3.81 \(\int (a+b x^2)^3 (A+B x+C x^2+D x^3) \, dx\)

Optimal. Leaf size=133 \[ \frac{1}{3} a^2 x^3 (a C+3 A b)+a^3 A x+\frac{1}{2} a^2 b D x^6+\frac{1}{4} a^3 D x^4+\frac{1}{7} b^2 x^7 (3 a C+A b)+\frac{3}{5} a b x^5 (a C+A b)+\frac{3}{8} a b^2 D x^8+\frac{B \left (a+b x^2\right )^4}{8 b}+\frac{1}{9} b^3 C x^9+\frac{1}{10} b^3 D x^{10} \]

[Out]

a^3*A*x + (a^2*(3*A*b + a*C)*x^3)/3 + (a^3*D*x^4)/4 + (3*a*b*(A*b + a*C)*x^5)/5 + (a^2*b*D*x^6)/2 + (b^2*(A*b
+ 3*a*C)*x^7)/7 + (3*a*b^2*D*x^8)/8 + (b^3*C*x^9)/9 + (b^3*D*x^10)/10 + (B*(a + b*x^2)^4)/(8*b)

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Rubi [A]  time = 0.0926051, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {1582, 1810} \[ \frac{1}{3} a^2 x^3 (a C+3 A b)+a^3 A x+\frac{1}{2} a^2 b D x^6+\frac{1}{4} a^3 D x^4+\frac{1}{7} b^2 x^7 (3 a C+A b)+\frac{3}{5} a b x^5 (a C+A b)+\frac{3}{8} a b^2 D x^8+\frac{B \left (a+b x^2\right )^4}{8 b}+\frac{1}{9} b^3 C x^9+\frac{1}{10} b^3 D x^{10} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]

[Out]

a^3*A*x + (a^2*(3*A*b + a*C)*x^3)/3 + (a^3*D*x^4)/4 + (3*a*b*(A*b + a*C)*x^5)/5 + (a^2*b*D*x^6)/2 + (b^2*(A*b
+ 3*a*C)*x^7)/7 + (3*a*b^2*D*x^8)/8 + (b^3*C*x^9)/9 + (b^3*D*x^10)/10 + (B*(a + b*x^2)^4)/(8*b)

Rule 1582

Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(Coeff[Px, x, n - 1]*(a + b*x^n)^(p + 1))/(b*n*(p +
1)), x] + Int[(Px - Coeff[Px, x, n - 1]*x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && I
GtQ[p, 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] &&  !MatchQ[P
x, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ[{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Co
eff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx &=\frac{B \left (a+b x^2\right )^4}{8 b}+\int \left (a+b x^2\right )^3 \left (A+C x^2+D x^3\right ) \, dx\\ &=\frac{B \left (a+b x^2\right )^4}{8 b}+\int \left (a^3 A+a^2 (3 A b+a C) x^2+a^3 D x^3+3 a b (A b+a C) x^4+3 a^2 b D x^5+b^2 (A b+3 a C) x^6+3 a b^2 D x^7+b^3 C x^8+b^3 D x^9\right ) \, dx\\ &=a^3 A x+\frac{1}{3} a^2 (3 A b+a C) x^3+\frac{1}{4} a^3 D x^4+\frac{3}{5} a b (A b+a C) x^5+\frac{1}{2} a^2 b D x^6+\frac{1}{7} b^2 (A b+3 a C) x^7+\frac{3}{8} a b^2 D x^8+\frac{1}{9} b^3 C x^9+\frac{1}{10} b^3 D x^{10}+\frac{B \left (a+b x^2\right )^4}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.0473235, size = 121, normalized size = 0.91 \[ \frac{126 a^2 b x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+210 a^3 x (12 A+x (6 B+x (4 C+3 D x)))+9 a b^2 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))+b^3 x^7 (360 A+7 x (45 B+4 x (10 C+9 D x)))}{2520} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]

[Out]

