3.14 \(\int (a+b x^2)^3 (c+d x^2)^3 \, dx\)

Optimal. Leaf size=154 \[ \frac{1}{3} b d x^9 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+\frac{1}{7} x^7 (a d+b c) \left (a^2 d^2+8 a b c d+b^2 c^2\right )+\frac{3}{5} a c x^5 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a^2 c^2 x^3 (a d+b c)+a^3 c^3 x+\frac{3}{11} b^2 d^2 x^{11} (a d+b c)+\frac{1}{13} b^3 d^3 x^{13} \]

[Out]

a^3*c^3*x + a^2*c^2*(b*c + a*d)*x^3 + (3*a*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^5)/5 + ((b*c + a*d)*(b^2*c^2 +
8*a*b*c*d + a^2*d^2)*x^7)/7 + (b*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^9)/3 + (3*b^2*d^2*(b*c + a*d)*x^11)/11 +
(b^3*d^3*x^13)/13

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Rubi [A]  time = 0.103268, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {373} \[ \frac{1}{3} b d x^9 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+\frac{1}{7} x^7 (a d+b c) \left (a^2 d^2+8 a b c d+b^2 c^2\right )+\frac{3}{5} a c x^5 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a^2 c^2 x^3 (a d+b c)+a^3 c^3 x+\frac{3}{11} b^2 d^2 x^{11} (a d+b c)+\frac{1}{13} b^3 d^3 x^{13} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3*(c + d*x^2)^3,x]

[Out]

a^3*c^3*x + a^2*c^2*(b*c + a*d)*x^3 + (3*a*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^5)/5 + ((b*c + a*d)*(b^2*c^2 +
8*a*b*c*d + a^2*d^2)*x^7)/7 + (b*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^9)/3 + (3*b^2*d^2*(b*c + a*d)*x^11)/11 +
(b^3*d^3*x^13)/13

Rule 373

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*} \int \left (a+b x^2\right )^3 \left (c+d x^2\right )^3 \, dx &=\int \left (a^3 c^3+3 a^2 c^2 (b c+a d) x^2+3 a c \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^4+(b c+a d) \left (b^2 c^2+8 a b c d+a^2 d^2\right ) x^6+3 b d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^8+3 b^2 d^2 (b c+a d) x^{10}+b^3 d^3 x^{12}\right ) \, dx\\ &=a^3 c^3 x+a^2 c^2 (b c+a d) x^3+\frac{3}{5} a c \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^5+\frac{1}{7} (b c+a d) \left (b^2 c^2+8 a b c d+a^2 d^2\right ) x^7+\frac{1}{3} b d \left (b^2 c^2+3 a b c d+a^2 d^2\right ) x^9+\frac{3}{11} b^2 d^2 (b c+a d) x^{11}+\frac{1}{13} b^3 d^3 x^{13}\\ \end{align*}

Mathematica [A]  time = 0.0284419, size = 161, normalized size = 1.05 \[ \frac{1}{3} b d x^9 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+\frac{1}{7} x^7 \left (9 a^2 b c d^2+a^3 d^3+9 a b^2 c^2 d+b^3 c^3\right )+\frac{3}{5} a c x^5 \left (a^2 d^2+3 a b c d+b^2 c^2\right )+a^2 c^2 x^3 (a d+b c)+a^3 c^3 x+\frac{3}{11} b^2 d^2 x^{11} (a d+b c)+\frac{1}{13} b^3 d^3 x^{13} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*c^3*x + a^2*c^2*(b*c + a*d)*x^3 + (3*a*c*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^5)/5 + ((b^3*c^3 + 9*a*b^2*c^2*
d + 9*a^2*b*c*d^2 + a^3*d^3)*x^7)/7 + (b*d*(b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^9)/3 + (3*b^2*d^2*(b*c + a*d)*x^1
1)/11 + (b^3*d^3*x^13)/13

