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

Optimal. Leaf size=44 \[ -\frac{f \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \sqrt{b^2-4 a c}} \]

[Out]

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

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Rubi [A]  time = 0.0631629, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1142, 1107, 618, 206} \[ -\frac{f \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((f*ArcTanh[(b + 2*c*(d + e*x)^2)/Sqrt[b^2 - 4*a*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 1107

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0172062, size = 47, normalized size = 1.07 \[ \frac{f \tan ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{4 a c-b^2}}\right )}{e \sqrt{4 a c-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.005, size = 130, normalized size = 3. \begin{align*}{\frac{f}{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}\,e+d \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)/(a+b*(e*x+d)^2+c*(e*x+d)^4),x)

[Out]

1/2*f/e*sum((_R*e+d)/(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{e f x + d f}{{\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)/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.55573, size = 603, normalized size = 13.7 \begin{align*} \left [\frac{f \log \left (\frac{2 \, c^{2} e^{4} x^{4} + 8 \, c^{2} d e^{3} x^{3} + 2 \, c^{2} d^{4} + 2 \,{\left (6 \, c^{2} d^{2} + b c\right )} e^{2} x^{2} + 2 \, b c d^{2} + 4 \,{\left (2 \, c^{2} d^{3} + b c d\right )} e x + b^{2} - 2 \, a c -{\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt{b^{2} - 4 \, a c}}{c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} +{\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \,{\left (2 \, c d^{3} + b d\right )} e x + a}\right )}{2 \, \sqrt{b^{2} - 4 \, a c} e}, -\frac{\sqrt{-b^{2} + 4 \, a c} f \arctan \left (-\frac{{\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{{\left (b^{2} - 4 \, a c\right )} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.3556, size = 189, normalized size = 4.3 \begin{align*} - \frac{f \sqrt{- \frac{1}{4 a c - b^{2}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 4 a c f \sqrt{- \frac{1}{4 a c - b^{2}}} + b^{2} f \sqrt{- \frac{1}{4 a c - b^{2}}} + b f + 2 c d^{2} f}{2 c e^{2} f} \right )}}{2 e} + \frac{f \sqrt{- \frac{1}{4 a c - b^{2}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{4 a c f \sqrt{- \frac{1}{4 a c - b^{2}}} - b^{2} f \sqrt{- \frac{1}{4 a c - b^{2}}} + b f + 2 c d^{2} f}{2 c e^{2} f} \right )}}{2 e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-f*sqrt(-1/(4*a*c - b**2))*log(2*d*x/e + x**2 + (-4*a*c*f*sqrt(-1/(4*a*c - b**2)) + b**2*f*sqrt(-1/(4*a*c - b*
*2)) + b*f + 2*c*d**2*f)/(2*c*e**2*f))/(2*e) + f*sqrt(-1/(4*a*c - b**2))*log(2*d*x/e + x**2 + (4*a*c*f*sqrt(-1
/(4*a*c - b**2)) - b**2*f*sqrt(-1/(4*a*c - b**2)) + b*f + 2*c*d**2*f)/(2*c*e**2*f))/(2*e)

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Giac [B]  time = 1.43048, size = 250, normalized size = 5.68 \begin{align*} \frac{\sqrt{b^{2} - 4 \, a c} f e \log \left ({\left |{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} + 2 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d x e +{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d^{2} + 2 \, a \right |}\right )}{2 \,{\left (b^{2} e^{2} - 4 \, a c e^{2}\right )}} - \frac{\sqrt{b^{2} - 4 \, a c} f e \log \left ({\left | -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} - 2 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d x e -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d^{2} - 2 \, a \right |}\right )}{2 \,{\left (b^{2} e^{2} - 4 \, a c e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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