\(\int x^3 (a+b x^2)^2 (c+d x^2)^3 \, dx\) [160]

Optimal result

Integrand size = 22, antiderivative size = 106 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=-\frac {c (b c-a d)^2 \left (c+d x^2\right )^4}{8 d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^5}{10 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^6}{12 d^4}+\frac {b^2 \left (c+d x^2\right )^7}{14 d^4} \]

[Out]

-1/8*c*(-a*d+b*c)^2*(d*x^2+c)^4/d^4+1/10*(-a*d+b*c)*(-a*d+3*b*c)*(d*x^2+c)^5/d^4-1/12*b*(-2*a*d+3*b*c)*(d*x^2+
c)^6/d^4+1/14*b^2*(d*x^2+c)^7/d^4

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=-\frac {b \left (c+d x^2\right )^6 (3 b c-2 a d)}{12 d^4}+\frac {\left (c+d x^2\right )^5 (b c-a d) (3 b c-a d)}{10 d^4}-\frac {c \left (c+d x^2\right )^4 (b c-a d)^2}{8 d^4}+\frac {b^2 \left (c+d x^2\right )^7}{14 d^4} \]

[In]

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

[Out]

-1/8*(c*(b*c - a*d)^2*(c + d*x^2)^4)/d^4 + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^5)/(10*d^4) - (b*(3*b*c - 2*
a*d)*(c + d*x^2)^6)/(12*d^4) + (b^2*(c + d*x^2)^7)/(14*d^4)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x (a+b x)^2 (c+d x)^3 \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {c (b c-a d)^2 (c+d x)^3}{d^3}+\frac {(b c-a d) (3 b c-a d) (c+d x)^4}{d^3}-\frac {b (3 b c-2 a d) (c+d x)^5}{d^3}+\frac {b^2 (c+d x)^6}{d^3}\right ) \, dx,x,x^2\right ) \\ & = -\frac {c (b c-a d)^2 \left (c+d x^2\right )^4}{8 d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^5}{10 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^6}{12 d^4}+\frac {b^2 \left (c+d x^2\right )^7}{14 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{840} x^4 \left (210 a^2 c^3+140 a c^2 (2 b c+3 a d) x^2+105 c \left (b^2 c^2+6 a b c d+3 a^2 d^2\right ) x^4+84 d \left (3 b^2 c^2+6 a b c d+a^2 d^2\right ) x^6+70 b d^2 (3 b c+2 a d) x^8+60 b^2 d^3 x^{10}\right ) \]

[In]

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

[Out]

(x^4*(210*a^2*c^3 + 140*a*c^2*(2*b*c + 3*a*d)*x^2 + 105*c*(b^2*c^2 + 6*a*b*c*d + 3*a^2*d^2)*x^4 + 84*d*(3*b^2*
c^2 + 6*a*b*c*d + a^2*d^2)*x^6 + 70*b*d^2*(3*b*c + 2*a*d)*x^8 + 60*b^2*d^3*x^10))/840

