\(\int (a+b x^4)^2 (c+d x^4) \, dx\) [156]

Optimal result

Integrand size = 17, antiderivative size = 50 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=a^2 c x+\frac {1}{5} a (2 b c+a d) x^5+\frac {1}{9} b (b c+2 a d) x^9+\frac {1}{13} b^2 d x^{13} \]

[Out]

a^2*c*x+1/5*a*(a*d+2*b*c)*x^5+1/9*b*(2*a*d+b*c)*x^9+1/13*b^2*d*x^13

Rubi [A] (verified)

Time = 0.03 (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} \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=a^2 c x+\frac {1}{9} b x^9 (2 a d+b c)+\frac {1}{5} a x^5 (a d+2 b c)+\frac {1}{13} b^2 d x^{13} \]

[In]

Int[(a + b*x^4)^2*(c + d*x^4),x]

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^5)/5 + (b*(b*c + 2*a*d)*x^9)/9 + (b^2*d*x^13)/13

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^4+b (b c+2 a d) x^8+b^2 d x^{12}\right ) \, dx \\ & = a^2 c x+\frac {1}{5} a (2 b c+a d) x^5+\frac {1}{9} b (b c+2 a d) x^9+\frac {1}{13} b^2 d x^{13} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=a^2 c x+\frac {1}{5} a (2 b c+a d) x^5+\frac {1}{9} b (b c+2 a d) x^9+\frac {1}{13} b^2 d x^{13} \]

[In]

Integrate[(a + b*x^4)^2*(c + d*x^4),x]

[Out]

a^2*c*x + (a*(2*b*c + a*d)*x^5)/5 + (b*(b*c + 2*a*d)*x^9)/9 + (b^2*d*x^13)/13

Maple [A] (verified)

Time = 3.95 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {b^{2} d \,x^{13}}{13}+\frac {\left (2 a b d +b^{2} c \right ) x^{9}}{9}+\frac {\left (a^{2} d +2 a b c \right ) x^{5}}{5}+a^{2} c x\)	\(49\)
norman	\(\frac {b^{2} d \,x^{13}}{13}+\left (\frac {2}{9} a b d +\frac {1}{9} b^{2} c \right ) x^{9}+\left (\frac {1}{5} a^{2} d +\frac {2}{5} a b c \right ) x^{5}+a^{2} c x\)	\(49\)
gosper	\(\frac {1}{13} b^{2} d \,x^{13}+\frac {2}{9} x^{9} a b d +\frac {1}{9} x^{9} b^{2} c +\frac {1}{5} x^{5} a^{2} d +\frac {2}{5} x^{5} a b c +a^{2} c x\)	\(51\)
risch	\(\frac {1}{13} b^{2} d \,x^{13}+\frac {2}{9} x^{9} a b d +\frac {1}{9} x^{9} b^{2} c +\frac {1}{5} x^{5} a^{2} d +\frac {2}{5} x^{5} a b c +a^{2} c x\)	\(51\)
parallelrisch	\(\frac {1}{13} b^{2} d \,x^{13}+\frac {2}{9} x^{9} a b d +\frac {1}{9} x^{9} b^{2} c +\frac {1}{5} x^{5} a^{2} d +\frac {2}{5} x^{5} a b c +a^{2} c x\)	\(51\)

[In]

int((b*x^4+a)^2*(d*x^4+c),x,method=_RETURNVERBOSE)

[Out]

1/13*b^2*d*x^13+1/9*(2*a*b*d+b^2*c)*x^9+1/5*(a^2*d+2*a*b*c)*x^5+a^2*c*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=\frac {1}{13} \, b^{2} d x^{13} + \frac {1}{9} \, {\left (b^{2} c + 2 \, a b d\right )} x^{9} + \frac {1}{5} \, {\left (2 \, a b c + a^{2} d\right )} x^{5} + a^{2} c x \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c),x, algorithm="fricas")

[Out]

1/13*b^2*d*x^13 + 1/9*(b^2*c + 2*a*b*d)*x^9 + 1/5*(2*a*b*c + a^2*d)*x^5 + a^2*c*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=a^{2} c x + \frac {b^{2} d x^{13}}{13} + x^{9} \cdot \left (\frac {2 a b d}{9} + \frac {b^{2} c}{9}\right ) + x^{5} \left (\frac {a^{2} d}{5} + \frac {2 a b c}{5}\right ) \]

[In]

integrate((b*x**4+a)**2*(d*x**4+c),x)

[Out]

a**2*c*x + b**2*d*x**13/13 + x**9*(2*a*b*d/9 + b**2*c/9) + x**5*(a**2*d/5 + 2*a*b*c/5)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=\frac {1}{13} \, b^{2} d x^{13} + \frac {1}{9} \, {\left (b^{2} c + 2 \, a b d\right )} x^{9} + \frac {1}{5} \, {\left (2 \, a b c + a^{2} d\right )} x^{5} + a^{2} c x \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c),x, algorithm="maxima")

[Out]

1/13*b^2*d*x^13 + 1/9*(b^2*c + 2*a*b*d)*x^9 + 1/5*(2*a*b*c + a^2*d)*x^5 + a^2*c*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=\frac {1}{13} \, b^{2} d x^{13} + \frac {1}{9} \, b^{2} c x^{9} + \frac {2}{9} \, a b d x^{9} + \frac {2}{5} \, a b c x^{5} + \frac {1}{5} \, a^{2} d x^{5} + a^{2} c x \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c),x, algorithm="giac")

[Out]

1/13*b^2*d*x^13 + 1/9*b^2*c*x^9 + 2/9*a*b*d*x^9 + 2/5*a*b*c*x^5 + 1/5*a^2*d*x^5 + a^2*c*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right ) \, dx=x^5\,\left (\frac {d\,a^2}{5}+\frac {2\,b\,c\,a}{5}\right )+x^9\,\left (\frac {c\,b^2}{9}+\frac {2\,a\,d\,b}{9}\right )+\frac {b^2\,d\,x^{13}}{13}+a^2\,c\,x \]

[In]

int((a + b*x^4)^2*(c + d*x^4),x)

[Out]

x^5*((a^2*d)/5 + (2*a*b*c)/5) + x^9*((b^2*c)/9 + (2*a*b*d)/9) + (b^2*d*x^13)/13 + a^2*c*x