(210*a^3*x*(12*A + x*(6*B + x*(4*C + 3*D*x))) + 126*a^2*b*x^3*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))) + 9*a*b^2*
x^5*(168*A + 5*x*(28*B + 3*x*(8*C + 7*D*x))) + b^3*x^7*(360*A + 7*x*(45*B + 4*x*(10*C + 9*D*x))))/2520

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Maple [A]  time = 0.003, size = 147, normalized size = 1.1 \begin{align*}{\frac{{b}^{3}D{x}^{10}}{10}}+{\frac{{b}^{3}C{x}^{9}}{9}}+{\frac{ \left ({b}^{3}B+3\,a{b}^{2}D \right ){x}^{8}}{8}}+{\frac{ \left ( A{b}^{3}+3\,a{b}^{2}C \right ){x}^{7}}{7}}+{\frac{ \left ( 3\,a{b}^{2}B+3\,{a}^{2}bD \right ){x}^{6}}{6}}+{\frac{ \left ( 3\,a{b}^{2}A+3\,{a}^{2}bC \right ){x}^{5}}{5}}+{\frac{ \left ( 3\,{a}^{2}bB+{a}^{3}D \right ){x}^{4}}{4}}+{\frac{ \left ( 3\,A{a}^{2}b+{a}^{3}C \right ){x}^{3}}{3}}+{\frac{B{x}^{2}{a}^{3}}{2}}+{a}^{3}Ax \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x)

[Out]

1/10*b^3*D*x^10+1/9*b^3*C*x^9+1/8*(B*b^3+3*D*a*b^2)*x^8+1/7*(A*b^3+3*C*a*b^2)*x^7+1/6*(3*B*a*b^2+3*D*a^2*b)*x^
6+1/5*(3*A*a*b^2+3*C*a^2*b)*x^5+1/4*(3*B*a^2*b+D*a^3)*x^4+1/3*(3*A*a^2*b+C*a^3)*x^3+1/2*B*x^2*a^3+a^3*A*x

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Maxima [A]  time = 1.01368, size = 192, normalized size = 1.44 \begin{align*} \frac{1}{10} \, D b^{3} x^{10} + \frac{1}{9} \, C b^{3} x^{9} + \frac{1}{8} \,{\left (3 \, D a b^{2} + B b^{3}\right )} x^{8} + \frac{1}{7} \,{\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac{1}{2} \,{\left (D a^{2} b + B a b^{2}\right )} x^{6} + \frac{1}{2} \, B a^{3} x^{2} + \frac{3}{5} \,{\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac{1}{4} \,{\left (D a^{3} + 3 \, B a^{2} b\right )} x^{4} + \frac{1}{3} \,{\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")

[Out]

1/10*D*b^3*x^10 + 1/9*C*b^3*x^9 + 1/8*(3*D*a*b^2 + B*b^3)*x^8 + 1/7*(3*C*a*b^2 + A*b^3)*x^7 + 1/2*(D*a^2*b + B
*a*b^2)*x^6 + 1/2*B*a^3*x^2 + 3/5*(C*a^2*b + A*a*b^2)*x^5 + A*a^3*x + 1/4*(D*a^3 + 3*B*a^2*b)*x^4 + 1/3*(C*a^3
+ 3*A*a^2*b)*x^3

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Fricas [A]  time = 1.35735, size = 355, normalized size = 2.67 \begin{align*} \frac{1}{10} x^{10} b^{3} D + \frac{1}{9} x^{9} b^{3} C + \frac{3}{8} x^{8} b^{2} a D + \frac{1}{8} x^{8} b^{3} B + \frac{3}{7} x^{7} b^{2} a C + \frac{1}{7} x^{7} b^{3} A + \frac{1}{2} x^{6} b a^{2} D + \frac{1}{2} x^{6} b^{2} a B + \frac{3}{5} x^{5} b a^{2} C + \frac{3}{5} x^{5} b^{2} a A + \frac{1}{4} x^{4} a^{3} D + \frac{3}{4} x^{4} b a^{2} B + \frac{1}{3} x^{3} a^{3} C + x^{3} b a^{2} A + \frac{1}{2} x^{2} a^{3} B + x a^{3} A \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")

