3.614 \(\int x^3 (a+b x^2)^2 (c+d x^2)^{3/2} \, dx\)

Optimal. Leaf size=114 \[ -\frac {b \left (c+d x^2\right )^{9/2} (3 b c-2 a d)}{9 d^4}+\frac {\left (c+d x^2\right )^{7/2} (b c-a d) (3 b c-a d)}{7 d^4}-\frac {c \left (c+d x^2\right )^{5/2} (b c-a d)^2}{5 d^4}+\frac {b^2 \left (c+d x^2\right )^{11/2}}{11 d^4} \]

[Out]

-1/5*c*(-a*d+b*c)^2*(d*x^2+c)^(5/2)/d^4+1/7*(-a*d+b*c)*(-a*d+3*b*c)*(d*x^2+c)^(7/2)/d^4-1/9*b*(-2*a*d+3*b*c)*(
d*x^2+c)^(9/2)/d^4+1/11*b^2*(d*x^2+c)^(11/2)/d^4

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Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {446, 77} \[ -\frac {b \left (c+d x^2\right )^{9/2} (3 b c-2 a d)}{9 d^4}+\frac {\left (c+d x^2\right )^{7/2} (b c-a d) (3 b c-a d)}{7 d^4}-\frac {c \left (c+d x^2\right )^{5/2} (b c-a d)^2}{5 d^4}+\frac {b^2 \left (c+d x^2\right )^{11/2}}{11 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(b*c - a*d)^2*(c + d*x^2)^(5/2))/(5*d^4) + ((b*c - a*d)*(3*b*c - a*d)*(c + d*x^2)^(7/2))/(7*d^4) - (b*(3*b
*c - 2*a*d)*(c + d*x^2)^(9/2))/(9*d^4) + (b^2*(c + d*x^2)^(11/2))/(11*d^4)

Rule 77

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 446

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*} \int x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x)^2 (c+d x)^{3/2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {c (b c-a d)^2 (c+d x)^{3/2}}{d^3}+\frac {(b c-a d) (3 b c-a d) (c+d x)^{5/2}}{d^3}-\frac {b (3 b c-2 a d) (c+d x)^{7/2}}{d^3}+\frac {b^2 (c+d x)^{9/2}}{d^3}\right ) \, dx,x,x^2\right )\\ &=-\frac {c (b c-a d)^2 \left (c+d x^2\right )^{5/2}}{5 d^4}+\frac {(b c-a d) (3 b c-a d) \left (c+d x^2\right )^{7/2}}{7 d^4}-\frac {b (3 b c-2 a d) \left (c+d x^2\right )^{9/2}}{9 d^4}+\frac {b^2 \left (c+d x^2\right )^{11/2}}{11 d^4}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 100, normalized size = 0.88 \[ \frac {\left (c+d x^2\right )^{5/2} \left (99 a^2 d^2 \left (5 d x^2-2 c\right )+22 a b d \left (8 c^2-20 c d x^2+35 d^2 x^4\right )-3 b^2 \left (16 c^3-40 c^2 d x^2+70 c d^2 x^4-105 d^3 x^6\right )\right )}{3465 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x^2)^(5/2)*(99*a^2*d^2*(-2*c + 5*d*x^2) + 22*a*b*d*(8*c^2 - 20*c*d*x^2 + 35*d^2*x^4) - 3*b^2*(16*c^3 -
40*c^2*d*x^2 + 70*c*d^2*x^4 - 105*d^3*x^6)))/(3465*d^4)

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fricas [A]  time = 0.77, size = 179, normalized size = 1.57 \[ \frac {{\left (315 \, b^{2} d^{5} x^{10} + 70 \, {\left (6 \, b^{2} c d^{4} + 11 \, a b d^{5}\right )} x^{8} - 48 \, b^{2} c^{5} + 176 \, a b c^{4} d - 198 \, a^{2} c^{3} d^{2} + 5 \, {\left (3 \, b^{2} c^{2} d^{3} + 220 \, a b c d^{4} + 99 \, a^{2} d^{5}\right )} x^{6} - 6 \, {\left (3 \, b^{2} c^{3} d^{2} - 11 \, a b c^{2} d^{3} - 132 \, a^{2} c d^{4}\right )} x^{4} + {\left (24 \, b^{2} c^{4} d - 88 \, a b c^{3} d^{2} + 99 \, a^{2} c^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3465 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(315*b^2*d^5*x^10 + 70*(6*b^2*c*d^4 + 11*a*b*d^5)*x^8 - 48*b^2*c^5 + 176*a*b*c^4*d - 198*a^2*c^3*d^2 +
5*(3*b^2*c^2*d^3 + 220*a*b*c*d^4 + 99*a^2*d^5)*x^6 - 6*(3*b^2*c^3*d^2 - 11*a*b*c^2*d^3 - 132*a^2*c*d^4)*x^4 +
(24*b^2*c^4*d - 88*a*b*c^3*d^2 + 99*a^2*c^2*d^3)*x^2)*sqrt(d*x^2 + c)/d^4

