3.623 \(\int \frac{(d+e x)^2}{(a+b (d+e x)^2+c (d+e x)^4)^2} \, dx\)

Optimal. Leaf size=254 \[ -\frac{(d+e x) \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac{\sqrt{c} \left (2 b-\sqrt{b^2-4 a c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} e \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\sqrt{c} \left (\sqrt{b^2-4 a c}+2 b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} e \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}} \]

[Out]

-((d + e*x)*(b + 2*c*(d + e*x)^2))/(2*(b^2 - 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (Sqrt[c]*(2*b - S
qrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2
)*Sqrt[b - Sqrt[b^2 - 4*a*c]]*e) - (Sqrt[c]*(2*b + Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[
b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]*e)

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Rubi [A]  time = 0.388064, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1142, 1119, 1166, 205} \[ -\frac{(d+e x) \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac{\sqrt{c} \left (2 b-\sqrt{b^2-4 a c}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} e \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\sqrt{c} \left (\sqrt{b^2-4 a c}+2 b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} e \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + e*x)*(b + 2*c*(d + e*x)^2))/(2*(b^2 - 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (Sqrt[c]*(2*b - S
qrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2
)*Sqrt[b - Sqrt[b^2 - 4*a*c]]*e) - (Sqrt[c]*(2*b + Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[
b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]*e)

Rule 1142

Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^2 + (c_.)*(v_)^4)^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m),
Subst[Int[x^m*(a + b*x^2 + c*x^(2*2))^p, x], x, v], x] /; FreeQ[{a, b, c, m, p}, x] && LinearPairQ[u, v, x]

Rule 1119

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(d*x)^(m - 1)*(b + 2*c*
x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*(p + 1)*(b^2 - 4*a*c)), x] - Dist[d^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(d*x
)^(m - 2)*(b*(m - 1) + 2*c*(m + 4*p + 5)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1166

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

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]

Rubi steps

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

Mathematica [A]  time = 1.08438, size = 247, normalized size = 0.97 \[ -\frac{\frac{b (d+e x)+2 c (d+e x)^3}{\left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac{\sqrt{2} \sqrt{c} \left (\sqrt{b^2-4 a c}-2 b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} \sqrt{c} \left (\sqrt{b^2-4 a c}+2 b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(d + e*x) + 2*c*(d + e*x)^3)/((b^2 - 4*a*c)*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) + (Sqrt[2]*Sqrt[c]*(-2*b
+ Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sq
rt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*Sqrt[c]*(2*b + Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqr
t[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(2*e)

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Maple [C]  time = 0.018, size = 319, normalized size = 1.3 \begin{align*}{\frac{1}{c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a} \left ({\frac{c{e}^{2}{x}^{3}}{4\,ac-{b}^{2}}}+3\,{\frac{cde{x}^{2}}{4\,ac-{b}^{2}}}+{\frac{ \left ( 6\,c{d}^{2}+b \right ) x}{8\,ac-2\,{b}^{2}}}+{\frac{d \left ( 2\,c{d}^{2}+b \right ) }{2\,e \left ( 4\,ac-{b}^{2} \right ) }} \right ) }+{\frac{1}{ \left ( 16\,ac-4\,{b}^{2} \right ) e}\sum _{{\it \_R}={\it RootOf} \left ( c{e}^{4}{{\it \_Z}}^{4}+4\,cd{e}^{3}{{\it \_Z}}^{3}+ \left ( 6\,c{d}^{2}{e}^{2}+b{e}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,c{d}^{3}e+2\,bde \right ){\it \_Z}+c{d}^{4}+b{d}^{2}+a \right ) }{\frac{ \left ( 2\,{{\it \_R}}^{2}c{e}^{2}+4\,{\it \_R}\,cde+2\,c{d}^{2}-b \right ) \ln \left ( x-{\it \_R} \right ) }{2\,c{e}^{3}{{\it \_R}}^{3}+6\,cd{e}^{2}{{\it \_R}}^{2}+6\,c{d}^{2}e{\it \_R}+2\,c{d}^{3}+be{\it \_R}+bd}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*e^2/(4*a*c-b^2)*x^3+3*d*c*e/(4*a*c-b^2)*x^2+1/2*(6*c*d^2+b)/(4*a*c-b^2)*x+1/2*d/e*(2*c*d^2+b)/(4*a*c-b^2))/
(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)+1/4/(4*a*c-b^2)/e*sum(
(2*_R^2*c*e^2+4*_R*c*d*e+2*c*d^2-b)/(2*_R^3*c*e^3+6*_R^2*c*d*e^2+6*_R*c*d^2*e+2*c*d^3+_R*b*e+b*d)*ln(x-_R),_R=
RootOf(c*e^4*_Z^4+4*c*d*e^3*_Z^3+(6*c*d^2*e^2+b*e^2)*_Z^2+(4*c*d^3*e+2*b*d*e)*_Z+c*d^4+b*d^2+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, c e^{3} x^{3} + 6 \, c d e^{2} x^{2} + 2 \, c d^{3} +{\left (6 \, c d^{2} + b\right )} e x + b d}{2 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{5} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{4} x^{3} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{3} x^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e^{2} x +{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} e\right )}} + \frac{1}{2} \, \int -\frac{2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} - b}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{4} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{3} x^{3} +{\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{2} x^{2} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*c*e^3*x^3 + 6*c*d*e^2*x^2 + 2*c*d^3 + (6*c*d^2 + b)*e*x + b*d)/((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c -
4*a*c^2)*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b^3 -
4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e) + 1/2*integrate(-(2*c*e
^2*x^2 + 4*c*d*e*x + 2*c*d^2 - b)/((b^2*c - 4*a*c^2)*e^4*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^3*x^3 + (b^2*c - 4*a*c^
2)*d^4 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^2*x^2 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2 + 2*(2*(b^2
*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e*x), x)

