3.2.32 \(\int \frac {x^2 (c+d x^2+e x^4+f x^6)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=167 \[ \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {x \left (11 a^3 f-7 a^2 b e+3 a b^2 d+b^3 c\right )}{8 a b^4 \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (35 a^3 f-15 a^2 b e+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac {x (b e-3 a f)}{b^4}+\frac {f x^3}{3 b^3} \]

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Rubi [A]  time = 0.26, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1804, 1585, 1257, 1153, 205} \begin {gather*} \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {x \left (-7 a^2 b e+11 a^3 f+3 a b^2 d+b^3 c\right )}{8 a b^4 \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^2 b e+35 a^3 f+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac {x (b e-3 a f)}{b^4}+\frac {f x^3}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x]

[Out]

((b*e - 3*a*f)*x)/b^4 + (f*x^3)/(3*b^3) + ((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^3)/(4*a*(a + b*x^2)^2) - ((
b^3*c + 3*a*b^2*d - 7*a^2*b*e + 11*a^3*f)*x)/(8*a*b^4*(a + b*x^2)) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3
*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))

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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

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^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x \left (-\left (\left (b c+3 a d-\frac {3 a^2 e}{b}+\frac {3 a^3 f}{b^2}\right ) x\right )-4 a \left (e-\frac {a f}{b}\right ) x^3-4 a f x^5\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^2 \left (-b c-3 a d+\frac {3 a^2 e}{b}-\frac {3 a^3 f}{b^2}-4 a \left (e-\frac {a f}{b}\right ) x^2-4 a f x^4\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\int \frac {b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f+8 a b (b e-2 a f) x^2+8 a b^2 f x^4}{a+b x^2} \, dx}{8 a b^4}\\ &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\int \left (8 a (b e-3 a f)+8 a b f x^2+\frac {b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f}{a+b x^2}\right ) \, dx}{8 a b^4}\\ &=\frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{8 a b^4}\\ &=\frac {(b e-3 a f) x}{b^4}+\frac {f x^3}{3 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac {\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{3/2} b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 156, normalized size = 0.93 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (35 a^3 f-15 a^2 b e+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac {x \left (-105 a^4 f+5 a^3 b \left (9 e-35 f x^2\right )+a^2 b^2 \left (-9 d+75 e x^2-56 f x^4\right )+a b^3 \left (-3 c-15 d x^2+24 e x^4+8 f x^6\right )+3 b^4 c x^2\right )}{24 a b^4 \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x]

[Out]

(x*(-105*a^4*f + 3*b^4*c*x^2 + 5*a^3*b*(9*e - 35*f*x^2) + a^2*b^2*(-9*d + 75*e*x^2 - 56*f*x^4) + a*b^3*(-3*c -
15*d*x^2 + 24*e*x^4 + 8*f*x^6)))/(24*a*b^4*(a + b*x^2)^2) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3*f)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x]

[Out]

IntegrateAlgebraic[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3, x]

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fricas [A]  time = 0.97, size = 555, normalized size = 3.32 \begin {gather*} \left [\frac {16 \, a^{2} b^{4} f x^{7} + 16 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + 2 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} - 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\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 (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{48 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, \frac {8 \, a^{2} b^{4} f x^{7} + 8 \, {\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} + 3 \, {\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f + {\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 3 \, {\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{24 \, {\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/48*(16*a^2*b^4*f*x^7 + 16*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + 2*(3*a*b^5*c - 15*a^2*b^4*d + 75*a^3*b^3*e - 17
5*a^4*b^2*f)*x^3 - 3*(a^2*b^3*c + 3*a^3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2*b^3*e + 35
*a^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b*f)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a
*b)*x - a)/(b*x^2 + a)) - 6*(a^2*b^4*c + 3*a^3*b^3*d - 15*a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*
x^2 + a^4*b^5), 1/24*(8*a^2*b^4*f*x^7 + 8*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + (3*a*b^5*c - 15*a^2*b^4*d + 75*a^3
*b^3*e - 175*a^4*b^2*f)*x^3 + 3*(a^2*b^3*c + 3*a^3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2
*b^3*e + 35*a^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b*f)*x^2)*sqrt(a*b)*arctan(sqrt(
a*b)*x/a) - 3*(a^2*b^4*c + 3*a^3*b^3*d - 15*a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5)
]

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giac [A]  time = 0.45, size = 173, normalized size = 1.04 \begin {gather*} \frac {{\left (b^{3} c + 3 \, a b^{2} d + 35 \, a^{3} f - 15 \, a^{2} b e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} + \frac {b^{4} c x^{3} - 5 \, a b^{3} d x^{3} - 13 \, a^{3} b f x^{3} + 9 \, a^{2} b^{2} x^{3} e - a b^{3} c x - 3 \, a^{2} b^{2} d x - 11 \, a^{4} f x + 7 \, a^{3} b x e}{8 \, {\left (b x^{2} + a\right )}^{2} a b^{4}} + \frac {b^{6} f x^{3} - 9 \, a b^{5} f x + 3 \, b^{6} x e}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/8*(b^3*c + 3*a*b^2*d + 35*a^3*f - 15*a^2*b*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^4) + 1/8*(b^4*c*x^3 - 5*a
*b^3*d*x^3 - 13*a^3*b*f*x^3 + 9*a^2*b^2*x^3*e - a*b^3*c*x - 3*a^2*b^2*d*x - 11*a^4*f*x + 7*a^3*b*x*e)/((b*x^2
+ a)^2*a*b^4) + 1/3*(b^6*f*x^3 - 9*a*b^5*f*x + 3*b^6*x*e)/b^9

