\(\int x^5 \sqrt {a+b x^2} (A+B x^2) \, dx\) [504]

Optimal result

Integrand size = 22, antiderivative size = 103 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {a^2 (A b-a B) \left (a+b x^2\right )^{3/2}}{3 b^4}-\frac {a (2 A b-3 a B) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{7/2}}{7 b^4}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 103, 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} \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {a^2 \left (a+b x^2\right )^{3/2} (A b-a B)}{3 b^4}+\frac {\left (a+b x^2\right )^{7/2} (A b-3 a B)}{7 b^4}-\frac {a \left (a+b x^2\right )^{5/2} (2 A b-3 a B)}{5 b^4}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^4} \]

[In]

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

[Out]

(a^2*(A*b - a*B)*(a + b*x^2)^(3/2))/(3*b^4) - (a*(2*A*b - 3*a*B)*(a + b*x^2)^(5/2))/(5*b^4) + ((A*b - 3*a*B)*(
a + b*x^2)^(7/2))/(7*b^4) + (B*(a + b*x^2)^(9/2))/(9*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 x^2 \sqrt {a+b x} (A+B x) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a^2 (-A b+a B) \sqrt {a+b x}}{b^3}+\frac {a (-2 A b+3 a B) (a+b x)^{3/2}}{b^3}+\frac {(A b-3 a B) (a+b x)^{5/2}}{b^3}+\frac {B (a+b x)^{7/2}}{b^3}\right ) \, dx,x,x^2\right ) \\ & = \frac {a^2 (A b-a B) \left (a+b x^2\right )^{3/2}}{3 b^4}-\frac {a (2 A b-3 a B) \left (a+b x^2\right )^{5/2}}{5 b^4}+\frac {(A b-3 a B) \left (a+b x^2\right )^{7/2}}{7 b^4}+\frac {B \left (a+b x^2\right )^{9/2}}{9 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.73 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {\left (a+b x^2\right )^{3/2} \left (-16 a^3 B+24 a^2 b \left (A+B x^2\right )-6 a b^2 x^2 \left (6 A+5 B x^2\right )+5 b^3 x^4 \left (9 A+7 B x^2\right )\right )}{315 b^4} \]

[In]

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

[Out]

((a + b*x^2)^(3/2)*(-16*a^3*B + 24*a^2*b*(A + B*x^2) - 6*a*b^2*x^2*(6*A + 5*B*x^2) + 5*b^3*x^4*(9*A + 7*B*x^2)
))/(315*b^4)

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {8 \left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (\frac {15 x^{4} \left (\frac {7 x^{2} B}{9}+A \right ) b^{3}}{8}-\frac {3 x^{2} \left (\frac {5 x^{2} B}{6}+A \right ) a \,b^{2}}{2}+a^{2} \left (x^{2} B +A \right ) b -\frac {2 a^{3} B}{3}\right )}{105 b^{4}}\)	\(67\)
gosper	\(\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (35 b^{3} B \,x^{6}+45 A \,b^{3} x^{4}-30 B a \,b^{2} x^{4}-36 a A \,b^{2} x^{2}+24 B \,a^{2} b \,x^{2}+24 a^{2} b A -16 a^{3} B \right )}{315 b^{4}}\)	\(77\)
trager	\(\frac {\left (35 B \,x^{8} b^{4}+45 A \,x^{6} b^{4}+5 B \,x^{6} a \,b^{3}+9 A a \,b^{3} x^{4}-6 B \,a^{2} b^{2} x^{4}-12 A \,a^{2} b^{2} x^{2}+8 B \,a^{3} b \,x^{2}+24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 b^{4}}\)	\(101\)
risch	\(\frac {\left (35 B \,x^{8} b^{4}+45 A \,x^{6} b^{4}+5 B \,x^{6} a \,b^{3}+9 A a \,b^{3} x^{4}-6 B \,a^{2} b^{2} x^{4}-12 A \,a^{2} b^{2} x^{2}+8 B \,a^{3} b \,x^{2}+24 A \,a^{3} b -16 B \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 b^{4}}\)	\(101\)
default	\(B \left (\frac {x^{6} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{9 b}-\frac {2 a \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{7 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )}{7 b}\right )}{3 b}\right )+A \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{7 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )}{7 b}\right )\)	\(144\)

[In]

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

[Out]

