3.62 $$\int \frac{x^5 (d+e x^2+f x^4)}{(a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=236 $-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (12 a^2 c^2 f-2 a c \left (6 b^2 f-3 b c e+2 c^2 d\right )+b^3 (-(c e-2 b f))\right )}{2 c^3 \left (b^2-4 a c\right )^{3/2}}+\frac{x^4 \left (x^2 \left (-\left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )-b (a f+c d)+2 a c e\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{x^2 \left (-c (6 a f+b e)+2 b^2 f+2 c^2 d\right )}{2 c^2 \left (b^2-4 a c\right )}+\frac{(c e-2 b f) \log \left (a+b x^2+c x^4\right )}{4 c^3}$

[Out]

((2*c^2*d + 2*b^2*f - c*(b*e + 6*a*f))*x^2)/(2*c^2*(b^2 - 4*a*c)) + (x^4*(2*a*c*e - b*(c*d + a*f) - (2*c^2*d -
b*c*e + b^2*f - 2*a*c*f)*x^2))/(2*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((12*a^2*c^2*f - b^3*(c*e - 2*b*f) -
2*a*c*(2*c^2*d - 3*b*c*e + 6*b^2*f))*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*(b^2 - 4*a*c)^(3/2)) +
((c*e - 2*b*f)*Log[a + b*x^2 + c*x^4])/(4*c^3)

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Rubi [A]  time = 0.440226, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.233, Rules used = {1663, 1644, 773, 634, 618, 206, 628} $-\frac{\tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right ) \left (12 a^2 c^2 f-2 a c \left (6 b^2 f-3 b c e+2 c^2 d\right )+b^3 (-(c e-2 b f))\right )}{2 c^3 \left (b^2-4 a c\right )^{3/2}}+\frac{x^4 \left (x^2 \left (-\left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )-b (a f+c d)+2 a c e\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{x^2 \left (-c (6 a f+b e)+2 b^2 f+2 c^2 d\right )}{2 c^2 \left (b^2-4 a c\right )}+\frac{(c e-2 b f) \log \left (a+b x^2+c x^4\right )}{4 c^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(x^5*(d + e*x^2 + f*x^4))/(a + b*x^2 + c*x^4)^2,x]

[Out]

((2*c^2*d + 2*b^2*f - c*(b*e + 6*a*f))*x^2)/(2*c^2*(b^2 - 4*a*c)) + (x^4*(2*a*c*e - b*(c*d + a*f) - (2*c^2*d -
b*c*e + b^2*f - 2*a*c*f)*x^2))/(2*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((12*a^2*c^2*f - b^3*(c*e - 2*b*f) -
2*a*c*(2*c^2*d - 3*b*c*e + 6*b^2*f))*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^3*(b^2 - 4*a*c)^(3/2)) +
((c*e - 2*b*f)*Log[a + b*x^2 + c*x^4])/(4*c^3)

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1644

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g +
(2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x
+ c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Q + g*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c
*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[p] ||  !IntegerQ[m
] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2,
0]))

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^5 \left (d+e x^2+f x^4\right )}{\left (a+b x^2+c x^4\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2 \left (d+e x+f x^2\right )}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{x^4 \left (2 a c e-b (c d+a f)-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{x \left (2 \left (2 a e-\frac{b (c d+a f)}{c}\right )-\frac{\left (2 c^2 d-b c e+2 b^2 f-6 a c f\right ) x}{c}\right )}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=\frac{\left (2 c^2 d+2 b^2 f-c (b e+6 a f)\right ) x^2}{2 c^2 \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a c e-b (c d+a f)-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a \left (2 c^2 d-b c e+2 b^2 f-6 a c f\right )}{c}+\left (\frac{b \left (2 c^2 d-b c e+2 b^2 f-6 a c f\right )}{c}+2 c \left (2 a e-\frac{b (c d+a f)}{c}\right )\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c \left (b^2-4 a c\right )}\\ &=\frac{\left (2 c^2 d+2 b^2 f-c (b e+6 a f)\right ) x^2}{2 c^2 \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a c e-b (c d+a f)-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{(c e-2 b f) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3}+\frac{\left (12 a^2 c^2 f-b^3 (c e-2 b f)-2 a c \left (2 c^2 d-3 b c e+6 b^2 f\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^3 \left (b^2-4 a c\right )}\\ &=\frac{\left (2 c^2 d+2 b^2 f-c (b e+6 a f)\right ) x^2}{2 c^2 \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a c e-b (c d+a f)-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{(c e-2 b f) \log \left (a+b x^2+c x^4\right )}{4 c^3}-\frac{\left (12 a^2 c^2 f-b^3 (c e-2 b f)-2 a c \left (2 c^2 d-3 b c e+6 b^2 f\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^3 \left (b^2-4 a c\right )}\\ &=\frac{\left (2 c^2 d+2 b^2 f-c (b e+6 a f)\right ) x^2}{2 c^2 \left (b^2-4 a c\right )}+\frac{x^4 \left (2 a c e-b (c d+a f)-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (12 a^2 c^2 f-b^3 (c e-2 b f)-2 a c \left (2 c^2 d-3 b c e+6 b^2 f\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^3 \left (b^2-4 a c\right )^{3/2}}+\frac{(c e-2 b f) \log \left (a+b x^2+c x^4\right )}{4 c^3}\\ \end{align*}

