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

[Out]

1/7*(A-a*(B*b^2-C*a*b+D*a^2)/b^3)*x^5/a/(b*x^2+a)^(7/2)+1/35*(2*A*b^3+a*(5*B*b^2-12*C*a*b+19*D*a^2))*x^5/a^2/b
^3/(b*x^2+a)^(5/2)+1/3*a*(C*b-3*D*a)*x/b^5/(b*x^2+a)^(3/2)+1/2*(2*C*b-9*D*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2)
)/b^(11/2)-1/3*(4*C*b-15*D*a)*x/b^5/(b*x^2+a)^(1/2)+1/2*D*x*(b*x^2+a)^(1/2)/b^5

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Rubi [A]  time = 0.39, antiderivative size = 210, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.50, size = 194, normalized size = 0.92 \[ \frac {105 a^{5/2} \sqrt {\frac {b x^2}{a}+1} \left (a+b x^2\right )^3 (2 b C-9 a D) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\sqrt {b} x \left (945 a^6 D-210 a^5 b \left (C-15 D x^2\right )+14 a^4 b^2 x^2 \left (261 D x^2-50 C\right )+4 a^3 b^3 x^4 \left (396 D x^2-203 C\right )+a^2 b^4 x^6 \left (105 D x^2-352 C\right )+6 a b^5 x^4 \left (7 A+5 B x^2\right )+12 A b^6 x^6\right )}{210 a^2 b^{11/2} \left (a+b x^2\right )^{7/2}} \]

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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fricas [A]  time = 0.96, size = 653, normalized size = 3.11 \[ \left [\frac {105 \, {\left ({\left (9 \, D a^{3} b^{4} - 2 \, C a^{2} b^{5}\right )} x^{8} + 9 \, D a^{7} - 2 \, C a^{6} b + 4 \, {\left (9 \, D a^{4} b^{3} - 2 \, C a^{3} b^{4}\right )} x^{6} + 6 \, {\left (9 \, D a^{5} b^{2} - 2 \, C a^{4} b^{3}\right )} x^{4} + 4 \, {\left (9 \, D a^{6} b - 2 \, C a^{5} b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (105 \, D a^{2} b^{5} x^{9} + 2 \, {\left (792 \, D a^{3} b^{4} - 176 \, C a^{2} b^{5} + 15 \, B a b^{6} + 6 \, A b^{7}\right )} x^{7} + 14 \, {\left (261 \, D a^{4} b^{3} - 58 \, C a^{3} b^{4} + 3 \, A a b^{6}\right )} x^{5} + 350 \, {\left (9 \, D a^{5} b^{2} - 2 \, C a^{4} b^{3}\right )} x^{3} + 105 \, {\left (9 \, D a^{6} b - 2 \, C a^{5} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{420 \, {\left (a^{2} b^{10} x^{8} + 4 \, a^{3} b^{9} x^{6} + 6 \, a^{4} b^{8} x^{4} + 4 \, a^{5} b^{7} x^{2} + a^{6} b^{6}\right )}}, \frac {105 \, {\left ({\left (9 \, D a^{3} b^{4} - 2 \, C a^{2} b^{5}\right )} x^{8} + 9 \, D a^{7} - 2 \, C a^{6} b + 4 \, {\left (9 \, D a^{4} b^{3} - 2 \, C a^{3} b^{4}\right )} x^{6} + 6 \, {\left (9 \, D a^{5} b^{2} - 2 \, C a^{4} b^{3}\right )} x^{4} + 4 \, {\left (9 \, D a^{6} b - 2 \, C a^{5} b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (105 \, D a^{2} b^{5} x^{9} + 2 \, {\left (792 \, D a^{3} b^{4} - 176 \, C a^{2} b^{5} + 15 \, B a b^{6} + 6 \, A b^{7}\right )} x^{7} + 14 \, {\left (261 \, D a^{4} b^{3} - 58 \, C a^{3} b^{4} + 3 \, A a b^{6}\right )} x^{5} + 350 \, {\left (9 \, D a^{5} b^{2} - 2 \, C a^{4} b^{3}\right )} x^{3} + 105 \, {\left (9 \, D a^{6} b - 2 \, C a^{5} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{210 \, {\left (a^{2} b^{10} x^{8} + 4 \, a^{3} b^{9} x^{6} + 6 \, a^{4} b^{8} x^{4} + 4 \, a^{5} b^{7} x^{2} + a^{6} b^{6}\right )}}\right ] \]

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]

[1/420*(105*((9*D*a^3*b^4 - 2*C*a^2*b^5)*x^8 + 9*D*a^7 - 2*C*a^6*b + 4*(9*D*a^4*b^3 - 2*C*a^3*b^4)*x^6 + 6*(9*
D*a^5*b^2 - 2*C*a^4*b^3)*x^4 + 4*(9*D*a^6*b - 2*C*a^5*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(
b)*x - a) + 2*(105*D*a^2*b^5*x^9 + 2*(792*D*a^3*b^4 - 176*C*a^2*b^5 + 15*B*a*b^6 + 6*A*b^7)*x^7 + 14*(261*D*a^
4*b^3 - 58*C*a^3*b^4 + 3*A*a*b^6)*x^5 + 350*(9*D*a^5*b^2 - 2*C*a^4*b^3)*x^3 + 105*(9*D*a^6*b - 2*C*a^5*b^2)*x)
*sqrt(b*x^2 + a))/(a^2*b^10*x^8 + 4*a^3*b^9*x^6 + 6*a^4*b^8*x^4 + 4*a^5*b^7*x^2 + a^6*b^6), 1/210*(105*((9*D*a
^3*b^4 - 2*C*a^2*b^5)*x^8 + 9*D*a^7 - 2*C*a^6*b + 4*(9*D*a^4*b^3 - 2*C*a^3*b^4)*x^6 + 6*(9*D*a^5*b^2 - 2*C*a^4
*b^3)*x^4 + 4*(9*D*a^6*b - 2*C*a^5*b^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (105*D*a^2*b^5*x^9
+ 2*(792*D*a^3*b^4 - 176*C*a^2*b^5 + 15*B*a*b^6 + 6*A*b^7)*x^7 + 14*(261*D*a^4*b^3 - 58*C*a^3*b^4 + 3*A*a*b^6)
*x^5 + 350*(9*D*a^5*b^2 - 2*C*a^4*b^3)*x^3 + 105*(9*D*a^6*b - 2*C*a^5*b^2)*x)*sqrt(b*x^2 + a))/(a^2*b^10*x^8 +
4*a^3*b^9*x^6 + 6*a^4*b^8*x^4 + 4*a^5*b^7*x^2 + a^6*b^6)]

