\(\int (a+b x^2)^2 (c+d x^2) \, dx\) [145]

Optimal result

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

[Out]

a^2*c*x+1/3*a*(a*d+2*b*c)*x^3+1/5*b*(2*a*d+b*c)*x^5+1/7*b^2*d*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} \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=a^2 c x+\frac {1}{5} b x^5 (2 a d+b c)+\frac {1}{3} a x^3 (a d+2 b c)+\frac {1}{7} b^2 d x^7 \]

[In]

Int[(a + b*x^2)^2*(c + d*x^2),x]

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^3)/3 + (b*(b*c + 2*a*d)*x^5)/5 + (b^2*d*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 c+a (2 b c+a d) x^2+b (b c+2 a d) x^4+b^2 d x^6\right ) \, dx \\ & = a^2 c x+\frac {1}{3} a (2 b c+a d) x^3+\frac {1}{5} b (b c+2 a d) x^5+\frac {1}{7} b^2 d x^7 \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[(a + b*x^2)^2*(c + d*x^2),x]

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^3)/3 + (b*(b*c + 2*a*d)*x^5)/5 + (b^2*d*x^7)/7

Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98

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

[In]

int((b*x^2+a)^2*(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

1/7*b^2*d*x^7+1/5*(2*a*b*d+b^2*c)*x^5+1/3*(a^2*d+2*a*b*c)*x^3+a^2*c*x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{7} \, b^{2} d x^{7} + \frac {1}{5} \, {\left (b^{2} c + 2 \, a b d\right )} x^{5} + a^{2} c x + \frac {1}{3} \, {\left (2 \, a b c + a^{2} d\right )} x^{3} \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c),x, algorithm="fricas")

[Out]

1/7*b^2*d*x^7 + 1/5*(b^2*c + 2*a*b*d)*x^5 + a^2*c*x + 1/3*(2*a*b*c + a^2*d)*x^3

Sympy [A] (verification not implemented)

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

[In]

integrate((b*x**2+a)**2*(d*x**2+c),x)

[Out]

a**2*c*x + b**2*d*x**7/7 + x**5*(2*a*b*d/5 + b**2*c/5) + x**3*(a**2*d/3 + 2*a*b*c/3)

Maxima [A] (verification not implemented)

none

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

[In]

integrate((b*x^2+a)^2*(d*x^2+c),x, algorithm="maxima")

[Out]

1/7*b^2*d*x^7 + 1/5*(b^2*c + 2*a*b*d)*x^5 + a^2*c*x + 1/3*(2*a*b*c + a^2*d)*x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=\frac {1}{7} \, b^{2} d x^{7} + \frac {1}{5} \, b^{2} c x^{5} + \frac {2}{5} \, a b d x^{5} + \frac {2}{3} \, a b c x^{3} + \frac {1}{3} \, a^{2} d x^{3} + a^{2} c x \]

[In]

integrate((b*x^2+a)^2*(d*x^2+c),x, algorithm="giac")

[Out]

1/7*b^2*d*x^7 + 1/5*b^2*c*x^5 + 2/5*a*b*d*x^5 + 2/3*a*b*c*x^3 + 1/3*a^2*d*x^3 + a^2*c*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right ) \, dx=x^3\,\left (\frac {d\,a^2}{3}+\frac {2\,b\,c\,a}{3}\right )+x^5\,\left (\frac {c\,b^2}{5}+\frac {2\,a\,d\,b}{5}\right )+\frac {b^2\,d\,x^7}{7}+a^2\,c\,x \]

[In]

int((a + b*x^2)^2*(c + d*x^2),x)

[Out]

x^3*((a^2*d)/3 + (2*a*b*c)/3) + x^5*((b^2*c)/5 + (2*a*b*d)/5) + (b^2*d*x^7)/7 + a^2*c*x