3.164 \(\int \frac {A+B x^2+C x^4+D x^6}{x^2 (a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=185 \[ -\frac {x^3 \left (48 A b^2-a (a C+6 b B)\right )}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {x (8 A b-a B)}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {2 b x^7 \left (4 b \left (48 A b^2-a (a C+6 b B)\right )-3 a^3 D\right )}{105 a^5 \left (a+b x^2\right )^{7/2}}-\frac {x^5 \left (4 b \left (48 A b^2-a (a C+6 b B)\right )-3 a^3 D\right )}{15 a^4 \left (a+b x^2\right )^{7/2}}-\frac {A}{a x \left (a+b x^2\right )^{7/2}} \]

[Out]

-A/a/x/(b*x^2+a)^(7/2)-(8*A*b-B*a)*x/a^2/(b*x^2+a)^(7/2)-1/3*(48*A*b^2-a*(6*B*b+C*a))*x^3/a^3/(b*x^2+a)^(7/2)-
1/15*(4*b*(48*A*b^2-a*(6*B*b+C*a))-3*a^3*D)*x^5/a^4/(b*x^2+a)^(7/2)-2/105*b*(4*b*(48*A*b^2-a*(6*B*b+C*a))-3*a^
3*D)*x^7/a^5/(b*x^2+a)^(7/2)

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Rubi [A]  time = 0.25, antiderivative size = 179, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1803, 1813, 12, 271, 264} \[ -\frac {2 b x^7 \left (-3 a^3 D-4 a b (a C+6 b B)+192 A b^3\right )}{105 a^5 \left (a+b x^2\right )^{7/2}}-\frac {x^5 \left (-3 a^3 D-4 a b (a C+6 b B)+192 A b^3\right )}{15 a^4 \left (a+b x^2\right )^{7/2}}-\frac {x^3 \left (48 A b^2-a (a C+6 b B)\right )}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {x (8 A b-a B)}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {A}{a x \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2 + C*x^4 + D*x^6)/(x^2*(a + b*x^2)^(9/2)),x]

[Out]

-(A/(a*x*(a + b*x^2)^(7/2))) - ((8*A*b - a*B)*x)/(a^2*(a + b*x^2)^(7/2)) - ((48*A*b^2 - a*(6*b*B + a*C))*x^3)/
(3*a^3*(a + b*x^2)^(7/2)) - ((192*A*b^3 - 4*a*b*(6*b*B + a*C) - 3*a^3*D)*x^5)/(15*a^4*(a + b*x^2)^(7/2)) - (2*
b*(192*A*b^3 - 4*a*b*(6*b*B + a*C) - 3*a^3*D)*x^7)/(105*a^5*(a + b*x^2)^(7/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 1803

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{A = Coeff[Pq, x, 0], Q = PolynomialQuotient
[Pq - Coeff[Pq, x, 0], x^2, x]}, Simp[(A*x^(m + 1)*(a + b*x^2)^(p + 1))/(a*(m + 1)), x] + Dist[1/(a*(m + 1)),
Int[x^(m + 2)*(a + b*x^2)^p*(a*(m + 1)*Q - A*b*(m + 2*(p + 1) + 1)), x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x^2] && IntegerQ[m/2] && ILtQ[(m + 1)/2 + p, 0] && LtQ[m + Expon[Pq, x] + 2*p + 1, 0]