Mathematica [A]  time = 0.385163, size = 236, normalized size = 1. $\frac{-\frac{2 \left (a^2 c \left (2 c \left (e+f x^2\right )-3 b f\right )+a \left (-b^2 c \left (e+4 f x^2\right )+b^3 f+b c^2 \left (d+3 e x^2\right )-2 c^3 d x^2\right )+b^2 x^2 \left (b^2 f-b c e+c^2 d\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{2 \tan ^{-1}\left (\frac{b+2 c x^2}{\sqrt{4 a c-b^2}}\right ) \left (12 a^2 c^2 f-2 a c \left (6 b^2 f-3 b c e+2 c^2 d\right )+b^3 (2 b f-c e)\right )}{\left (4 a c-b^2\right )^{3/2}}+(c e-2 b f) \log \left (a+b x^2+c x^4\right )+2 c f x^2}{4 c^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(x^5*(d + e*x^2 + f*x^4))/(a + b*x^2 + c*x^4)^2,x]

[Out]

(2*c*f*x^2 - (2*(b^2*(c^2*d - b*c*e + b^2*f)*x^2 + a^2*c*(-3*b*f + 2*c*(e + f*x^2)) + a*(b^3*f - 2*c^3*d*x^2 +
b*c^2*(d + 3*e*x^2) - b^2*c*(e + 4*f*x^2))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (2*(12*a^2*c^2*f + b^3*(-(
c*e) + 2*b*f) - 2*a*c*(2*c^2*d - 3*b*c*e + 6*b^2*f))*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^
(3/2) + (c*e - 2*b*f)*Log[a + b*x^2 + c*x^4])/(4*c^3)

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Maple [B]  time = 0.017, size = 832, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a)^2,x)

[Out]

1/2*f*x^2/c^2+1/c/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a^2*f-2/c^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a*b^2*f+3/2/c/(c
*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a*b*e-1/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*a*d+1/2/c^3/(c*x^4+b*x^2+a)/(4*a*c-b^2)*
x^2*b^4*f-1/2/c^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*b^3*e+1/2/c/(c*x^4+b*x^2+a)/(4*a*c-b^2)*x^2*b^2*d-3/2/c^2/(c
*x^4+b*x^2+a)*a^2/(4*a*c-b^2)*b*f+1/c/(c*x^4+b*x^2+a)*a^2/(4*a*c-b^2)*e+1/2/c^3/(c*x^4+b*x^2+a)*a/(4*a*c-b^2)*
b^3*f-1/2/c^2/(c*x^4+b*x^2+a)*a/(4*a*c-b^2)*b^2*e+1/2/c/(c*x^4+b*x^2+a)*a/(4*a*c-b^2)*b*d-2/c^2/(4*a*c-b^2)*ln
(c*x^4+b*x^2+a)*a*b*f+1/c/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*a*e+1/2/c^3/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*b^3*f-1/4/c^
2/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*b^2*e-6/c/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a^2*f+6/c^2/
(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*b^2*f-3/c/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*
c-b^2)^(1/2))*a*b*e+2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*d-1/c^3/(4*a*c-b^2)^(3/2)*arct
an((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^4*f+1/2/c^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^3*e

