\(\int (a+b x^2) (A+B x^2) \, dx\) [3]

Optimal result

Integrand size = 15, antiderivative size = 28 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=a A x+\frac {1}{3} (A b+a B) x^3+\frac {1}{5} b B x^5 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {380} \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{3} x^3 (a B+A b)+a A x+\frac {1}{5} b B x^5 \]

[In]

Int[(a + b*x^2)*(A + B*x^2),x]

[Out]

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

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 A+(A b+a B) x^2+b B x^4\right ) \, dx \\ & = a A x+\frac {1}{3} (A b+a B) x^3+\frac {1}{5} b B x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=a A x+\frac {1}{3} (A b+a B) x^3+\frac {1}{5} b B x^5 \]

[In]

Integrate[(a + b*x^2)*(A + B*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
default	\(a A x +\frac {\left (A b +B a \right ) x^{3}}{3}+\frac {b B \,x^{5}}{5}\)	\(25\)
norman	\(\frac {b B \,x^{5}}{5}+\left (\frac {A b}{3}+\frac {B a}{3}\right ) x^{3}+a A x\)	\(26\)
gosper	\(\frac {1}{5} b B \,x^{5}+\frac {1}{3} x^{3} A b +\frac {1}{3} x^{3} B a +a A x\)	\(27\)
risch	\(\frac {1}{5} b B \,x^{5}+\frac {1}{3} x^{3} A b +\frac {1}{3} x^{3} B a +a A x\)	\(27\)
parallelrisch	\(\frac {1}{5} b B \,x^{5}+\frac {1}{3} x^{3} A b +\frac {1}{3} x^{3} B a +a A x\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, {\left (B a + A b\right )} x^{3} + A a x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=A a x + \frac {B b x^{5}}{5} + x^{3} \left (\frac {A b}{3} + \frac {B a}{3}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, {\left (B a + A b\right )} x^{3} + A a x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, B a x^{3} + \frac {1}{3} \, A b x^{3} + A a x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {B\,b\,x^5}{5}+\left (\frac {A\,b}{3}+\frac {B\,a}{3}\right )\,x^3+A\,a\,x \]

[In]

int((A + B*x^2)*(a + b*x^2),x)

[Out]

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