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giac [A]  time = 0.37, size = 150, normalized size = 1.32 \[ \frac {315 \, {\left (d x^{2} + c\right )}^{\frac {11}{2}} b^{2} - 1155 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} b^{2} c + 1485 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c^{2} - 693 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{3} + 770 \, {\left (d x^{2} + c\right )}^{\frac {9}{2}} a b d - 1980 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b c d + 1386 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c^{2} d + 495 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d^{2} - 693 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} c d^{2}}{3465 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(315*(d*x^2 + c)^(11/2)*b^2 - 1155*(d*x^2 + c)^(9/2)*b^2*c + 1485*(d*x^2 + c)^(7/2)*b^2*c^2 - 693*(d*x^
2 + c)^(5/2)*b^2*c^3 + 770*(d*x^2 + c)^(9/2)*a*b*d - 1980*(d*x^2 + c)^(7/2)*a*b*c*d + 1386*(d*x^2 + c)^(5/2)*a
*b*c^2*d + 495*(d*x^2 + c)^(7/2)*a^2*d^2 - 693*(d*x^2 + c)^(5/2)*a^2*c*d^2)/d^4

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maple [A]  time = 0.01, size = 108, normalized size = 0.95 \[ -\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} \left (-315 b^{2} x^{6} d^{3}-770 a b \,d^{3} x^{4}+210 b^{2} c \,d^{2} x^{4}-495 a^{2} d^{3} x^{2}+440 a b c \,d^{2} x^{2}-120 b^{2} c^{2} d \,x^{2}+198 a^{2} c \,d^{2}-176 a b \,c^{2} d +48 b^{2} c^{3}\right )}{3465 d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3465*(d*x^2+c)^(5/2)*(-315*b^2*d^3*x^6-770*a*b*d^3*x^4+210*b^2*c*d^2*x^4-495*a^2*d^3*x^2+440*a*b*c*d^2*x^2-
120*b^2*c^2*d*x^2+198*a^2*c*d^2-176*a*b*c^2*d+48*b^2*c^3)/d^4

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maxima [A]  time = 1.11, size = 181, normalized size = 1.59 \[ \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x^{6}}{11 \, d} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c x^{4}}{33 \, d^{2}} + \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b x^{4}}{9 \, d} + \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} x^{2}}{231 \, d^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c x^{2}}{63 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} x^{2}}{7 \, d} - \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{3}}{1155 \, d^{4}} + \frac {16 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c^{2}}{315 \, d^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} c}{35 \, d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11*(d*x^2 + c)^(5/2)*b^2*x^6/d - 2/33*(d*x^2 + c)^(5/2)*b^2*c*x^4/d^2 + 2/9*(d*x^2 + c)^(5/2)*a*b*x^4/d + 8/
231*(d*x^2 + c)^(5/2)*b^2*c^2*x^2/d^3 - 8/63*(d*x^2 + c)^(5/2)*a*b*c*x^2/d^2 + 1/7*(d*x^2 + c)^(5/2)*a^2*x^2/d
- 16/1155*(d*x^2 + c)^(5/2)*b^2*c^3/d^4 + 16/315*(d*x^2 + c)^(5/2)*a*b*c^2/d^3 - 2/35*(d*x^2 + c)^(5/2)*a^2*c
/d^2