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Fricas [B]  time = 2.2738, size = 5304, normalized size = 20.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*c*e^3*x^3 + 12*c*d*e^2*x^2 + 4*c*d^3 + 2*(6*c*d^2 + b)*e*x - sqrt(1/2)*((b^2*c - 4*a*c^2)*e^5*x^4 + 4*
(b^2*c - 4*a*c^2)*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 +
(b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e)*sqrt(-((a*b^6 -
12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3
)*e^4)) + b^3 + 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))*log((3*b^2*c + 4*a*c^2)*
e*x + (3*b^2*c + 4*a*c^2)*d + 1/2*sqrt(1/2)*((a*b^8 - 8*a^2*b^6*c + 128*a^4*b^2*c^3 - 256*a^5*c^4)*e^3*sqrt(1/
((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(-((a*
b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^
5*c^3)*e^4)) + b^3 + 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))) + sqrt(1/2)*((b^2*
c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(
2*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c
)*d^2)*e)*sqrt(-((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48
*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) + b^3 + 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))
*log((3*b^2*c + 4*a*c^2)*e*x + (3*b^2*c + 4*a*c^2)*d - 1/2*sqrt(1/2)*((a*b^8 - 8*a^2*b^6*c + 128*a^4*b^2*c^3 -
256*a^5*c^4)*e^3*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - (b^5 - 8*a*b^3*c + 16
*a^2*b*c^2)*e)*sqrt(-((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c
+ 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) + b^3 + 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*
e^2))) + sqrt(1/2)*((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*
a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 -
4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e)*sqrt(((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/((a^2
*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - b^3 - 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2
*c^2 - 64*a^4*c^3)*e^2))*log((3*b^2*c + 4*a*c^2)*e*x + (3*b^2*c + 4*a*c^2)*d + 1/2*sqrt(1/2)*((a*b^8 - 8*a^2*b
^6*c + 128*a^4*b^2*c^3 - 256*a^5*c^4)*e^3*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4))
+ (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2*sqrt(1/(
(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - b^3 - 12*a*b*c)/((a*b^6 - 12*a^2*b^4*c + 48*a^3
*b^2*c^2 - 64*a^4*c^3)*e^2))) - sqrt(1/2)*((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)*d*e^4*x^3 + (b^3 -
4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*d)*e^2*x + ((b^2*c -
4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e)*sqrt(((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^
4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - b^3 - 12*a*b*c)/((a*b^6 - 12
*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))*log((3*b^2*c + 4*a*c^2)*e*x + (3*b^2*c + 4*a*c^2)*d - 1/2*sqrt
(1/2)*((a*b^8 - 8*a^2*b^6*c + 128*a^4*b^2*c^3 - 256*a^5*c^4)*e^3*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*
c^2 - 64*a^5*c^3)*e^4)) + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e)*sqrt(((a*b^6 - 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 6
4*a^4*c^3)*e^2*sqrt(1/((a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3)*e^4)) - b^3 - 12*a*b*c)/((a*b^6
- 12*a^2*b^4*c + 48*a^3*b^2*c^2 - 64*a^4*c^3)*e^2))) + 2*b*d)/((b^2*c - 4*a*c^2)*e^5*x^4 + 4*(b^2*c - 4*a*c^2)
*d*e^4*x^3 + (b^3 - 4*a*b*c + 6*(b^2*c - 4*a*c^2)*d^2)*e^3*x^2 + 2*(2*(b^2*c - 4*a*c^2)*d^3 + (b^3 - 4*a*b*c)*
d)*e^2*x + ((b^2*c - 4*a*c^2)*d^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*d^2)*e)

