\(\int \frac {x^3 (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [557]

Optimal result

Integrand size = 22, antiderivative size = 71 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (A b-a B) \sqrt {a+b x^2}}{b^3}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, 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} \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\left (a+b x^2\right )^{3/2} (A b-2 a B)}{3 b^3}-\frac {a \sqrt {a+b x^2} (A b-a B)}{b^3}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^3} \]

[In]

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

[Out]

-((a*(A*b - a*B)*Sqrt[a + b*x^2])/b^3) + ((A*b - 2*a*B)*(a + b*x^2)^(3/2))/(3*b^3) + (B*(a + b*x^2)^(5/2))/(5*
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)}{\sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {a (-A b+a B)}{b^2 \sqrt {a+b x}}+\frac {(A b-2 a B) \sqrt {a+b x}}{b^2}+\frac {B (a+b x)^{3/2}}{b^2}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a (A b-a B) \sqrt {a+b x^2}}{b^3}+\frac {(A b-2 a B) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {B \left (a+b x^2\right )^{5/2}}{5 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-10 a A b+8 a^2 B+5 A b^2 x^2-4 a b B x^2+3 b^2 B x^4\right )}{15 b^3} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-10*a*A*b + 8*a^2*B + 5*A*b^2*x^2 - 4*a*b*B*x^2 + 3*b^2*B*x^4))/(15*b^3)

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-\frac {x^{2} \left (\frac {3 x^{2} B}{5}+A \right ) b^{2}}{2}+\left (\frac {2 x^{2} B}{5}+A \right ) a b -\frac {4 a^{2} B}{5}\right )}{3 b^{3}}\)	\(49\)
gosper	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+4 B a b \,x^{2}+10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(53\)
trager	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+4 B a b \,x^{2}+10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(53\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 b^{2} B \,x^{4}-5 A \,b^{2} x^{2}+4 B a b \,x^{2}+10 a b A -8 a^{2} B \right )}{15 b^{3}}\)	\(53\)
default	\(B \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )+A \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )\)	\(96\)

[In]

int(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {{\left (3 \, B b^{2} x^{4} + 8 \, B a^{2} - 10 \, A a b - {\left (4 \, B a b - 5 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, b^{3}} \]

[In]

integrate(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/15*(3*B*b^2*x^4 + 8*B*a^2 - 10*A*a*b - (4*B*a*b - 5*A*b^2)*x^2)*sqrt(b*x^2 + a)/b^3

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.70 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {2 A a \sqrt {a + b x^{2}}}{3 b^{2}} + \frac {A x^{2} \sqrt {a + b x^{2}}}{3 b} + \frac {8 B a^{2} \sqrt {a + b x^{2}}}{15 b^{3}} - \frac {4 B a x^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {B x^{4} \sqrt {a + b x^{2}}}{5 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{6}}{6}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(B*x**2+A)/(b*x**2+a)**(1/2),x)

[Out]

Piecewise((-2*A*a*sqrt(a + b*x**2)/(3*b**2) + A*x**2*sqrt(a + b*x**2)/(3*b) + 8*B*a**2*sqrt(a + b*x**2)/(15*b*
*3) - 4*B*a*x**2*sqrt(a + b*x**2)/(15*b**2) + B*x**4*sqrt(a + b*x**2)/(5*b), Ne(b, 0)), ((A*x**4/4 + B*x**6/6)
/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{4}}{5 \, b} - \frac {4 \, \sqrt {b x^{2} + a} B a x^{2}}{15 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{2}}{3 \, b} + \frac {8 \, \sqrt {b x^{2} + a} B a^{2}}{15 \, b^{3}} - \frac {2 \, \sqrt {b x^{2} + a} A a}{3 \, b^{2}} \]

[In]

integrate(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(b*x^2 + a)*B*x^4/b - 4/15*sqrt(b*x^2 + a)*B*a*x^2/b^2 + 1/3*sqrt(b*x^2 + a)*A*x^2/b + 8/15*sqrt(b*x^2
+ a)*B*a^2/b^3 - 2/3*sqrt(b*x^2 + a)*A*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.97 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {{\left (B a^{2} - A a b\right )} \sqrt {b x^{2} + a}}{b^{3}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B - 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{15 \, b^{3}} \]

[In]

integrate(x^3*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

(B*a^2 - A*a*b)*sqrt(b*x^2 + a)/b^3 + 1/15*(3*(b*x^2 + a)^(5/2)*B - 10*(b*x^2 + a)^(3/2)*B*a + 5*(b*x^2 + a)^(
3/2)*A*b)/b^3

Mupad [B] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\sqrt {b\,x^2+a}\,\left (\frac {8\,B\,a^2-10\,A\,a\,b}{15\,b^3}+\frac {x^2\,\left (5\,A\,b^2-4\,B\,a\,b\right )}{15\,b^3}+\frac {B\,x^4}{5\,b}\right ) \]

[In]

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

[Out]

(a + b*x^2)^(1/2)*((8*B*a^2 - 10*A*a*b)/(15*b^3) + (x^2*(5*A*b^2 - 4*B*a*b))/(15*b^3) + (B*x^4)/(5*b))