\(\int x^2 (a+b x^2)^5 (A+B x^2) \, dx\) [30]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{3} a^5 A x^3+\frac {1}{5} a^4 (5 A b+a B) x^5+\frac {5}{7} a^3 b (2 A b+a B) x^7+\frac {10}{9} a^2 b^2 (A b+a B) x^9+\frac {5}{11} a b^3 (A b+2 a B) x^{11}+\frac {1}{13} b^4 (A b+5 a B) x^{13}+\frac {1}{15} b^5 B x^{15} \]

[Out]

1/3*a^5*A*x^3+1/5*a^4*(5*A*b+B*a)*x^5+5/7*a^3*b*(2*A*b+B*a)*x^7+10/9*a^2*b^2*(A*b+B*a)*x^9+5/11*a*b^3*(A*b+2*B
*a)*x^11+1/13*b^4*(A*b+5*B*a)*x^13+1/15*b^5*B*x^15

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {459} \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{3} a^5 A x^3+\frac {1}{5} a^4 x^5 (a B+5 A b)+\frac {5}{7} a^3 b x^7 (a B+2 A b)+\frac {10}{9} a^2 b^2 x^9 (a B+A b)+\frac {1}{13} b^4 x^{13} (5 a B+A b)+\frac {5}{11} a b^3 x^{11} (2 a B+A b)+\frac {1}{15} b^5 B x^{15} \]

[In]

Int[x^2*(a + b*x^2)^5*(A + B*x^2),x]

[Out]

(a^5*A*x^3)/3 + (a^4*(5*A*b + a*B)*x^5)/5 + (5*a^3*b*(2*A*b + a*B)*x^7)/7 + (10*a^2*b^2*(A*b + a*B)*x^9)/9 + (
5*a*b^3*(A*b + 2*a*B)*x^11)/11 + (b^4*(A*b + 5*a*B)*x^13)/13 + (b^5*B*x^15)/15

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^5 A x^2+a^4 (5 A b+a B) x^4+5 a^3 b (2 A b+a B) x^6+10 a^2 b^2 (A b+a B) x^8+5 a b^3 (A b+2 a B) x^{10}+b^4 (A b+5 a B) x^{12}+b^5 B x^{14}\right ) \, dx \\ & = \frac {1}{3} a^5 A x^3+\frac {1}{5} a^4 (5 A b+a B) x^5+\frac {5}{7} a^3 b (2 A b+a B) x^7+\frac {10}{9} a^2 b^2 (A b+a B) x^9+\frac {5}{11} a b^3 (A b+2 a B) x^{11}+\frac {1}{13} b^4 (A b+5 a B) x^{13}+\frac {1}{15} b^5 B x^{15} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{3} a^5 A x^3+\frac {1}{5} a^4 (5 A b+a B) x^5+\frac {5}{7} a^3 b (2 A b+a B) x^7+\frac {10}{9} a^2 b^2 (A b+a B) x^9+\frac {5}{11} a b^3 (A b+2 a B) x^{11}+\frac {1}{13} b^4 (A b+5 a B) x^{13}+\frac {1}{15} b^5 B x^{15} \]

[In]

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

[Out]

(a^5*A*x^3)/3 + (a^4*(5*A*b + a*B)*x^5)/5 + (5*a^3*b*(2*A*b + a*B)*x^7)/7 + (10*a^2*b^2*(A*b + a*B)*x^9)/9 + (
5*a*b^3*(A*b + 2*a*B)*x^11)/11 + (b^4*(A*b + 5*a*B)*x^13)/13 + (b^5*B*x^15)/15

