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

Optimal. Leaf size=108 \[ -\frac {b \left (c+d x^2\right )^{3/2} (3 b c-2 a d)}{3 d^4}+\frac {\sqrt {c+d x^2} (b c-a d) (3 b c-a d)}{d^4}+\frac {c (b c-a d)^2}{d^4 \sqrt {c+d x^2}}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^4} \]

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Rubi [A]  time = 0.09, antiderivative size = 108, 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} \begin {gather*} -\frac {b \left (c+d x^2\right )^{3/2} (3 b c-2 a d)}{3 d^4}+\frac {\sqrt {c+d x^2} (b c-a d) (3 b c-a d)}{d^4}+\frac {c (b c-a d)^2}{d^4 \sqrt {c+d x^2}}+\frac {b^2 \left (c+d x^2\right )^{5/2}}{5 d^4} \end {gather*}

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

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Mathematica [A]  time = 0.06, size = 97, normalized size = 0.90 \begin {gather*} \frac {15 a^2 d^2 \left (2 c+d x^2\right )+10 a b d \left (-8 c^2-4 c d x^2+d^2 x^4\right )+3 b^2 \left (16 c^3+8 c^2 d x^2-2 c d^2 x^4+d^3 x^6\right )}{15 d^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*d^2*(2*c + d*x^2) + 10*a*b*d*(-8*c^2 - 4*c*d*x^2 + d^2*x^4) + 3*b^2*(16*c^3 + 8*c^2*d*x^2 - 2*c*d^2*x^
4 + d^3*x^6))/(15*d^4*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.06, size = 111, normalized size = 1.03 \begin {gather*} \frac {30 a^2 c d^2+15 a^2 d^3 x^2-80 a b c^2 d-40 a b c d^2 x^2+10 a b d^3 x^4+48 b^2 c^3+24 b^2 c^2 d x^2-6 b^2 c d^2 x^4+3 b^2 d^3 x^6}{15 d^4 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(48*b^2*c^3 - 80*a*b*c^2*d + 30*a^2*c*d^2 + 24*b^2*c^2*d*x^2 - 40*a*b*c*d^2*x^2 + 15*a^2*d^3*x^2 - 6*b^2*c*d^2
*x^4 + 10*a*b*d^3*x^4 + 3*b^2*d^3*x^6)/(15*d^4*Sqrt[c + d*x^2])

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fricas [A]  time = 0.75, size = 115, normalized size = 1.06 \begin {gather*} \frac {{\left (3 \, b^{2} d^{3} x^{6} + 48 \, b^{2} c^{3} - 80 \, a b c^{2} d + 30 \, a^{2} c d^{2} - 2 \, {\left (3 \, b^{2} c d^{2} - 5 \, a b d^{3}\right )} x^{4} + {\left (24 \, b^{2} c^{2} d - 40 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (d^{5} x^{2} + c d^{4}\right )}} \end {gather*}

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]

1/15*(3*b^2*d^3*x^6 + 48*b^2*c^3 - 80*a*b*c^2*d + 30*a^2*c*d^2 - 2*(3*b^2*c*d^2 - 5*a*b*d^3)*x^4 + (24*b^2*c^2
*d - 40*a*b*c*d^2 + 15*a^2*d^3)*x^2)*sqrt(d*x^2 + c)/(d^5*x^2 + c*d^4)

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giac [A]  time = 0.41, size = 149, normalized size = 1.38 \begin {gather*} \frac {b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}}{\sqrt {d x^{2} + c} d^{4}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d^{16} - 15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c d^{16} + 45 \, \sqrt {d x^{2} + c} b^{2} c^{2} d^{16} + 10 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{17} - 60 \, \sqrt {d x^{2} + c} a b c d^{17} + 15 \, \sqrt {d x^{2} + c} a^{2} d^{18}}{15 \, d^{20}} \end {gather*}

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]

(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)/(sqrt(d*x^2 + c)*d^4) + 1/15*(3*(d*x^2 + c)^(5/2)*b^2*d^16 - 15*(d*x^2 + c
)^(3/2)*b^2*c*d^16 + 45*sqrt(d*x^2 + c)*b^2*c^2*d^16 + 10*(d*x^2 + c)^(3/2)*a*b*d^17 - 60*sqrt(d*x^2 + c)*a*b*
c*d^17 + 15*sqrt(d*x^2 + c)*a^2*d^18)/d^20

