3.1.98 \(\int \frac {x^4 (A+B x+C x^2+D x^3)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=185 \[ \frac {3 (A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}-\frac {3 x (A b-5 a C)}{8 a b^3}+\frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}+\frac {(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}-\frac {x^2 (b B-3 a D)}{2 a b^3} \]

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Rubi [A]  time = 0.34, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1804, 801, 635, 205, 260} \begin {gather*} \frac {x^3 (4 x (b B-2 a D)-5 a C+A b)}{8 a b^2 \left (a+b x^2\right )}-\frac {3 x (A b-5 a C)}{8 a b^3}+\frac {3 (A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-x (A b-a C)\right )}{4 a b \left (a+b x^2\right )^2}-\frac {x^2 (b B-3 a D)}{2 a b^3}+\frac {(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3,x]

[Out]

(-3*(A*b - 5*a*C)*x)/(8*a*b^3) - ((b*B - 3*a*D)*x^2)/(2*a*b^3) - (x^4*(a*(B - (a*D)/b) - (A*b - a*C)*x))/(4*a*
b*(a + b*x^2)^2) + (x^3*(A*b - 5*a*C + 4*(b*B - 2*a*D)*x))/(8*a*b^2*(a + b*x^2)) + (3*(A*b - 5*a*C)*ArcTan[(Sq
rt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2)) + ((b*B - 3*a*D)*Log[a + b*x^2])/(2*b^4)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx &=-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}-\frac {\int \frac {x^3 \left (-4 a \left (B-\frac {a D}{b}\right )+(A b-5 a C) x-4 a D x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {\int \frac {x^2 (-3 a (A b-5 a C)-8 a (b B-3 a D) x)}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {\int \left (-\frac {3 a (A b-5 a C)}{b}-\frac {8 a (b B-3 a D) x}{b}+\frac {3 a^2 (A b-5 a C)+8 a^2 (b B-3 a D) x}{b \left (a+b x^2\right )}\right ) \, dx}{8 a^2 b^2}\\ &=-\frac {3 (A b-5 a C) x}{8 a b^3}-\frac {(b B-3 a D) x^2}{2 a b^3}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {\int \frac {3 a^2 (A b-5 a C)+8 a^2 (b B-3 a D) x}{a+b x^2} \, dx}{8 a^2 b^3}\\ &=-\frac {3 (A b-5 a C) x}{8 a b^3}-\frac {(b B-3 a D) x^2}{2 a b^3}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {(3 (A b-5 a C)) \int \frac {1}{a+b x^2} \, dx}{8 b^3}+\frac {(b B-3 a D) \int \frac {x}{a+b x^2} \, dx}{b^3}\\ &=-\frac {3 (A b-5 a C) x}{8 a b^3}-\frac {(b B-3 a D) x^2}{2 a b^3}-\frac {x^4 \left (a \left (B-\frac {a D}{b}\right )-(A b-a C) x\right )}{4 a b \left (a+b x^2\right )^2}+\frac {x^3 (A b-5 a C+4 (b B-2 a D) x)}{8 a b^2 \left (a+b x^2\right )}+\frac {3 (A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {(b B-3 a D) \log \left (a+b x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 139, normalized size = 0.75 \begin {gather*} \frac {\frac {2 a \left (a^2 D-a b (B+C x)+A b^2 x\right )}{\left (a+b x^2\right )^2}+\frac {-12 a^2 D+8 a b B+9 a b C x-5 A b^2 x}{a+b x^2}+\frac {3 \sqrt {b} (A b-5 a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}}+4 (b B-3 a D) \log \left (a+b x^2\right )+8 b C x+4 b D x^2}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3,x]

[Out]

(8*b*C*x + 4*b*D*x^2 + (8*a*b*B - 12*a^2*D - 5*A*b^2*x + 9*a*b*C*x)/(a + b*x^2) + (2*a*(a^2*D + A*b^2*x - a*b*
(B + C*x)))/(a + b*x^2)^2 + (3*Sqrt[b]*(A*b - 5*a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[a] + 4*(b*B - 3*a*D)*Lo
g[a + b*x^2])/(8*b^4)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{\left (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3,x]

[Out]

IntegrateAlgebraic[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2)^3, x]

