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

Optimal. Leaf size=202 \[ -\frac{f^4 \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} e \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{f^4 \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} c^{3/2} e \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{f^4 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.361944, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {1142, 1122, 1166, 205} \[ -\frac{f^4 \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} e \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{f^4 \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} c^{3/2} e \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{f^4 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^4*x)/c - ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*f^4*ArcTan[(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b - Sqrt[b^2 -
4*a*c]]])/(Sqrt[2]*c^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]*e) - ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*f^4*ArcTan[
(Sqrt[2]*Sqrt[c]*(d + e*x))/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*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 1122

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d^3*(d*x)^(m - 3)*(a + b*
x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 1)), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && 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 f+e f x)^4}{a+b (d+e x)^2+c (d+e x)^4} \, dx &=\frac{f^4 \operatorname{Subst}\left (\int \frac{x^4}{a+b x^2+c x^4} \, dx,x,d+e x\right )}{e}\\ &=\frac{f^4 x}{c}-\frac{f^4 \operatorname{Subst}\left (\int \frac{a+b x^2}{a+b x^2+c x^4} \, dx,x,d+e x\right )}{c e}\\ &=\frac{f^4 x}{c}-\frac{\left (\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) f^4\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 c e}-\frac{\left (\left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) f^4\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 c e}\\ &=\frac{f^4 x}{c}-\frac{\left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) f^4 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}} e}-\frac{\left (b+\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) f^4 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}} e}\\ \end{align*}

Mathematica [A]  time = 0.14723, size = 222, normalized size = 1.1 \[ \frac{f^4 \left (-\frac{\sqrt{2} \left (b \sqrt{b^2-4 a c}+2 a c-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\sqrt{2} \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} (d+e x)}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}+2 \sqrt{c} (d+e x)\right )}{2 c^{3/2} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.003, size = 164, normalized size = 0.8 \begin{align*}{\frac{{f}^{4}x}{c}}+{\frac{{f}^{4}}{2\,ce}\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}b{e}^{2}-2\,{\it \_R}\,bde-b{d}^{2}-a \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)^4/(a+b*(e*x+d)^2+c*(e*x+d)^4),x)

[Out]

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

[Out]

Timed out

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Fricas [B]  time = 1.65602, size = 2772, normalized size = 13.72 \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)^4/(a+b*(e*x+d)^2+c*(e*x+d)^4),x, algorithm="fricas")

[Out]

1/2*(2*f^4*x - sqrt(1/2)*c*sqrt(-((b^3 - 3*a*b*c)*f^8 + sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*
c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))*log(-2*(a*b^2 - a^2*c)*e*f^12*x - 2*(a*b^2 - a^
2*c)*d*f^12 + sqrt(1/2)*((b^4 - 5*a*b^2*c + 4*a^2*c^2)*e*f^8 - sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6
- 4*a*c^7)*e^4))*(b^3*c^3 - 4*a*b*c^4)*e^3)*sqrt(-((b^3 - 3*a*b*c)*f^8 + sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^1
6/((b^2*c^6 - 4*a*c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))) + sqrt(1/2)*c*sqrt(-((b^3 -
3*a*b*c)*f^8 + sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2
*c^3 - 4*a*c^4)*e^2))*log(-2*(a*b^2 - a^2*c)*e*f^12*x - 2*(a*b^2 - a^2*c)*d*f^12 - sqrt(1/2)*((b^4 - 5*a*b^2*c
+ 4*a^2*c^2)*e*f^8 - sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^3*c^3 - 4*a*b*c^4)*e
^3)*sqrt(-((b^3 - 3*a*b*c)*f^8 + sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^2*c^3 - 4
*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))) - sqrt(1/2)*c*sqrt(-((b^3 - 3*a*b*c)*f^8 - sqrt((b^4 - 2*a*b^2*c + a^
2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))*log(-2*(a*b^2 - a^2
*c)*e*f^12*x - 2*(a*b^2 - a^2*c)*d*f^12 + sqrt(1/2)*((b^4 - 5*a*b^2*c + 4*a^2*c^2)*e*f^8 + sqrt((b^4 - 2*a*b^2
*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^3*c^3 - 4*a*b*c^4)*e^3)*sqrt(-((b^3 - 3*a*b*c)*f^8 - sqrt((b^
4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2)))
+ sqrt(1/2)*c*sqrt(-((b^3 - 3*a*b*c)*f^8 - sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e^4))*(b
^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))*log(-2*(a*b^2 - a^2*c)*e*f^12*x - 2*(a*b^2 - a^2*c)*d*f^12 -
sqrt(1/2)*((b^4 - 5*a*b^2*c + 4*a^2*c^2)*e*f^8 + sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 - 4*a*c^7)*e
^4))*(b^3*c^3 - 4*a*b*c^4)*e^3)*sqrt(-((b^3 - 3*a*b*c)*f^8 - sqrt((b^4 - 2*a*b^2*c + a^2*c^2)*f^16/((b^2*c^6 -
4*a*c^7)*e^4))*(b^2*c^3 - 4*a*c^4)*e^2)/((b^2*c^3 - 4*a*c^4)*e^2))))/c

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Sympy [A]  time = 3.70435, size = 219, normalized size = 1.08 \begin{align*} \operatorname{RootSum}{\left (t^{4} \left (256 a^{2} c^{5} e^{4} - 128 a b^{2} c^{4} e^{4} + 16 b^{4} c^{3} e^{4}\right ) + t^{2} \left (48 a^{2} b c^{2} e^{2} f^{8} - 28 a b^{3} c e^{2} f^{8} + 4 b^{5} e^{2} f^{8}\right ) + a^{3} f^{16}, \left ( t \mapsto t \log{\left (x + \frac{32 t^{3} a b c^{4} e^{3} - 8 t^{3} b^{3} c^{3} e^{3} - 4 t a^{2} c^{2} e f^{8} + 8 t a b^{2} c e f^{8} - 2 t b^{4} e f^{8} + a^{2} c d f^{12} - a b^{2} d f^{12}}{a^{2} c e f^{12} - a b^{2} e f^{12}} \right )} \right )\right )} + \frac{f^{4} x}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(_t**4*(256*a**2*c**5*e**4 - 128*a*b**2*c**4*e**4 + 16*b**4*c**3*e**4) + _t**2*(48*a**2*b*c**2*e**2*f**
8 - 28*a*b**3*c*e**2*f**8 + 4*b**5*e**2*f**8) + a**3*f**16, Lambda(_t, _t*log(x + (32*_t**3*a*b*c**4*e**3 - 8*
_t**3*b**3*c**3*e**3 - 4*_t*a**2*c**2*e*f**8 + 8*_t*a*b**2*c*e*f**8 - 2*_t*b**4*e*f**8 + a**2*c*d*f**12 - a*b*
*2*d*f**12)/(a**2*c*e*f**12 - a*b**2*e*f**12)))) + f**4*x/c

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

[Out]

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