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

Optimal. Leaf size=210 \[ \frac{x^5 \left (a \left (19 a^2 D-12 a b C+5 b^2 B\right )+2 A b^3\right )}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{x^5 \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/2}}-\frac{x (4 b C-15 a D)}{3 b^5 \sqrt{a+b x^2}}+\frac{a x (b C-3 a D)}{3 b^5 \left (a+b x^2\right )^{3/2}}+\frac{(2 b C-9 a D) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{11/2}}+\frac{D x \sqrt{a+b x^2}}{2 b^5} \]

[Out]

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x^5)/(7*a*(a + b*x^2)^(7/2)) + ((2*A*b^3 + a*(5*b^2*B - 12*a*b*C + 19*a
^2*D))*x^5)/(35*a^2*b^3*(a + b*x^2)^(5/2)) + (a*(b*C - 3*a*D)*x)/(3*b^5*(a + b*x^2)^(3/2)) - ((4*b*C - 15*a*D)
*x)/(3*b^5*Sqrt[a + b*x^2]) + (D*x*Sqrt[a + b*x^2])/(2*b^5) + ((2*b*C - 9*a*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*
x^2]])/(2*b^(11/2))

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Rubi [A]  time = 0.387121, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {1804, 1585, 1263, 1584, 455, 1157, 388, 217, 206} \[ \frac{x^5 \left (a \left (19 a^2 D-12 a b C+5 b^2 B\right )+2 A b^3\right )}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{x^5 \left (A-\frac{a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{7 a \left (a+b x^2\right )^{7/2}}-\frac{x (4 b C-15 a D)}{3 b^5 \sqrt{a+b x^2}}+\frac{a x (b C-3 a D)}{3 b^5 \left (a+b x^2\right )^{3/2}}+\frac{(2 b C-9 a D) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{11/2}}+\frac{D x \sqrt{a+b x^2}}{2 b^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(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^5)/(7*a*(a + b*x^2)^(7/2)) + ((2*A*b^3 + a*(5*b^2*B - 12*a*b*C + 19*a
^2*D))*x^5)/(35*a^2*b^3*(a + b*x^2)^(5/2)) + (a*(b*C - 3*a*D)*x)/(3*b^5*(a + b*x^2)^(3/2)) - ((4*b*C - 15*a*D)
*x)/(3*b^5*Sqrt[a + b*x^2]) + (D*x*Sqrt[a + b*x^2])/(2*b^5) + ((2*b*C - 9*a*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*
x^2]])/(2*b^(11/2))

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 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(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*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

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^4 \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^5}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{x^3 \left (-\left (2 A b+\frac{5 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^5}{7 a \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{x^4 \left (-2 A b-\frac{5 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^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{x^3 \left (\frac{35 a^2 (b C-2 a D) x}{b^2}+\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^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{x^4 \left (\frac{35 a^2 (b C-2 a D)}{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^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{a (b C-3 a D) x}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{\frac{35 a^3 (b C-3 a D)}{b}-105 a^2 (b C-3 a D) x^2-105 a^2 b D x^4}{\left (a+b x^2\right )^{3/2}} \, dx}{105 a^2 b^4}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{a (b C-3 a D) x}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{(4 b C-15 a D) x}{3 b^5 \sqrt{a+b x^2}}+\frac{\int \frac{\frac{105 a^3 (b C-4 a D)}{b}+105 a^3 D x^2}{\sqrt{a+b x^2}} \, dx}{105 a^3 b^4}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{a (b C-3 a D) x}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{(4 b C-15 a D) x}{3 b^5 \sqrt{a+b x^2}}+\frac{D x \sqrt{a+b x^2}}{2 b^5}+\frac{(2 b C-9 a D) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{2 b^5}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{a (b C-3 a D) x}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{(4 b C-15 a D) x}{3 b^5 \sqrt{a+b x^2}}+\frac{D x \sqrt{a+b x^2}}{2 b^5}+\frac{(2 b C-9 a D) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 b^5}\\ &=\frac{\left (A-\frac{a \left (b^2 B-a b C+a^2 D\right )}{b^3}\right ) x^5}{7 a \left (a+b x^2\right )^{7/2}}+\frac{\left (2 A b^3+a \left (5 b^2 B-12 a b C+19 a^2 D\right )\right ) x^5}{35 a^2 b^3 \left (a+b x^2\right )^{5/2}}+\frac{a (b C-3 a D) x}{3 b^5 \left (a+b x^2\right )^{3/2}}-\frac{(4 b C-15 a D) x}{3 b^5 \sqrt{a+b x^2}}+\frac{D x \sqrt{a+b x^2}}{2 b^5}+\frac{(2 b C-9 a D) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.440886, size = 194, normalized size = 0.92 \[ \frac{\sqrt{b} x \left (14 a^4 b^2 x^2 \left (261 D x^2-50 C\right )+4 a^3 b^3 x^4 \left (396 D x^2-203 C\right )+a^2 b^4 x^6 \left (105 D x^2-352 C\right )-210 a^5 b \left (C-15 D x^2\right )+945 a^6 D+6 a b^5 x^4 \left (7 A+5 B x^2\right )+12 A b^6 x^6\right )+105 a^{5/2} \sqrt{\frac{b x^2}{a}+1} \left (a+b x^2\right )^3 (2 b C-9 a D) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{210 a^2 b^{11/2} \left (a+b x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*(945*a^6*D + 12*A*b^6*x^6 + 6*a*b^5*x^4*(7*A + 5*B*x^2) - 210*a^5*b*(C - 15*D*x^2) + a^2*b^4*x^6*(-
352*C + 105*D*x^2) + 14*a^4*b^2*x^2*(-50*C + 261*D*x^2) + 4*a^3*b^3*x^4*(-203*C + 396*D*x^2)) + 105*a^(5/2)*(2
*b*C - 9*a*D)*(a + b*x^2)^3*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(210*a^2*b^(11/2)*(a + b*x^2)^(7
/2))

