3.159 \(\int \frac{x^8 (A+B x^2+C x^4+D x^6)}{(a+b x^2)^{9/2}} \, dx\)

Optimal. Leaf size=381 \[ -\frac{x^9 \left (2 A b^3-a \left (23 a^2 D-16 a b C+9 b^2 B\right )\right )}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{x^9 \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/2}}-\frac{x^7 \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}-\frac{x^5 \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{30 a^2 b^5 \sqrt{a+b x^2}}+\frac{x^3 \sqrt{a+b x^2} \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{24 a^2 b^6}-\frac{x \sqrt{a+b x^2} \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{16 a b^7}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \left (198 a^2 b C-429 a^3 D-72 a b^2 B+16 A b^3\right )}{16 b^{15/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}} \]

[Out]

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x^9)/(7*a*(a + b*x^2)^(7/2)) - ((2*A*b^3 - a*(9*b^2*B - 16*a*b*C + 23*a
^2*D))*x^9)/(35*a^2*b^3*(a + b*x^2)^(5/2)) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x^7)/(210*a^2
*b^4*(a + b*x^2)^(3/2)) + (D*x^9)/(6*b^3*(a + b*x^2)^(3/2)) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*
D))*x^5)/(30*a^2*b^5*Sqrt[a + b*x^2]) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x*Sqrt[a + b*x^2])
/(16*a*b^7) + ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x^3*Sqrt[a + b*x^2])/(24*a^2*b^6) + ((16*A*b
^3 - 72*a*b^2*B + 198*a^2*b*C - 429*a^3*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(15/2))

________________________________________________________________________________________

Rubi [A]  time = 0.662335, antiderivative size = 381, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {1804, 1585, 1263, 1584, 459, 288, 321, 217, 206} \[ -\frac{x^9 \left (2 A b^3-a \left (23 a^2 D-16 a b C+9 b^2 B\right )\right )}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{x^9 \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/2}}-\frac{x^7 \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}-\frac{x^5 \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{30 a^2 b^5 \sqrt{a+b x^2}}+\frac{x^3 \sqrt{a+b x^2} \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{24 a^2 b^6}-\frac{x \sqrt{a+b x^2} \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{16 a b^7}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \left (198 a^2 b C-429 a^3 D-72 a b^2 B+16 A b^3\right )}{16 b^{15/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(9/2),x]

[Out]

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x^9)/(7*a*(a + b*x^2)^(7/2)) - ((2*A*b^3 - a*(9*b^2*B - 16*a*b*C + 23*a
^2*D))*x^9)/(35*a^2*b^3*(a + b*x^2)^(5/2)) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x^7)/(210*a^2
*b^4*(a + b*x^2)^(3/2)) + (D*x^9)/(6*b^3*(a + b*x^2)^(3/2)) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*
D))*x^5)/(30*a^2*b^5*Sqrt[a + b*x^2]) - ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x*Sqrt[a + b*x^2])
/(16*a*b^7) + ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*x^3*Sqrt[a + b*x^2])/(24*a^2*b^6) + ((16*A*b
^3 - 72*a*b^2*B + 198*a^2*b*C - 429*a^3*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(15/2))

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1263

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*
x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(2*d*f*(q + 1)), x] + Dist[f/(2*d*(
q + 1)), Int[(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*x*Qx + R*(m + 2*q + 3)*x, x], x], x]] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[q, -1] && GtQ[m, 0]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 459

