\(\int \frac {x^3 (a+b x^2)^2}{\sqrt {c+d x^2}} \, dx\) [637]

Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {c (b c-a d)^2 \sqrt {c+d x^2}}{d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{3/2}}{3 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^{5/2}}{5 d^4}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 112, 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} \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {b \left (c+d x^2\right )^{5/2} (3 b c-2 a d)}{5 d^4}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d) (3 b c-a d)}{3 d^4}-\frac {c \sqrt {c+d x^2} (b c-a d)^2}{d^4}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^4} \]

[In]

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

[Out]

-((c*(b*c - a*d)^2*Sqrt[c + d*x^2])/d^4) + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^(3/2))/(3*d^4) - (b*(3*b*c -
2*a*d)*(c + d*x^2)^(5/2))/(5*d^4) + (b^2*(c + d*x^2)^(7/2))/(7*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 \frac {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}{d^3 \sqrt {c+d x}}+\frac {(b c-a d) (3 b c-a d) \sqrt {c+d x}}{d^3}-\frac {b (3 b c-2 a d) (c+d x)^{3/2}}{d^3}+\frac {b^2 (c+d x)^{5/2}}{d^3}\right ) \, dx,x,x^2\right ) \\ & = -\frac {c (b c-a d)^2 \sqrt {c+d x^2}}{d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{3/2}}{3 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^{5/2}}{5 d^4}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {c+d x^2} \left (35 a^2 d^2 \left (-2 c+d x^2\right )+14 a b d \left (8 c^2-4 c d x^2+3 d^2 x^4\right )-3 b^2 \left (16 c^3-8 c^2 d x^2+6 c d^2 x^4-5 d^3 x^6\right )\right )}{105 d^4} \]

[In]

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

[Out]

(Sqrt[c + d*x^2]*(35*a^2*d^2*(-2*c + d*x^2) + 14*a*b*d*(8*c^2 - 4*c*d*x^2 + 3*d^2*x^4) - 3*b^2*(16*c^3 - 8*c^2
*d*x^2 + 6*c*d^2*x^4 - 5*d^3*x^6)))/(105*d^4)

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {3}{14} b^{2} x^{6}-\frac {3}{5} a b \,x^{4}-\frac {1}{2} a^{2} x^{2}\right ) d^{3}+c \left (\frac {9}{35} b^{2} x^{4}+\frac {4}{5} a b \,x^{2}+a^{2}\right ) d^{2}-\frac {8 \left (\frac {3 b \,x^{2}}{14}+a \right ) b \,c^{2} d}{5}+\frac {24 b^{2} c^{3}}{35}\right ) \sqrt {d \,x^{2}+c}}{3 d^{4}}\)	\(91\)
gosper	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-15 b^{2} d^{3} x^{6}-42 a b \,d^{3} x^{4}+18 b^{2} c \,d^{2} x^{4}-35 a^{2} d^{3} x^{2}+56 a b c \,d^{2} x^{2}-24 b^{2} c^{2} d \,x^{2}+70 c \,a^{2} d^{2}-112 a b \,c^{2} d +48 b^{2} c^{3}\right )}{105 d^{4}}\)	\(108\)
trager	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-15 b^{2} d^{3} x^{6}-42 a b \,d^{3} x^{4}+18 b^{2} c \,d^{2} x^{4}-35 a^{2} d^{3} x^{2}+56 a b c \,d^{2} x^{2}-24 b^{2} c^{2} d \,x^{2}+70 c \,a^{2} d^{2}-112 a b \,c^{2} d +48 b^{2} c^{3}\right )}{105 d^{4}}\)	\(108\)
risch	\(-\frac {\sqrt {d \,x^{2}+c}\, \left (-15 b^{2} d^{3} x^{6}-42 a b \,d^{3} x^{4}+18 b^{2} c \,d^{2} x^{4}-35 a^{2} d^{3} x^{2}+56 a b c \,d^{2} x^{2}-24 b^{2} c^{2} d \,x^{2}+70 c \,a^{2} d^{2}-112 a b \,c^{2} d +48 b^{2} c^{3}\right )}{105 d^{4}}\)	\(108\)
default	\(b^{2} \left (\frac {x^{6} \sqrt {d \,x^{2}+c}}{7 d}-\frac {6 c \left (\frac {x^{4} \sqrt {d \,x^{2}+c}}{5 d}-\frac {4 c \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )}{5 d}\right )}{7 d}\right )+a^{2} \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )+2 a b \left (\frac {x^{4} \sqrt {d \,x^{2}+c}}{5 d}-\frac {4 c \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )}{5 d}\right )\)	\(185\)

[In]

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

[Out]

-2/3*((-3/14*b^2*x^6-3/5*a*b*x^4-1/2*a^2*x^2)*d^3+c*(9/35*b^2*x^4+4/5*a*b*x^2+a^2)*d^2-8/5*(3/14*b*x^2+a)*b*c^
2*d+24/35*b^2*c^3)*(d*x^2+c)^(1/2)/d^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {{\left (15 \, b^{2} d^{3} x^{6} - 48 \, b^{2} c^{3} + 112 \, a b c^{2} d - 70 \, a^{2} c d^{2} - 6 \, {\left (3 \, b^{2} c d^{2} - 7 \, a b d^{3}\right )} x^{4} + {\left (24 \, b^{2} c^{2} d - 56 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{105 \, d^{4}} \]

[In]

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

[Out]

1/105*(15*b^2*d^3*x^6 - 48*b^2*c^3 + 112*a*b*c^2*d - 70*a^2*c*d^2 - 6*(3*b^2*c*d^2 - 7*a*b*d^3)*x^4 + (24*b^2*
c^2*d - 56*a*b*c*d^2 + 35*a^2*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. 240 vs. \(2 (100) = 200\).

