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

[Out]

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

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Rubi [A]  time = 0.41, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.51, size = 197, normalized size = 0.92 \[ \frac {x \left (-105 a^7 F-350 a^6 b F x^2-406 a^5 b^2 F x^4-176 a^4 b^3 F x^6+a^3 b^4 \left (105 A+35 B x^2+21 C x^4+15 D x^6\right )+2 a^2 b^5 x^2 \left (105 A+14 B x^2+3 C x^4\right )+8 a b^6 x^4 \left (21 A+B x^2\right )+48 A b^7 x^6\right )}{105 a^4 b^4 \left (a+b x^2\right )^{7/2}}+\frac {\sqrt {a} F \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{9/2} \sqrt {a+b x^2}} \]

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])

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fricas [A]  time = 1.17, size = 567, normalized size = 2.65 \[ \left [\frac {105 \, {\left (F a^{4} b^{4} x^{8} + 4 \, F a^{5} b^{3} x^{6} + 6 \, F a^{6} b^{2} x^{4} + 4 \, F a^{7} b x^{2} + F a^{8}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (176 \, F a^{4} b^{4} - 15 \, D a^{3} b^{5} - 6 \, C a^{2} b^{6} - 8 \, B a b^{7} - 48 \, A b^{8}\right )} x^{7} + 7 \, {\left (58 \, F a^{5} b^{3} - 3 \, C a^{3} b^{5} - 4 \, B a^{2} b^{6} - 24 \, A a b^{7}\right )} x^{5} + 35 \, {\left (10 \, F a^{6} b^{2} - B a^{3} b^{5} - 6 \, A a^{2} b^{6}\right )} x^{3} + 105 \, {\left (F a^{7} b - A a^{3} b^{5}\right )} x\right )} \sqrt {b x^{2} + a}}{210 \, {\left (a^{4} b^{9} x^{8} + 4 \, a^{5} b^{8} x^{6} + 6 \, a^{6} b^{7} x^{4} + 4 \, a^{7} b^{6} x^{2} + a^{8} b^{5}\right )}}, -\frac {105 \, {\left (F a^{4} b^{4} x^{8} + 4 \, F a^{5} b^{3} x^{6} + 6 \, F a^{6} b^{2} x^{4} + 4 \, F a^{7} b x^{2} + F a^{8}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (176 \, F a^{4} b^{4} - 15 \, D a^{3} b^{5} - 6 \, C a^{2} b^{6} - 8 \, B a b^{7} - 48 \, A b^{8}\right )} x^{7} + 7 \, {\left (58 \, F a^{5} b^{3} - 3 \, C a^{3} b^{5} - 4 \, B a^{2} b^{6} - 24 \, A a b^{7}\right )} x^{5} + 35 \, {\left (10 \, F a^{6} b^{2} - B a^{3} b^{5} - 6 \, A a^{2} b^{6}\right )} x^{3} + 105 \, {\left (F a^{7} b - A a^{3} b^{5}\right )} x\right )} \sqrt {b x^{2} + a}}{105 \, {\left (a^{4} b^{9} x^{8} + 4 \, a^{5} b^{8} x^{6} + 6 \, a^{6} b^{7} x^{4} + 4 \, a^{7} b^{6} x^{2} + a^{8} b^{5}\right )}}\right ] \]

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]

[1/210*(105*(F*a^4*b^4*x^8 + 4*F*a^5*b^3*x^6 + 6*F*a^6*b^2*x^4 + 4*F*a^7*b*x^2 + F*a^8)*sqrt(b)*log(-2*b*x^2 -
2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*((176*F*a^4*b^4 - 15*D*a^3*b^5 - 6*C*a^2*b^6 - 8*B*a*b^7 - 48*A*b^8)*x^7
+ 7*(58*F*a^5*b^3 - 3*C*a^3*b^5 - 4*B*a^2*b^6 - 24*A*a*b^7)*x^5 + 35*(10*F*a^6*b^2 - B*a^3*b^5 - 6*A*a^2*b^6)
*x^3 + 105*(F*a^7*b - A*a^3*b^5)*x)*sqrt(b*x^2 + a))/(a^4*b^9*x^8 + 4*a^5*b^8*x^6 + 6*a^6*b^7*x^4 + 4*a^7*b^6*
x^2 + a^8*b^5), -1/105*(105*(F*a^4*b^4*x^8 + 4*F*a^5*b^3*x^6 + 6*F*a^6*b^2*x^4 + 4*F*a^7*b*x^2 + F*a^8)*sqrt(-
b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + ((176*F*a^4*b^4 - 15*D*a^3*b^5 - 6*C*a^2*b^6 - 8*B*a*b^7 - 48*A*b^8)*x
^7 + 7*(58*F*a^5*b^3 - 3*C*a^3*b^5 - 4*B*a^2*b^6 - 24*A*a*b^7)*x^5 + 35*(10*F*a^6*b^2 - B*a^3*b^5 - 6*A*a^2*b^
6)*x^3 + 105*(F*a^7*b - A*a^3*b^5)*x)*sqrt(b*x^2 + a))/(a^4*b^9*x^8 + 4*a^5*b^8*x^6 + 6*a^6*b^7*x^4 + 4*a^7*b^
6*x^2 + a^8*b^5)]

