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

Optimal. Leaf size=214 \[ \frac{x^7 \left (a \left (15 a^2 b D-176 a^3 F+6 a b^2 C+8 b^3 B\right )+48 A b^4\right )}{105 a^4 b \left (a+b x^2\right )^{7/2}}+\frac{x^5 \left (a \left (-58 a^3 F+3 a b^2 C+4 b^3 B\right )+24 A b^4\right )}{15 a^3 b^2 \left (a+b x^2\right )^{7/2}}+\frac{x^3 \left (-10 a^4 F+a b^3 B+6 A b^4\right )}{3 a^2 b^3 \left (a+b x^2\right )^{7/2}}+\frac{x \left (A b^4-a^4 F\right )}{a b^4 \left (a+b x^2\right )^{7/2}}+\frac{F \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{9/2}} \]

[Out]

((A*b^4 - a^4*F)*x)/(a*b^4*(a + b*x^2)^(7/2)) + ((6*A*b^4 + a*b^3*B - 10*a^4*F)*x^3)/(3*a^2*b^3*(a + b*x^2)^(7
/2)) + ((24*A*b^4 + a*(4*b^3*B + 3*a*b^2*C - 58*a^3*F))*x^5)/(15*a^3*b^2*(a + b*x^2)^(7/2)) + ((48*A*b^4 + a*(
8*b^3*B + 6*a*b^2*C + 15*a^2*b*D - 176*a^3*F))*x^7)/(105*a^4*b*(a + b*x^2)^(7/2)) + (F*ArcTanh[(Sqrt[b]*x)/Sqr
t[a + b*x^2]])/b^(9/2)

________________________________________________________________________________________

Rubi [A]  time = 0.409623, antiderivative size = 250, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {1814, 1157, 385, 217, 206} \[ \frac{x \left (\frac{15 a^2 b D-176 a^3 F+6 a b^2 C+8 b^3 B}{b^4}+\frac{48 A}{a}\right )}{105 a^3 \sqrt{a+b x^2}}+\frac{x \left (a \left (-45 a^2 b D+122 a^3 F+3 a b^2 C+4 b^3 B\right )+24 A b^4\right )}{105 a^3 b^4 \left (a+b x^2\right )^{3/2}}+\frac{x \left (\frac{15 a^2 b D-22 a^3 F-8 a b^2 C+b^3 B}{b^4}+\frac{6 A}{a}\right )}{35 a \left (a+b x^2\right )^{5/2}}+\frac{x \left (\frac{A}{a}-\frac{a^2 b D+a^3 (-F)-a b^2 C+b^3 B}{b^4}\right )}{7 \left (a+b x^2\right )^{7/2}}+\frac{F \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A/a - (b^3*B - a*b^2*C + a^2*b*D - a^3*F)/b^4)*x)/(7*(a + b*x^2)^(7/2)) + (((6*A)/a + (b^3*B - 8*a*b^2*C + 1
5*a^2*b*D - 22*a^3*F)/b^4)*x)/(35*a*(a + b*x^2)^(5/2)) + ((24*A*b^4 + a*(4*b^3*B + 3*a*b^2*C - 45*a^2*b*D + 12
2*a^3*F))*x)/(105*a^3*b^4*(a + b*x^2)^(3/2)) + (((48*A)/a + (8*b^3*B + 6*a*b^2*C + 15*a^2*b*D - 176*a^3*F)/b^4
)*x)/(105*a^3*Sqrt[a + b*x^2]) + (F*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/b^(9/2)