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.19

method	result	size
norman	\(\frac {a^{2} c^{3} x^{4}}{4}+\left (\frac {1}{2} a^{2} c^{2} d +\frac {1}{3} a b \,c^{3}\right ) x^{6}+\left (\frac {3}{8} c \,a^{2} d^{2}+\frac {3}{4} a b \,c^{2} d +\frac {1}{8} b^{2} c^{3}\right ) x^{8}+\left (\frac {1}{10} a^{2} d^{3}+\frac {3}{5} a b c \,d^{2}+\frac {3}{10} b^{2} c^{2} d \right ) x^{10}+\left (\frac {1}{6} a b \,d^{3}+\frac {1}{4} b^{2} c \,d^{2}\right ) x^{12}+\frac {b^{2} d^{3} x^{14}}{14}\)	\(126\)
default	\(\frac {b^{2} d^{3} x^{14}}{14}+\frac {\left (2 a b \,d^{3}+3 b^{2} c \,d^{2}\right ) x^{12}}{12}+\frac {\left (a^{2} d^{3}+6 a b c \,d^{2}+3 b^{2} c^{2} d \right ) x^{10}}{10}+\frac {\left (3 c \,a^{2} d^{2}+6 a b \,c^{2} d +b^{2} c^{3}\right ) x^{8}}{8}+\frac {\left (3 a^{2} c^{2} d +2 a b \,c^{3}\right ) x^{6}}{6}+\frac {a^{2} c^{3} x^{4}}{4}\)	\(128\)
gosper	\(\frac {1}{4} a^{2} c^{3} x^{4}+\frac {1}{2} x^{6} a^{2} c^{2} d +\frac {1}{3} x^{6} a b \,c^{3}+\frac {3}{8} x^{8} c \,a^{2} d^{2}+\frac {3}{4} x^{8} a b \,c^{2} d +\frac {1}{8} x^{8} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {1}{6} x^{12} a b \,d^{3}+\frac {1}{4} x^{12} b^{2} c \,d^{2}+\frac {1}{14} b^{2} d^{3} x^{14}\)	\(136\)
risch	\(\frac {1}{4} a^{2} c^{3} x^{4}+\frac {1}{2} x^{6} a^{2} c^{2} d +\frac {1}{3} x^{6} a b \,c^{3}+\frac {3}{8} x^{8} c \,a^{2} d^{2}+\frac {3}{4} x^{8} a b \,c^{2} d +\frac {1}{8} x^{8} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {1}{6} x^{12} a b \,d^{3}+\frac {1}{4} x^{12} b^{2} c \,d^{2}+\frac {1}{14} b^{2} d^{3} x^{14}\)	\(136\)
parallelrisch	\(\frac {1}{4} a^{2} c^{3} x^{4}+\frac {1}{2} x^{6} a^{2} c^{2} d +\frac {1}{3} x^{6} a b \,c^{3}+\frac {3}{8} x^{8} c \,a^{2} d^{2}+\frac {3}{4} x^{8} a b \,c^{2} d +\frac {1}{8} x^{8} b^{2} c^{3}+\frac {1}{10} x^{10} a^{2} d^{3}+\frac {3}{5} x^{10} a b c \,d^{2}+\frac {3}{10} x^{10} b^{2} c^{2} d +\frac {1}{6} x^{12} a b \,d^{3}+\frac {1}{4} x^{12} b^{2} c \,d^{2}+\frac {1}{14} b^{2} d^{3} x^{14}\)	\(136\)

[In]

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

[Out]

1/4*a^2*c^3*x^4+(1/2*a^2*c^2*d+1/3*a*b*c^3)*x^6+(3/8*c*a^2*d^2+3/4*a*b*c^2*d+1/8*b^2*c^3)*x^8+(1/10*a^2*d^3+3/
5*a*b*c*d^2+3/10*b^2*c^2*d)*x^10+(1/6*a*b*d^3+1/4*b^2*c*d^2)*x^12+1/14*b^2*d^3*x^14

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{14} \, b^{2} d^{3} x^{14} + \frac {1}{12} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{12} + \frac {1}{10} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{10} + \frac {1}{4} \, a^{2} c^{3} x^{4} + \frac {1}{8} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{8} + \frac {1}{6} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{6} \]

[In]

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

[Out]

1/14*b^2*d^3*x^14 + 1/12*(3*b^2*c*d^2 + 2*a*b*d^3)*x^12 + 1/10*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^10 + 1/
4*a^2*c^3*x^4 + 1/8*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^8 + 1/6*(2*a*b*c^3 + 3*a^2*c^2*d)*x^6

