\(\int (a+b x^2)^3 (A+B x+C x^2+D x^3) \, dx\) [81]

Optimal result

Integrand size = 25, antiderivative size = 133 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A b+a C) x^3+\frac {1}{4} a^3 D x^4+\frac {3}{5} a b (A b+a C) x^5+\frac {1}{2} a^2 b D x^6+\frac {1}{7} b^2 (A b+3 a C) x^7+\frac {3}{8} a b^2 D x^8+\frac {1}{9} b^3 C x^9+\frac {1}{10} b^3 D x^{10}+\frac {B \left (a+b x^2\right )^4}{8 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1596, 1824} \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=a^3 A x+\frac {1}{4} a^3 D x^4+\frac {1}{3} a^2 x^3 (a C+3 A b)+\frac {1}{2} a^2 b D x^6+\frac {1}{7} b^2 x^7 (3 a C+A b)+\frac {3}{5} a b x^5 (a C+A b)+\frac {3}{8} a b^2 D x^8+\frac {B \left (a+b x^2\right )^4}{8 b}+\frac {1}{9} b^3 C x^9+\frac {1}{10} b^3 D x^{10} \]

[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 1596

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 1824

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*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {210 a^3 x (12 A+x (6 B+x (4 C+3 D x)))+126 a^2 b x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+9 a b^2 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))+b^3 x^7 (360 A+7 x (45 B+4 x (10 C+9 D x)))}{2520} \]

[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

Maple [A] (verified)

Time = 3.45 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08

method	result	size
norman	\(\frac {b^{3} D x^{10}}{10}+\frac {b^{3} C \,x^{9}}{9}+\left (\frac {1}{8} B \,b^{3}+\frac {3}{8} a \,b^{2} D\right ) x^{8}+\left (\frac {1}{7} b^{3} A +\frac {3}{7} C \,b^{2} a \right ) x^{7}+\left (\frac {1}{2} a \,b^{2} B +\frac {1}{2} D a^{2} b \right ) x^{6}+\left (\frac {3}{5} a \,b^{2} A +\frac {3}{5} C \,a^{2} b \right ) x^{5}+\left (\frac {3}{4} a^{2} b B +\frac {1}{4} D a^{3}\right ) x^{4}+\left (a^{2} b A +\frac {1}{3} C \,a^{3}\right ) x^{3}+\frac {a^{3} B \,x^{2}}{2}+a^{3} A x\)	\(144\)
default	\(\frac {b^{3} D x^{10}}{10}+\frac {b^{3} C \,x^{9}}{9}+\frac {\left (B \,b^{3}+3 a \,b^{2} D\right ) x^{8}}{8}+\frac {\left (b^{3} A +3 C \,b^{2} a \right ) x^{7}}{7}+\frac {\left (3 a \,b^{2} B +3 D a^{2} b \right ) x^{6}}{6}+\frac {\left (3 a \,b^{2} A +3 C \,a^{2} b \right ) x^{5}}{5}+\frac {\left (3 a^{2} b B +D a^{3}\right ) x^{4}}{4}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{3}}{3}+\frac {a^{3} B \,x^{2}}{2}+a^{3} A x\)	\(147\)
gosper	\(\frac {1}{10} b^{3} D x^{10}+\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {3}{8} a \,b^{2} D x^{8}+\frac {1}{7} x^{7} b^{3} A +\frac {3}{7} x^{7} C \,b^{2} a +\frac {1}{2} x^{6} a \,b^{2} B +\frac {1}{2} a^{2} b D x^{6}+\frac {3}{5} x^{5} a \,b^{2} A +\frac {3}{5} x^{5} C \,a^{2} b +\frac {3}{4} x^{4} a^{2} b B +\frac {1}{4} a^{3} D x^{4}+x^{3} a^{2} b A +\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} a^{3} B \,x^{2}+a^{3} A x\)	\(150\)
parallelrisch	\(\frac {1}{10} b^{3} D x^{10}+\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {3}{8} a \,b^{2} D x^{8}+\frac {1}{7} x^{7} b^{3} A +\frac {3}{7} x^{7} C \,b^{2} a +\frac {1}{2} x^{6} a \,b^{2} B +\frac {1}{2} a^{2} b D x^{6}+\frac {3}{5} x^{5} a \,b^{2} A +\frac {3}{5} x^{5} C \,a^{2} b +\frac {3}{4} x^{4} a^{2} b B +\frac {1}{4} a^{3} D x^{4}+x^{3} a^{2} b A +\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} a^{3} B \,x^{2}+a^{3} A x\)	\(150\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{8} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (D a^{2} b + B a b^{2}\right )} x^{6} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{4} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{4} + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \]

[In]

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

[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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.19 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=A a^{3} x + \frac {B a^{3} x^{2}}{2} + \frac {C b^{3} x^{9}}{9} + \frac {D b^{3} x^{10}}{10} + x^{8} \left (\frac {B b^{3}}{8} + \frac {3 D a b^{2}}{8}\right ) + x^{7} \left (\frac {A b^{3}}{7} + \frac {3 C a b^{2}}{7}\right ) + x^{6} \left (\frac {B a b^{2}}{2} + \frac {D a^{2} b}{2}\right ) + x^{5} \cdot \left (\frac {3 A a b^{2}}{5} + \frac {3 C a^{2} b}{5}\right ) + x^{4} \cdot \left (\frac {3 B a^{2} b}{4} + \frac {D a^{3}}{4}\right ) + x^{3} \left (A a^{2} b + \frac {C a^{3}}{3}\right ) \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{8} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (D a^{2} b + B a b^{2}\right )} x^{6} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{4} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{4} + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {3}{8} \, D a b^{2} x^{8} + \frac {1}{8} \, B b^{3} x^{8} + \frac {3}{7} \, C a b^{2} x^{7} + \frac {1}{7} \, A b^{3} x^{7} + \frac {1}{2} \, D a^{2} b x^{6} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {3}{5} \, C a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {1}{4} \, D a^{3} x^{4} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {1}{3} \, C a^{3} x^{3} + A a^{2} b x^{3} + \frac {1}{2} \, B a^{3} x^{2} + A a^{3} x \]

[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

Mupad [B] (verification not implemented)

Time = 5.77 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {B\,a^3\,x^2}{2}+\frac {A\,b^3\,x^7}{7}+\frac {C\,a^3\,x^3}{3}+\frac {B\,b^3\,x^8}{8}+\frac {C\,b^3\,x^9}{9}+\frac {a^3\,x^4\,D}{4}+\frac {b^3\,x^{10}\,D}{10}+A\,a^3\,x+\frac {a^2\,b\,x^6\,D}{2}+\frac {3\,a\,b^2\,x^8\,D}{8}+A\,a^2\,b\,x^3+\frac {3\,A\,a\,b^2\,x^5}{5}+\frac {3\,B\,a^2\,b\,x^4}{4}+\frac {B\,a\,b^2\,x^6}{2}+\frac {3\,C\,a^2\,b\,x^5}{5}+\frac {3\,C\,a\,b^2\,x^7}{7} \]

[In]

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

[Out]

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