Rule 1814

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

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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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{A+B x^2+C x^4+D x^6+F x^8}{\left (a+b x^2\right )^{9/2}} \, dx &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}-\frac{\int \frac{-6 A-\frac{a \left (b^3 B-a b^2 C+a^2 b D-a^3 F\right )}{b^4}-\frac{7 a \left (b^2 C-a b D+a^2 F\right ) x^2}{b^3}-\frac{7 a (b D-a F) x^4}{b^2}-\frac{7 a F x^6}{b}}{\left (a+b x^2\right )^{7/2}} \, dx}{7 a}\\ &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}+\frac{\left (\frac{6 A}{a}+\frac{b^3 B-8 a b^2 C+15 a^2 b D-22 a^3 F}{b^4}\right ) x}{35 a \left (a+b x^2\right )^{5/2}}+\frac{\int \frac{\frac{24 A b^4+4 a b^3 B+3 a^2 b^2 C-10 a^3 b D+17 a^4 F}{b^4}+\frac{35 a^2 (b D-2 a F) x^2}{b^3}+\frac{35 a^2 F x^4}{b^2}}{\left (a+b x^2\right )^{5/2}} \, dx}{35 a^2}\\ &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}+\frac{\left (\frac{6 A}{a}+\frac{b^3 B-8 a b^2 C+15 a^2 b D-22 a^3 F}{b^4}\right ) x}{35 a \left (a+b x^2\right )^{5/2}}+\frac{\left (24 A b^4+a \left (4 b^3 B+3 a b^2 C-45 a^2 b D+122 a^3 F\right )\right ) x}{105 a^3 b^4 \left (a+b x^2\right )^{3/2}}-\frac{\int \frac{-\frac{48 A b^4+8 a b^3 B+6 a^2 b^2 C+15 a^3 b D-71 a^4 F}{b^4}-\frac{105 a^3 F x^2}{b^3}}{\left (a+b x^2\right )^{3/2}} \, dx}{105 a^3}\\ &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}+\frac{\left (\frac{6 A}{a}+\frac{b^3 B-8 a b^2 C+15 a^2 b D-22 a^3 F}{b^4}\right ) x}{35 a \left (a+b x^2\right )^{5/2}}+\frac{\left (24 A b^4+a \left (4 b^3 B+3 a b^2 C-45 a^2 b D+122 a^3 F\right )\right ) x}{105 a^3 b^4 \left (a+b x^2\right )^{3/2}}+\frac{\left (\frac{48 A}{a}+\frac{8 b^3 B+6 a b^2 C+15 a^2 b D-176 a^3 F}{b^4}\right ) x}{105 a^3 \sqrt{a+b x^2}}+\frac{F \int \frac{1}{\sqrt{a+b x^2}} \, dx}{b^4}\\ &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}+\frac{\left (\frac{6 A}{a}+\frac{b^3 B-8 a b^2 C+15 a^2 b D-22 a^3 F}{b^4}\right ) x}{35 a \left (a+b x^2\right )^{5/2}}+\frac{\left (24 A b^4+a \left (4 b^3 B+3 a b^2 C-45 a^2 b D+122 a^3 F\right )\right ) x}{105 a^3 b^4 \left (a+b x^2\right )^{3/2}}+\frac{\left (\frac{48 A}{a}+\frac{8 b^3 B+6 a b^2 C+15 a^2 b D-176 a^3 F}{b^4}\right ) x}{105 a^3 \sqrt{a+b x^2}}+\frac{F \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{b^4}\\ &=\frac{\left (\frac{A}{a}-\frac{b^3 B-a b^2 C+a^2 b D-a^3 F}{b^4}\right ) x}{7 \left (a+b x^2\right )^{7/2}}+\frac{\left (\frac{6 A}{a}+\frac{b^3 B-8 a b^2 C+15 a^2 b D-22 a^3 F}{b^4}\right ) x}{35 a \left (a+b x^2\right )^{5/2}}+\frac{\left (24 A b^4+a \left (4 b^3 B+3 a b^2 C-45 a^2 b D+122 a^3 F\right )\right ) x}{105 a^3 b^4 \left (a+b x^2\right )^{3/2}}+\frac{\left (\frac{48 A}{a}+\frac{8 b^3 B+6 a b^2 C+15 a^2 b D-176 a^3 F}{b^4}\right ) x}{105 a^3 \sqrt{a+b x^2}}+\frac{F \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.445792, size = 197, normalized size = 0.92 \[ \frac{x \left (a^3 b^4 \left (105 A+35 B x^2+21 C x^4+15 D x^6\right )+2 a^2 b^5 x^2 \left (105 A+14 B x^2+3 C x^4\right )-406 a^5 b^2 F x^4-176 a^4 b^3 F x^6-350 a^6 b F x^2-105 a^7 F+8 a b^6 x^4 \left (21 A+B x^2\right )+48 A b^7 x^6\right )}{105 a^4 b^4 \left (a+b x^2\right )^{7/2}}+\frac{\sqrt{a} F \sqrt{\frac{b x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{9/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-105*a^7*F - 350*a^6*b*F*x^2 - 406*a^5*b^2*F*x^4 + 48*A*b^7*x^6 - 176*a^4*b^3*F*x^6 + 8*a*b^6*x^4*(21*A +
B*x^2) + 2*a^2*b^5*x^2*(105*A + 14*B*x^2 + 3*C*x^4) + a^3*b^4*(105*A + 35*B*x^2 + 21*C*x^4 + 15*D*x^6)))/(105*
a^4*b^4*(a + b*x^2)^(7/2)) + (Sqrt[a]*F*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(b^(9/2)*Sqrt[a + b*
x^2])