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Maple [A]  time = 0.001, size = 177, normalized size = 1.2 \begin{align*}{\frac{{b}^{3}{d}^{3}{x}^{13}}{13}}+{\frac{ \left ( 3\,a{b}^{2}{d}^{3}+3\,{b}^{3}c{d}^{2} \right ){x}^{11}}{11}}+{\frac{ \left ( 3\,{a}^{2}b{d}^{3}+9\,a{b}^{2}c{d}^{2}+3\,{b}^{3}{c}^{2}d \right ){x}^{9}}{9}}+{\frac{ \left ({a}^{3}{d}^{3}+9\,{a}^{2}bc{d}^{2}+9\,a{b}^{2}{c}^{2}d+{b}^{3}{c}^{3} \right ){x}^{7}}{7}}+{\frac{ \left ( 3\,{a}^{3}c{d}^{2}+9\,{a}^{2}b{c}^{2}d+3\,a{b}^{2}{c}^{3} \right ){x}^{5}}{5}}+{\frac{ \left ( 3\,{a}^{3}{c}^{2}d+3\,{a}^{2}b{c}^{3} \right ){x}^{3}}{3}}+{a}^{3}{c}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^3*d^3*x^13+1/11*(3*a*b^2*d^3+3*b^3*c*d^2)*x^11+1/9*(3*a^2*b*d^3+9*a*b^2*c*d^2+3*b^3*c^2*d)*x^9+1/7*(a^3
*d^3+9*a^2*b*c*d^2+9*a*b^2*c^2*d+b^3*c^3)*x^7+1/5*(3*a^3*c*d^2+9*a^2*b*c^2*d+3*a*b^2*c^3)*x^5+1/3*(3*a^3*c^2*d
+3*a^2*b*c^3)*x^3+a^3*c^3*x

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Maxima [A]  time = 1.08075, size = 225, normalized size = 1.46 \begin{align*} \frac{1}{13} \, b^{3} d^{3} x^{13} + \frac{3}{11} \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{11} + \frac{1}{3} \,{\left (b^{3} c^{2} d + 3 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{9} + \frac{1}{7} \,{\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} x^{7} + a^{3} c^{3} x + \frac{3}{5} \,{\left (a b^{2} c^{3} + 3 \, a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{5} +{\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^3*d^3*x^13 + 3/11*(b^3*c*d^2 + a*b^2*d^3)*x^11 + 1/3*(b^3*c^2*d + 3*a*b^2*c*d^2 + a^2*b*d^3)*x^9 + 1/7*
(b^3*c^3 + 9*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*x^7 + a^3*c^3*x + 3/5*(a*b^2*c^3 + 3*a^2*b*c^2*d + a^3*c*d
^2)*x^5 + (a^2*b*c^3 + a^3*c^2*d)*x^3

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Fricas [A]  time = 1.50852, size = 414, normalized size = 2.69 \begin{align*} \frac{1}{13} x^{13} d^{3} b^{3} + \frac{3}{11} x^{11} d^{2} c b^{3} + \frac{3}{11} x^{11} d^{3} b^{2} a + \frac{1}{3} x^{9} d c^{2} b^{3} + x^{9} d^{2} c b^{2} a + \frac{1}{3} x^{9} d^{3} b a^{2} + \frac{1}{7} x^{7} c^{3} b^{3} + \frac{9}{7} x^{7} d c^{2} b^{2} a + \frac{9}{7} x^{7} d^{2} c b a^{2} + \frac{1}{7} x^{7} d^{3} a^{3} + \frac{3}{5} x^{5} c^{3} b^{2} a + \frac{9}{5} x^{5} d c^{2} b a^{2} + \frac{3}{5} x^{5} d^{2} c a^{3} + x^{3} c^{3} b a^{2} + x^{3} d c^{2} a^{3} + x c^{3} a^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*x^13*d^3*b^3 + 3/11*x^11*d^2*c*b^3 + 3/11*x^11*d^3*b^2*a + 1/3*x^9*d*c^2*b^3 + x^9*d^2*c*b^2*a + 1/3*x^9*
d^3*b*a^2 + 1/7*x^7*c^3*b^3 + 9/7*x^7*d*c^2*b^2*a + 9/7*x^7*d^2*c*b*a^2 + 1/7*x^7*d^3*a^3 + 3/5*x^5*c^3*b^2*a
+ 9/5*x^5*d*c^2*b*a^2 + 3/5*x^5*d^2*c*a^3 + x^3*c^3*b*a^2 + x^3*d*c^2*a^3 + x*c^3*a^3