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

Rule 288

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^8 \left (A+B x^2+C x^4+D x^6\right )}{\left (a+b x^2\right )^{9/2}} \, dx &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{x^7 \left (\left (2 A b-\frac{9 a \left (b^2 B-a b C+a^2 D\right )}{b^2}\right ) x-7 a \left (C-\frac{a D}{b}\right ) x^3-7 a D x^5\right )}{\left (a+b x^2\right )^{7/2}} \, dx}{7 a b}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{x^8 \left (2 A b-\frac{9 a \left (b^2 B-a b C+a^2 D\right )}{b^2}-7 a \left (C-\frac{a D}{b}\right ) x^2-7 a D x^4\right )}{\left (a+b x^2\right )^{7/2}} \, dx}{7 a b}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{x^7 \left (\left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}\right ) x+\frac{35 a^2 D x^3}{b}\right )}{\left (a+b x^2\right )^{5/2}} \, dx}{35 a^2 b}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{x^8 \left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}+\frac{35 a^2 D x^2}{b}\right )}{\left (a+b x^2\right )^{5/2}} \, dx}{35 a^2 b}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (\frac{315 a^3 D}{b}-6 b \left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}\right )\right ) \int \frac{x^8}{\left (a+b x^2\right )^{5/2}} \, dx}{210 a^2 b^2}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (\frac{315 a^3 D}{b}-6 b \left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}\right )\right ) \int \frac{x^6}{\left (a+b x^2\right )^{3/2}} \, dx}{90 a^2 b^3}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^5}{30 a^2 b^5 \sqrt{a+b x^2}}-\frac{\left (\frac{315 a^3 D}{b}-6 b \left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}\right )\right ) \int \frac{x^4}{\sqrt{a+b x^2}} \, dx}{18 a^2 b^4}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^5}{30 a^2 b^5 \sqrt{a+b x^2}}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^3 \sqrt{a+b x^2}}{24 a^2 b^6}+\frac{\left (\frac{315 a^3 D}{b}-6 b \left (8 A b-\frac{9 a \left (4 b^2 B-11 a b C+18 a^2 D\right )}{b^2}\right )\right ) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{24 a b^5}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^5}{30 a^2 b^5 \sqrt{a+b x^2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x \sqrt{a+b x^2}}{16 a b^7}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^3 \sqrt{a+b x^2}}{24 a^2 b^6}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{16 b^7}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^5}{30 a^2 b^5 \sqrt{a+b x^2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x \sqrt{a+b x^2}}{16 a b^7}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^3 \sqrt{a+b x^2}}{24 a^2 b^6}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 b^7}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^9}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\left (2 A b^3-a \left (9 b^2 B-16 a b C+23 a^2 D\right )\right ) x^9}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^7}{210 a^2 b^4 \left (a+b x^2\right )^{3/2}}+\frac{D x^9}{6 b^3 \left (a+b x^2\right )^{3/2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^5}{30 a^2 b^5 \sqrt{a+b x^2}}-\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x \sqrt{a+b x^2}}{16 a b^7}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) x^3 \sqrt{a+b x^2}}{24 a^2 b^6}+\frac{\left (16 A b^3-72 a b^2 B+198 a^2 b C-429 a^3 D\right ) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{15/2}}\\ \end{align*}

Mathematica [A]  time = 0.533611, size = 273, normalized size = 0.72 \[ \frac{x \left (-12 a^3 b^3 \left (140 A-2100 B x^2+6699 C x^4-6292 D x^6\right )+a^2 b^4 x^2 \left (-5600 A+29232 B x^2-34848 C x^4+5005 D x^6\right )+42 a^4 b^2 \left (180 B-1650 C x^2+4147 D x^4\right )-2310 a^5 b \left (9 C-65 D x^2\right )+45045 a^6 D-2 a b^5 x^4 \left (3248 A-6336 B x^2+1155 C x^4+455 D x^6\right )+4 b^6 x^6 \left (35 \left (6 B x^2+3 C x^4+2 D x^6\right )-704 A\right )\right )}{1680 b^7 \left (a+b x^2\right )^{7/2}}+\frac{\sqrt{a+b x^2} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (16 A b^3-3 a \left (143 a^2 D-66 a b C+24 b^2 B\right )\right )}{16 \sqrt{a} b^{15/2} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2 + C*x^4 + D*x^6))/(a + b*x^2)^(9/2),x]

[Out]

(x*(45045*a^6*D - 2310*a^5*b*(9*C - 65*D*x^2) + 42*a^4*b^2*(180*B - 1650*C*x^2 + 4147*D*x^4) - 12*a^3*b^3*(140
*A - 2100*B*x^2 + 6699*C*x^4 - 6292*D*x^6) - 2*a*b^5*x^4*(3248*A - 6336*B*x^2 + 1155*C*x^4 + 455*D*x^6) + a^2*
b^4*x^2*(-5600*A + 29232*B*x^2 - 34848*C*x^4 + 5005*D*x^6) + 4*b^6*x^6*(-704*A + 35*(6*B*x^2 + 3*C*x^4 + 2*D*x
^6))))/(1680*b^7*(a + b*x^2)^(7/2)) + ((16*A*b^3 - 3*a*(24*b^2*B - 66*a*b*C + 143*a^2*D))*Sqrt[a + b*x^2]*ArcS
inh[(Sqrt[b]*x)/Sqrt[a]])/(16*Sqrt[a]*b^(15/2)*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.301, size = 517, normalized size = 1.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(9/2),x)

[Out]