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maple [A]  time = 0.01, size = 108, normalized size = 1.00 \begin {gather*} \frac {3 b^{2} x^{6} d^{3}+10 a b \,d^{3} x^{4}-6 b^{2} c \,d^{2} x^{4}+15 a^{2} d^{3} x^{2}-40 a b c \,d^{2} x^{2}+24 b^{2} c^{2} d \,x^{2}+30 a^{2} c \,d^{2}-80 a b \,c^{2} d +48 b^{2} c^{3}}{15 \sqrt {d \,x^{2}+c}\, d^{4}} \end {gather*}

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]

1/15*(3*b^2*d^3*x^6+10*a*b*d^3*x^4-6*b^2*c*d^2*x^4+15*a^2*d^3*x^2-40*a*b*c*d^2*x^2+24*b^2*c^2*d*x^2+30*a^2*c*d
^2-80*a*b*c^2*d+48*b^2*c^3)/(d*x^2+c)^(1/2)/d^4

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maxima [A]  time = 0.92, size = 180, normalized size = 1.67 \begin {gather*} \frac {b^{2} x^{6}}{5 \, \sqrt {d x^{2} + c} d} - \frac {2 \, b^{2} c x^{4}}{5 \, \sqrt {d x^{2} + c} d^{2}} + \frac {2 \, a b x^{4}}{3 \, \sqrt {d x^{2} + c} d} + \frac {8 \, b^{2} c^{2} x^{2}}{5 \, \sqrt {d x^{2} + c} d^{3}} - \frac {8 \, a b c x^{2}}{3 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a^{2} x^{2}}{\sqrt {d x^{2} + c} d} + \frac {16 \, b^{2} c^{3}}{5 \, \sqrt {d x^{2} + c} d^{4}} - \frac {16 \, a b c^{2}}{3 \, \sqrt {d x^{2} + c} d^{3}} + \frac {2 \, a^{2} c}{\sqrt {d x^{2} + c} d^{2}} \end {gather*}

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]

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

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mupad [B]  time = 0.88, size = 107, normalized size = 0.99 \begin {gather*} \frac {30\,a^2\,c\,d^2+15\,a^2\,d^3\,x^2-80\,a\,b\,c^2\,d-40\,a\,b\,c\,d^2\,x^2+10\,a\,b\,d^3\,x^4+48\,b^2\,c^3+24\,b^2\,c^2\,d\,x^2-6\,b^2\,c\,d^2\,x^4+3\,b^2\,d^3\,x^6}{15\,d^4\,\sqrt {d\,x^2+c}} \end {gather*}

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]

(48*b^2*c^3 + 30*a^2*c*d^2 + 15*a^2*d^3*x^2 + 3*b^2*d^3*x^6 + 24*b^2*c^2*d*x^2 - 6*b^2*c*d^2*x^4 - 80*a*b*c^2*
d + 10*a*b*d^3*x^4 - 40*a*b*c*d^2*x^2)/(15*d^4*(c + d*x^2)^(1/2))

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sympy [A]  time = 1.99, size = 236, normalized size = 2.19 \begin {gather*} \begin {cases} \frac {2 a^{2} c}{d^{2} \sqrt {c + d x^{2}}} + \frac {a^{2} x^{2}}{d \sqrt {c + d x^{2}}} - \frac {16 a b c^{2}}{3 d^{3} \sqrt {c + d x^{2}}} - \frac {8 a b c x^{2}}{3 d^{2} \sqrt {c + d x^{2}}} + \frac {2 a b x^{4}}{3 d \sqrt {c + d x^{2}}} + \frac {16 b^{2} c^{3}}{5 d^{4} \sqrt {c + d x^{2}}} + \frac {8 b^{2} c^{2} x^{2}}{5 d^{3} \sqrt {c + d x^{2}}} - \frac {2 b^{2} c x^{4}}{5 d^{2} \sqrt {c + d x^{2}}} + \frac {b^{2} x^{6}}{5 d \sqrt {c + d x^{2}}} & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b^{2} x^{8}}{8}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

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/(d**2*sqrt(c + d*x**2)) + a**2*x**2/(d*sqrt(c + d*x**2)) - 16*a*b*c**2/(3*d**3*sqrt(c + d*
x**2)) - 8*a*b*c*x**2/(3*d**2*sqrt(c + d*x**2)) + 2*a*b*x**4/(3*d*sqrt(c + d*x**2)) + 16*b**2*c**3/(5*d**4*sqr
t(c + d*x**2)) + 8*b**2*c**2*x**2/(5*d**3*sqrt(c + d*x**2)) - 2*b**2*c*x**4/(5*d**2*sqrt(c + d*x**2)) + b**2*x
**6/(5*d*sqrt(c + d*x**2)), Ne(d, 0)), ((a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8)/c**(3/2), True))

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