Rule 1813

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{A = Coeff[Pq, x, 0], Q = PolynomialQuotient[Pq - Coef
f[Pq, x, 0], x^2, x]}, Simp[(A*x*(a + b*x^2)^(p + 1))/a, x] + Dist[1/a, Int[x^2*(a + b*x^2)^p*(a*Q - A*b*(2*p
+ 3)), x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x^2] && ILtQ[p + 1/2, 0] && LtQ[Expon[Pq, x] + 2*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x^2+C x^4+D x^6}{x^2 \left (a+b x^2\right )^{9/2}} \, dx &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {8 A b-a \left (B+C x^2+D x^4\right )}{\left (a+b x^2\right )^{9/2}} \, dx}{a}\\ &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {x^2 \left (6 b (8 A b-a B)+a \left (-a C-a D x^2\right )\right )}{\left (a+b x^2\right )^{9/2}} \, dx}{a^2}\\ &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\left (48 A b^2-a (6 b B+a C)\right ) x^3}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {\int \frac {\left (4 b \left (48 A b^2-6 a b B-a^2 C\right )-3 a^3 D\right ) x^4}{\left (a+b x^2\right )^{9/2}} \, dx}{3 a^3}\\ &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\left (48 A b^2-a (6 b B+a C)\right ) x^3}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {\left (192 A b^3-4 a b (6 b B+a C)-3 a^3 D\right ) \int \frac {x^4}{\left (a+b x^2\right )^{9/2}} \, dx}{3 a^3}\\ &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\left (48 A b^2-a (6 b B+a C)\right ) x^3}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {\left (192 A b^3-4 a b (6 b B+a C)-3 a^3 D\right ) x^5}{15 a^4 \left (a+b x^2\right )^{7/2}}-\frac {\left (2 b \left (192 A b^3-4 a b (6 b B+a C)-3 a^3 D\right )\right ) \int \frac {x^6}{\left (a+b x^2\right )^{9/2}} \, dx}{15 a^4}\\ &=-\frac {A}{a x \left (a+b x^2\right )^{7/2}}-\frac {(8 A b-a B) x}{a^2 \left (a+b x^2\right )^{7/2}}-\frac {\left (48 A b^2-a (6 b B+a C)\right ) x^3}{3 a^3 \left (a+b x^2\right )^{7/2}}-\frac {\left (192 A b^3-4 a b (6 b B+a C)-3 a^3 D\right ) x^5}{15 a^4 \left (a+b x^2\right )^{7/2}}-\frac {2 b \left (192 A b^3-4 a b (6 b B+a C)-3 a^3 D\right ) x^7}{105 a^5 \left (a+b x^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 133, normalized size = 0.72 \[ \frac {-7 a^4 \left (15 A-15 B x^2-5 C x^4-3 D x^6\right )+2 a^3 b x^2 \left (-420 A+105 B x^2+14 C x^4+3 D x^6\right )+8 a^2 b^2 x^4 \left (-210 A+21 B x^2+C x^4\right )+48 a b^3 x^6 \left (B x^2-28 A\right )-384 A b^4 x^8}{105 a^5 x \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(x^2*(a + b*x^2)^(9/2)),x]

[Out]

(-384*A*b^4*x^8 + 48*a*b^3*x^6*(-28*A + B*x^2) + 8*a^2*b^2*x^4*(-210*A + 21*B*x^2 + C*x^4) - 7*a^4*(15*A - 15*
B*x^2 - 5*C*x^4 - 3*D*x^6) + 2*a^3*b*x^2*(-420*A + 105*B*x^2 + 14*C*x^4 + 3*D*x^6))/(105*a^5*x*(a + b*x^2)^(7/
2))

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fricas [A]  time = 0.94, size = 182, normalized size = 0.98 \[ \frac {{\left (2 \, {\left (3 \, D a^{3} b + 4 \, C a^{2} b^{2} + 24 \, B a b^{3} - 192 \, A b^{4}\right )} x^{8} + 7 \, {\left (3 \, D a^{4} + 4 \, C a^{3} b + 24 \, B a^{2} b^{2} - 192 \, A a b^{3}\right )} x^{6} - 105 \, A a^{4} + 35 \, {\left (C a^{4} + 6 \, B a^{3} b - 48 \, A a^{2} b^{2}\right )} x^{4} + 105 \, {\left (B a^{4} - 8 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

1/105*(2*(3*D*a^3*b + 4*C*a^2*b^2 + 24*B*a*b^3 - 192*A*b^4)*x^8 + 7*(3*D*a^4 + 4*C*a^3*b + 24*B*a^2*b^2 - 192*
A*a*b^3)*x^6 - 105*A*a^4 + 35*(C*a^4 + 6*B*a^3*b - 48*A*a^2*b^2)*x^4 + 105*(B*a^4 - 8*A*a^3*b)*x^2)*sqrt(b*x^2
+ a)/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x)

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giac [A]  time = 0.58, size = 211, normalized size = 1.14 \[ \frac {{\left ({\left (x^{2} {\left (\frac {{\left (6 \, D a^{12} b^{4} + 8 \, C a^{11} b^{5} + 48 \, B a^{10} b^{6} - 279 \, A a^{9} b^{7}\right )} x^{2}}{a^{14} b^{3}} + \frac {7 \, {\left (3 \, D a^{13} b^{3} + 4 \, C a^{12} b^{4} + 24 \, B a^{11} b^{5} - 132 \, A a^{10} b^{6}\right )}}{a^{14} b^{3}}\right )} + \frac {35 \, {\left (C a^{13} b^{3} + 6 \, B a^{12} b^{4} - 30 \, A a^{11} b^{5}\right )}}{a^{14} b^{3}}\right )} x^{2} + \frac {105 \, {\left (B a^{13} b^{3} - 4 \, A a^{12} b^{4}\right )}}{a^{14} b^{3}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {2 \, A \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

