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

Optimal. Leaf size=174 \[ -\frac {3 (5 b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {(3 A b-a C) \log \left (a+b x^2\right )}{2 a^4}-\frac {\log (x) (3 A b-a C)}{a^4}-\frac {4 (2 A b-a C)+x (7 b B-3 a D)}{8 a^3 \left (a+b x^2\right )}-\frac {A}{2 a^3 x^2}-\frac {B}{a^3 x}-\frac {\frac {A b}{a}+x \left (\frac {b B}{a}-D\right )-C}{4 a \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 174, 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 = {1805, 1802, 635, 205, 260} \begin {gather*} -\frac {4 (2 A b-a C)+x (7 b B-3 a D)}{8 a^3 \left (a+b x^2\right )}+\frac {(3 A b-a C) \log \left (a+b x^2\right )}{2 a^4}-\frac {\log (x) (3 A b-a C)}{a^4}-\frac {A}{2 a^3 x^2}-\frac {3 (5 b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {B}{a^3 x}-\frac {\frac {A b}{a}+x \left (\frac {b B}{a}-D\right )-C}{4 a \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(2*a^3*x^2) - B/(a^3*x) - ((A*b)/a - C + ((b*B)/a - D)*x)/(4*a*(a + b*x^2)^2) - (4*(2*A*b - a*C) + (7*b*B -
3*a*D)*x)/(8*a^3*(a + b*x^2)) - (3*(5*b*B - a*D)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]) - ((3*A*b -
a*C)*Log[x])/a^4 + ((3*A*b - a*C)*Log[a + b*x^2])/(2*a^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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*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[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 147, normalized size = 0.84 \begin {gather*} \frac {\frac {2 a^2 (a (C+D x)-A b-b B x)}{\left (a+b x^2\right )^2}+\frac {a (4 a C+3 a D x-8 A b-7 b B x)}{a+b x^2}+4 (3 A b-a C) \log \left (a+b x^2\right )+8 \log (x) (a C-3 A b)-\frac {4 a A}{x^2}+\frac {3 \sqrt {a} (a D-5 b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {8 a B}{x}}{8 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.83, size = 696, normalized size = 4.00 \begin {gather*} \left [-\frac {16 \, B a^{3} b x - 6 \, {\left (D a^{2} b^{2} - 5 \, B a b^{3}\right )} x^{5} + 8 \, A a^{3} b - 8 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} - 10 \, {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x^{3} - 12 \, {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (D a b^{2} - 5 \, B b^{3}\right )} x^{6} + 2 \, {\left (D a^{2} b - 5 \, B a b^{2}\right )} x^{4} + {\left (D a^{3} - 5 \, B a^{2} b\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 8 \, {\left ({\left (C a b^{3} - 3 \, A b^{4}\right )} x^{6} + 2 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 16 \, {\left ({\left (C a b^{3} - 3 \, A b^{4}\right )} x^{6} + 2 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \relax (x)}{16 \, {\left (a^{4} b^{3} x^{6} + 2 \, a^{5} b^{2} x^{4} + a^{6} b x^{2}\right )}}, -\frac {8 \, B a^{3} b x - 3 \, {\left (D a^{2} b^{2} - 5 \, B a b^{3}\right )} x^{5} + 4 \, A a^{3} b - 4 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} - 5 \, {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x^{3} - 6 \, {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (D a b^{2} - 5 \, B b^{3}\right )} x^{6} + 2 \, {\left (D a^{2} b - 5 \, B a b^{2}\right )} x^{4} + {\left (D a^{3} - 5 \, B a^{2} b\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 4 \, {\left ({\left (C a b^{3} - 3 \, A b^{4}\right )} x^{6} + 2 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 8 \, {\left ({\left (C a b^{3} - 3 \, A b^{4}\right )} x^{6} + 2 \, {\left (C a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + {\left (C a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \relax (x)}{8 \, {\left (a^{4} b^{3} x^{6} + 2 \, a^{5} b^{2} x^{4} + a^{6} b x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 162, normalized size = 0.93 \begin {gather*} \frac {3 \, {\left (D a - 5 