8/105*(b*x^2+a)^(3/2)*(15/8*x^4*(7/9*x^2*B+A)*b^3-3/2*x^2*(5/6*x^2*B+A)*a*b^2+a^2*(B*x^2+A)*b-2/3*a^3*B)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {{\left (35 \, B b^{4} x^{8} + 5 \, {\left (B a b^{3} + 9 \, A b^{4}\right )} x^{6} - 16 \, B a^{4} + 24 \, A a^{3} b - 3 \, {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 4 \, {\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{315 \, b^{4}} \]

[In]

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

[Out]

1/315*(35*B*b^4*x^8 + 5*(B*a*b^3 + 9*A*b^4)*x^6 - 16*B*a^4 + 24*A*a^3*b - 3*(2*B*a^2*b^2 - 3*A*a*b^3)*x^4 + 4*
(2*B*a^3*b - 3*A*a^2*b^2)*x^2)*sqrt(b*x^2 + a)/b^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (94) = 188\).

Time = 0.31 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.06 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\begin {cases} \frac {8 A a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 A a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {A a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {A x^{6} \sqrt {a + b x^{2}}}{7} - \frac {16 B a^{4} \sqrt {a + b x^{2}}}{315 b^{4}} + \frac {8 B a^{3} x^{2} \sqrt {a + b x^{2}}}{315 b^{3}} - \frac {2 B a^{2} x^{4} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {B a x^{6} \sqrt {a + b x^{2}}}{63 b} + \frac {B x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{6}}{6} + \frac {B x^{8}}{8}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((8*A*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*A*a**2*x**2*sqrt(a + b*x**2)/(105*b**2) + A*a*x**4*sqrt(a
+ b*x**2)/(35*b) + A*x**6*sqrt(a + b*x**2)/7 - 16*B*a**4*sqrt(a + b*x**2)/(315*b**4) + 8*B*a**3*x**2*sqrt(a +
b*x**2)/(315*b**3) - 2*B*a**2*x**4*sqrt(a + b*x**2)/(105*b**2) + B*a*x**6*sqrt(a + b*x**2)/(63*b) + B*x**8*sqr
t(a + b*x**2)/9, Ne(b, 0)), (sqrt(a)*(A*x**6/6 + B*x**8/8), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.28 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{6}}{9 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{4}}{21 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{4}}{7 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x^{2}}{105 \, b^{3}} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x^{2}}{35 \, b^{2}} - \frac {16 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3}}{315 \, b^{4}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2}}{105 \, b^{3}} \]

[In]

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

[Out]

1/9*(b*x^2 + a)^(3/2)*B*x^6/b - 2/21*(b*x^2 + a)^(3/2)*B*a*x^4/b^2 + 1/7*(b*x^2 + a)^(3/2)*A*x^4/b + 8/105*(b*
x^2 + a)^(3/2)*B*a^2*x^2/b^3 - 4/35*(b*x^2 + a)^(3/2)*A*a*x^2/b^2 - 16/315*(b*x^2 + a)^(3/2)*B*a^3/b^4 + 8/105
*(b*x^2 + a)^(3/2)*A*a^2/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {35 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B - 135 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a + 189 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} - 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} + 45 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b - 126 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b + 105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} b}{315 \, b^{4}} \]

[In]

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

[Out]

1/315*(35*(b*x^2 + a)^(9/2)*B - 135*(b*x^2 + a)^(7/2)*B*a + 189*(b*x^2 + a)^(5/2)*B*a^2 - 105*(b*x^2 + a)^(3/2
)*B*a^3 + 45*(b*x^2 + a)^(7/2)*A*b - 126*(b*x^2 + a)^(5/2)*A*a*b + 105*(b*x^2 + a)^(3/2)*A*a^2*b)/b^4

Mupad [B] (verification not implemented)

Time = 5.83 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.93 \[ \int x^5 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\sqrt {b\,x^2+a}\,\left (\frac {B\,x^8}{9}-\frac {16\,B\,a^4-24\,A\,a^3\,b}{315\,b^4}+\frac {x^6\,\left (45\,A\,b^4+5\,B\,a\,b^3\right )}{315\,b^4}-\frac {4\,a^2\,x^2\,\left (3\,A\,b-2\,B\,a\right )}{315\,b^3}+\frac {a\,x^4\,\left (3\,A\,b-2\,B\,a\right )}{105\,b^2}\right ) \]

[In]

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

[Out]

(a + b*x^2)^(1/2)*((B*x^8)/9 - (16*B*a^4 - 24*A*a^3*b)/(315*b^4) + (x^6*(45*A*b^4 + 5*B*a*b^3))/(315*b^4) - (4
*a^2*x^2*(3*A*b - 2*B*a))/(315*b^3) + (a*x^4*(3*A*b - 2*B*a))/(105*b^2))