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fricas [A]  time = 0.84, size = 574, normalized size = 3.10 \begin {gather*} \left [\frac {8 \, D a b^{3} x^{6} + 16 \, C a b^{3} x^{5} + 16 \, D a^{2} b^{2} x^{4} - 20 \, D a^{4} + 12 \, B a^{3} b + 10 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} - 16 \, {\left (D a^{3} b - B a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (5 \, C a b^{2} - A b^{3}\right )} x^{4} + 5 \, C a^{3} - A a^{2} b + 2 \, {\left (5 \, C a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, C a^{3} b - A a^{2} b^{2}\right )} x - 8 \, {\left (3 \, D a^{4} - B a^{3} b + {\left (3 \, D a^{2} b^{2} - B a b^{3}\right )} x^{4} + 2 \, {\left (3 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {4 \, D a b^{3} x^{6} + 8 \, C a b^{3} x^{5} + 8 \, D a^{2} b^{2} x^{4} - 10 \, D a^{4} + 6 \, B a^{3} b + 5 \, {\left (5 \, C a^{2} b^{2} - A a b^{3}\right )} x^{3} - 8 \, {\left (D a^{3} b - B a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (5 \, C a b^{2} - A b^{3}\right )} x^{4} + 5 \, C a^{3} - A a^{2} b + 2 \, {\left (5 \, C a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, C a^{3} b - A a^{2} b^{2}\right )} x - 4 \, {\left (3 \, D a^{4} - B a^{3} b + {\left (3 \, D a^{2} b^{2} - B a b^{3}\right )} x^{4} + 2 \, {\left (3 \, D a^{3} b - B a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(8*D*a*b^3*x^6 + 16*C*a*b^3*x^5 + 16*D*a^2*b^2*x^4 - 20*D*a^4 + 12*B*a^3*b + 10*(5*C*a^2*b^2 - A*a*b^3)*
x^3 - 16*(D*a^3*b - B*a^2*b^2)*x^2 + 3*((5*C*a*b^2 - A*b^3)*x^4 + 5*C*a^3 - A*a^2*b + 2*(5*C*a^2*b - A*a*b^2)*
x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*C*a^3*b - A*a^2*b^2)*x - 8*(3*D*a^4 - B*a
^3*b + (3*D*a^2*b^2 - B*a*b^3)*x^4 + 2*(3*D*a^3*b - B*a^2*b^2)*x^2)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2
+ a^3*b^4), 1/8*(4*D*a*b^3*x^6 + 8*C*a*b^3*x^5 + 8*D*a^2*b^2*x^4 - 10*D*a^4 + 6*B*a^3*b + 5*(5*C*a^2*b^2 - A*
a*b^3)*x^3 - 8*(D*a^3*b - B*a^2*b^2)*x^2 - 3*((5*C*a*b^2 - A*b^3)*x^4 + 5*C*a^3 - A*a^2*b + 2*(5*C*a^2*b - A*a
*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(5*C*a^3*b - A*a^2*b^2)*x - 4*(3*D*a^4 - B*a^3*b + (3*D*a^2*b^2
- B*a*b^3)*x^4 + 2*(3*D*a^3*b - B*a^2*b^2)*x^2)*log(b*x^2 + a))/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4)]

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giac [A]  time = 0.41, size = 157, normalized size = 0.85 \begin {gather*} -\frac {3 \, {\left (5 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} - \frac {{\left (3 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {D b^{3} x^{2} + 2 \, C b^{3} x}{2 \, b^{6}} - \frac {10 \, D a^{3} - 6 \, B a^{2} b - {\left (9 \, C a b^{2} - 5 \, A b^{3}\right )} x^{3} + 4 \, {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2} - {\left (7 \, C a^{2} b - 3 \, A a b^{2}\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

-3/8*(5*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(3*D*a - B*b)*log(b*x^2 + a)/b^4 + 1/2*(D*b^3*x
^2 + 2*C*b^3*x)/b^6 - 1/8*(10*D*a^3 - 6*B*a^2*b - (9*C*a*b^2 - 5*A*b^3)*x^3 + 4*(3*D*a^2*b - 2*B*a*b^2)*x^2 -
(7*C*a^2*b - 3*A*a*b^2)*x)/((b*x^2 + a)^2*b^4)

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maple [A]  time = 0.01, size = 235, normalized size = 1.27 \begin {gather*} -\frac {5 A \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b}+\frac {9 C a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {B a \,x^{2}}{\left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {3 D a^{2} x^{2}}{2 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {3 A a x}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {7 C \,a^{2} x}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}+\frac {3 A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}+\frac {3 B \,a^{2}}{4 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {15 C a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}-\frac {5 D a^{3}}{4 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {D x^{2}}{2 b^{3}}+\frac {B \ln \left (b \,x^{2}+a \right )}{2 b^{3}}+\frac {C x}{b^{3}}-\frac {3 D a \ln \left (b \,x^{2}+a \right )}{2 b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x)

[Out]

1/2/b^3*D*x^2+1/b^3*C*x-5/8/(b*x^2+a)^2*A/b*x^3+9/8/b^2/(b*x^2+a)^2*C*x^3*a+1/b^2/(b*x^2+a)^2*B*x^2*a-3/2/b^3/
(b*x^2+a)^2*D*x^2*a^2-3/8/(b*x^2+a)^2*A*a/b^2*x+7/8/b^3/(b*x^2+a)^2*a^2*C*x+3/4/(b*x^2+a)^2*B*a^2/b^3-5/4/b^4/
(b*x^2+a)^2*a^3*D+1/2*B/b^3*ln(b*x^2+a)-3/2/b^4*ln(b*x^2+a)*a*D+3/8/(a*b)^(1/2)*A/b^2*arctan(1/(a*b)^(1/2)*b*x
)-15/8/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*a*C

