\(\int (a+b x^4)^2 (c+d x^4)^3 \, dx\) [154]

Optimal result

Integrand size = 19, antiderivative size = 122 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=a^2 c^3 x+\frac {1}{5} a c^2 (2 b c+3 a d) x^5+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{13} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{13}+\frac {1}{17} b d^2 (3 b c+2 a d) x^{17}+\frac {1}{21} b^2 d^3 x^{21} \]

[Out]

a^2*c^3*x+1/5*a*c^2*(3*a*d+2*b*c)*x^5+1/9*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^9+1/13*d*(a^2*d^2+6*a*b*c*d+3*b^2*
c^2)*x^13+1/17*b*d^2*(2*a*d+3*b*c)*x^17+1/21*b^2*d^3*x^21

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {380} \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=\frac {1}{13} d x^{13} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {1}{9} c x^9 \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+a^2 c^3 x+\frac {1}{5} a c^2 x^5 (3 a d+2 b c)+\frac {1}{17} b d^2 x^{17} (2 a d+3 b c)+\frac {1}{21} b^2 d^3 x^{21} \]

[In]

Int[(a + b*x^4)^2*(c + d*x^4)^3,x]

[Out]

a^2*c^3*x + (a*c^2*(2*b*c + 3*a*d)*x^5)/5 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^9)/9 + (d*(3*b^2*c^2 + 6*a*
b*c*d + a^2*d^2)*x^13)/13 + (b*d^2*(3*b*c + 2*a*d)*x^17)/17 + (b^2*d^3*x^21)/21

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^3+a c^2 (2 b c+3 a d) x^4+c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^8+d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{12}+b d^2 (3 b c+2 a d) x^{16}+b^2 d^3 x^{20}\right ) \, dx \\ & = a^2 c^3 x+\frac {1}{5} a c^2 (2 b c+3 a d) x^5+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{13} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{13}+\frac {1}{17} b d^2 (3 b c+2 a d) x^{17}+\frac {1}{21} b^2 d^3 x^{21} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=a^2 c^3 x+\frac {1}{5} a c^2 (2 b c+3 a d) x^5+\frac {1}{9} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^9+\frac {1}{13} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{13}+\frac {1}{17} b d^2 (3 b c+2 a d) x^{17}+\frac {1}{21} b^2 d^3 x^{21} \]

[In]

Integrate[(a + b*x^4)^2*(c + d*x^4)^3,x]

[Out]

a^2*c^3*x + (a*c^2*(2*b*c + 3*a*d)*x^5)/5 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^9)/9 + (d*(3*b^2*c^2 + 6*a*
b*c*d + a^2*d^2)*x^13)/13 + (b*d^2*(3*b*c + 2*a*d)*x^17)/17 + (b^2*d^3*x^21)/21

Maple [A] (verified)