________________________________________________________________________________________

Maple [B]  time = 0.008, size = 427, normalized size = 2. \begin{align*} -{\frac{F{x}^{7}}{7\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{F{x}^{5}}{5\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}-{\frac{F{x}^{3}}{3\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{Fx}{{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{F\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{9}{2}}}}-{\frac{D{x}^{5}}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{5\,D{x}^{3}a}{8\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{15\,{a}^{2}Dx}{56\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{3\,aDx}{56\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{Dx}{14\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{Dx}{7\,{b}^{3}a}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{C{x}^{3}}{4\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}-{\frac{3\,aCx}{28\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{3\,Cx}{140\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{Cx}{35\,a{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,Cx}{35\,{b}^{2}{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{Bx}{7\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{Bx}{35\,ab} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{4\,Bx}{105\,{a}^{2}b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,Bx}{105\,{a}^{3}b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{Ax}{7\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{2}}}}+{\frac{6\,Ax}{35\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{2}}}}+{\frac{8\,Ax}{35\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{16\,Ax}{35\,{a}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*F*x^7/b/(b*x^2+a)^(7/2)-1/5*F/b^2*x^5/(b*x^2+a)^(5/2)-1/3*F/b^3*x^3/(b*x^2+a)^(3/2)-F/b^4*x/(b*x^2+a)^(1/
2)+F/b^(9/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-1/2*D*x^5/b/(b*x^2+a)^(7/2)-5/8*D/b^2*a*x^3/(b*x^2+a)^(7/2)-15/56*D
/b^3*a^2*x/(b*x^2+a)^(7/2)+3/56*D/b^3*a*x/(b*x^2+a)^(5/2)+1/14*D/b^3*x/(b*x^2+a)^(3/2)+1/7*D/b^3/a*x/(b*x^2+a)
^(1/2)-1/4*C*x^3/b/(b*x^2+a)^(7/2)-3/28*C/b^2*a*x/(b*x^2+a)^(7/2)+3/140*C/b^2*x/(b*x^2+a)^(5/2)+1/35*C/b^2/a*x
/(b*x^2+a)^(3/2)+2/35*C/b^2/a^2*x/(b*x^2+a)^(1/2)-1/7*B/b*x/(b*x^2+a)^(7/2)+1/35*B/b/a*x/(b*x^2+a)^(5/2)+4/105
*B*x/a^2/b/(b*x^2+a)^(3/2)+8/105*B*x/a^3/b/(b*x^2+a)^(1/2)+1/7*A*x/a/(b*x^2+a)^(7/2)+6/35*A/a^2*x/(b*x^2+a)^(5
/2)+8/35*A/a^3*x/(b*x^2+a)^(3/2)+16/35*A/a^4*x/(b*x^2+a)^(1/2)

________________________________________________________________________________________

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((F*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

________________________________________________________________________________________

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((F*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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.24285, size = 275, normalized size = 1.29 \begin{align*} -\frac{{\left ({\left (x^{2}{\left (\frac{{\left (176 \, F a^{4} b^{6} - 15 \, D a^{3} b^{7} - 6 \, C a^{2} b^{8} - 8 \, B a b^{9} - 48 \, A b^{10}\right )} x^{2}}{a^{4} b^{7}} + \frac{7 \,{\left (58 \, F a^{5} b^{5} - 3 \, C a^{3} b^{7} - 4 \, B a^{2} b^{8} - 24 \, A a b^{9}\right )}}{a^{4} b^{7}}\right )} + \frac{35 \,{\left (10 \, F a^{6} b^{4} - B a^{3} b^{7} - 6 \, A a^{2} b^{8}\right )}}{a^{4} b^{7}}\right )} x^{2} + \frac{105 \,{\left (F a^{7} b^{3} - A a^{3} b^{7}\right )}}{a^{4} b^{7}}\right )} x}{105 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}}} - \frac{F \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{b^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*((x^2*((176*F*a^4*b^6 - 15*D*a^3*b^7 - 6*C*a^2*b^8 - 8*B*a*b^9 - 48*A*b^10)*x^2/(a^4*b^7) + 7*(58*F*a^5
*b^5 - 3*C*a^3*b^7 - 4*B*a^2*b^8 - 24*A*a*b^9)/(a^4*b^7)) + 35*(10*F*a^6*b^4 - B*a^3*b^7 - 6*A*a^2*b^8)/(a^4*b
^7))*x^2 + 105*(F*a^7*b^3 - A*a^3*b^7)/(a^4*b^7))*x/(b*x^2 + a)^(7/2) - F*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)
))/b^(9/2)