Integrand size = 22, antiderivative size = 73 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=-\frac {a (A b-a B) \left (a+b x^2\right )^{7/2}}{7 b^3}+\frac {(A b-2 a B) \left (a+b x^2\right )^{9/2}}{9 b^3}+\frac {B \left (a+b x^2\right )^{11/2}}{11 b^3} \] Output:
-1/7*a*(A*b-B*a)*(b*x^2+a)^(7/2)/b^3+1/9*(A*b-2*B*a)*(b*x^2+a)^(9/2)/b^3+1 /11*B*(b*x^2+a)^(11/2)/b^3
Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {\left (a+b x^2\right )^{7/2} \left (-22 a A b+8 a^2 B+77 A b^2 x^2-28 a b B x^2+63 b^2 B x^4\right )}{693 b^3} \] Input:
Integrate[x^3*(a + b*x^2)^(5/2)*(A + B*x^2),x]
Output:
((a + b*x^2)^(7/2)*(-22*a*A*b + 8*a^2*B + 77*A*b^2*x^2 - 28*a*b*B*x^2 + 63 *b^2*B*x^4))/(693*b^3)
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {354, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \left (b x^2+a\right )^{5/2} \left (B x^2+A\right )dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B \left (b x^2+a\right )^{9/2}}{b^2}+\frac {(A b-2 a B) \left (b x^2+a\right )^{7/2}}{b^2}+\frac {a (a B-A b) \left (b x^2+a\right )^{5/2}}{b^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (a+b x^2\right )^{9/2} (A b-2 a B)}{9 b^3}-\frac {2 a \left (a+b x^2\right )^{7/2} (A b-a B)}{7 b^3}+\frac {2 B \left (a+b x^2\right )^{11/2}}{11 b^3}\right )\) |
Input:
Int[x^3*(a + b*x^2)^(5/2)*(A + B*x^2),x]
Output:
((-2*a*(A*b - a*B)*(a + b*x^2)^(7/2))/(7*b^3) + (2*(A*b - 2*a*B)*(a + b*x^ 2)^(9/2))/(9*b^3) + (2*B*(a + b*x^2)^(11/2))/(11*b^3))/2
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((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]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.44 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-\frac {7 \left (\frac {9 x^{2} B}{11}+A \right ) x^{2} b^{2}}{2}+a \left (\frac {14 x^{2} B}{11}+A \right ) b -\frac {4 a^{2} B}{11}\right ) \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{3}}\) | \(49\) |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-63 b^{2} B \,x^{4}-77 A \,b^{2} x^{2}+28 B a b \,x^{2}+22 a b A -8 a^{2} B \right )}{693 b^{3}}\) | \(53\) |
orering | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-63 b^{2} B \,x^{4}-77 A \,b^{2} x^{2}+28 B a b \,x^{2}+22 a b A -8 a^{2} B \right )}{693 b^{3}}\) | \(53\) |
default | \(A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{9 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{2}}\right )+B \left (\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{11 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{9 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{2}}\right )}{11 b}\right )\) | \(96\) |
trager | \(-\frac {\left (-63 b^{5} B \,x^{10}-77 A \,x^{8} b^{5}-161 a \,b^{4} B \,x^{8}-209 a \,b^{4} A \,x^{6}-113 a^{2} b^{3} B \,x^{6}-165 a^{2} b^{3} A \,x^{4}-3 B \,a^{3} b^{2} x^{4}-11 a^{3} A \,b^{2} x^{2}+4 B \,a^{4} b \,x^{2}+22 a^{4} b A -8 a^{5} B \right ) \sqrt {b \,x^{2}+a}}{693 b^{3}}\) | \(125\) |
risch | \(-\frac {\left (-63 b^{5} B \,x^{10}-77 A \,x^{8} b^{5}-161 a \,b^{4} B \,x^{8}-209 a \,b^{4} A \,x^{6}-113 a^{2} b^{3} B \,x^{6}-165 a^{2} b^{3} A \,x^{4}-3 B \,a^{3} b^{2} x^{4}-11 a^{3} A \,b^{2} x^{2}+4 B \,a^{4} b \,x^{2}+22 a^{4} b A -8 a^{5} B \right ) \sqrt {b \,x^{2}+a}}{693 b^{3}}\) | \(125\) |
Input:
int(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x,method=_RETURNVERBOSE)
Output:
-2/63*(-7/2*(9/11*x^2*B+A)*x^2*b^2+a*(14/11*x^2*B+A)*b-4/11*a^2*B)*(b*x^2+ a)^(7/2)/b^3
Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {{\left (63 \, B b^{5} x^{10} + 7 \, {\left (23 \, B a b^{4} + 11 \, A b^{5}\right )} x^{8} + {\left (113 \, B a^{2} b^{3} + 209 \, A a b^{4}\right )} x^{6} + 8 \, B a^{5} - 22 \, A a^{4} b + 3 \, {\left (B a^{3} b^{2} + 55 \, A a^{2} b^{3}\right )} x^{4} - {\left (4 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{693 \, b^{3}} \] Input:
integrate(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="fricas")
Output:
1/693*(63*B*b^5*x^10 + 7*(23*B*a*b^4 + 11*A*b^5)*x^8 + (113*B*a^2*b^3 + 20 9*A*a*b^4)*x^6 + 8*B*a^5 - 22*A*a^4*b + 3*(B*a^3*b^2 + 55*A*a^2*b^3)*x^4 - (4*B*a^4*b - 11*A*a^3*b^2)*x^2)*sqrt(b*x^2 + a)/b^3
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (65) = 130\).
