\(\int \frac {x^3 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [572]

Optimal result

Integrand size = 22, antiderivative size = 67 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {a (A b-a B)}{b^3 \sqrt {a+b x^2}}+\frac {(A b-2 a B) \sqrt {a+b x^2}}{b^3}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\sqrt {a+b x^2} (A b-2 a B)}{b^3}+\frac {a (A b-a B)}{b^3 \sqrt {a+b x^2}}+\frac {B \left (a+b x^2\right )^{3/2}}{3 b^3} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {6 a A b-8 a^2 B+3 A b^2 x^2-4 a b B x^2+b^2 B x^4}{3 b^3 \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {x^{2} \left (\frac {x^{2} B}{3}+A \right ) b^{2}+2 a \left (-\frac {2 x^{2} B}{3}+A \right ) b -\frac {8 a^{2} B}{3}}{\sqrt {b \,x^{2}+a}\, b^{3}}\)	\(49\)
gosper	\(\frac {b^{2} B \,x^{4}+3 A \,b^{2} x^{2}-4 B a b \,x^{2}+6 a b A -8 a^{2} B}{3 \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(52\)
trager	\(\frac {b^{2} B \,x^{4}+3 A \,b^{2} x^{2}-4 B a b \,x^{2}+6 a b A -8 a^{2} B}{3 \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(52\)
risch	\(\frac {\left (b B \,x^{2}+3 A b -5 B a \right ) \sqrt {b \,x^{2}+a}}{3 b^{3}}+\frac {a \left (A b -B a \right )}{b^{3} \sqrt {b \,x^{2}+a}}\)	\(53\)
default	\(B \left (\frac {x^{4}}{3 b \sqrt {b \,x^{2}+a}}-\frac {4 a \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )}{3 b}\right )+A \left (\frac {x^{2}}{\sqrt {b \,x^{2}+a}\, b}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (B b^{2} x^{4} - 8 \, B a^{2} + 6 \, A a b - {\left (4 \, B a b - 3 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.75 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\begin {cases} \frac {2 A a}{b^{2} \sqrt {a + b x^{2}}} + \frac {A x^{2}}{b \sqrt {a + b x^{2}}} - \frac {8 B a^{2}}{3 b^{3} \sqrt {a + b x^{2}}} - \frac {4 B a x^{2}}{3 b^{2} \sqrt {a + b x^{2}}} + \frac {B x^{4}}{3 b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{6}}{6}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*A*a/(b**2*sqrt(a + b*x**2)) + A*x**2/(b*sqrt(a + b*x**2)) - 8*B*a**2/(3*b**3*sqrt(a + b*x**2)) -
4*B*a*x**2/(3*b**2*sqrt(a + b*x**2)) + B*x**4/(3*b*sqrt(a + b*x**2)), Ne(b, 0)), ((A*x**4/4 + B*x**6/6)/a**(3/
2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.33 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{4}}{3 \, \sqrt {b x^{2} + a} b} - \frac {4 \, B a x^{2}}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{2}}{\sqrt {b x^{2} + a} b} - \frac {8 \, B a^{2}}{3 \, \sqrt {b x^{2} + a} b^{3}} + \frac {2 \, A a}{\sqrt {b x^{2} + a} b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {B a^{2} - A a b}{\sqrt {b x^{2} + a} b^{3}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{6} - 6 \, \sqrt {b x^{2} + a} B a b^{6} + 3 \, \sqrt {b x^{2} + a} A b^{7}}{3 \, b^{9}} \]

[In]

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

[Out]

-(B*a^2 - A*a*b)/(sqrt(b*x^2 + a)*b^3) + 1/3*((b*x^2 + a)^(3/2)*B*b^6 - 6*sqrt(b*x^2 + a)*B*a*b^6 + 3*sqrt(b*x
^2 + a)*A*b^7)/b^9

Mupad [B] (verification not implemented)

Time = 5.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B\,{\left (b\,x^2+a\right )}^2-3\,B\,a^2+3\,A\,b\,\left (b\,x^2+a\right )-6\,B\,a\,\left (b\,x^2+a\right )+3\,A\,a\,b}{3\,b^3\,\sqrt {b\,x^2+a}} \]

[In]

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

[Out]

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