\(\int \frac {x^5 (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [555]

Optimal result

Integrand size = 22, antiderivative size = 100 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {a^2 (A b-a B) \sqrt {a+b x^2}}{b^4}-\frac {a (2 A b-3 a B) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[Out]

-1/3*a*(2*A*b-3*B*a)*(b*x^2+a)^(3/2)/b^4+1/5*(A*b-3*B*a)*(b*x^2+a)^(5/2)/b^4+1/7*B*(b*x^2+a)^(7/2)/b^4+a^2*(A*
b-B*a)*(b*x^2+a)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 100, 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} \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {a^2 \sqrt {a+b x^2} (A b-a B)}{b^4}+\frac {\left (a+b x^2\right )^{5/2} (A b-3 a B)}{5 b^4}-\frac {a \left (a+b x^2\right )^{3/2} (2 A b-3 a B)}{3 b^4}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b^4} \]

[In]

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

[Out]

(a^2*(A*b - a*B)*Sqrt[a + b*x^2])/b^4 - (a*(2*A*b - 3*a*B)*(a + b*x^2)^(3/2))/(3*b^4) + ((A*b - 3*a*B)*(a + b*
x^2)^(5/2))/(5*b^4) + (B*(a + b*x^2)^(7/2))/(7*b^4)

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^2 (A+B x)}{\sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a^2 (-A b+a B)}{b^3 \sqrt {a+b x}}+\frac {a (-2 A b+3 a B) \sqrt {a+b x}}{b^3}+\frac {(A b-3 a B) (a+b x)^{3/2}}{b^3}+\frac {B (a+b x)^{5/2}}{b^3}\right ) \, dx,x,x^2\right ) \\ & = \frac {a^2 (A b-a B) \sqrt {a+b x^2}}{b^4}-\frac {a (2 A b-3 a B) \left (a+b x^2\right )^{3/2}}{3 b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (56 a^2 A b-48 a^3 B-28 a A b^2 x^2+24 a^2 b B x^2+21 A b^3 x^4-18 a b^2 B x^4+15 b^3 B x^6\right )}{105 b^4} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(56*a^2*A*b - 48*a^3*B - 28*a*A*b^2*x^2 + 24*a^2*b*B*x^2 + 21*A*b^3*x^4 - 18*a*b^2*B*x^4 + 15
*b^3*B*x^6))/(105*b^4)

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {8 \left (\frac {3 x^{4} \left (\frac {5 x^{2} B}{7}+A \right ) b^{3}}{8}-\frac {x^{2} a \left (\frac {9 x^{2} B}{14}+A \right ) b^{2}}{2}+a^{2} \left (\frac {3 x^{2} B}{7}+A \right ) b -\frac {6 a^{3} B}{7}\right ) \sqrt {b \,x^{2}+a}}{15 b^{4}}\)	\(68\)
gosper	\(\frac {\sqrt {b \,x^{2}+a}\, \left (15 b^{3} B \,x^{6}+21 A \,b^{3} x^{4}-18 B a \,b^{2} x^{4}-28 a A \,b^{2} x^{2}+24 B \,a^{2} b \,x^{2}+56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(77\)
trager	\(\frac {\sqrt {b \,x^{2}+a}\, \left (15 b^{3} B \,x^{6}+21 A \,b^{3} x^{4}-18 B a \,b^{2} x^{4}-28 a A \,b^{2} x^{2}+24 B \,a^{2} b \,x^{2}+56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(77\)
risch	\(\frac {\sqrt {b \,x^{2}+a}\, \left (15 b^{3} B \,x^{6}+21 A \,b^{3} x^{4}-18 B a \,b^{2} x^{4}-28 a A \,b^{2} x^{2}+24 B \,a^{2} b \,x^{2}+56 a^{2} b A -48 a^{3} B \right )}{105 b^{4}}\)	\(77\)
default	\(B \left (\frac {x^{6} \sqrt {b \,x^{2}+a}}{7 b}-\frac {6 a \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )}{7 b}\right )+A \left (\frac {x^{4} \sqrt {b \,x^{2}+a}}{5 b}-\frac {4 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )}{5 b}\right )\)	\(144\)

[In]

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

[Out]

8/15*(3/8*x^4*(5/7*x^2*B+A)*b^3-1/2*x^2*a*(9/14*x^2*B+A)*b^2+a^2*(3/7*x^2*B+A)*b-6/7*a^3*B)*(b*x^2+a)^(1/2)/b^
4