A/b^(9/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/4*C*x^11/b/(b*x^2+a)^(7/2)+99/8*C/b^(13/2)*a^2*ln(x*b^(1/2)+(b*x^2+a
)^(1/2))-1/7*A*x^7/b/(b*x^2+a)^(7/2)-1/5*A/b^2*x^5/(b*x^2+a)^(5/2)-1/3*A/b^3*x^3/(b*x^2+a)^(3/2)-A/b^4*x/(b*x^
2+a)^(1/2)+1/6*D*x^13/b/(b*x^2+a)^(7/2)-429/16*D/b^(15/2)*a^3*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/2*B*x^9/b/(b*x^2
+a)^(7/2)-9/2*B/b^(11/2)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+429/16*D/b^7*a^3*x/(b*x^2+a)^(1/2)-11/8*C/b^2*a*x^9/(
b*x^2+a)^(7/2)-99/56*C/b^3*a^2*x^7/(b*x^2+a)^(7/2)-99/40*C/b^4*a^2*x^5/(b*x^2+a)^(5/2)-33/8*C/b^5*a^2*x^3/(b*x
^2+a)^(3/2)-99/8*C/b^6*a^2*x/(b*x^2+a)^(1/2)+9/14*B/b^2*a*x^7/(b*x^2+a)^(7/2)+9/10*B/b^3*a*x^5/(b*x^2+a)^(5/2)
+3/2*B/b^4*a*x^3/(b*x^2+a)^(3/2)+9/2*B/b^5*a*x/(b*x^2+a)^(1/2)+429/112*D/b^4*a^3*x^7/(b*x^2+a)^(7/2)+429/80*D/
b^5*a^3*x^5/(b*x^2+a)^(5/2)+143/16*D/b^6*a^3*x^3/(b*x^2+a)^(3/2)-13/24*D/b^2*a*x^11/(b*x^2+a)^(7/2)+143/48*D/b
^3*a^2*x^9/(b*x^2+a)^(7/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(9/2),x)

[Out]

Timed out

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Giac [A]  time = 1.24753, size = 462, normalized size = 1.21 \begin{align*} \frac{{\left ({\left ({\left ({\left (35 \,{\left (2 \,{\left (\frac{4 \, D x^{2}}{b} - \frac{13 \, D a^{4} b^{11} - 6 \, C a^{3} b^{12}}{a^{3} b^{13}}\right )} x^{2} + \frac{143 \, D a^{5} b^{10} - 66 \, C a^{4} b^{11} + 24 \, B a^{3} b^{12}}{a^{3} b^{13}}\right )} x^{2} + \frac{176 \,{\left (429 \, D a^{6} b^{9} - 198 \, C a^{5} b^{10} + 72 \, B a^{4} b^{11} - 16 \, A a^{3} b^{12}\right )}}{a^{3} b^{13}}\right )} x^{2} + \frac{406 \,{\left (429 \, D a^{7} b^{8} - 198 \, C a^{6} b^{9} + 72 \, B a^{5} b^{10} - 16 \, A a^{4} b^{11}\right )}}{a^{3} b^{13}}\right )} x^{2} + \frac{350 \,{\left (429 \, D a^{8} b^{7} - 198 \, C a^{7} b^{8} + 72 \, B a^{6} b^{9} - 16 \, A a^{5} b^{10}\right )}}{a^{3} b^{13}}\right )} x^{2} + \frac{105 \,{\left (429 \, D a^{9} b^{6} - 198 \, C a^{8} b^{7} + 72 \, B a^{7} b^{8} - 16 \, A a^{6} b^{9}\right )}}{a^{3} b^{13}}\right )} x}{1680 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} + \frac{{\left (429 \, D a^{3} - 198 \, C a^{2} b + 72 \, B a b^{2} - 16 \, A b^{3}\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{15}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

1/1680*((((35*(2*(4*D*x^2/b - (13*D*a^4*b^11 - 6*C*a^3*b^12)/(a^3*b^13))*x^2 + (143*D*a^5*b^10 - 66*C*a^4*b^11
+ 24*B*a^3*b^12)/(a^3*b^13))*x^2 + 176*(429*D*a^6*b^9 - 198*C*a^5*b^10 + 72*B*a^4*b^11 - 16*A*a^3*b^12)/(a^3*
b^13))*x^2 + 406*(429*D*a^7*b^8 - 198*C*a^6*b^9 + 72*B*a^5*b^10 - 16*A*a^4*b^11)/(a^3*b^13))*x^2 + 350*(429*D*
a^8*b^7 - 198*C*a^7*b^8 + 72*B*a^6*b^9 - 16*A*a^5*b^10)/(a^3*b^13))*x^2 + 105*(429*D*a^9*b^6 - 198*C*a^8*b^7 +
72*B*a^7*b^8 - 16*A*a^6*b^9)/(a^3*b^13))*x/(b*x^2 + a)^(7/2) + 1/16*(429*D*a^3 - 198*C*a^2*b + 72*B*a*b^2 - 1
6*A*b^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(15/2)