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Sympy [A]  time = 0.086279, size = 189, normalized size = 1.23 \begin{align*} a^{3} c^{3} x + \frac{b^{3} d^{3} x^{13}}{13} + x^{11} \left (\frac{3 a b^{2} d^{3}}{11} + \frac{3 b^{3} c d^{2}}{11}\right ) + x^{9} \left (\frac{a^{2} b d^{3}}{3} + a b^{2} c d^{2} + \frac{b^{3} c^{2} d}{3}\right ) + x^{7} \left (\frac{a^{3} d^{3}}{7} + \frac{9 a^{2} b c d^{2}}{7} + \frac{9 a b^{2} c^{2} d}{7} + \frac{b^{3} c^{3}}{7}\right ) + x^{5} \left (\frac{3 a^{3} c d^{2}}{5} + \frac{9 a^{2} b c^{2} d}{5} + \frac{3 a b^{2} c^{3}}{5}\right ) + x^{3} \left (a^{3} c^{2} d + a^{2} b c^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c**3*x + b**3*d**3*x**13/13 + x**11*(3*a*b**2*d**3/11 + 3*b**3*c*d**2/11) + x**9*(a**2*b*d**3/3 + a*b**2*
c*d**2 + b**3*c**2*d/3) + x**7*(a**3*d**3/7 + 9*a**2*b*c*d**2/7 + 9*a*b**2*c**2*d/7 + b**3*c**3/7) + x**5*(3*a
**3*c*d**2/5 + 9*a**2*b*c**2*d/5 + 3*a*b**2*c**3/5) + x**3*(a**3*c**2*d + a**2*b*c**3)

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Giac [A]  time = 1.09512, size = 252, normalized size = 1.64 \begin{align*} \frac{1}{13} \, b^{3} d^{3} x^{13} + \frac{3}{11} \, b^{3} c d^{2} x^{11} + \frac{3}{11} \, a b^{2} d^{3} x^{11} + \frac{1}{3} \, b^{3} c^{2} d x^{9} + a b^{2} c d^{2} x^{9} + \frac{1}{3} \, a^{2} b d^{3} x^{9} + \frac{1}{7} \, b^{3} c^{3} x^{7} + \frac{9}{7} \, a b^{2} c^{2} d x^{7} + \frac{9}{7} \, a^{2} b c d^{2} x^{7} + \frac{1}{7} \, a^{3} d^{3} x^{7} + \frac{3}{5} \, a b^{2} c^{3} x^{5} + \frac{9}{5} \, a^{2} b c^{2} d x^{5} + \frac{3}{5} \, a^{3} c d^{2} x^{5} + a^{2} b c^{3} x^{3} + a^{3} c^{2} d x^{3} + a^{3} c^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13*b^3*d^3*x^13 + 3/11*b^3*c*d^2*x^11 + 3/11*a*b^2*d^3*x^11 + 1/3*b^3*c^2*d*x^9 + a*b^2*c*d^2*x^9 + 1/3*a^2*
b*d^3*x^9 + 1/7*b^3*c^3*x^7 + 9/7*a*b^2*c^2*d*x^7 + 9/7*a^2*b*c*d^2*x^7 + 1/7*a^3*d^3*x^7 + 3/5*a*b^2*c^3*x^5
+ 9/5*a^2*b*c^2*d*x^5 + 3/5*a^3*c*d^2*x^5 + a^2*b*c^3*x^3 + a^3*c^2*d*x^3 + a^3*c^3*x