2}{x^4 (a+b x^2)^{3/2}} \, dx\) [579]

Optimal result

Integrand size = 22, antiderivative size = 82 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {A}{3 a x^3 \sqrt {a+b x^2}}+\frac {4 A b-3 a B}{3 a^2 x \sqrt {a+b x^2}}+\frac {2 b (4 A b-3 a B) x}{3 a^3 \sqrt {a+b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {464, 277, 197} \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\frac {2 b x (4 A b-3 a B)}{3 a^3 \sqrt {a+b x^2}}+\frac {4 A b-3 a B}{3 a^2 x \sqrt {a+b x^2}}-\frac {A}{3 a x^3 \sqrt {a+b x^2}} \]

[In]

Int[(A + B*x^2)/(x^4*(a + b*x^2)^(3/2)),x]

[Out]

-1/3*A/(a*x^3*Sqrt[a + b*x^2]) + (4*A*b - 3*a*B)/(3*a^2*x*Sqrt[a + b*x^2]) + (2*b*(4*A*b - 3*a*B)*x)/(3*a^3*Sq
rt[a + b*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\frac {-a^2 A+4 a A b x^2-3 a^2 B x^2+8 A b^2 x^4-6 a b B x^4}{3 a^3 x^3 \sqrt {a+b x^2}} \]

[In]

Integrate[(A + B*x^2)/(x^4*(a + b*x^2)^(3/2)),x]

[Out]

(-(a^2*A) + 4*a*A*b*x^2 - 3*a^2*B*x^2 + 8*A*b^2*x^4 - 6*a*b*B*x^4)/(3*a^3*x^3*Sqrt[a + b*x^2])

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(-\frac {\left (3 x^{2} B +A \right ) a^{2}-4 x^{2} \left (-\frac {3 x^{2} B}{2}+A \right ) b a -8 A \,b^{2} x^{4}}{3 \sqrt {b \,x^{2}+a}\, x^{3} a^{3}}\)	\(55\)
gosper	\(-\frac {-8 A \,b^{2} x^{4}+6 B a b \,x^{4}-4 a A b \,x^{2}+3 a^{2} B \,x^{2}+a^{2} A}{3 x^{3} \sqrt {b \,x^{2}+a}\, a^{3}}\)	\(58\)
trager	\(-\frac {-8 A \,b^{2} x^{4}+6 B a b \,x^{4}-4 a A b \,x^{2}+3 a^{2} B \,x^{2}+a^{2} A}{3 x^{3} \sqrt {b \,x^{2}+a}\, a^{3}}\)	\(58\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-5 A b \,x^{2}+3 B a \,x^{2}+A a \right )}{3 a^{3} x^{3}}+\frac {x \left (A b -B a \right ) b}{\sqrt {b \,x^{2}+a}\, a^{3}}\)	\(60\)
default	\(A \left (-\frac {1}{3 a \,x^{3} \sqrt {b \,x^{2}+a}}-\frac {4 b \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )}{3 a}\right )+B \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(98\)

[In]

int((B*x^2+A)/x^4/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {{\left (2 \, {\left (3 \, B a b - 4 \, A b^{2}\right )} x^{4} + A a^{2} + {\left (3 \, B a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}} \]

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (73) = 146\).

Time = 3.36 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.46 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=A \left (- \frac {a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {3 a^{2} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {12 a b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {8 b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}}\right ) + B \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) \]

[In]

integrate((B*x**2+A)/x**4/(b*x**2+a)**(3/2),x)

[Out]

A*(-a**3*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x**6) + 3*a**2*b**(1
1/2)*x**2*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x**6) + 12*a*b**(13/2)*x**4*
sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x**6) + 8*b**(15/2)*x**6*sqrt(a/(b*x**
2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x**6)) + B*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)
) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1)))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {2 \, B b x}{\sqrt {b x^{2} + a} a^{2}} + \frac {8 \, A b^{2} x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {B}{\sqrt {b x^{2} + a} a x} + \frac {4 \, A b}{3 \, \sqrt {b x^{2} + a} a^{2} x} - \frac {A}{3 \, \sqrt {b x^{2} + a} a x^{3}} \]

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (70) = 140\).

Time = 0.31 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.21 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {{\left (B a b - A b^{2}\right )} x}{\sqrt {b x^{2} + a} a^{3}} + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a \sqrt {b} - 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a b^{\frac {3}{2}} + 3 \, B a^{3} \sqrt {b} - 5 \, A a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{2}} \]

[In]

integrate((B*x^2+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-(B*a*b - A*b^2)*x/(sqrt(b*x^2 + a)*a^3) + 2/3*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a*sqrt(b) - 3*(sqrt(b)*x -
sqrt(b*x^2 + a))^4*A*b^(3/2) - 6*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*sqrt(b) + 12*(sqrt(b)*x - sqrt(b*x^2 +
a))^2*A*a*b^(3/2) + 3*B*a^3*sqrt(b) - 5*A*a^2*b^(3/2))/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3*a^2)

Mupad [B] (verification not implemented)

Time = 5.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {3\,B\,a^2\,x^2+A\,a^2+6\,B\,a\,b\,x^4-4\,A\,a\,b\,x^2-8\,A\,b^2\,x^4}{3\,a^3\,x^3\,\sqrt {b\,x^2+a}} \]

[In]

int((A + B*x^2)/(x^4*(a + b*x^2)^(3/2)),x)

[Out]

-(A*a^2 + 3*B*a^2*x^2 - 8*A*b^2*x^4 - 4*A*a*b*x^2 + 6*B*a*b*x^4)/(3*a^3*x^3*(a + b*x^2)^(1/2))