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maple [A]  time = 0.01, size = 259, normalized size = 1.55 \begin {gather*} -\frac {13 a^{2} f \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}+\frac {9 a e \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {c \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a}-\frac {5 d \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b}-\frac {11 a^{3} f x}{8 \left (b \,x^{2}+a \right )^{2} b^{4}}+\frac {7 a^{2} e x}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}-\frac {3 a d x}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {c x}{8 \left (b \,x^{2}+a \right )^{2} b}+\frac {f \,x^{3}}{3 b^{3}}+\frac {35 a^{2} f \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{4}}-\frac {15 a e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}+\frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a b}+\frac {3 d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}-\frac {3 a f x}{b^{4}}+\frac {e x}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x)

[Out]

1/3*f*x^3/b^3-3/b^4*a*f*x+1/b^3*e*x-13/8/b^3/(b*x^2+a)^2*x^3*a^2*f+9/8/b^2/(b*x^2+a)^2*x^3*a*e-5/8/b/(b*x^2+a)
^2*x^3*d+1/8/(b*x^2+a)^2/a*x^3*c-11/8/b^4/(b*x^2+a)^2*a^3*f*x+7/8/b^3/(b*x^2+a)^2*a^2*e*x-3/8/b^2/(b*x^2+a)^2*
a*d*x-1/8/b/(b*x^2+a)^2*c*x+35/8/b^4*a^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*f-15/8/b^3*a/(a*b)^(1/2)*arctan
(1/(a*b)^(1/2)*b*x)*e+3/8/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*d+1/8/b/a/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)
*b*x)*c

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maxima [A]  time = 3.02, size = 169, normalized size = 1.01 \begin {gather*} \frac {{\left (b^{4} c - 5 \, a b^{3} d + 9 \, a^{2} b^{2} e - 13 \, a^{3} b f\right )} x^{3} - {\left (a b^{3} c + 3 \, a^{2} b^{2} d - 7 \, a^{3} b e + 11 \, a^{4} f\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} + \frac {b f x^{3} + 3 \, {\left (b e - 3 \, a f\right )} x}{3 \, b^{4}} + \frac {{\left (b^{3} c + 3 \, a b^{2} d - 15 \, a^{2} b e + 35 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/8*((b^4*c - 5*a*b^3*d + 9*a^2*b^2*e - 13*a^3*b*f)*x^3 - (a*b^3*c + 3*a^2*b^2*d - 7*a^3*b*e + 11*a^4*f)*x)/(a
*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4) + 1/3*(b*f*x^3 + 3*(b*e - 3*a*f)*x)/b^4 + 1/8*(b^3*c + 3*a*b^2*d - 15*a^2*
b*e + 35*a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^4)

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mupad [B]  time = 1.02, size = 163, normalized size = 0.98 \begin {gather*} x\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )-\frac {x\,\left (\frac {11\,f\,a^3}{8}-\frac {7\,e\,a^2\,b}{8}+\frac {3\,d\,a\,b^2}{8}+\frac {c\,b^3}{8}\right )-\frac {x^3\,\left (-13\,f\,a^3\,b+9\,e\,a^2\,b^2-5\,d\,a\,b^3+c\,b^4\right )}{8\,a}}{a^2\,b^4+2\,a\,b^5\,x^2+b^6\,x^4}+\frac {f\,x^3}{3\,b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (35\,f\,a^3-15\,e\,a^2\,b+3\,d\,a\,b^2+c\,b^3\right )}{8\,a^{3/2}\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x)

[Out]

x*(e/b^3 - (3*a*f)/b^4) - (x*((b^3*c)/8 + (11*a^3*f)/8 + (3*a*b^2*d)/8 - (7*a^2*b*e)/8) - (x^3*(b^4*c + 9*a^2*
b^2*e - 5*a*b^3*d - 13*a^3*b*f))/(8*a))/(a^2*b^4 + b^6*x^4 + 2*a*b^5*x^2) + (f*x^3)/(3*b^3) + (atan((b^(1/2)*x
)/a^(1/2))*(b^3*c + 35*a^3*f + 3*a*b^2*d - 15*a^2*b*e))/(8*a^(3/2)*b^(9/2))

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sympy [A]  time = 13.07, size = 260, normalized size = 1.56 \begin {gather*} x \left (- \frac {3 a f}{b^{4}} + \frac {e}{b^{3}}\right ) - \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (- a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{3} b^{9}}} \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log {\left (a^{2} b^{4} \sqrt {- \frac {1}{a^{3} b^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 13 a^{3} b f + 9 a^{2} b^{2} e - 5 a b^{3} d + b^{4} c\right ) + x \left (- 11 a^{4} f + 7 a^{3} b e - 3 a^{2} b^{2} d - a b^{3} c\right )}{8 a^{3} b^{4} + 16 a^{2} b^{5} x^{2} + 8 a b^{6} x^{4}} + \frac {f x^{3}}{3 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**3,x)

[Out]

x*(-3*a*f/b**4 + e/b**3) - sqrt(-1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(-a**2*b**4
*sqrt(-1/(a**3*b**9)) + x)/16 + sqrt(-1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(a**2*
b**4*sqrt(-1/(a**3*b**9)) + x)/16 + (x**3*(-13*a**3*b*f + 9*a**2*b**2*e - 5*a*b**3*d + b**4*c) + x*(-11*a**4*f
+ 7*a**3*b*e - 3*a**2*b**2*d - a*b**3*c))/(8*a**3*b**4 + 16*a**2*b**5*x**2 + 8*a*b**6*x**4) + f*x**3/(3*b**3)

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