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {{\left (15 \, B b^{3} x^{6} - 3 \, {\left (6 \, B a b^{2} - 7 \, A b^{3}\right )} x^{4} - 48 \, B a^{3} + 56 \, A a^{2} b + 4 \, {\left (6 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, b^{4}} \]

[In]

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

[Out]

1/105*(15*B*b^3*x^6 - 3*(6*B*a*b^2 - 7*A*b^3)*x^4 - 48*B*a^3 + 56*A*a^2*b + 4*(6*B*a^2*b - 7*A*a*b^2)*x^2)*sqr
t(b*x^2 + a)/b^4

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.72 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {8 A a^{2} \sqrt {a + b x^{2}}}{15 b^{3}} - \frac {4 A a x^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {A x^{4} \sqrt {a + b x^{2}}}{5 b} - \frac {16 B a^{3} \sqrt {a + b x^{2}}}{35 b^{4}} + \frac {8 B a^{2} x^{2} \sqrt {a + b x^{2}}}{35 b^{3}} - \frac {6 B a x^{4} \sqrt {a + b x^{2}}}{35 b^{2}} + \frac {B x^{6} \sqrt {a + b x^{2}}}{7 b} & \text {for}\: b \neq 0 \\\frac {\frac {A x^{6}}{6} + \frac {B x^{8}}{8}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((8*A*a**2*sqrt(a + b*x**2)/(15*b**3) - 4*A*a*x**2*sqrt(a + b*x**2)/(15*b**2) + A*x**4*sqrt(a + b*x**
2)/(5*b) - 16*B*a**3*sqrt(a + b*x**2)/(35*b**4) + 8*B*a**2*x**2*sqrt(a + b*x**2)/(35*b**3) - 6*B*a*x**4*sqrt(a
+ b*x**2)/(35*b**2) + B*x**6*sqrt(a + b*x**2)/(7*b), Ne(b, 0)), ((A*x**6/6 + B*x**8/8)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.32 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{6}}{7 \, b} - \frac {6 \, \sqrt {b x^{2} + a} B a x^{4}}{35 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{4}}{5 \, b} + \frac {8 \, \sqrt {b x^{2} + a} B a^{2} x^{2}}{35 \, b^{3}} - \frac {4 \, \sqrt {b x^{2} + a} A a x^{2}}{15 \, b^{2}} - \frac {16 \, \sqrt {b x^{2} + a} B a^{3}}{35 \, b^{4}} + \frac {8 \, \sqrt {b x^{2} + a} A a^{2}}{15 \, b^{3}} \]

[In]

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

[Out]

1/7*sqrt(b*x^2 + a)*B*x^6/b - 6/35*sqrt(b*x^2 + a)*B*a*x^4/b^2 + 1/5*sqrt(b*x^2 + a)*A*x^4/b + 8/35*sqrt(b*x^2
+ a)*B*a^2*x^2/b^3 - 4/15*sqrt(b*x^2 + a)*A*a*x^2/b^2 - 16/35*sqrt(b*x^2 + a)*B*a^3/b^4 + 8/15*sqrt(b*x^2 + a
)*A*a^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {{\left (B a^{3} - A a^{2} b\right )} \sqrt {b x^{2} + a}}{b^{4}} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B - 63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a + 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} + 21 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b - 70 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b}{105 \, b^{4}} \]

[In]

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

[Out]

-(B*a^3 - A*a^2*b)*sqrt(b*x^2 + a)/b^4 + 1/105*(15*(b*x^2 + a)^(7/2)*B - 63*(b*x^2 + a)^(5/2)*B*a + 105*(b*x^2
+ a)^(3/2)*B*a^2 + 21*(b*x^2 + a)^(5/2)*A*b - 70*(b*x^2 + a)^(3/2)*A*a*b)/b^4

Mupad [B] (verification not implemented)

Time = 5.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {x^5 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\sqrt {b\,x^2+a}\,\left (\frac {48\,B\,a^3-56\,A\,a^2\,b}{105\,b^4}-\frac {B\,x^6}{7\,b}-\frac {x^4\,\left (21\,A\,b^3-18\,B\,a\,b^2\right )}{105\,b^4}+\frac {4\,a\,x^2\,\left (7\,A\,b-6\,B\,a\right )}{105\,b^3}\right ) \]

[In]

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

[Out]

-(a + b*x^2)^(1/2)*((48*B*a^3 - 56*A*a^2*b)/(105*b^4) - (B*x^6)/(7*b) - (x^4*(21*A*b^3 - 18*B*a*b^2))/(105*b^4
) + (4*a*x^2*(7*A*b - 6*B*a))/(105*b^3))