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giac [A]  time = 0.60, size = 203, normalized size = 0.97 \[ \frac {{\left ({\left ({\left ({\left (\frac {105 \, D x^{2}}{b} + \frac {2 \, {\left (792 \, D a^{4} b^{7} - 176 \, C a^{3} b^{8} + 15 \, B a^{2} b^{9} + 6 \, A a b^{10}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac {14 \, {\left (261 \, D a^{5} b^{6} - 58 \, C a^{4} b^{7} + 3 \, A a^{2} b^{9}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac {350 \, {\left (9 \, D a^{6} b^{5} - 2 \, C a^{5} b^{6}\right )}}{a^{3} b^{9}}\right )} x^{2} + \frac {105 \, {\left (9 \, D a^{7} b^{4} - 2 \, C a^{6} b^{5}\right )}}{a^{3} b^{9}}\right )} x}{210 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}}} + \frac {{\left (9 \, D a - 2 \, C b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {11}{2}}} \]

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)

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maple [B]  time = 0.01, size = 405, normalized size = 1.93 \[ \frac {D x^{9}}{2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {C \,x^{7}}{7 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}+\frac {9 D a \,x^{7}}{14 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}-\frac {B \,x^{5}}{2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {C \,x^{5}}{5 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2}}+\frac {9 D a \,x^{5}}{10 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3}}-\frac {A \,x^{3}}{4 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b}-\frac {5 B a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}-\frac {3 A a x}{28 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}-\frac {15 B \,a^{2} x}{56 \left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{3}}-\frac {C \,x^{3}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}+\frac {3 D a \,x^{3}}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{4}}+\frac {3 A x}{140 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{2}}+\frac {3 B a x}{56 \left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3}}+\frac {A x}{35 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,b^{2}}+\frac {B x}{14 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}+\frac {2 A x}{35 \sqrt {b \,x^{2}+a}\, a^{2} b^{2}}+\frac {B x}{7 \sqrt {b \,x^{2}+a}\, a \,b^{3}}-\frac {C x}{\sqrt {b \,x^{2}+a}\, b^{4}}+\frac {9 D a x}{2 \sqrt {b \,x^{2}+a}\, b^{5}}+\frac {C \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {9}{2}}}-\frac {9 D a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {11}{2}}} \]

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

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maxima [B]  time = 1.69, size = 753, normalized size = 3.59 \[ \text {result too large to display} \]

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]

1/2*D*x^9/((b*x^2 + a)^(7/2)*b) - 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))*C*x + 9/70*(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))*D*a*x/b
+ 3/10*D*a*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^2 - 1/15*C*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*B*x^5/((b*x^2 + a)^(7/2)*b) + 3/2*D*a*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)
)/b^3 - 1/3*C*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b^2 + 9/2*D*a^2*x^3/((b*x^2 + a)^(
5/2)*b^4) - C*a*x^3/((b*x^2 + a)^(5/2)*b^3) - 5/8*B*a*x^3/((b*x^2 + a)^(7/2)*b^2) - 1/4*A*x^3/((b*x^2 + a)^(7/
2)*b) - 417/70*D*a*x/(sqrt(b*x^2 + a)*b^5) - 51/70*D*a^2*x/((b*x^2 + a)^(3/2)*b^5) + 261/70*D*a^3*x/((b*x^2 +
a)^(5/2)*b^5) + 139/105*C*x/(sqrt(b*x^2 + a)*b^4) + 17/105*C*a*x/((b*x^2 + a)^(3/2)*b^4) - 29/35*C*a^2*x/((b*x
^2 + a)^(5/2)*b^4) + 1/14*B*x/((b*x^2 + a)^(3/2)*b^3) + 1/7*B*x/(sqrt(b*x^2 + a)*a*b^3) + 3/56*B*a*x/((b*x^2 +
a)^(5/2)*b^3) - 15/56*B*a^2*x/((b*x^2 + a)^(7/2)*b^3) + 3/140*A*x/((b*x^2 + a)^(5/2)*b^2) + 2/35*A*x/(sqrt(b*
x^2 + a)*a^2*b^2) + 1/35*A*x/((b*x^2 + a)^(3/2)*a*b^2) - 3/28*A*a*x/((b*x^2 + a)^(7/2)*b^2) - 9/2*D*a*arcsinh(
b*x/sqrt(a*b))/b^(11/2) + C*arcsinh(b*x/sqrt(a*b))/b^(9/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^4*(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 \[ \text {Timed out} \]

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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