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

Optimal. Leaf size=87 \[ \frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{2 c e \sqrt{b^2-4 a c}}+\frac{f^3 \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 c e} \]

[Out]

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

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Rubi [A]  time = 0.127477, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1142, 1114, 634, 618, 206, 628} \[ \frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{2 c e \sqrt{b^2-4 a c}}+\frac{f^3 \log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f^3*ArcTanh[(b + 2*c*(d + e*x)^2)/Sqrt[b^2 - 4*a*c]])/(2*c*Sqrt[b^2 - 4*a*c]*e) + (f^3*Log[a + b*(d + e*x)^
2 + c*(d + e*x)^4])/(4*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 1114

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

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

Mathematica [A]  time = 0.0418863, size = 80, normalized size = 0.92 \[ \frac{f^3 \left (\log \left (a+b (d+e x)^2+c (d+e x)^4\right )-\frac{2 b \tan ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{4 a c-b^2}}\right )}{\sqrt{4 a c-b^2}}\right )}{4 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.004, size = 154, normalized size = 1.8 \begin{align*}{\frac{{f}^{3}}{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}}^{3}{e}^{3}+3\,{{\it \_R}}^{2}d{e}^{2}+3\,{\it \_R}\,{d}^{2}e+{d}^{3} \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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x)

[Out]

1/2*f^3/e*sum((_R^3*e^3+3*_R^2*d*e^2+3*_R*d^2*e+d^3)/(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 )}^{3}}{{\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)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56381, size = 977, normalized size = 11.23 \begin{align*} \left [\frac{\sqrt{b^{2} - 4 \, a c} b f^{3} \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 ) +{\left (b^{2} - 4 \, a c\right )} f^{3} \log \left (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 )}{4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e}, \frac{2 \, \sqrt{-b^{2} + 4 \, a c} b f^{3} \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 )} f^{3} \log \left (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 )}{4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.15366, size = 332, normalized size = 3.82 \begin{align*} \left (- \frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 8 a c e \left (- \frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) + 2 a f^{3} + 2 b^{2} e \left (- \frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) + b d^{2} f^{3}}{b e^{2} f^{3}} \right )} + \left (\frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 8 a c e \left (\frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) + 2 a f^{3} + 2 b^{2} e \left (\frac{b f^{3} \sqrt{- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac{f^{3}}{4 c e}\right ) + b d^{2} f^{3}}{b e^{2} f^{3}} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*f**3*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + f**3/(4*c*e))*log(2*d*x/e + x**2 + (-8*a*c*e*(-b*f**3*sq
rt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + f**3/(4*c*e)) + 2*a*f**3 + 2*b**2*e*(-b*f**3*sqrt(-4*a*c + b**2)/(4
*c*e*(4*a*c - b**2)) + f**3/(4*c*e)) + b*d**2*f**3)/(b*e**2*f**3)) + (b*f**3*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c
- b**2)) + f**3/(4*c*e))*log(2*d*x/e + x**2 + (-8*a*c*e*(b*f**3*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) +
f**3/(4*c*e)) + 2*a*f**3 + 2*b**2*e*(b*f**3*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + f**3/(4*c*e)) + b*d**
2*f**3)/(b*e**2*f**3))

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Giac [B]  time = 1.47233, size = 387, normalized size = 4.45 \begin{align*} -\frac{\sqrt{b^{2} - 4 \, a c} b c f^{3} e \log \left ({\left | 2 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{6} + 4 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d x e^{5} + 2 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d^{2} e^{4} + 4 \, a e^{4} \right |}\right )}{4 \,{\left (b^{2} c^{2} e^{2} - 4 \, a c^{3} e^{2}\right )}} + \frac{\sqrt{b^{2} - 4 \, a c} b c f^{3} e \log \left ({\left | -2 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{6} - 4 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d x e^{5} - 2 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d^{2} e^{4} - 4 \, a e^{4} \right |}\right )}{4 \,{\left (b^{2} c^{2} e^{2} - 4 \, a c^{3} e^{2}\right )}} + \frac{f^{3} e^{\left (-1\right )} \log \left ({\left | c x^{4} e^{4} + 4 \, c d x^{3} e^{3} + 6 \, c d^{2} x^{2} e^{2} + 4 \, c d^{3} x e + c d^{4} + b x^{2} e^{2} + 2 \, b d x e + b d^{2} + a \right |}\right )}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(b^2 - 4*a*c)*b*c*f^3*e*log(abs(2*(b + sqrt(b^2 - 4*a*c))*x^2*e^6 + 4*(b + sqrt(b^2 - 4*a*c))*d*x*e^5
+ 2*(b + sqrt(b^2 - 4*a*c))*d^2*e^4 + 4*a*e^4))/(b^2*c^2*e^2 - 4*a*c^3*e^2) + 1/4*sqrt(b^2 - 4*a*c)*b*c*f^3*e
*log(abs(-2*(b - sqrt(b^2 - 4*a*c))*x^2*e^6 - 4*(b - sqrt(b^2 - 4*a*c))*d*x*e^5 - 2*(b - sqrt(b^2 - 4*a*c))*d^
2*e^4 - 4*a*e^4))/(b^2*c^2*e^2 - 4*a*c^3*e^2) + 1/4*f^3*e^(-1)*log(abs(c*x^4*e^4 + 4*c*d*x^3*e^3 + 6*c*d^2*x^2
*e^2 + 4*c*d^3*x*e + c*d^4 + b*x^2*e^2 + 2*b*d*x*e + b*d^2 + a))/c