\(\int x^3 (a+b x^2)^2 \sqrt {c+d x^2} \, dx\) [599]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=-\frac {c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}{3 d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{5/2}}{5 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^4}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^4} \]

[Out]

-1/3*c*(-a*d+b*c)^2*(d*x^2+c)^(3/2)/d^4+1/5*(-a*d+b*c)*(-a*d+3*b*c)*(d*x^2+c)^(5/2)/d^4-1/7*b*(-2*a*d+3*b*c)*(
d*x^2+c)^(7/2)/d^4+1/9*b^2*(d*x^2+c)^(9/2)/d^4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {457, 78} \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=-\frac {b \left (c+d x^2\right )^{7/2} (3 b c-2 a d)}{7 d^4}+\frac {\left (c+d x^2\right )^{5/2} (b c-a d) (3 b c-a d)}{5 d^4}-\frac {c \left (c+d x^2\right )^{3/2} (b c-a d)^2}{3 d^4}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^4} \]

[In]

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

[Out]

-1/3*(c*(b*c - a*d)^2*(c + d*x^2)^(3/2))/d^4 + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^(5/2))/(5*d^4) - (b*(3*b
*c - 2*a*d)*(c + d*x^2)^(7/2))/(7*d^4) + (b^2*(c + d*x^2)^(9/2))/(9*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 \sqrt {c+d x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {c (b c-a d)^2 \sqrt {c+d x}}{d^3}+\frac {(b c-a d) (3 b c-a d) (c+d x)^{3/2}}{d^3}-\frac {b (3 b c-2 a d) (c+d x)^{5/2}}{d^3}+\frac {b^2 (c+d x)^{7/2}}{d^3}\right ) \, dx,x,x^2\right ) \\ & = -\frac {c (b c-a d)^2 \left (c+d x^2\right )^{3/2}}{3 d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{5/2}}{5 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^{7/2}}{7 d^4}+\frac {b^2 \left (c+d x^2\right )^{9/2}}{9 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.87 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\left (c+d x^2\right )^{3/2} \left (21 a^2 d^2 \left (-2 c+3 d x^2\right )+6 a b d \left (8 c^2-12 c d x^2+15 d^2 x^4\right )+b^2 \left (-16 c^3+24 c^2 d x^2-30 c d^2 x^4+35 d^3 x^6\right )\right )}{315 d^4} \]

[In]

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

[Out]

((c + d*x^2)^(3/2)*(21*a^2*d^2*(-2*c + 3*d*x^2) + 6*a*b*d*(8*c^2 - 12*c*d*x^2 + 15*d^2*x^4) + b^2*(-16*c^3 + 2
4*c^2*d*x^2 - 30*c*d^2*x^4 + 35*d^3*x^6)))/(315*d^4)