1/105*((x^2*((6*D*a^12*b^4 + 8*C*a^11*b^5 + 48*B*a^10*b^6 - 279*A*a^9*b^7)*x^2/(a^14*b^3) + 7*(3*D*a^13*b^3 +
4*C*a^12*b^4 + 24*B*a^11*b^5 - 132*A*a^10*b^6)/(a^14*b^3)) + 35*(C*a^13*b^3 + 6*B*a^12*b^4 - 30*A*a^11*b^5)/(a
^14*b^3))*x^2 + 105*(B*a^13*b^3 - 4*A*a^12*b^4)/(a^14*b^3))*x/(b*x^2 + a)^(7/2) + 2*A*sqrt(b)/(((sqrt(b)*x - s
qrt(b*x^2 + a))^2 - a)*a^4)

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maple [A]  time = 0.01, size = 157, normalized size = 0.85 \[ -\frac {384 A \,b^{4} x^{8}-48 B a \,b^{3} x^{8}-8 C \,a^{2} b^{2} x^{8}-6 D a^{3} b \,x^{8}+1344 A a \,b^{3} x^{6}-168 B \,a^{2} b^{2} x^{6}-28 C \,a^{3} b \,x^{6}-21 D a^{4} x^{6}+1680 A \,a^{2} b^{2} x^{4}-210 B \,a^{3} b \,x^{4}-35 C \,a^{4} x^{4}+840 A \,a^{3} b \,x^{2}-105 B \,a^{4} x^{2}+105 A \,a^{4}}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x)

[Out]

-1/105*(384*A*b^4*x^8-48*B*a*b^3*x^8-8*C*a^2*b^2*x^8-6*D*a^3*b*x^8+1344*A*a*b^3*x^6-168*B*a^2*b^2*x^6-28*C*a^3
*b*x^6-21*D*a^4*x^6+1680*A*a^2*b^2*x^4-210*B*a^3*b*x^4-35*C*a^4*x^4+840*A*a^3*b*x^2-105*B*a^4*x^2+105*A*a^4)/(
b*x^2+a)^(7/2)/x/a^5

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maxima [A]  time = 1.47, size = 313, normalized size = 1.69 \[ -\frac {D x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {16 \, B x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {B x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} + \frac {3 \, D x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, D x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {D x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, D a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {C x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, C x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, C x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} - \frac {128 \, A b x}{35 \, \sqrt {b x^{2} + a} a^{5}} - \frac {64 \, A b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} - \frac {48 \, A b x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{3}} - \frac {8 \, A b x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a^{2}} - \frac {A}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x^6+C*x^4+B*x^2+A)/x^2/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

-1/4*D*x^3/((b*x^2 + a)^(7/2)*b) + 16/35*B*x/(sqrt(b*x^2 + a)*a^4) + 8/35*B*x/((b*x^2 + a)^(3/2)*a^3) + 6/35*B
*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*B*x/((b*x^2 + a)^(7/2)*a) + 3/140*D*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*D*x/(sqr
t(b*x^2 + a)*a^2*b^2) + 1/35*D*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*D*a*x/((b*x^2 + a)^(7/2)*b^2) - 1/7*C*x/((b*
x^2 + a)^(7/2)*b) + 8/105*C*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*C*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*C*x/((b*x^2
+ a)^(5/2)*a*b) - 128/35*A*b*x/(sqrt(b*x^2 + a)*a^5) - 64/35*A*b*x/((b*x^2 + a)^(3/2)*a^4) - 48/35*A*b*x/((b*
x^2 + a)^(5/2)*a^3) - 8/7*A*b*x/((b*x^2 + a)^(7/2)*a^2) - A/((b*x^2 + a)^(7/2)*a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x^2+C\,x^4+x^6\,D}{x^2\,{\left (b\,x^2+a\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2 + C*x^4 + x^6*D)/(x^2*(a + b*x^2)^(9/2)),x)

[Out]

int((A + B*x^2 + C*x^4 + x^6*D)/(x^2*(a + b*x^2)^(9/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((D*x**6+C*x**4+B*x**2+A)/x**2/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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