\(\int (a+b x^3)^2 (c+d x^3)^3 \, dx\) [8]

Optimal result

Integrand size = 19, antiderivative size = 122 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=a^2 c^3 x+\frac {1}{4} a c^2 (2 b c+3 a d) x^4+\frac {1}{7} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^7+\frac {1}{10} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{10}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{16} b^2 d^3 x^{16} \]

[Out]

a^2*c^3*x+1/4*a*c^2*(3*a*d+2*b*c)*x^4+1/7*c*(3*a^2*d^2+6*a*b*c*d+b^2*c^2)*x^7+1/10*d*(a^2*d^2+6*a*b*c*d+3*b^2*
c^2)*x^10+1/13*b*d^2*(2*a*d+3*b*c)*x^13+1/16*b^2*d^3*x^16

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} \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=\frac {1}{10} d x^{10} \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {1}{7} c x^7 \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+a^2 c^3 x+\frac {1}{4} a c^2 x^4 (3 a d+2 b c)+\frac {1}{13} b d^2 x^{13} (2 a d+3 b c)+\frac {1}{16} b^2 d^3 x^{16} \]

[In]

Int[(a + b*x^3)^2*(c + d*x^3)^3,x]

[Out]

a^2*c^3*x + (a*c^2*(2*b*c + 3*a*d)*x^4)/4 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^7)/7 + (d*(3*b^2*c^2 + 6*a*
b*c*d + a^2*d^2)*x^10)/10 + (b*d^2*(3*b*c + 2*a*d)*x^13)/13 + (b^2*d^3*x^16)/16

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^3+c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^6+d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^9+b d^2 (3 b c+2 a d) x^{12}+b^2 d^3 x^{15}\right ) \, dx \\ & = a^2 c^3 x+\frac {1}{4} a c^2 (2 b c+3 a d) x^4+\frac {1}{7} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^7+\frac {1}{10} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{10}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{16} b^2 d^3 x^{16} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=a^2 c^3 x+\frac {1}{4} a c^2 (2 b c+3 a d) x^4+\frac {1}{7} c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^7+\frac {1}{10} d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^{10}+\frac {1}{13} b d^2 (3 b c+2 a d) x^{13}+\frac {1}{16} b^2 d^3 x^{16} \]

[In]

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

[Out]

a^2*c^3*x + (a*c^2*(2*b*c + 3*a*d)*x^4)/4 + (c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^7)/7 + (d*(3*b^2*c^2 + 6*a*
b*c*d + a^2*d^2)*x^10)/10 + (b*d^2*(3*b*c + 2*a*d)*x^13)/13 + (b^2*d^3*x^16)/16

Maple [A] (verified)

Time = 3.92 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.01

method	result	size
norman	\(a^{2} c^{3} x +\left (\frac {3}{4} a^{2} c^{2} d +\frac {1}{2} a b \,c^{3}\right ) x^{4}+\left (\frac {3}{7} c \,a^{2} d^{2}+\frac {6}{7} a b \,c^{2} d +\frac {1}{7} b^{2} c^{3}\right ) x^{7}+\left (\frac {1}{10} a^{2} d^{3}+\frac {3}{5} a b c \,d^{2}+\frac {3}{10} b^{2} c^{2} d \right ) x^{10}+\left (\frac {2}{13} a b \,d^{3}+\frac {3}{13} b^{2} c \,d^{2}\right ) x^{13}+\frac {b^{2} d^{3} x^{16}}{16}\)	\(123\)
default	\(\frac {b^{2} d^{3} x^{16}}{16}+\frac {\left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{13}}{13}+\frac {\left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{10}}{10}+\frac {\left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{7}}{7}+\frac {\left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{4}}{4}+a^{2} c^{3} x\)	\(125\)
gosper	\(a^{2} c^{3} x +\frac {3}{4} x^{4} a^{2} c^{2} d +\frac {1}{2} x^{4} a b \,c^{3}+\frac {3}{7} x^{7} c \,a^{2} d^{2}+\frac {6}{7} x^{7} a b \,c^{2} d +\frac {1}{7} x^{7} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{16} b^{2} d^{3} x^{16}\)	\(133\)
risch	\(a^{2} c^{3} x +\frac {3}{4} x^{4} a^{2} c^{2} d +\frac {1}{2} x^{4} a b \,c^{3}+\frac {3}{7} x^{7} c \,a^{2} d^{2}+\frac {6}{7} x^{7} a b \,c^{2} d +\frac {1}{7} x^{7} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{16} b^{2} d^{3} x^{16}\)	\(133\)
parallelrisch	\(a^{2} c^{3} x +\frac {3}{4} x^{4} a^{2} c^{2} d +\frac {1}{2} x^{4} a b \,c^{3}+\frac {3}{7} x^{7} c \,a^{2} d^{2}+\frac {6}{7} x^{7} a b \,c^{2} d +\frac {1}{7} x^{7} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {2}{13} x^{13} a b \,d^{3}+\frac {3}{13} x^{13} b^{2} c \,d^{2}+\frac {1}{16} b^{2} d^{3} x^{16}\)	\(133\)

[In]

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

[Out]

a^2*c^3*x+(3/4*a^2*c^2*d+1/2*a*b*c^3)*x^4+(3/7*c*a^2*d^2+6/7*a*b*c^2*d+1/7*b^2*c^3)*x^7+(1/10*a^2*d^3+3/5*a*b*
c*d^2+3/10*b^2*c^2*d)*x^10+(2/13*a*b*d^3+3/13*b^2*c*d^2)*x^13+1/16*b^2*d^3*x^16