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03

method	result	size
norman	\(\frac {a^{5} A \,x^{3}}{3}+\left (a^{4} b A +\frac {1}{5} a^{5} B \right ) x^{5}+\left (\frac {10}{7} a^{3} b^{2} A +\frac {5}{7} a^{4} b B \right ) x^{7}+\left (\frac {10}{9} a^{2} b^{3} A +\frac {10}{9} a^{3} b^{2} B \right ) x^{9}+\left (\frac {5}{11} a \,b^{4} A +\frac {10}{11} a^{2} b^{3} B \right ) x^{11}+\left (\frac {1}{13} b^{5} A +\frac {5}{13} a \,b^{4} B \right ) x^{13}+\frac {b^{5} B \,x^{15}}{15}\)	\(120\)
default	\(\frac {b^{5} B \,x^{15}}{15}+\frac {\left (b^{5} A +5 a \,b^{4} B \right ) x^{13}}{13}+\frac {\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) x^{11}}{11}+\frac {\left (10 a^{2} b^{3} A +10 a^{3} b^{2} B \right ) x^{9}}{9}+\frac {\left (10 a^{3} b^{2} A +5 a^{4} b B \right ) x^{7}}{7}+\frac {\left (5 a^{4} b A +a^{5} B \right ) x^{5}}{5}+\frac {a^{5} A \,x^{3}}{3}\)	\(124\)
gosper	\(\frac {1}{3} a^{5} A \,x^{3}+x^{5} a^{4} b A +\frac {1}{5} x^{5} a^{5} B +\frac {10}{7} x^{7} a^{3} b^{2} A +\frac {5}{7} x^{7} a^{4} b B +\frac {10}{9} x^{9} a^{2} b^{3} A +\frac {10}{9} x^{9} a^{3} b^{2} B +\frac {5}{11} x^{11} a \,b^{4} A +\frac {10}{11} x^{11} a^{2} b^{3} B +\frac {1}{13} x^{13} b^{5} A +\frac {5}{13} x^{13} a \,b^{4} B +\frac {1}{15} b^{5} B \,x^{15}\)	\(125\)
risch	\(\frac {1}{3} a^{5} A \,x^{3}+x^{5} a^{4} b A +\frac {1}{5} x^{5} a^{5} B +\frac {10}{7} x^{7} a^{3} b^{2} A +\frac {5}{7} x^{7} a^{4} b B +\frac {10}{9} x^{9} a^{2} b^{3} A +\frac {10}{9} x^{9} a^{3} b^{2} B +\frac {5}{11} x^{11} a \,b^{4} A +\frac {10}{11} x^{11} a^{2} b^{3} B +\frac {1}{13} x^{13} b^{5} A +\frac {5}{13} x^{13} a \,b^{4} B +\frac {1}{15} b^{5} B \,x^{15}\)	\(125\)
parallelrisch	\(\frac {1}{3} a^{5} A \,x^{3}+x^{5} a^{4} b A +\frac {1}{5} x^{5} a^{5} B +\frac {10}{7} x^{7} a^{3} b^{2} A +\frac {5}{7} x^{7} a^{4} b B +\frac {10}{9} x^{9} a^{2} b^{3} A +\frac {10}{9} x^{9} a^{3} b^{2} B +\frac {5}{11} x^{11} a \,b^{4} A +\frac {10}{11} x^{11} a^{2} b^{3} B +\frac {1}{13} x^{13} b^{5} A +\frac {5}{13} x^{13} a \,b^{4} B +\frac {1}{15} b^{5} B \,x^{15}\)	\(125\)

[In]

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

[Out]

1/3*a^5*A*x^3+(a^4*b*A+1/5*a^5*B)*x^5+(10/7*a^3*b^2*A+5/7*a^4*b*B)*x^7+(10/9*a^2*b^3*A+10/9*a^3*b^2*B)*x^9+(5/
11*a*b^4*A+10/11*a^2*b^3*B)*x^11+(1/13*b^5*A+5/13*a*b^4*B)*x^13+1/15*b^5*B*x^15

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{15} \, B b^{5} x^{15} + \frac {1}{13} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{13} + \frac {5}{11} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{11} + \frac {10}{9} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + \frac {1}{3} \, A a^{5} x^{3} + \frac {5}{7} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{7} + \frac {1}{5} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{5} \]

[In]

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

[Out]

1/15*B*b^5*x^15 + 1/13*(5*B*a*b^4 + A*b^5)*x^13 + 5/11*(2*B*a^2*b^3 + A*a*b^4)*x^11 + 10/9*(B*a^3*b^2 + A*a^2*
b^3)*x^9 + 1/3*A*a^5*x^3 + 5/7*(B*a^4*b + 2*A*a^3*b^2)*x^7 + 1/5*(B*a^5 + 5*A*a^4*b)*x^5