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {5}{6} b^{2} x^{6}-\frac {15}{7} a b \,x^{4}-\frac {3}{2} a^{2} x^{2}\right ) d^{3}+\left (b \,x^{2}+a \right ) \left (\frac {5 b \,x^{2}}{7}+a \right ) c \,d^{2}-\frac {8 \left (\frac {b \,x^{2}}{2}+a \right ) b \,c^{2} d}{7}+\frac {8 b^{2} c^{3}}{21}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{4}}\)	\(87\)
gosper	\(-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \left (-35 b^{2} d^{3} x^{6}-90 a b \,d^{3} x^{4}+30 b^{2} c \,d^{2} x^{4}-63 a^{2} d^{3} x^{2}+72 a b c \,d^{2} x^{2}-24 b^{2} c^{2} d \,x^{2}+42 c \,a^{2} d^{2}-48 a b \,c^{2} d +16 b^{2} c^{3}\right )}{315 d^{4}}\)	\(108\)
trager	\(-\frac {\left (-35 b^{2} x^{8} d^{4}-90 a b \,d^{4} x^{6}-5 b^{2} c \,d^{3} x^{6}-63 a^{2} d^{4} x^{4}-18 c a b \,x^{4} d^{3}+6 b^{2} c^{2} d^{2} x^{4}-21 a^{2} c \,d^{3} x^{2}+24 a b \,c^{2} d^{2} x^{2}-8 b^{2} c^{3} d \,x^{2}+42 a^{2} c^{2} d^{2}-48 a b \,c^{3} d +16 b^{2} c^{4}\right ) \sqrt {d \,x^{2}+c}}{315 d^{4}}\)	\(149\)
risch	\(-\frac {\left (-35 b^{2} x^{8} d^{4}-90 a b \,d^{4} x^{6}-5 b^{2} c \,d^{3} x^{6}-63 a^{2} d^{4} x^{4}-18 c a b \,x^{4} d^{3}+6 b^{2} c^{2} d^{2} x^{4}-21 a^{2} c \,d^{3} x^{2}+24 a b \,c^{2} d^{2} x^{2}-8 b^{2} c^{3} d \,x^{2}+42 a^{2} c^{2} d^{2}-48 a b \,c^{3} d +16 b^{2} c^{4}\right ) \sqrt {d \,x^{2}+c}}{315 d^{4}}\)	\(149\)
default	\(b^{2} \left (\frac {x^{6} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{9 d}-\frac {2 c \left (\frac {x^{4} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )}{3 d}\right )+a^{2} \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )+2 a b \left (\frac {x^{4} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{7 d}-\frac {4 c \left (\frac {x^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 d}-\frac {2 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 d^{2}}\right )}{7 d}\right )\)	\(185\)

[In]

int(x^3*(b*x^2+a)^2*(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*((-5/6*b^2*x^6-15/7*a*b*x^4-3/2*a^2*x^2)*d^3+(b*x^2+a)*(5/7*b*x^2+a)*c*d^2-8/7*(1/2*b*x^2+a)*b*c^2*d+8/2
1*b^2*c^3)*(d*x^2+c)^(3/2)/d^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.23 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (35 \, b^{2} d^{4} x^{8} + 5 \, {\left (b^{2} c d^{3} + 18 \, a b d^{4}\right )} x^{6} - 16 \, b^{2} c^{4} + 48 \, a b c^{3} d - 42 \, a^{2} c^{2} d^{2} - 3 \, {\left (2 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} - 21 \, a^{2} d^{4}\right )} x^{4} + {\left (8 \, b^{2} c^{3} d - 24 \, a b c^{2} d^{2} + 21 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{315 \, d^{4}} \]

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

1/315*(35*b^2*d^4*x^8 + 5*(b^2*c*d^3 + 18*a*b*d^4)*x^6 - 16*b^2*c^4 + 48*a*b*c^3*d - 42*a^2*c^2*d^2 - 3*(2*b^2
*c^2*d^2 - 6*a*b*c*d^3 - 21*a^2*d^4)*x^4 + (8*b^2*c^3*d - 24*a*b*c^2*d^2 + 21*a^2*c*d^3)*x^2)*sqrt(d*x^2 + c)/
d^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (102) = 204\).

Time = 0.31 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.70 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} - \frac {2 a^{2} c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {a^{2} c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {a^{2} x^{4} \sqrt {c + d x^{2}}}{5} + \frac {16 a b c^{3} \sqrt {c + d x^{2}}}{105 d^{3}} - \frac {8 a b c^{2} x^{2} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {2 a b c x^{4} \sqrt {c + d x^{2}}}{35 d} + \frac {2 a b x^{6} \sqrt {c + d x^{2}}}{7} - \frac {16 b^{2} c^{4} \sqrt {c + d x^{2}}}{315 d^{4}} + \frac {8 b^{2} c^{3} x^{2} \sqrt {c + d x^{2}}}{315 d^{3}} - \frac {2 b^{2} c^{2} x^{4} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {b^{2} c x^{6} \sqrt {c + d x^{2}}}{63 d} + \frac {b^{2} x^{8} \sqrt {c + d x^{2}}}{9} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b^{2} x^{8}}{8}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(b*x**2+a)**2*(d*x**2+c)**(1/2),x)

[Out]