Time = 0.50 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.56 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\begin {cases} - \frac {2 A a^{4} \sqrt {a + b x^{2}}}{63 b^{2}} + \frac {A a^{3} x^{2} \sqrt {a + b x^{2}}}{63 b} + \frac {5 A a^{2} x^{4} \sqrt {a + b x^{2}}}{21} + \frac {19 A a b x^{6} \sqrt {a + b x^{2}}}{63} + \frac {A b^{2} x^{8} \sqrt {a + b x^{2}}}{9} + \frac {8 B a^{5} \sqrt {a + b x^{2}}}{693 b^{3}} - \frac {4 B a^{4} x^{2} \sqrt {a + b x^{2}}}{693 b^{2}} + \frac {B a^{3} x^{4} \sqrt {a + b x^{2}}}{231 b} + \frac {113 B a^{2} x^{6} \sqrt {a + b x^{2}}}{693} + \frac {23 B a b x^{8} \sqrt {a + b x^{2}}}{99} + \frac {B b^{2} x^{10} \sqrt {a + b x^{2}}}{11} & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{4}}{4} + \frac {B x^{6}}{6}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(b*x**2+a)**(5/2)*(B*x**2+A),x)
Output:
Piecewise((-2*A*a**4*sqrt(a + b*x**2)/(63*b**2) + A*a**3*x**2*sqrt(a + b*x **2)/(63*b) + 5*A*a**2*x**4*sqrt(a + b*x**2)/21 + 19*A*a*b*x**6*sqrt(a + b *x**2)/63 + A*b**2*x**8*sqrt(a + b*x**2)/9 + 8*B*a**5*sqrt(a + b*x**2)/(69 3*b**3) - 4*B*a**4*x**2*sqrt(a + b*x**2)/(693*b**2) + B*a**3*x**4*sqrt(a + b*x**2)/(231*b) + 113*B*a**2*x**6*sqrt(a + b*x**2)/693 + 23*B*a*b*x**8*sq rt(a + b*x**2)/99 + B*b**2*x**10*sqrt(a + b*x**2)/11, Ne(b, 0)), (a**(5/2) *(A*x**4/4 + B*x**6/6), True))
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{4}}{11 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x^{2}}{99 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x^{2}}{9 \, b} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2}}{693 \, b^{3}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a}{63 \, b^{2}} \] Input:
integrate(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="maxima")
Output:
1/11*(b*x^2 + a)^(7/2)*B*x^4/b - 4/99*(b*x^2 + a)^(7/2)*B*a*x^2/b^2 + 1/9* (b*x^2 + a)^(7/2)*A*x^2/b + 8/693*(b*x^2 + a)^(7/2)*B*a^2/b^3 - 2/63*(b*x^ 2 + a)^(7/2)*A*a/b^2
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {63 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} B - 154 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a + 99 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} + 77 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b - 99 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b}{693 \, b^{3}} \] Input:
integrate(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="giac")
Output:
1/693*(63*(b*x^2 + a)^(11/2)*B - 154*(b*x^2 + a)^(9/2)*B*a + 99*(b*x^2 + a )^(7/2)*B*a^2 + 77*(b*x^2 + a)^(9/2)*A*b - 99*(b*x^2 + a)^(7/2)*A*a*b)/b^3
Time = 0.53 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.58 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\sqrt {b\,x^2+a}\,\left (\frac {8\,B\,a^5-22\,A\,a^4\,b}{693\,b^3}+\frac {B\,b^2\,x^{10}}{11}+\frac {x^8\,\left (77\,A\,b^5+161\,B\,a\,b^4\right )}{693\,b^3}+\frac {a\,x^6\,\left (209\,A\,b+113\,B\,a\right )}{693}+\frac {a^3\,x^2\,\left (11\,A\,b-4\,B\,a\right )}{693\,b^2}+\frac {a^2\,x^4\,\left (55\,A\,b+B\,a\right )}{231\,b}\right ) \] Input:
int(x^3*(A + B*x^2)*(a + b*x^2)^(5/2),x)
Output:
(a + b*x^2)^(1/2)*((8*B*a^5 - 22*A*a^4*b)/(693*b^3) + (B*b^2*x^10)/11 + (x ^8*(77*A*b^5 + 161*B*a*b^4))/(693*b^3) + (a*x^6*(209*A*b + 113*B*a))/693 + (a^3*x^2*(11*A*b - 4*B*a))/(693*b^2) + (a^2*x^4*(55*A*b + B*a))/(231*b))
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (9 b^{5} x^{10}+34 a \,b^{4} x^{8}+46 a^{2} b^{3} x^{6}+24 a^{3} b^{2} x^{4}+a^{4} b \,x^{2}-2 a^{5}\right )}{99 b^{2}} \] Input:
int(x^3*(b*x^2+a)^(5/2)*(B*x^2+A),x)
Output:
(sqrt(a + b*x**2)*( - 2*a**5 + a**4*b*x**2 + 24*a**3*b**2*x**4 + 46*a**2*b **3*x**6 + 34*a*b**4*x**8 + 9*b**5*x**10))/(99*b**2)