3.7.43 \(\int \frac {x^3 (a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {b \sqrt {c+d x^2} (3 b c-2 a d)}{d^4}-\frac {(b c-a d) (3 b c-a d)}{d^4 \sqrt {c+d x^2}}+\frac {c (b c-a d)^2}{3 d^4 \left (c+d x^2\right )^{3/2}}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^4} \]

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Rubi [A]  time = 0.09, antiderivative size = 110, 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 \sqrt {c+d x^2} (3 b c-2 a d)}{d^4}-\frac {(b c-a d) (3 b c-a d)}{d^4 \sqrt {c+d x^2}}+\frac {c (b c-a d)^2}{3 d^4 \left (c+d x^2\right )^{3/2}}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.06, size = 98, normalized size = 0.89 \begin {gather*} \frac {-a^2 d^2 \left (2 c+3 d x^2\right )+2 a b d \left (8 c^2+12 c d x^2+3 d^2 x^4\right )+b^2 \left (-16 c^3-24 c^2 d x^2-6 c d^2 x^4+d^3 x^6\right )}{3 d^4 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

(-(a^2*d^2*(2*c + 3*d*x^2)) + 2*a*b*d*(8*c^2 + 12*c*d*x^2 + 3*d^2*x^4) + b^2*(-16*c^3 - 24*c^2*d*x^2 - 6*c*d^2
*x^4 + d^3*x^6))/(3*d^4*(c + d*x^2)^(3/2))

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IntegrateAlgebraic [A]  time = 0.06, size = 110, normalized size = 1.00 \begin {gather*} \frac {-2 a^2 c d^2-3 a^2 d^3 x^2+16 a b c^2 d+24 a b c d^2 x^2+6 a b d^3 x^4-16 b^2 c^3-24 b^2 c^2 d x^2-6 b^2 c d^2 x^4+b^2 d^3 x^6}{3 d^4 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

(-16*b^2*c^3 + 16*a*b*c^2*d - 2*a^2*c*d^2 - 24*b^2*c^2*d*x^2 + 24*a*b*c*d^2*x^2 - 3*a^2*d^3*x^2 - 6*b^2*c*d^2*
x^4 + 6*a*b*d^3*x^4 + b^2*d^3*x^6)/(3*d^4*(c + d*x^2)^(3/2))

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fricas [A]  time = 1.52, size = 124, normalized size = 1.13 \begin {gather*} \frac {{\left (b^{2} d^{3} x^{6} - 16 \, b^{2} c^{3} + 16 \, a b c^{2} d - 2 \, a^{2} c d^{2} - 6 \, {\left (b^{2} c d^{2} - a b d^{3}\right )} x^{4} - 3 \, {\left (8 \, b^{2} c^{2} d - 8 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{3 \, {\left (d^{6} x^{4} + 2 \, c d^{5} x^{2} + c^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.39, size = 140, normalized size = 1.27 \begin {gather*} -\frac {9 \, {\left (d x^{2} + c\right )} b^{2} c^{2} - b^{2} c^{3} - 12 \, {\left (d x^{2} + c\right )} a b c d + 2 \, a b c^{2} d + 3 \, {\left (d x^{2} + c\right )} a^{2} d^{2} - a^{2} c d^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{4}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{8} - 9 \, \sqrt {d x^{2} + c} b^{2} c d^{8} + 6 \, \sqrt {d x^{2} + c} a b d^{9}}{3 \, d^{12}} \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)^(5/2),x, algorithm="giac")

[Out]

-1/3*(9*(d*x^2 + c)*b^2*c^2 - b^2*c^3 - 12*(d*x^2 + c)*a*b*c*d + 2*a*b*c^2*d + 3*(d*x^2 + c)*a^2*d^2 - a^2*c*d
^2)/((d*x^2 + c)^(3/2)*d^4) + 1/3*((d*x^2 + c)^(3/2)*b^2*d^8 - 9*sqrt(d*x^2 + c)*b^2*c*d^8 + 6*sqrt(d*x^2 + c)
*a*b*d^9)/d^12