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.30 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {a^{2} c^{3} x^{4}}{4} + \frac {b^{2} d^{3} x^{14}}{14} + x^{12} \left (\frac {a b d^{3}}{6} + \frac {b^{2} c d^{2}}{4}\right ) + x^{10} \left (\frac {a^{2} d^{3}}{10} + \frac {3 a b c d^{2}}{5} + \frac {3 b^{2} c^{2} d}{10}\right ) + x^{8} \cdot \left (\frac {3 a^{2} c d^{2}}{8} + \frac {3 a b c^{2} d}{4} + \frac {b^{2} c^{3}}{8}\right ) + x^{6} \left (\frac {a^{2} c^{2} d}{2} + \frac {a b c^{3}}{3}\right ) \]

[In]

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

[Out]

a**2*c**3*x**4/4 + b**2*d**3*x**14/14 + x**12*(a*b*d**3/6 + b**2*c*d**2/4) + x**10*(a**2*d**3/10 + 3*a*b*c*d**
2/5 + 3*b**2*c**2*d/10) + x**8*(3*a**2*c*d**2/8 + 3*a*b*c**2*d/4 + b**2*c**3/8) + x**6*(a**2*c**2*d/2 + a*b*c*
*3/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{14} \, b^{2} d^{3} x^{14} + \frac {1}{12} \, {\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{12} + \frac {1}{10} \, {\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{10} + \frac {1}{4} \, a^{2} c^{3} x^{4} + \frac {1}{8} \, {\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{8} + \frac {1}{6} \, {\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{6} \]

[In]

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

[Out]

1/14*b^2*d^3*x^14 + 1/12*(3*b^2*c*d^2 + 2*a*b*d^3)*x^12 + 1/10*(3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^10 + 1/
4*a^2*c^3*x^4 + 1/8*(b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^8 + 1/6*(2*a*b*c^3 + 3*a^2*c^2*d)*x^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.27 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=\frac {1}{14} \, b^{2} d^{3} x^{14} + \frac {1}{4} \, b^{2} c d^{2} x^{12} + \frac {1}{6} \, a b d^{3} x^{12} + \frac {3}{10} \, b^{2} c^{2} d x^{10} + \frac {3}{5} \, a b c d^{2} x^{10} + \frac {1}{10} \, a^{2} d^{3} x^{10} + \frac {1}{8} \, b^{2} c^{3} x^{8} + \frac {3}{4} \, a b c^{2} d x^{8} + \frac {3}{8} \, a^{2} c d^{2} x^{8} + \frac {1}{3} \, a b c^{3} x^{6} + \frac {1}{2} \, a^{2} c^{2} d x^{6} + \frac {1}{4} \, a^{2} c^{3} x^{4} \]

[In]

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

[Out]

1/14*b^2*d^3*x^14 + 1/4*b^2*c*d^2*x^12 + 1/6*a*b*d^3*x^12 + 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 + 1/8*b^2*c^3*x^8 + 3/4*a*b*c^2*d*x^8 + 3/8*a^2*c*d^2*x^8 + 1/3*a*b*c^3*x^6 + 1/2*a^2*c^2*d*x^6 + 1
/4*a^2*c^3*x^4

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^3 \, dx=x^8\,\left (\frac {3\,a^2\,c\,d^2}{8}+\frac {3\,a\,b\,c^2\,d}{4}+\frac {b^2\,c^3}{8}\right )+x^{10}\,\left (\frac {a^2\,d^3}{10}+\frac {3\,a\,b\,c\,d^2}{5}+\frac {3\,b^2\,c^2\,d}{10}\right )+\frac {a^2\,c^3\,x^4}{4}+\frac {b^2\,d^3\,x^{14}}{14}+\frac {a\,c^2\,x^6\,\left (3\,a\,d+2\,b\,c\right )}{6}+\frac {b\,d^2\,x^{12}\,\left (2\,a\,d+3\,b\,c\right )}{12} \]

[In]

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

[Out]

x^8*((b^2*c^3)/8 + (3*a^2*c*d^2)/8 + (3*a*b*c^2*d)/4) + 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^4)/4 + (b^2*d^3*x^14)/14 + (a*c^2*x^6*(3*a*d + 2*b*c))/6 + (b*d^2*x^12*(2*a*d + 3*b*c))/12