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Sympy [B]  time = 35.2598, size = 578, normalized size = 2.28 \begin{align*} \frac{b d + 2 c d^{3} + 6 c d e^{2} x^{2} + 2 c e^{3} x^{3} + x \left (b e + 6 c d^{2} e\right )}{8 a^{2} c e - 2 a b^{2} e + 8 a b c d^{2} e + 8 a c^{2} d^{4} e - 2 b^{3} d^{2} e - 2 b^{2} c d^{4} e + x^{4} \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \left (8 a b c e^{3} + 48 a c^{2} d^{2} e^{3} - 2 b^{3} e^{3} - 12 b^{2} c d^{2} e^{3}\right ) + x \left (16 a b c d e^{2} + 32 a c^{2} d^{3} e^{2} - 4 b^{3} d e^{2} - 8 b^{2} c d^{3} e^{2}\right )} + \operatorname{RootSum}{\left (t^{4} \left (1048576 a^{7} c^{6} e^{4} - 1572864 a^{6} b^{2} c^{5} e^{4} + 983040 a^{5} b^{4} c^{4} e^{4} - 327680 a^{4} b^{6} c^{3} e^{4} + 61440 a^{3} b^{8} c^{2} e^{4} - 6144 a^{2} b^{10} c e^{4} + 256 a b^{12} e^{4}\right ) + t^{2} \left (- 12288 a^{4} b c^{4} e^{2} + 8192 a^{3} b^{3} c^{3} e^{2} - 1536 a^{2} b^{5} c^{2} e^{2} + 16 b^{9} e^{2}\right ) + 16 a^{2} c^{3} + 24 a b^{2} c^{2} + 9 b^{4} c, \left ( t \mapsto t \log{\left (x + \frac{16384 t^{3} a^{5} c^{4} e^{3} - 8192 t^{3} a^{4} b^{2} c^{3} e^{3} + 512 t^{3} a^{2} b^{6} c e^{3} - 64 t^{3} a b^{8} e^{3} - 128 t a^{2} b c^{2} e - 16 t a b^{3} c e - 4 t b^{5} e + 4 a c^{2} d + 3 b^{2} c d}{4 a c^{2} e + 3 b^{2} c e} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d + 2*c*d**3 + 6*c*d*e**2*x**2 + 2*c*e**3*x**3 + x*(b*e + 6*c*d**2*e))/(8*a**2*c*e - 2*a*b**2*e + 8*a*b*c*d
**2*e + 8*a*c**2*d**4*e - 2*b**3*d**2*e - 2*b**2*c*d**4*e + x**4*(8*a*c**2*e**5 - 2*b**2*c*e**5) + x**3*(32*a*
c**2*d*e**4 - 8*b**2*c*d*e**4) + x**2*(8*a*b*c*e**3 + 48*a*c**2*d**2*e**3 - 2*b**3*e**3 - 12*b**2*c*d**2*e**3)
+ x*(16*a*b*c*d*e**2 + 32*a*c**2*d**3*e**2 - 4*b**3*d*e**2 - 8*b**2*c*d**3*e**2)) + RootSum(_t**4*(1048576*a*
*7*c**6*e**4 - 1572864*a**6*b**2*c**5*e**4 + 983040*a**5*b**4*c**4*e**4 - 327680*a**4*b**6*c**3*e**4 + 61440*a
**3*b**8*c**2*e**4 - 6144*a**2*b**10*c*e**4 + 256*a*b**12*e**4) + _t**2*(-12288*a**4*b*c**4*e**2 + 8192*a**3*b
**3*c**3*e**2 - 1536*a**2*b**5*c**2*e**2 + 16*b**9*e**2) + 16*a**2*c**3 + 24*a*b**2*c**2 + 9*b**4*c, Lambda(_t
, _t*log(x + (16384*_t**3*a**5*c**4*e**3 - 8192*_t**3*a**4*b**2*c**3*e**3 + 512*_t**3*a**2*b**6*c*e**3 - 64*_t
**3*a*b**8*e**3 - 128*_t*a**2*b*c**2*e - 16*_t*a*b**3*c*e - 4*_t*b**5*e + 4*a*c**2*d + 3*b**2*c*d)/(4*a*c**2*e
+ 3*b**2*c*e))))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{2}}{{\left ({\left (e x + d\right )}^{4} c +{\left (e x + d\right )}^{2} b + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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