[Out]

1/10*x^10*b^3*D + 1/9*x^9*b^3*C + 3/8*x^8*b^2*a*D + 1/8*x^8*b^3*B + 3/7*x^7*b^2*a*C + 1/7*x^7*b^3*A + 1/2*x^6*
b*a^2*D + 1/2*x^6*b^2*a*B + 3/5*x^5*b*a^2*C + 3/5*x^5*b^2*a*A + 1/4*x^4*a^3*D + 3/4*x^4*b*a^2*B + 1/3*x^3*a^3*
C + x^3*b*a^2*A + 1/2*x^2*a^3*B + x*a^3*A

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Sympy [A]  time = 0.085419, size = 158, normalized size = 1.19 \begin{align*} A a^{3} x + \frac{B a^{3} x^{2}}{2} + \frac{C b^{3} x^{9}}{9} + \frac{D b^{3} x^{10}}{10} + x^{8} \left (\frac{B b^{3}}{8} + \frac{3 D a b^{2}}{8}\right ) + x^{7} \left (\frac{A b^{3}}{7} + \frac{3 C a b^{2}}{7}\right ) + x^{6} \left (\frac{B a b^{2}}{2} + \frac{D a^{2} b}{2}\right ) + x^{5} \left (\frac{3 A a b^{2}}{5} + \frac{3 C a^{2} b}{5}\right ) + x^{4} \left (\frac{3 B a^{2} b}{4} + \frac{D a^{3}}{4}\right ) + x^{3} \left (A a^{2} b + \frac{C a^{3}}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3*(D*x**3+C*x**2+B*x+A),x)

[Out]

A*a**3*x + B*a**3*x**2/2 + C*b**3*x**9/9 + D*b**3*x**10/10 + x**8*(B*b**3/8 + 3*D*a*b**2/8) + x**7*(A*b**3/7 +
3*C*a*b**2/7) + x**6*(B*a*b**2/2 + D*a**2*b/2) + x**5*(3*A*a*b**2/5 + 3*C*a**2*b/5) + x**4*(3*B*a**2*b/4 + D*
a**3/4) + x**3*(A*a**2*b + C*a**3/3)

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Giac [A]  time = 1.12483, size = 201, normalized size = 1.51 \begin{align*} \frac{1}{10} \, D b^{3} x^{10} + \frac{1}{9} \, C b^{3} x^{9} + \frac{3}{8} \, D a b^{2} x^{8} + \frac{1}{8} \, B b^{3} x^{8} + \frac{3}{7} \, C a b^{2} x^{7} + \frac{1}{7} \, A b^{3} x^{7} + \frac{1}{2} \, D a^{2} b x^{6} + \frac{1}{2} \, B a b^{2} x^{6} + \frac{3}{5} \, C a^{2} b x^{5} + \frac{3}{5} \, A a b^{2} x^{5} + \frac{1}{4} \, D a^{3} x^{4} + \frac{3}{4} \, B a^{2} b x^{4} + \frac{1}{3} \, C a^{3} x^{3} + A a^{2} b x^{3} + \frac{1}{2} \, B a^{3} x^{2} + A a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")

[Out]

1/10*D*b^3*x^10 + 1/9*C*b^3*x^9 + 3/8*D*a*b^2*x^8 + 1/8*B*b^3*x^8 + 3/7*C*a*b^2*x^7 + 1/7*A*b^3*x^7 + 1/2*D*a^
2*b*x^6 + 1/2*B*a*b^2*x^6 + 3/5*C*a^2*b*x^5 + 3/5*A*a*b^2*x^5 + 1/4*D*a^3*x^4 + 3/4*B*a^2*b*x^4 + 1/3*C*a^3*x^
3 + A*a^2*b*x^3 + 1/2*B*a^3*x^2 + A*a^3*x