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maple [A]  time = 0.01, size = 108, normalized size = 0.98 \begin {gather*} -\frac {-b^{2} x^{6} d^{3}-6 a b \,d^{3} x^{4}+6 b^{2} c \,d^{2} x^{4}+3 a^{2} d^{3} x^{2}-24 a b c \,d^{2} x^{2}+24 b^{2} c^{2} d \,x^{2}+2 a^{2} c \,d^{2}-16 a b \,c^{2} d +16 b^{2} c^{3}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} 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)^(5/2),x)

[Out]

-1/3*(-b^2*d^3*x^6-6*a*b*d^3*x^4+6*b^2*c*d^2*x^4+3*a^2*d^3*x^2-24*a*b*c*d^2*x^2+24*b^2*c^2*d*x^2+2*a^2*c*d^2-1
6*a*b*c^2*d+16*b^2*c^3)/(d*x^2+c)^(3/2)/d^4

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maxima [A]  time = 0.91, size = 181, normalized size = 1.65 \begin {gather*} \frac {b^{2} x^{6}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {2 \, b^{2} c x^{4}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} + \frac {2 \, a b x^{4}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {8 \, b^{2} c^{2} x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} + \frac {8 \, a b c x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} - \frac {a^{2} x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {16 \, b^{2} c^{3}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{4}} + \frac {16 \, a b c^{2}}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{3}} - \frac {2 \, a^{2} c}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} 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)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.78, size = 107, normalized size = 0.97 \begin {gather*} -\frac {2\,a^2\,c\,d^2+3\,a^2\,d^3\,x^2-16\,a\,b\,c^2\,d-24\,a\,b\,c\,d^2\,x^2-6\,a\,b\,d^3\,x^4+16\,b^2\,c^3+24\,b^2\,c^2\,d\,x^2+6\,b^2\,c\,d^2\,x^4-b^2\,d^3\,x^6}{3\,d^4\,{\left (d\,x^2+c\right )}^{3/2}} \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)^(5/2),x)

[Out]

-(16*b^2*c^3 + 2*a^2*c*d^2 + 3*a^2*d^3*x^2 - b^2*d^3*x^6 + 24*b^2*c^2*d*x^2 + 6*b^2*c*d^2*x^4 - 16*a*b*c^2*d -
6*a*b*d^3*x^4 - 24*a*b*c*d^2*x^2)/(3*d^4*(c + d*x^2)^(3/2))

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sympy [A]  time = 2.95, size = 454, normalized size = 4.13 \begin {gather*} \begin {cases} - \frac {2 a^{2} c d^{2}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} - \frac {3 a^{2} d^{3} x^{2}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} + \frac {16 a b c^{2} d}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} + \frac {24 a b c d^{2} x^{2}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} + \frac {6 a b d^{3} x^{4}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} - \frac {16 b^{2} c^{3}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} - \frac {24 b^{2} c^{2} d x^{2}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} - \frac {6 b^{2} c d^{2} x^{4}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \sqrt {c + d x^{2}}} + \frac {b^{2} d^{3} x^{6}}{3 c d^{4} \sqrt {c + d x^{2}} + 3 d^{5} x^{2} \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 {5}{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)**(5/2),x)

[Out]

Piecewise((-2*a**2*c*d**2/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)) - 3*a**2*d**3*x**2/(3*c*d
**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)) + 16*a*b*c**2*d/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*
sqrt(c + d*x**2)) + 24*a*b*c*d**2*x**2/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)) + 6*a*b*d**3
*x**4/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)) - 16*b**2*c**3/(3*c*d**4*sqrt(c + d*x**2) + 3
*d**5*x**2*sqrt(c + d*x**2)) - 24*b**2*c**2*d*x**2/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2))
- 6*b**2*c*d**2*x**4/(3*c*d**4*sqrt(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)) + b**2*d**3*x**6/(3*c*d**4*sqr
t(c + d*x**2) + 3*d**5*x**2*sqrt(c + d*x**2)), Ne(d, 0)), ((a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8)/c**(5/2),
True))

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