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Maple [B]  time = 0.012, size = 405, normalized size = 1.9 \begin{align*}{\frac{D{x}^{9}}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{9\,aD{x}^{7}}{14\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{9\,aD{x}^{5}}{10\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{3\,D{x}^{3}a}{2\,{b}^{4}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{9\,aDx}{2\,{b}^{5}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{9\,aD}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{11}{2}}}}-{\frac{C{x}^{7}}{7\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{C{x}^{5}}{5\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}-{\frac{C{x}^{3}}{3\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{Cx}{{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{C\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{9}{2}}}}-{\frac{B{x}^{5}}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{5\,aB{x}^{3}}{8\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{15\,{a}^{2}Bx}{56\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{3\,Bax}{56\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{Bx}{14\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{Bx}{7\,{b}^{3}a}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{A{x}^{3}}{4\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{3\,aAx}{28\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{3\,Ax}{140\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{Ax}{35\,{b}^{2}a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Ax}{35\,{b}^{2}{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*D*x^9/b/(b*x^2+a)^(7/2)+9/14*D/b^2*a*x^7/(b*x^2+a)^(7/2)+9/10*D/b^3*a*x^5/(b*x^2+a)^(5/2)+3/2*D/b^4*a*x^3/
(b*x^2+a)^(3/2)+9/2*D/b^5*a*x/(b*x^2+a)^(1/2)-9/2*D/b^(11/2)*a*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-1/7*C*x^7/b/(b*x^
2+a)^(7/2)-1/5*C/b^2*x^5/(b*x^2+a)^(5/2)-1/3*C/b^3*x^3/(b*x^2+a)^(3/2)-C/b^4*x/(b*x^2+a)^(1/2)+C/b^(9/2)*ln(x*
b^(1/2)+(b*x^2+a)^(1/2))-1/2*B*x^5/b/(b*x^2+a)^(7/2)-5/8*B/b^2*a*x^3/(b*x^2+a)^(7/2)-15/56*B/b^3*a^2*x/(b*x^2+
a)^(7/2)+3/56*B/b^3*a*x/(b*x^2+a)^(5/2)+1/14*B/b^3*x/(b*x^2+a)^(3/2)+1/7*B/b^3/a*x/(b*x^2+a)^(1/2)-1/4*A*x^3/b
/(b*x^2+a)^(7/2)-3/28*A/b^2*a*x/(b*x^2+a)^(7/2)+3/140*A/b^2*x/(b*x^2+a)^(5/2)+1/35*A/b^2/a*x/(b*x^2+a)^(3/2)+2
/35*A/b^2/a^2*x/(b*x^2+a)^(1/2)

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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^4*(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^4*(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**4*(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 = 1.20251, size = 274, normalized size = 1.3 \begin{align*} \frac{{\left ({\left ({\left ({\left (\frac{105 \, D x^{2}}{b} + \frac{2 \,{\left (792 \, D a^{4} b^{7} - 176 \, C a^{3} b^{8} + 15 \, B a^{2} b^{9} + 6 \, A a b^{10}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac{14 \,{\left (261 \, D a^{5} b^{6} - 58 \, C a^{4} b^{7} + 3 \, A a^{2} b^{9}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac{350 \,{\left (9 \, D a^{6} b^{5} - 2 \, C a^{5} b^{6}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac{105 \,{\left (9 \, D a^{7} b^{4} - 2 \, C a^{6} b^{5}\right )}}{a^{3} b^{9}}\right )} x}{210 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} + \frac{{\left (9 \, D a - 2 \, C b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*((((105*D*x^2/b + 2*(792*D*a^4*b^7 - 176*C*a^3*b^8 + 15*B*a^2*b^9 + 6*A*a*b^10)/(a^3*b^9))*x^2 + 14*(261
*D*a^5*b^6 - 58*C*a^4*b^7 + 3*A*a^2*b^9)/(a^3*b^9))*x^2 + 350*(9*D*a^6*b^5 - 2*C*a^5*b^6)/(a^3*b^9))*x^2 + 105
*(9*D*a^7*b^4 - 2*C*a^6*b^5)/(a^3*b^9))*x/(b*x^2 + a)^(7/2) + 1/2*(9*D*a - 2*C*b)*log(abs(-sqrt(b)*x + sqrt(b*
x^2 + a)))/b^(11/2)