Time = 3.88 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.01

method	result	size
norman	\(a^{2} c^{3} x +\left (\frac {3}{5} a^{2} c^{2} d +\frac {2}{5} a b \,c^{3}\right ) x^{5}+\left (\frac {1}{3} c \,a^{2} d^{2}+\frac {2}{3} a b \,c^{2} d +\frac {1}{9} b^{2} c^{3}\right ) x^{9}+\left (\frac {1}{13} a^{2} d^{3}+\frac {6}{13} a b c \,d^{2}+\frac {3}{13} b^{2} c^{2} d \right ) x^{13}+\left (\frac {2}{17} a b \,d^{3}+\frac {3}{17} b^{2} c \,d^{2}\right ) x^{17}+\frac {b^{2} d^{3} x^{21}}{21}\)	\(123\)
default	\(\frac {b^{2} d^{3} x^{21}}{21}+\frac {\left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{17}}{17}+\frac {\left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{13}}{13}+\frac {\left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{9}}{9}+\frac {\left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{5}}{5}+a^{2} c^{3} x\)	\(125\)
gosper	\(a^{2} c^{3} x +\frac {3}{5} x^{5} a^{2} c^{2} d +\frac {2}{5} x^{5} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{13} x^{13} a^{2} d^{3}+\frac {6}{13} x^{13} a b c \,d^{2}+\frac {3}{13} x^{13} b^{2} c^{2} d +\frac {2}{17} x^{17} a b \,d^{3}+\frac {3}{17} x^{17} b^{2} c \,d^{2}+\frac {1}{21} b^{2} d^{3} x^{21}\)	\(133\)
risch	\(a^{2} c^{3} x +\frac {3}{5} x^{5} a^{2} c^{2} d +\frac {2}{5} x^{5} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{13} x^{13} a^{2} d^{3}+\frac {6}{13} x^{13} a b c \,d^{2}+\frac {3}{13} x^{13} b^{2} c^{2} d +\frac {2}{17} x^{17} a b \,d^{3}+\frac {3}{17} x^{17} b^{2} c \,d^{2}+\frac {1}{21} b^{2} d^{3} x^{21}\)	\(133\)
parallelrisch	\(a^{2} c^{3} x +\frac {3}{5} x^{5} a^{2} c^{2} d +\frac {2}{5} x^{5} a b \,c^{3}+\frac {1}{3} x^{9} c \,a^{2} d^{2}+\frac {2}{3} x^{9} a b \,c^{2} d +\frac {1}{9} x^{9} b^{2} c^{3}+\frac {1}{13} x^{13} a^{2} d^{3}+\frac {6}{13} x^{13} a b c \,d^{2}+\frac {3}{13} x^{13} b^{2} c^{2} d +\frac {2}{17} x^{17} a b \,d^{3}+\frac {3}{17} x^{17} b^{2} c \,d^{2}+\frac {1}{21} b^{2} d^{3} x^{21}\)	\(133\)

[In]

int((b*x^4+a)^2*(d*x^4+c)^3,x,method=_RETURNVERBOSE)

[Out]

a^2*c^3*x+(3/5*a^2*c^2*d+2/5*a*b*c^3)*x^5+(1/3*c*a^2*d^2+2/3*a*b*c^2*d+1/9*b^2*c^3)*x^9+(1/13*a^2*d^3+6/13*a*b
*c*d^2+3/13*b^2*c^2*d)*x^13+(2/17*a*b*d^3+3/17*b^2*c*d^2)*x^17+1/21*b^2*d^3*x^21

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=\frac {1}{21} \, b^{2} d^{3} x^{21} + \frac {1}{17} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{17} + \frac {1}{13} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{13} + \frac {1}{9} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{9} + a^{2} c^{3} x + \frac {1}{5} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{5} \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^3,x, algorithm="fricas")

[Out]

1/21*b^2*d^3*x^21 + 1/17*(3*b^2*c*d^2 + 2*a*b*d^3)*x^17 + 1/13*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^13 + 1/
9*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^9 + a^2*c^3*x + 1/5*(2*a*b*c^3 + 3*a^2*c^2*d)*x^5

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=a^{2} c^{3} x + \frac {b^{2} d^{3} x^{21}}{21} + x^{17} \cdot \left (\frac {2 a b d^{3}}{17} + \frac {3 b^{2} c d^{2}}{17}\right ) + x^{13} \left (\frac {a^{2} d^{3}}{13} + \frac {6 a b c d^{2}}{13} + \frac {3 b^{2} c^{2} d}{13}\right ) + x^{9} \left (\frac {a^{2} c d^{2}}{3} + \frac {2 a b c^{2} d}{3} + \frac {b^{2} c^{3}}{9}\right ) + x^{5} \cdot \left (\frac {3 a^{2} c^{2} d}{5} + \frac {2 a b c^{3}}{5}\right ) \]

[In]

integrate((b*x**4+a)**2*(d*x**4+c)**3,x)