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=\frac {1}{16} \, b^{2} d^{3} x^{16} + \frac {1}{13} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{13} + \frac {1}{10} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{10} + \frac {1}{7} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{7} + a^{2} c^{3} x + \frac {1}{4} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{4} \]

[In]

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

[Out]

1/16*b^2*d^3*x^16 + 1/13*(3*b^2*c*d^2 + 2*a*b*d^3)*x^13 + 1/10*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^10 + 1/
7*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^7 + a^2*c^3*x + 1/4*(2*a*b*c^3 + 3*a^2*c^2*d)*x^4

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.14 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=a^{2} c^{3} x + \frac {b^{2} d^{3} x^{16}}{16} + x^{13} \cdot \left (\frac {2 a b d^{3}}{13} + \frac {3 b^{2} c d^{2}}{13}\right ) + x^{10} \left (\frac {a^{2} d^{3}}{10} + \frac {3 a b c d^{2}}{5} + \frac {3 b^{2} c^{2} d}{10}\right ) + x^{7} \cdot \left (\frac {3 a^{2} c d^{2}}{7} + \frac {6 a b c^{2} d}{7} + \frac {b^{2} c^{3}}{7}\right ) + x^{4} \cdot \left (\frac {3 a^{2} c^{2} d}{4} + \frac {a b c^{3}}{2}\right ) \]

[In]

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

[Out]

a**2*c**3*x + b**2*d**3*x**16/16 + x**13*(2*a*b*d**3/13 + 3*b**2*c*d**2/13) + x**10*(a**2*d**3/10 + 3*a*b*c*d*
*2/5 + 3*b**2*c**2*d/10) + x**7*(3*a**2*c*d**2/7 + 6*a*b*c**2*d/7 + b**2*c**3/7) + x**4*(3*a**2*c**2*d/4 + a*b
*c**3/2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=\frac {1}{16} \, b^{2} d^{3} x^{16} + \frac {1}{13} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{13} + \frac {1}{10} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{10} + \frac {1}{7} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{7} + a^{2} c^{3} x + \frac {1}{4} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{4} \]

[In]

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

[Out]

1/16*b^2*d^3*x^16 + 1/13*(3*b^2*c*d^2 + 2*a*b*d^3)*x^13 + 1/10*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^10 + 1/
7*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^7 + a^2*c^3*x + 1/4*(2*a*b*c^3 + 3*a^2*c^2*d)*x^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=\frac {1}{16} \, b^{2} d^{3} x^{16} + \frac {3}{13} \, b^{2} c d^{2} x^{13} + \frac {2}{13} \, a b d^{3} x^{13} + \frac {3}{10} \, b^{2} c^{2} d x^{10} + \frac {3}{5} \, a b c d^{2} x^{10} + \frac {1}{10} \, a^{2} d^{3} x^{10} + \frac {1}{7} \, b^{2} c^{3} x^{7} + \frac {6}{7} \, a b c^{2} d x^{7} + \frac {3}{7} \, a^{2} c d^{2} x^{7} + \frac {1}{2} \, a b c^{3} x^{4} + \frac {3}{4} \, a^{2} c^{2} d x^{4} + a^{2} c^{3} x \]

[In]

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

[Out]

1/16*b^2*d^3*x^16 + 3/13*b^2*c*d^2*x^13 + 2/13*a*b*d^3*x^13 + 3/10*b^2*c^2*d*x^10 + 3/5*a*b*c*d^2*x^10 + 1/10*
a^2*d^3*x^10 + 1/7*b^2*c^3*x^7 + 6/7*a*b*c^2*d*x^7 + 3/7*a^2*c*d^2*x^7 + 1/2*a*b*c^3*x^4 + 3/4*a^2*c^2*d*x^4 +
a^2*c^3*x

Mupad [B] (verification not implemented)

Time = 5.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.95 \[ \int \left (a+b x^3\right )^2 \left (c+d x^3\right )^3 \, dx=x^7\,\left (\frac {3\,a^2\,c\,d^2}{7}+\frac {6\,a\,b\,c^2\,d}{7}+\frac {b^2\,c^3}{7}\right )+x^{10}\,\left (\frac {a^2\,d^3}{10}+\frac {3\,a\,b\,c\,d^2}{5}+\frac {3\,b^2\,c^2\,d}{10}\right )+a^2\,c^3\,x+\frac {b^2\,d^3\,x^{16}}{16}+\frac {a\,c^2\,x^4\,\left (3\,a\,d+2\,b\,c\right )}{4}+\frac {b\,d^2\,x^{13}\,\left (2\,a\,d+3\,b\,c\right )}{13} \]

[In]

int((a + b*x^3)^2*(c + d*x^3)^3,x)

[Out]

x^7*((b^2*c^3)/7 + (3*a^2*c*d^2)/7 + (6*a*b*c^2*d)/7) + x^10*((a^2*d^3)/10 + (3*b^2*c^2*d)/10 + (3*a*b*c*d^2)/
5) + a^2*c^3*x + (b^2*d^3*x^16)/16 + (a*c^2*x^4*(3*a*d + 2*b*c))/4 + (b*d^2*x^13*(2*a*d + 3*b*c))/13