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {{\left (C a - 3 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {{\left (C a - 3 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {3 \, D a b x^{5} - 15 \, B b^{2} x^{5} + 4 \, C a b x^{4} - 12 \, A b^{2} x^{4} + 5 \, D a^{2} x^{3} - 25 \, B a b x^{3} + 6 \, C a^{2} x^{2} - 18 \, A a b x^{2} - 8 \, B a^{2} x - 4 \, A a^{2}}{8 \, {\left (b x^{3} + a x\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(D*a - 5*B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/2*(C*a - 3*A*b)*log(b*x^2 + a)/a^4 + (C*a - 3*A*b)
*log(abs(x))/a^4 + 1/8*(3*D*a*b*x^5 - 15*B*b^2*x^5 + 4*C*a*b*x^4 - 12*A*b^2*x^4 + 5*D*a^2*x^3 - 25*B*a*b*x^3 +
6*C*a^2*x^2 - 18*A*a*b*x^2 - 8*B*a^2*x - 4*A*a^2)/((b*x^3 + a*x)^2*a^3)

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maple [A]  time = 0.02, size = 250, normalized size = 1.44 \begin {gather*} -\frac {7 B \,b^{2} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {3 D b \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {A \,b^{2} x^{2}}{\left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {C b \,x^{2}}{2 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {9 B b x}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {5 D x}{8 \left (b \,x^{2}+a \right )^{2} a}-\frac {5 A b}{4 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {15 B b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}+\frac {3 C}{4 \left (b \,x^{2}+a \right )^{2} a}+\frac {3 D \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{2}}-\frac {3 A b \ln \relax (x )}{a^{4}}+\frac {3 A b \ln \left (b \,x^{2}+a \right )}{2 a^{4}}+\frac {C \ln \relax (x )}{a^{3}}-\frac {C \ln \left (b \,x^{2}+a \right )}{2 a^{3}}-\frac {B}{a^{3} x}-\frac {A}{2 a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.99, size = 172, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left (D a b - 5 \, B b^{2}\right )} x^{5} + 4 \, {\left (C a b - 3 \, A b^{2}\right )} x^{4} - 8 \, B a^{2} x + 5 \, {\left (D a^{2} - 5 \, B a b\right )} x^{3} - 4 \, A a^{2} + 6 \, {\left (C a^{2} - 3 \, A a b\right )} x^{2}}{8 \, {\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}\right )}} + \frac {3 \, {\left (D a - 5 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {{\left (C a - 3 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {{\left (C a - 3 \, A b\right )} \log \relax (x)}{a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.46, size = 229, normalized size = 1.32 \begin {gather*} \frac {\frac {3\,C}{4\,a}+\frac {C\,b\,x^2}{2\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {A}{2\,a}+\frac {9\,A\,b\,x^2}{4\,a^2}+\frac {3\,A\,b^2\,x^4}{2\,a^3}}{a^2\,x^2+2\,a\,b\,x^4+b^2\,x^6}-\frac {\frac {B}{a}+\frac {25\,B\,b\,x^2}{8\,a^2}+\frac {15\,B\,b^2\,x^4}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {C\,\ln \left (b\,x^2+a\right )}{2\,a^3}+\frac {C\,\ln \relax (x)}{a^3}+\frac {3\,A\,b\,\ln \left (b\,x^2+a\right )}{2\,a^4}-\frac {3\,A\,b\,\ln \relax (x)}{a^4}+\frac {x\,D\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},3;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{a^3}-\frac {15\,B\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*C)/(4*a) + (C*b*x^2)/(2*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) - (A/(2*a) + (9*A*b*x^2)/(4*a^2) + (3*A*b^2*x^4)
/(2*a^3))/(a^2*x^2 + b^2*x^6 + 2*a*b*x^4) - (B/a + (25*B*b*x^2)/(8*a^2) + (15*B*b^2*x^4)/(8*a^3))/(a^2*x + b^2
*x^5 + 2*a*b*x^3) - (C*log(a + b*x^2))/(2*a^3) + (C*log(x))/a^3 + (3*A*b*log(a + b*x^2))/(2*a^4) - (3*A*b*log(
x))/a^4 + (x*D*hypergeom([1/2, 3], 3/2, -(b*x^2)/a))/a^3 - (15*B*b^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(7/2)
)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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