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mupad [B]  time = 0.82, size = 170, normalized size = 1.49 \[ \sqrt {d\,x^2+c}\,\left (\frac {x^6\,\left (495\,a^2\,d^5+1100\,a\,b\,c\,d^4+15\,b^2\,c^2\,d^3\right )}{3465\,d^4}-\frac {198\,a^2\,c^3\,d^2-176\,a\,b\,c^4\,d+48\,b^2\,c^5}{3465\,d^4}+\frac {2\,b\,x^8\,\left (11\,a\,d+6\,b\,c\right )}{99}+\frac {b^2\,d\,x^{10}}{11}+\frac {2\,c\,x^4\,\left (132\,a^2\,d^2+11\,a\,b\,c\,d-3\,b^2\,c^2\right )}{1155\,d^2}+\frac {c^2\,x^2\,\left (99\,a^2\,d^2-88\,a\,b\,c\,d+24\,b^2\,c^2\right )}{3465\,d^3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(1/2)*((x^6*(495*a^2*d^5 + 15*b^2*c^2*d^3 + 1100*a*b*c*d^4))/(3465*d^4) - (48*b^2*c^5 + 198*a^2*c^
3*d^2 - 176*a*b*c^4*d)/(3465*d^4) + (2*b*x^8*(11*a*d + 6*b*c))/99 + (b^2*d*x^10)/11 + (2*c*x^4*(132*a^2*d^2 -
3*b^2*c^2 + 11*a*b*c*d))/(1155*d^2) + (c^2*x^2*(99*a^2*d^2 + 24*b^2*c^2 - 88*a*b*c*d))/(3465*d^3))

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sympy [A]  time = 8.40, size = 384, normalized size = 3.37 \[ \begin {cases} - \frac {2 a^{2} c^{3} \sqrt {c + d x^{2}}}{35 d^{2}} + \frac {a^{2} c^{2} x^{2} \sqrt {c + d x^{2}}}{35 d} + \frac {8 a^{2} c x^{4} \sqrt {c + d x^{2}}}{35} + \frac {a^{2} d x^{6} \sqrt {c + d x^{2}}}{7} + \frac {16 a b c^{4} \sqrt {c + d x^{2}}}{315 d^{3}} - \frac {8 a b c^{3} x^{2} \sqrt {c + d x^{2}}}{315 d^{2}} + \frac {2 a b c^{2} x^{4} \sqrt {c + d x^{2}}}{105 d} + \frac {20 a b c x^{6} \sqrt {c + d x^{2}}}{63} + \frac {2 a b d x^{8} \sqrt {c + d x^{2}}}{9} - \frac {16 b^{2} c^{5} \sqrt {c + d x^{2}}}{1155 d^{4}} + \frac {8 b^{2} c^{4} x^{2} \sqrt {c + d x^{2}}}{1155 d^{3}} - \frac {2 b^{2} c^{3} x^{4} \sqrt {c + d x^{2}}}{385 d^{2}} + \frac {b^{2} c^{2} x^{6} \sqrt {c + d x^{2}}}{231 d} + \frac {4 b^{2} c x^{8} \sqrt {c + d x^{2}}}{33} + \frac {b^{2} d x^{10} \sqrt {c + d x^{2}}}{11} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b^{2} x^{8}}{8}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*a**2*c**3*sqrt(c + d*x**2)/(35*d**2) + a**2*c**2*x**2*sqrt(c + d*x**2)/(35*d) + 8*a**2*c*x**4*sq
rt(c + d*x**2)/35 + a**2*d*x**6*sqrt(c + d*x**2)/7 + 16*a*b*c**4*sqrt(c + d*x**2)/(315*d**3) - 8*a*b*c**3*x**2
*sqrt(c + d*x**2)/(315*d**2) + 2*a*b*c**2*x**4*sqrt(c + d*x**2)/(105*d) + 20*a*b*c*x**6*sqrt(c + d*x**2)/63 +
2*a*b*d*x**8*sqrt(c + d*x**2)/9 - 16*b**2*c**5*sqrt(c + d*x**2)/(1155*d**4) + 8*b**2*c**4*x**2*sqrt(c + d*x**2
)/(1155*d**3) - 2*b**2*c**3*x**4*sqrt(c + d*x**2)/(385*d**2) + b**2*c**2*x**6*sqrt(c + d*x**2)/(231*d) + 4*b**
2*c*x**8*sqrt(c + d*x**2)/33 + b**2*d*x**10*sqrt(c + d*x**2)/11, Ne(d, 0)), (c**(3/2)*(a**2*x**4/4 + a*b*x**6/
3 + b**2*x**8/8), True))

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