Time = 0.30 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.14 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\begin {cases} - \frac {2 a^{2} c \sqrt {c + d x^{2}}}{3 d^{2}} + \frac {a^{2} x^{2} \sqrt {c + d x^{2}}}{3 d} + \frac {16 a b c^{2} \sqrt {c + d x^{2}}}{15 d^{3}} - \frac {8 a b c x^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {2 a b x^{4} \sqrt {c + d x^{2}}}{5 d} - \frac {16 b^{2} c^{3} \sqrt {c + d x^{2}}}{35 d^{4}} + \frac {8 b^{2} c^{2} x^{2} \sqrt {c + d x^{2}}}{35 d^{3}} - \frac {6 b^{2} c x^{4} \sqrt {c + d x^{2}}}{35 d^{2}} + \frac {b^{2} x^{6} \sqrt {c + d x^{2}}}{7 d} & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b^{2} x^{8}}{8}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*a**2*c*sqrt(c + d*x**2)/(3*d**2) + a**2*x**2*sqrt(c + d*x**2)/(3*d) + 16*a*b*c**2*sqrt(c + d*x**
2)/(15*d**3) - 8*a*b*c*x**2*sqrt(c + d*x**2)/(15*d**2) + 2*a*b*x**4*sqrt(c + d*x**2)/(5*d) - 16*b**2*c**3*sqrt
(c + d*x**2)/(35*d**4) + 8*b**2*c**2*x**2*sqrt(c + d*x**2)/(35*d**3) - 6*b**2*c*x**4*sqrt(c + d*x**2)/(35*d**2
) + b**2*x**6*sqrt(c + d*x**2)/(7*d), Ne(d, 0)), ((a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8)/sqrt(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.62 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{6}}{7 \, d} - \frac {6 \, \sqrt {d x^{2} + c} b^{2} c x^{4}}{35 \, d^{2}} + \frac {2 \, \sqrt {d x^{2} + c} a b x^{4}}{5 \, d} + \frac {8 \, \sqrt {d x^{2} + c} b^{2} c^{2} x^{2}}{35 \, d^{3}} - \frac {8 \, \sqrt {d x^{2} + c} a b c x^{2}}{15 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} x^{2}}{3 \, d} - \frac {16 \, \sqrt {d x^{2} + c} b^{2} c^{3}}{35 \, d^{4}} + \frac {16 \, \sqrt {d x^{2} + c} a b c^{2}}{15 \, d^{3}} - \frac {2 \, \sqrt {d x^{2} + c} a^{2} c}{3 \, d^{2}} \]

[In]

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

[Out]

1/7*sqrt(d*x^2 + c)*b^2*x^6/d - 6/35*sqrt(d*x^2 + c)*b^2*c*x^4/d^2 + 2/5*sqrt(d*x^2 + c)*a*b*x^4/d + 8/35*sqrt
(d*x^2 + c)*b^2*c^2*x^2/d^3 - 8/15*sqrt(d*x^2 + c)*a*b*c*x^2/d^2 + 1/3*sqrt(d*x^2 + c)*a^2*x^2/d - 16/35*sqrt(
d*x^2 + c)*b^2*c^3/d^4 + 16/15*sqrt(d*x^2 + c)*a*b*c^2/d^3 - 2/3*sqrt(d*x^2 + c)*a^2*c/d^2

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.22 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \sqrt {d x^{2} + c}}{d^{4}} + \frac {15 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} - 63 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c + 105 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} + 42 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d - 140 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d + 35 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{105 \, d^{4}} \]

[In]

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

[Out]

-(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*sqrt(d*x^2 + c)/d^4 + 1/105*(15*(d*x^2 + c)^(7/2)*b^2 - 63*(d*x^2 + c)^(5
/2)*b^2*c + 105*(d*x^2 + c)^(3/2)*b^2*c^2 + 42*(d*x^2 + c)^(5/2)*a*b*d - 140*(d*x^2 + c)^(3/2)*a*b*c*d + 35*(d
*x^2 + c)^(3/2)*a^2*d^2)/d^4

Mupad [B] (verification not implemented)

Time = 5.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\sqrt {d\,x^2+c}\,\left (\frac {b^2\,x^6}{7\,d}-\frac {70\,a^2\,c\,d^2-112\,a\,b\,c^2\,d+48\,b^2\,c^3}{105\,d^4}+\frac {x^2\,\left (35\,a^2\,d^3-56\,a\,b\,c\,d^2+24\,b^2\,c^2\,d\right )}{105\,d^4}+\frac {2\,b\,x^4\,\left (7\,a\,d-3\,b\,c\right )}{35\,d^2}\right ) \]

[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^6)/(7*d) - (48*b^2*c^3 + 70*a^2*c*d^2 - 112*a*b*c^2*d)/(105*d^4) + (x^2*(35*a^2*d^3
+ 24*b^2*c^2*d - 56*a*b*c*d^2))/(105*d^4) + (2*b*x^4*(7*a*d - 3*b*c))/(35*d^2))