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30547, size = 3043, normalized size = 12.89 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/4*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f*x^6 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f*x^4 - 2*((b^4*c^
2 - 6*a*b^2*c^3 + 8*a^2*c^4)*d - (b^5*c - 7*a*b^3*c^2 + 12*a^2*b*c^3)*e + (b^6 - 9*a*b^4*c + 26*a^2*b^2*c^2 -
24*a^3*c^3)*f)*x^2 + (4*a^2*c^3*d + (4*a*c^4*d + (b^3*c^2 - 6*a*b*c^3)*e - 2*(b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)
*f)*x^4 + (4*a*b*c^3*d + (b^4*c - 6*a*b^2*c^2)*e - 2*(b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*f)*x^2 + (a*b^3*c - 6*a^2
*b*c^2)*e - 2*(a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*f)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^2 + b^2 - 2*a*c
+ (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) - 2*(a*b^3*c^2 - 4*a^2*b*c^3)*d + 2*(a*b^4*c - 6*a^2*b
^2*c^2 + 8*a^3*c^3)*e - 2*(a*b^5 - 7*a^2*b^3*c + 12*a^3*b*c^2)*f + (((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*e -
2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f)*x^4 + ((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*e - 2*(b^6 - 8*a*b^4*c +
16*a^2*b^2*c^2)*f)*x^2 + (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e - 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*f)
*log(c*x^4 + b*x^2 + a))/(a*b^4*c^3 - 8*a^2*b^2*c^4 + 16*a^3*c^5 + (b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^4 +
(b^5*c^3 - 8*a*b^3*c^4 + 16*a^2*b*c^5)*x^2), 1/4*(2*(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f*x^6 + 2*(b^5*c - 8*
a*b^3*c^2 + 16*a^2*b*c^3)*f*x^4 - 2*((b^4*c^2 - 6*a*b^2*c^3 + 8*a^2*c^4)*d - (b^5*c - 7*a*b^3*c^2 + 12*a^2*b*c
^3)*e + (b^6 - 9*a*b^4*c + 26*a^2*b^2*c^2 - 24*a^3*c^3)*f)*x^2 + 2*(4*a^2*c^3*d + (4*a*c^4*d + (b^3*c^2 - 6*a*
b*c^3)*e - 2*(b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*f)*x^4 + (4*a*b*c^3*d + (b^4*c - 6*a*b^2*c^2)*e - 2*(b^5 - 6*a*
b^3*c + 6*a^2*b*c^2)*f)*x^2 + (a*b^3*c - 6*a^2*b*c^2)*e - 2*(a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*f)*sqrt(-b^2 + 4
*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - 2*(a*b^3*c^2 - 4*a^2*b*c^3)*d + 2*(a*b^4*c - 6
*a^2*b^2*c^2 + 8*a^3*c^3)*e - 2*(a*b^5 - 7*a^2*b^3*c + 12*a^3*b*c^2)*f + (((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4
)*e - 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f)*x^4 + ((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*e - 2*(b^6 - 8*a*b
^4*c + 16*a^2*b^2*c^2)*f)*x^2 + (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*e - 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c
^2)*f)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^3 - 8*a^2*b^2*c^4 + 16*a^3*c^5 + (b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*
x^4 + (b^5*c^3 - 8*a*b^3*c^4 + 16*a^2*b*c^5)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(f*x**4+e*x**2+d)/(c*x**4+b*x**2+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 19.2062, size = 377, normalized size = 1.6 \begin{align*} \frac{f x^{2}}{2 \, c^{2}} - \frac{{\left (4 \, a c^{3} d - 2 \, b^{4} f + 12 \, a b^{2} c f - 12 \, a^{2} c^{2} f + b^{3} c e - 6 \, a b c^{2} e\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (b^{2} c^{3} - 4 \, a c^{4}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{2 \, b^{3} f x^{4} - 8 \, a b c f x^{4} - b^{2} c x^{4} e + 4 \, a c^{2} x^{4} e - 2 \, b^{2} c d x^{2} + 4 \, a c^{2} d x^{2} - 4 \, a^{2} c f x^{2} + b^{3} x^{2} e - 2 \, a b c x^{2} e - 2 \, a b c d - 2 \, a^{2} b f + a b^{2} e}{4 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} - \frac{{\left (2 \, b f - c e\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(f*x^4+e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*f*x^2/c^2 - 1/2*(4*a*c^3*d - 2*b^4*f + 12*a*b^2*c*f - 12*a^2*c^2*f + b^3*c*e - 6*a*b*c^2*e)*arctan((2*c*x^
2 + b)/sqrt(-b^2 + 4*a*c))/((b^2*c^3 - 4*a*c^4)*sqrt(-b^2 + 4*a*c)) + 1/4*(2*b^3*f*x^4 - 8*a*b*c*f*x^4 - b^2*c
*x^4*e + 4*a*c^2*x^4*e - 2*b^2*c*d*x^2 + 4*a*c^2*d*x^2 - 4*a^2*c*f*x^2 + b^3*x^2*e - 2*a*b*c*x^2*e - 2*a*b*c*d
- 2*a^2*b*f + a*b^2*e)/((c*x^4 + b*x^2 + a)*(b^2*c^2 - 4*a*c^3)) - 1/4*(2*b*f - c*e)*log(c*x^4 + b*x^2 + a)/c
^3