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giac [A]  time = 0.59, size = 204, normalized size = 0.95 \[ -\frac {{\left ({\left (x^{2} {\left (\frac {{\left (176 \, F a^{4} b^{6} - 15 \, D a^{3} b^{7} - 6 \, C a^{2} b^{8} - 8 \, B a b^{9} - 48 \, A b^{10}\right )} x^{2}}{a^{4} b^{7}} + \frac {7 \, {\left (58 \, F a^{5} b^{5} - 3 \, C a^{3} b^{7} - 4 \, B a^{2} b^{8} - 24 \, A a b^{9}\right )}}{a^{4} b^{7}}\right )} + \frac {35 \, {\left (10 \, F a^{6} b^{4} - B a^{3} b^{7} - 6 \, A a^{2} b^{8}\right )}}{a^{4} b^{7}}\right )} x^{2} + \frac {105 \, {\left (F a^{7} b^{3} - A a^{3} b^{7}\right )}}{a^{4} b^{7}}\right )} x}{105 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} - \frac {F \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {9}{2}}} \]

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)

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maple [B]  time = 0.01, size = 427, normalized size = 2.00 \[ -\frac {F \,x^{7}}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {D x^{5}}{2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {F \,x^{5}}{5 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2}}-\frac {C \,x^{3}}{4 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {5 D a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}+\frac {A x}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a}-\frac {B x}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {3 C a x}{28 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}-\frac {15 D a^{2} x}{56 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{3}}-\frac {F \,x^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}+\frac {6 A x}{35 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{2}}+\frac {B x}{35 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a b}+\frac {3 C x}{140 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2}}+\frac {3 D a x}{56 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3}}+\frac {8 A x}{35 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3}}+\frac {4 B x}{105 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} b}+\frac {C x}{35 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,b^{2}}+\frac {D x}{14 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}+\frac {16 A x}{35 \sqrt {b \,x^{2}+a}\, a^{4}}+\frac {8 B x}{105 \sqrt {b \,x^{2}+a}\, a^{3} b}+\frac {2 C x}{35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}}+\frac {D x}{7 \sqrt {b \,x^{2}+a}\, a \,b^{3}}-\frac {F x}{\sqrt {b \,x^{2}+a}\, b^{4}}+\frac {F \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {9}{2}}} \]

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

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maxima [B]  time = 1.74, size = 597, normalized size = 2.79 \[ -\frac {1}{35} \, {\left (\frac {35 \, x^{6}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b} + \frac {70 \, a x^{4}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}} + \frac {56 \, a^{2} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{3}} + \frac {16 \, a^{3}}{{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{4}}\right )} F x - \frac {F x {\left (\frac {15 \, x^{4}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {20 \, a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}\right )}}{15 \, b} - \frac {D x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b} - \frac {F x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{3 \, b^{2}} - \frac {F a x^{3}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}} - \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 {139 \, F x}{105 \, \sqrt {b x^{2} + a} b^{4}} + \frac {17 \, F a x}{105 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}} - \frac {29 \, F a^{2} x}{35 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4}} + \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} + \frac {F \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {9}{2}}} \]

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]

-1/35*(35*x^6/((b*x^2 + a)^(7/2)*b) + 70*a*x^4/((b*x^2 + a)^(7/2)*b^2) + 56*a^2*x^2/((b*x^2 + a)^(7/2)*b^3) +
16*a^3/((b*x^2 + a)^(7/2)*b^4))*F*x - 1/15*F*x*(15*x^4/((b*x^2 + a)^(5/2)*b) + 20*a*x^2/((b*x^2 + a)^(5/2)*b^2
) + 8*a^2/((b*x^2 + a)^(5/2)*b^3))/b - 1/2*D*x^5/((b*x^2 + a)^(7/2)*b) - 1/3*F*x*(3*x^2/((b*x^2 + a)^(3/2)*b)
+ 2*a/((b*x^2 + a)^(3/2)*b^2))/b^2 - F*a*x^3/((b*x^2 + a)^(5/2)*b^3) - 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) + 139/105*F*x/(sqrt(b*x^2 + a)*b^4) + 17/105*F*a*x/((
b*x^2 + a)^(3/2)*b^4) - 29/35*F*a^2*x/((b*x^2 + a)^(5/2)*b^4) + 1/14*D*x/((b*x^2 + a)^(3/2)*b^3) + 1/7*D*x/(sq
rt(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) + F*arcsinh(b*x/sqrt(a*b))/b^(9/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x^2+C\,x^4+F\,x^8+x^6\,D}{{\left (b\,x^2+a\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*x^2 + C*x^4 + F*x^8 + 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 \[ \text {Timed out} \]

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

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