[Out]

a**2*c**3*x + b**2*d**3*x**21/21 + x**17*(2*a*b*d**3/17 + 3*b**2*c*d**2/17) + x**13*(a**2*d**3/13 + 6*a*b*c*d*
*2/13 + 3*b**2*c**2*d/13) + x**9*(a**2*c*d**2/3 + 2*a*b*c**2*d/3 + b**2*c**3/9) + x**5*(3*a**2*c**2*d/5 + 2*a*
b*c**3/5)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=\frac {1}{21} \, b^{2} d^{3} x^{21} + \frac {1}{17} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{17} + \frac {1}{13} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{13} + \frac {1}{9} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{9} + a^{2} c^{3} x + \frac {1}{5} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{5} \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^3,x, algorithm="maxima")

[Out]

1/21*b^2*d^3*x^21 + 1/17*(3*b^2*c*d^2 + 2*a*b*d^3)*x^17 + 1/13*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^13 + 1/
9*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^9 + a^2*c^3*x + 1/5*(2*a*b*c^3 + 3*a^2*c^2*d)*x^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=\frac {1}{21} \, b^{2} d^{3} x^{21} + \frac {3}{17} \, b^{2} c d^{2} x^{17} + \frac {2}{17} \, a b d^{3} x^{17} + \frac {3}{13} \, b^{2} c^{2} d x^{13} + \frac {6}{13} \, a b c d^{2} x^{13} + \frac {1}{13} \, a^{2} d^{3} x^{13} + \frac {1}{9} \, b^{2} c^{3} x^{9} + \frac {2}{3} \, a b c^{2} d x^{9} + \frac {1}{3} \, a^{2} c d^{2} x^{9} + \frac {2}{5} \, a b c^{3} x^{5} + \frac {3}{5} \, a^{2} c^{2} d x^{5} + a^{2} c^{3} x \]

[In]

integrate((b*x^4+a)^2*(d*x^4+c)^3,x, algorithm="giac")

[Out]

1/21*b^2*d^3*x^21 + 3/17*b^2*c*d^2*x^17 + 2/17*a*b*d^3*x^17 + 3/13*b^2*c^2*d*x^13 + 6/13*a*b*c*d^2*x^13 + 1/13
*a^2*d^3*x^13 + 1/9*b^2*c^3*x^9 + 2/3*a*b*c^2*d*x^9 + 1/3*a^2*c*d^2*x^9 + 2/5*a*b*c^3*x^5 + 3/5*a^2*c^2*d*x^5
+ a^2*c^3*x

Mupad [B] (verification not implemented)

Time = 5.59 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.95 \[ \int \left (a+b x^4\right )^2 \left (c+d x^4\right )^3 \, dx=x^9\,\left (\frac {a^2\,c\,d^2}{3}+\frac {2\,a\,b\,c^2\,d}{3}+\frac {b^2\,c^3}{9}\right )+x^{13}\,\left (\frac {a^2\,d^3}{13}+\frac {6\,a\,b\,c\,d^2}{13}+\frac {3\,b^2\,c^2\,d}{13}\right )+a^2\,c^3\,x+\frac {b^2\,d^3\,x^{21}}{21}+\frac {a\,c^2\,x^5\,\left (3\,a\,d+2\,b\,c\right )}{5}+\frac {b\,d^2\,x^{17}\,\left (2\,a\,d+3\,b\,c\right )}{17} \]

[In]

int((a + b*x^4)^2*(c + d*x^4)^3,x)

[Out]

x^9*((b^2*c^3)/9 + (a^2*c*d^2)/3 + (2*a*b*c^2*d)/3) + x^13*((a^2*d^3)/13 + (3*b^2*c^2*d)/13 + (6*a*b*c*d^2)/13
) + a^2*c^3*x + (b^2*d^3*x^21)/21 + (a*c^2*x^5*(3*a*d + 2*b*c))/5 + (b*d^2*x^17*(2*a*d + 3*b*c))/17