Piecewise((-2*a**2*c**2*sqrt(c + d*x**2)/(15*d**2) + a**2*c*x**2*sqrt(c + d*x**2)/(15*d) + a**2*x**4*sqrt(c +
d*x**2)/5 + 16*a*b*c**3*sqrt(c + d*x**2)/(105*d**3) - 8*a*b*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + 2*a*b*c*x*
*4*sqrt(c + d*x**2)/(35*d) + 2*a*b*x**6*sqrt(c + d*x**2)/7 - 16*b**2*c**4*sqrt(c + d*x**2)/(315*d**4) + 8*b**2
*c**3*x**2*sqrt(c + d*x**2)/(315*d**3) - 2*b**2*c**2*x**4*sqrt(c + d*x**2)/(105*d**2) + b**2*c*x**6*sqrt(c + d
*x**2)/(63*d) + b**2*x**8*sqrt(c + d*x**2)/9, Ne(d, 0)), (sqrt(c)*(a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8), Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.59 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{6}}{9 \, d} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{4}}{21 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{4}}{7 \, d} + \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x^{2}}{105 \, d^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x^{2}}{35 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x^{2}}{5 \, d} - \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3}}{315 \, d^{4}} + \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2}}{105 \, d^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c}{15 \, d^{2}} \]

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

1/9*(d*x^2 + c)^(3/2)*b^2*x^6/d - 2/21*(d*x^2 + c)^(3/2)*b^2*c*x^4/d^2 + 2/7*(d*x^2 + c)^(3/2)*a*b*x^4/d + 8/1
05*(d*x^2 + c)^(3/2)*b^2*c^2*x^2/d^3 - 8/35*(d*x^2 + c)^(3/2)*a*b*c*x^2/d^2 + 1/5*(d*x^2 + c)^(3/2)*a^2*x^2/d
- 16/315*(d*x^2 + c)^(3/2)*b^2*c^3/d^4 + 16/105*(d*x^2 + c)^(3/2)*a*b*c^2/d^3 - 2/15*(d*x^2 + c)^(3/2)*a^2*c/d
^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.32 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {35 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} - 135 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c + 189 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} - 105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} + 90 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d - 252 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d + 210 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d + 63 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{2} - 105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{2}}{315 \, d^{4}} \]

[In]

integrate(x^3*(b*x^2+a)^2*(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/315*(35*(d*x^2 + c)^(9/2)*b^2 - 135*(d*x^2 + c)^(7/2)*b^2*c + 189*(d*x^2 + c)^(5/2)*b^2*c^2 - 105*(d*x^2 + c
)^(3/2)*b^2*c^3 + 90*(d*x^2 + c)^(7/2)*a*b*d - 252*(d*x^2 + c)^(5/2)*a*b*c*d + 210*(d*x^2 + c)^(3/2)*a*b*c^2*d
+ 63*(d*x^2 + c)^(5/2)*a^2*d^2 - 105*(d*x^2 + c)^(3/2)*a^2*c*d^2)/d^4

Mupad [B] (verification not implemented)

Time = 5.14 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int x^3 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {b^2\,x^8}{9}-\frac {42\,a^2\,c^2\,d^2-48\,a\,b\,c^3\,d+16\,b^2\,c^4}{315\,d^4}+\frac {x^4\,\left (63\,a^2\,d^4+18\,a\,b\,c\,d^3-6\,b^2\,c^2\,d^2\right )}{315\,d^4}+\frac {b\,x^6\,\left (18\,a\,d+b\,c\right )}{63\,d}+\frac {c\,x^2\,\left (21\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{315\,d^3}\right ) \]

[In]

int(x^3*(a + b*x^2)^2*(c + d*x^2)^(1/2),x)

[Out]

(c + d*x^2)^(1/2)*((b^2*x^8)/9 - (16*b^2*c^4 + 42*a^2*c^2*d^2 - 48*a*b*c^3*d)/(315*d^4) + (x^4*(63*a^2*d^4 - 6
*b^2*c^2*d^2 + 18*a*b*c*d^3))/(315*d^4) + (b*x^6*(18*a*d + b*c))/(63*d) + (c*x^2*(21*a^2*d^2 + 8*b^2*c^2 - 24*
a*b*c*d))/(315*d^3))