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.15 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {A a^{5} x^{3}}{3} + \frac {B b^{5} x^{15}}{15} + x^{13} \left (\frac {A b^{5}}{13} + \frac {5 B a b^{4}}{13}\right ) + x^{11} \cdot \left (\frac {5 A a b^{4}}{11} + \frac {10 B a^{2} b^{3}}{11}\right ) + x^{9} \cdot \left (\frac {10 A a^{2} b^{3}}{9} + \frac {10 B a^{3} b^{2}}{9}\right ) + x^{7} \cdot \left (\frac {10 A a^{3} b^{2}}{7} + \frac {5 B a^{4} b}{7}\right ) + x^{5} \left (A a^{4} b + \frac {B a^{5}}{5}\right ) \]

[In]

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

[Out]

A*a**5*x**3/3 + B*b**5*x**15/15 + x**13*(A*b**5/13 + 5*B*a*b**4/13) + x**11*(5*A*a*b**4/11 + 10*B*a**2*b**3/11
) + x**9*(10*A*a**2*b**3/9 + 10*B*a**3*b**2/9) + x**7*(10*A*a**3*b**2/7 + 5*B*a**4*b/7) + x**5*(A*a**4*b + B*a
**5/5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{15} \, B b^{5} x^{15} + \frac {1}{13} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{13} + \frac {5}{11} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{11} + \frac {10}{9} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + \frac {1}{3} \, A a^{5} x^{3} + \frac {5}{7} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{7} + \frac {1}{5} \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{5} \]

[In]

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

[Out]

1/15*B*b^5*x^15 + 1/13*(5*B*a*b^4 + A*b^5)*x^13 + 5/11*(2*B*a^2*b^3 + A*a*b^4)*x^11 + 10/9*(B*a^3*b^2 + A*a^2*
b^3)*x^9 + 1/3*A*a^5*x^3 + 5/7*(B*a^4*b + 2*A*a^3*b^2)*x^7 + 1/5*(B*a^5 + 5*A*a^4*b)*x^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=\frac {1}{15} \, B b^{5} x^{15} + \frac {5}{13} \, B a b^{4} x^{13} + \frac {1}{13} \, A b^{5} x^{13} + \frac {10}{11} \, B a^{2} b^{3} x^{11} + \frac {5}{11} \, A a b^{4} x^{11} + \frac {10}{9} \, B a^{3} b^{2} x^{9} + \frac {10}{9} \, A a^{2} b^{3} x^{9} + \frac {5}{7} \, B a^{4} b x^{7} + \frac {10}{7} \, A a^{3} b^{2} x^{7} + \frac {1}{5} \, B a^{5} x^{5} + A a^{4} b x^{5} + \frac {1}{3} \, A a^{5} x^{3} \]

[In]

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

[Out]

1/15*B*b^5*x^15 + 5/13*B*a*b^4*x^13 + 1/13*A*b^5*x^13 + 10/11*B*a^2*b^3*x^11 + 5/11*A*a*b^4*x^11 + 10/9*B*a^3*
b^2*x^9 + 10/9*A*a^2*b^3*x^9 + 5/7*B*a^4*b*x^7 + 10/7*A*a^3*b^2*x^7 + 1/5*B*a^5*x^5 + A*a^4*b*x^5 + 1/3*A*a^5*
x^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^2\right )^5 \left (A+B x^2\right ) \, dx=x^5\,\left (\frac {B\,a^5}{5}+A\,b\,a^4\right )+x^{13}\,\left (\frac {A\,b^5}{13}+\frac {5\,B\,a\,b^4}{13}\right )+\frac {A\,a^5\,x^3}{3}+\frac {B\,b^5\,x^{15}}{15}+\frac {10\,a^2\,b^2\,x^9\,\left (A\,b+B\,a\right )}{9}+\frac {5\,a^3\,b\,x^7\,\left (2\,A\,b+B\,a\right )}{7}+\frac {5\,a\,b^3\,x^{11}\,\left (A\,b+2\,B\,a\right )}{11} \]

[In]

int(x^2*(A + B*x^2)*(a + b*x^2)^5,x)

[Out]

x^5*((B*a^5)/5 + A*a^4*b) + x^13*((A*b^5)/13 + (5*B*a*b^4)/13) + (A*a^5*x^3)/3 + (B*b^5*x^15)/15 + (10*a^2*b^2
*x^9*(A*b + B*a))/9 + (5*a^3*b*x^7*(2*A*b + B*a))/7 + (5*a*b^3*x^11*(A*b + 2*B*a))/11