\(\int (a+b x^2)^2 (A+B x^2) \, dx\) [13]

Optimal result

Integrand size = 17, antiderivative size = 50 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=a^2 A x+\frac {1}{3} a (2 A b+a B) x^3+\frac {1}{5} b (A b+2 a B) x^5+\frac {1}{7} b^2 B x^7 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {380} \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=a^2 A x+\frac {1}{5} b x^5 (2 a B+A b)+\frac {1}{3} a x^3 (a B+2 A b)+\frac {1}{7} b^2 B x^7 \]

[In]

Int[(a + b*x^2)^2*(A + B*x^2),x]

[Out]

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

Rule 380

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=a^2 A x+\frac {1}{3} a (2 A b+a B) x^3+\frac {1}{5} b (A b+2 a B) x^5+\frac {1}{7} b^2 B x^7 \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {b^{2} B \,x^{7}}{7}+\frac {\left (b^{2} A +2 a b B \right ) x^{5}}{5}+\frac {\left (2 a b A +a^{2} B \right ) x^{3}}{3}+a^{2} A x\)	\(49\)
norman	\(\frac {b^{2} B \,x^{7}}{7}+\left (\frac {1}{5} b^{2} A +\frac {2}{5} a b B \right ) x^{5}+\left (\frac {2}{3} a b A +\frac {1}{3} a^{2} B \right ) x^{3}+a^{2} A x\)	\(49\)
gosper	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{5} x^{5} b^{2} A +\frac {2}{5} x^{5} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +a^{2} A x\)	\(51\)
risch	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{5} x^{5} b^{2} A +\frac {2}{5} x^{5} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +a^{2} A x\)	\(51\)
parallelrisch	\(\frac {1}{7} b^{2} B \,x^{7}+\frac {1}{5} x^{5} b^{2} A +\frac {2}{5} x^{5} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +a^{2} A x\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {1}{5} \, {\left (2 \, B a b + A b^{2}\right )} x^{5} + A a^{2} x + \frac {1}{3} \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=A a^{2} x + \frac {B b^{2} x^{7}}{7} + x^{5} \left (\frac {A b^{2}}{5} + \frac {2 B a b}{5}\right ) + x^{3} \cdot \left (\frac {2 A a b}{3} + \frac {B a^{2}}{3}\right ) \]

[In]

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

[Out]

A*a**2*x + B*b**2*x**7/7 + x**5*(A*b**2/5 + 2*B*a*b/5) + x**3*(2*A*a*b/3 + B*a**2/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {1}{5} \, {\left (2 \, B a b + A b^{2}\right )} x^{5} + A a^{2} x + \frac {1}{3} \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=\frac {1}{7} \, B b^{2} x^{7} + \frac {2}{5} \, B a b x^{5} + \frac {1}{5} \, A b^{2} x^{5} + \frac {1}{3} \, B a^{2} x^{3} + \frac {2}{3} \, A a b x^{3} + A a^{2} x \]

[In]

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

[Out]

1/7*B*b^2*x^7 + 2/5*B*a*b*x^5 + 1/5*A*b^2*x^5 + 1/3*B*a^2*x^3 + 2/3*A*a*b*x^3 + A*a^2*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx=x^3\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+x^5\,\left (\frac {A\,b^2}{5}+\frac {2\,B\,a\,b}{5}\right )+\frac {B\,b^2\,x^7}{7}+A\,a^2\,x \]

[In]

int((A + B*x^2)*(a + b*x^2)^2,x)

[Out]

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