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

Optimal. Leaf size=170 \[ \frac{f^2 \sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{c} e \sqrt{b^2-4 a c}}-\frac{f^2 \sqrt{b-\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} e \sqrt{b^2-4 a c}} \]

[Out]

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

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Rubi [A]  time = 0.152551, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1142, 1130, 205} \[ \frac{f^2 \sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{c} e \sqrt{b^2-4 a c}}-\frac{f^2 \sqrt{b-\sqrt{b^2-4 a c}} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} e \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b - Sqrt[b^2 - 4*a*c]]*f^2*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*S
qrt[c]*Sqrt[b^2 - 4*a*c]*e)) + (Sqrt[b + Sqrt[b^2 - 4*a*c]]*f^2*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + Sq
rt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*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 1130

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

Rubi steps

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

Mathematica [A]  time = 0.094886, size = 178, normalized size = 1.05 \[ \frac{f^2 \left (\left (\sqrt{b^2-4 a c}-b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )+\sqrt{b-\sqrt{b^2-4 a c}} \sqrt{\sqrt{b^2-4 a c}+b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )\right )}{\sqrt{2} \sqrt{c} e \sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^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[b - Sqrt
[b^2 - 4*a*c]]*Sqrt[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]*Sqrt[c]*Sqrt[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*e)

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Maple [C]  time = 0.003, size = 143, normalized size = 0.8 \begin{align*}{\frac{{f}^{2}}{2\,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 ({{\it \_R}}^{2}{e}^{2}+2\,{\it \_R}\,de+{d}^{2} \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*f*x+d*f)^2/(a+b*(e*x+d)^2+c*(e*x+d)^4),x)

[Out]

1/2*f^2/e*sum((_R^2*e^2+2*_R*d*e+d^2)/(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*} \int \frac{{\left (e f x + d f\right )}^{2}}{{\left (e x + d\right )}^{4} c +{\left (e x + d\right )}^{2} b + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.6106, size = 1646, normalized size = 9.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.91699, size = 124, normalized size = 0.73 \begin{align*} \operatorname{RootSum}{\left (t^{4} \left (256 a^{2} c^{3} e^{4} - 128 a b^{2} c^{2} e^{4} + 16 b^{4} c e^{4}\right ) + t^{2} \left (- 16 a b c e^{2} f^{4} + 4 b^{3} e^{2} f^{4}\right ) + a f^{8}, \left ( t \mapsto t \log{\left (x + \frac{64 t^{3} a c^{2} e^{3} - 16 t^{3} b^{2} c e^{3} - 2 t b e f^{4} + d f^{6}}{e f^{6}} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_t**4*(256*a**2*c**3*e**4 - 128*a*b**2*c**2*e**4 + 16*b**4*c*e**4) + _t**2*(-16*a*b*c*e**2*f**4 + 4*b*
*3*e**2*f**4) + a*f**8, Lambda(_t, _t*log(x + (64*_t**3*a*c**2*e**3 - 16*_t**3*b**2*c*e**3 - 2*_t*b*e*f**4 + d
*f**6)/(e*f**6))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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