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

Optimal. Leaf size=134 \[ \frac {x^5 \left (a (3 a C+4 b B)+24 A b^2\right )}{15 a^3 \left (a+b x^2\right )^{7/2}}+\frac {x^3 (a B+6 A b)}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {x^7 \left (a \left (15 a^2 D+6 a b C+8 b^2 B\right )+48 A b^3\right )}{105 a^4 \left (a+b x^2\right )^{7/2}}+\frac {A x}{a \left (a+b x^2\right )^{7/2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1813, 1803, 12, 264} \begin {gather*} \frac {x^7 \left (a \left (15 a^2 D+6 a b C+8 b^2 B\right )+48 A b^3\right )}{105 a^4 \left (a+b x^2\right )^{7/2}}+\frac {x^5 \left (a (3 a C+4 b B)+24 A b^2\right )}{15 a^3 \left (a+b x^2\right )^{7/2}}+\frac {x^3 (a B+6 A b)}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {A x}{a \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*x)/(a*(a + b*x^2)^(7/2)) + ((6*A*b + a*B)*x^3)/(3*a^2*(a + b*x^2)^(7/2)) + ((24*A*b^2 + a*(4*b*B + 3*a*C))*
x^5)/(15*a^3*(a + b*x^2)^(7/2)) + ((48*A*b^3 + a*(8*b^2*B + 6*a*b*C + 15*a^2*D))*x^7)/(105*a^4*(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 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}{\left (a+b x^2\right )^{9/2}} \, dx &=\frac {A x}{a \left (a+b x^2\right )^{7/2}}+\frac {\int \frac {x^2 \left (6 A b+a \left (B+C x^2+D x^4\right )\right )}{\left (a+b x^2\right )^{9/2}} \, dx}{a}\\ &=\frac {A x}{a \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a B) x^3}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {\int \frac {x^4 \left (4 b (6 A b+a B)+3 a \left (a C+a D x^2\right )\right )}{\left (a+b x^2\right )^{9/2}} \, dx}{3 a^2}\\ &=\frac {A x}{a \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a B) x^3}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {\left (24 A b^2+a (4 b B+3 a C)\right ) x^5}{15 a^3 \left (a+b x^2\right )^{7/2}}+\frac {\int \frac {\left (2 b \left (24 A b^2+4 a b B+3 a^2 C\right )+15 a^3 D\right ) x^6}{\left (a+b x^2\right )^{9/2}} \, dx}{15 a^3}\\ &=\frac {A x}{a \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a B) x^3}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {\left (24 A b^2+a (4 b B+3 a C)\right ) x^5}{15 a^3 \left (a+b x^2\right )^{7/2}}+\frac {\left (48 A b^3+a \left (8 b^2 B+6 a b C+15 a^2 D\right )\right ) \int \frac {x^6}{\left (a+b x^2\right )^{9/2}} \, dx}{15 a^3}\\ &=\frac {A x}{a \left (a+b x^2\right )^{7/2}}+\frac {(6 A b+a B) x^3}{3 a^2 \left (a+b x^2\right )^{7/2}}+\frac {\left (24 A b^2+a (4 b B+3 a C)\right ) x^5}{15 a^3 \left (a+b x^2\right )^{7/2}}+\frac {\left (48 A b^3+a \left (8 b^2 B+6 a b C+15 a^2 D\right )\right ) x^7}{105 a^4 \left (a+b x^2\right )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 98, normalized size = 0.73 \begin {gather*} \frac {a^3 \left (105 A x+35 B x^3+21 C x^5+15 D x^7\right )+2 a^2 b x^3 \left (105 A+14 B x^2+3 C x^4\right )+8 a b^2 x^5 \left (21 A+B x^2\right )+48 A b^3 x^7}{105 a^4 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(48*A*b^3*x^7 + 8*a*b^2*x^5*(21*A + B*x^2) + 2*a^2*b*x^3*(105*A + 14*B*x^2 + 3*C*x^4) + a^3*(105*A*x + 35*B*x^
3 + 21*C*x^5 + 15*D*x^7))/(105*a^4*(a + b*x^2)^(7/2))

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IntegrateAlgebraic [A]  time = 0.46, size = 112, normalized size = 0.84 \begin {gather*} \frac {105 a^3 A x+35 a^3 B x^3+21 a^3 C x^5+15 a^3 D x^7+210 a^2 A b x^3+28 a^2 b B x^5+6 a^2 b C x^7+168 a A b^2 x^5+8 a b^2 B x^7+48 A b^3 x^7}{105 a^4 \left (a+b x^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(105*a^3*A*x + 210*a^2*A*b*x^3 + 35*a^3*B*x^3 + 168*a*A*b^2*x^5 + 28*a^2*b*B*x^5 + 21*a^3*C*x^5 + 48*A*b^3*x^7
+ 8*a*b^2*B*x^7 + 6*a^2*b*C*x^7 + 15*a^3*D*x^7)/(105*a^4*(a + b*x^2)^(7/2))