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maxima [A]  time = 3.07, size = 165, normalized size = 0.89 \begin {gather*} -\frac {10 \, D a^{3} - 6 \, B a^{2} b - {\left (9 \, C a b^{2} - 5 \, A b^{3}\right )} x^{3} + 4 \, {\left (3 \, D a^{2} b - 2 \, B a b^{2}\right )} x^{2} - {\left (7 \, C a^{2} b - 3 \, A a b^{2}\right )} x}{8 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} - \frac {3 \, {\left (5 \, C a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {D x^{2} + 2 \, C x}{2 \, b^{3}} - \frac {{\left (3 \, D a - B b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/8*(10*D*a^3 - 6*B*a^2*b - (9*C*a*b^2 - 5*A*b^3)*x^3 + 4*(3*D*a^2*b - 2*B*a*b^2)*x^2 - (7*C*a^2*b - 3*A*a*b^
2)*x)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4) - 3/8*(5*C*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*(D*x^2
+ 2*C*x)/b^3 - 1/2*(3*D*a - B*b)*log(b*x^2 + a)/b^4

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mupad [B]  time = 1.56, size = 232, normalized size = 1.25 \begin {gather*} \frac {\frac {7\,C\,a^2\,x}{8}+\frac {9\,C\,b\,a\,x^3}{8}}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}-\frac {\frac {5\,A\,x^3}{8\,b}+\frac {3\,A\,a\,x}{8\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {3\,B\,a^2}{4\,b^3}+\frac {B\,a\,x^2}{b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {D\,\left (3\,a\,\ln \left (b\,x^2+a\right )-b\,x^2+\frac {3\,a^2}{b\,x^2+a}-\frac {a^3}{2\,{\left (b\,x^2+a\right )}^2}\right )}{2\,b^4}+\frac {B\,\ln \left (b\,x^2+a\right )}{2\,b^3}+\frac {C\,x}{b^3}+\frac {3\,A\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,b^{5/2}}-\frac {15\,C\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2)^3,x)

[Out]

((7*C*a^2*x)/8 + (9*C*a*b*x^3)/8)/(a^2*b^3 + b^5*x^4 + 2*a*b^4*x^2) - ((5*A*x^3)/(8*b) + (3*A*a*x)/(8*b^2))/(a
^2 + b^2*x^4 + 2*a*b*x^2) + ((3*B*a^2)/(4*b^3) + (B*a*x^2)/b^2)/(a^2 + b^2*x^4 + 2*a*b*x^2) - (D*(3*a*log(a +
b*x^2) - b*x^2 + (3*a^2)/(a + b*x^2) - a^3/(2*(a + b*x^2)^2)))/(2*b^4) + (B*log(a + b*x^2))/(2*b^3) + (C*x)/b^
3 + (3*A*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(1/2)*b^(5/2)) - (15*C*a^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*b^(7/2))

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sympy [B]  time = 29.59, size = 357, normalized size = 1.93 \begin {gather*} \frac {C x}{b^{3}} + \frac {D x^{2}}{2 b^{3}} + \left (- \frac {- B b + 3 D a}{2 b^{4}} - \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log {\left (x + \frac {8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac {- B b + 3 D a}{2 b^{4}} - \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \left (- \frac {- B b + 3 D a}{2 b^{4}} + \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right ) \log {\left (x + \frac {8 B a b - 24 D a^{2} - 16 a b^{4} \left (- \frac {- B b + 3 D a}{2 b^{4}} + \frac {3 \sqrt {- a b^{9}} \left (- A b + 5 C a\right )}{16 a b^{8}}\right )}{- 3 A b^{2} + 15 C a b} \right )} + \frac {6 B a^{2} b - 10 D a^{3} + x^{3} \left (- 5 A b^{3} + 9 C a b^{2}\right ) + x^{2} \left (8 B a b^{2} - 12 D a^{2} b\right ) + x \left (- 3 A a b^{2} + 7 C a^{2} b\right )}{8 a^{2} b^{4} + 16 a b^{5} x^{2} + 8 b^{6} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(D*x**3+C*x**2+B*x+A)/(b*x**2+a)**3,x)

[Out]

C*x/b**3 + D*x**2/(2*b**3) + (-(-B*b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8))*log(x + (
8*B*a*b - 24*D*a**2 - 16*a*b**4*(-(-B*b + 3*D*a)/(2*b**4) - 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A
*b**2 + 15*C*a*b)) + (-(-B*b + 3*D*a)/(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8))*log(x + (8*B*a*b
- 24*D*a**2 - 16*a*b**4*(-(-B*b + 3*D*a)/(2*b**4) + 3*sqrt(-a*b**9)*(-A*b + 5*C*a)/(16*a*b**8)))/(-3*A*b**2 +
15*C*a*b)) + (6*B*a**2*b - 10*D*a**3 + x**3*(-5*A*b**3 + 9*C*a*b**2) + x**2*(8*B*a*b**2 - 12*D*a**2*b) + x*(-3
*A*a*b**2 + 7*C*a**2*b))/(8*a**2*b**4 + 16*a*b**5*x**2 + 8*b**6*x**4)

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