\(\int \frac {x (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [574]

Optimal result

Integrand size = 20, antiderivative size = 41 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {A b-a B}{b^2 \sqrt {a+b x^2}}+\frac {B \sqrt {a+b x^2}}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {455, 45} \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B \sqrt {a+b x^2}}{b^2}-\frac {A b-a B}{b^2 \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-A b+2 a B+b B x^2}{b^2 \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.82 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {-b B \,x^{2}+A b -2 B a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(30\)
trager	\(-\frac {-b B \,x^{2}+A b -2 B a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(30\)
pseudoelliptic	\(\frac {\left (x^{2} B -A \right ) b +2 B a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(30\)
risch	\(\frac {B \sqrt {b \,x^{2}+a}}{b^{2}}-\frac {A b -B a}{\sqrt {b \,x^{2}+a}\, b^{2}}\)	\(38\)
default	\(B \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )-\frac {A}{b \sqrt {b \,x^{2}+a}}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.98 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (B b x^{2} + 2 \, B a - A b\right )} \sqrt {b x^{2} + a}}{b^{3} x^{2} + a b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.61 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\begin {cases} - \frac {A}{b \sqrt {a + b x^{2}}} + \frac {2 B a}{b^{2} \sqrt {a + b x^{2}}} + \frac {B x^{2}}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{2}}{2} + \frac {B x^{4}}{4}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-A/(b*sqrt(a + b*x**2)) + 2*B*a/(b**2*sqrt(a + b*x**2)) + B*x**2/(b*sqrt(a + b*x**2)), Ne(b, 0)), (
(A*x**2/2 + B*x**4/4)/a**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{2}}{\sqrt {b x^{2} + a} b} + \frac {2 \, B a}{\sqrt {b x^{2} + a} b^{2}} - \frac {A}{\sqrt {b x^{2} + a} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {b x^{2} + a} B}{b^{2}} + \frac {B a - A b}{\sqrt {b x^{2} + a} b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.73 \[ \int \frac {x \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B\,a-A\,b+B\,\left (b\,x^2+a\right )}{b^2\,\sqrt {b\,x^2+a}} \]

[In]

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

[Out]

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