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fricas [A]  time = 1.01, size = 141, normalized size = 1.05 \begin {gather*} \frac {{\left ({\left (15 \, D a^{3} + 6 \, C a^{2} b + 8 \, B a b^{2} + 48 \, A b^{3}\right )} x^{7} + 7 \, {\left (3 \, C a^{3} + 4 \, B a^{2} b + 24 \, A a b^{2}\right )} x^{5} + 105 \, A a^{3} x + 35 \, {\left (B a^{3} + 6 \, A a^{2} b\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((15*D*a^3 + 6*C*a^2*b + 8*B*a*b^2 + 48*A*b^3)*x^7 + 7*(3*C*a^3 + 4*B*a^2*b + 24*A*a*b^2)*x^5 + 105*A*a^
3*x + 35*(B*a^3 + 6*A*a^2*b)*x^3)*sqrt(b*x^2 + a)/(a^4*b^4*x^8 + 4*a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 +
a^8)

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giac [A]  time = 0.53, size = 131, normalized size = 0.98 \begin {gather*} \frac {{\left ({\left (x^{2} {\left (\frac {{\left (15 \, D a^{3} b^{3} + 6 \, C a^{2} b^{4} + 8 \, B a b^{5} + 48 \, A b^{6}\right )} x^{2}}{a^{4} b^{3}} + \frac {7 \, {\left (3 \, C a^{3} b^{3} + 4 \, B a^{2} b^{4} + 24 \, A a b^{5}\right )}}{a^{4} b^{3}}\right )} + \frac {35 \, {\left (B a^{3} b^{3} + 6 \, A a^{2} b^{4}\right )}}{a^{4} b^{3}}\right )} x^{2} + \frac {105 \, A}{a}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*((x^2*((15*D*a^3*b^3 + 6*C*a^2*b^4 + 8*B*a*b^5 + 48*A*b^6)*x^2/(a^4*b^3) + 7*(3*C*a^3*b^3 + 4*B*a^2*b^4
+ 24*A*a*b^5)/(a^4*b^3)) + 35*(B*a^3*b^3 + 6*A*a^2*b^4)/(a^4*b^3))*x^2 + 105*A/a)*x/(b*x^2 + a)^(7/2)

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maple [A]  time = 0.01, size = 109, normalized size = 0.81 \begin {gather*} \frac {\left (48 A \,b^{3} x^{6}+8 B a \,b^{2} x^{6}+6 a^{2} b C \,x^{6}+15 D a^{3} x^{6}+168 A a \,b^{2} x^{4}+28 B \,a^{2} b \,x^{4}+21 a^{3} C \,x^{4}+210 A \,a^{2} b \,x^{2}+35 B \,a^{3} x^{2}+105 A \,a^{3}\right ) x}{105 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*x*(48*A*b^3*x^6+8*B*a*b^2*x^6+6*C*a^2*b*x^6+15*D*a^3*x^6+168*A*a*b^2*x^4+28*B*a^2*b*x^4+21*C*a^3*x^4+210
*A*a^2*b*x^2+35*B*a^3*x^2+105*A*a^3)/(b*x^2+a)^(7/2)/a^4

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maxima [B]  time = 1.42, size = 335, normalized size = 2.50 \begin {gather*} -\frac {D x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {5 \, D a x^{3}}{8 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {C x^{3}}{4 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {16 \, A x}{35 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {6 \, A x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}} + \frac {A x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a} + \frac {D x}{14 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {D x}{7 \, \sqrt {b x^{2} + a} a b^{3}} + \frac {3 \, D a x}{56 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}} - \frac {15 \, D a^{2} x}{56 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} + \frac {3 \, C x}{140 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {2 \, C x}{35 \, \sqrt {b x^{2} + a} a^{2} b^{2}} + \frac {C x}{35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{2}} - \frac {3 \, C a x}{28 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} - \frac {B x}{7 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {8 \, B x}{105 \, \sqrt {b x^{2} + a} a^{3} b} + \frac {4 \, B x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} b} + \frac {B x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*D*x^5/((b*x^2 + a)^(7/2)*b) - 5/8*D*a*x^3/((b*x^2 + a)^(7/2)*b^2) - 1/4*C*x^3/((b*x^2 + a)^(7/2)*b) + 16/
35*A*x/(sqrt(b*x^2 + a)*a^4) + 8/35*A*x/((b*x^2 + a)^(3/2)*a^3) + 6/35*A*x/((b*x^2 + a)^(5/2)*a^2) + 1/7*A*x/(
(b*x^2 + a)^(7/2)*a) + 1/14*D*x/((b*x^2 + a)^(3/2)*b^3) + 1/7*D*x/(sqrt(b*x^2 + a)*a*b^3) + 3/56*D*a*x/((b*x^2
+ a)^(5/2)*b^3) - 15/56*D*a^2*x/((b*x^2 + a)^(7/2)*b^3) + 3/140*C*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*C*x/(sqrt(
b*x^2 + a)*a^2*b^2) + 1/35*C*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*C*a*x/((b*x^2 + a)^(7/2)*b^2) - 1/7*B*x/((b*x^
2 + a)^(7/2)*b) + 8/105*B*x/(sqrt(b*x^2 + a)*a^3*b) + 4/105*B*x/((b*x^2 + a)^(3/2)*a^2*b) + 1/35*B*x/((b*x^2 +
a)^(5/2)*a*b)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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