\(\int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx\) [641]

Optimal result

Integrand size = 31, antiderivative size = 44 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=-\frac {f \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1156, 1121, 632, 212} \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=-\frac {f \text {arctanh}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{e \sqrt {b^2-4 a c}} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 1121

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 1156

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]

Rubi steps \begin{align*} \text {integral}& = \frac {f \text {Subst}\left (\int \frac {x}{a+b x^2+c x^4} \, dx,x,d+e x\right )}{e} \\ & = \frac {f \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,(d+e x)^2\right )}{2 e} \\ & = -\frac {f \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f \arctan \left (\frac {b+2 c (d+e x)^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c} e} \]

[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)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.95

method	result	size
default	\(\frac {f \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,e^{4} \textit {\_Z}^{4}+4 c d \,e^{3} \textit {\_Z}^{3}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 d^{3} e c +2 b d e \right ) \textit {\_Z} +d^{4} c +b \,d^{2}+a \right )}{\sum }\frac {\left (\textit {\_R} e +d \right ) \ln \left (x -\textit {\_R} \right )}{2 e^{3} c \,\textit {\_R}^{3}+6 c d \,e^{2} \textit {\_R}^{2}+6 c \,d^{2} e \textit {\_R} +2 d^{3} c +b e \textit {\_R} +b d}\right )}{2 e}\)	\(130\)
risch	\(-\frac {f \ln \left (\left (e^{2} \sqrt {-4 a c +b^{2}}-b \,e^{2}\right ) x^{2}+\left (2 e d \sqrt {-4 a c +b^{2}}-2 b d e \right ) x +d^{2} \sqrt {-4 a c +b^{2}}-b \,d^{2}-2 a \right )}{2 \sqrt {-4 a c +b^{2}}\, e}+\frac {f \ln \left (\left (e^{2} \sqrt {-4 a c +b^{2}}+b \,e^{2}\right ) x^{2}+\left (2 e d \sqrt {-4 a c +b^{2}}+2 b d e \right ) x +d^{2} \sqrt {-4 a c +b^{2}}+b \,d^{2}+2 a \right )}{2 \sqrt {-4 a c +b^{2}}\, e}\)	\(176\)

[In]

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

[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+d^4*c+b*d^2+a))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 274, normalized size of antiderivative = 6.23 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\left [\frac {f \log \left (\frac {2 \, c^{2} e^{4} x^{4} + 8 \, c^{2} d e^{3} x^{3} + 2 \, c^{2} d^{4} + 2 \, {\left (6 \, c^{2} d^{2} + b c\right )} e^{2} x^{2} + 2 \, b c d^{2} + 4 \, {\left (2 \, c^{2} d^{3} + b c d\right )} e x + b^{2} - 2 \, a c - {\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a}\right )}{2 \, \sqrt {b^{2} - 4 \, a c} e}, -\frac {\sqrt {-b^{2} + 4 \, a c} f \arctan \left (-\frac {{\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{{\left (b^{2} - 4 \, a c\right )} e}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (41) = 82\).

Time = 0.59 (sec) , antiderivative size = 189, normalized size of antiderivative = 4.30 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=- \frac {f \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {- 4 a c f \sqrt {- \frac {1}{4 a c - b^{2}}} + b^{2} f \sqrt {- \frac {1}{4 a c - b^{2}}} + b f + 2 c d^{2} f}{2 c e^{2} f} \right )}}{2 e} + \frac {f \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {4 a c f \sqrt {- \frac {1}{4 a c - b^{2}}} - b^{2} f \sqrt {- \frac {1}{4 a c - b^{2}}} + b f + 2 c d^{2} f}{2 c e^{2} f} \right )}}{2 e} \]

[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)

Maxima [F]

\[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\int { \frac {e f x + d f}{{\left (e x + d\right )}^{4} c + {\left (e x + d\right )}^{2} b + a} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f \arctan \left (\frac {2 \, c d^{2} f + 2 \, {\left (e f x^{2} + 2 \, d f x\right )} c e + b f}{\sqrt {-b^{2} + 4 \, a c} f}\right )}{\sqrt {-b^{2} + 4 \, a c} e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 477, normalized size of antiderivative = 10.84 \[ \int \frac {d f+e f x}{a+b (d+e x)^2+c (d+e x)^4} \, dx=\frac {f\,\mathrm {atan}\left (\frac {\frac {f\,\left (4\,c^2\,d^2\,e^7\,f+4\,c^2\,e^9\,f\,x^2-\frac {f\,\left (8\,b\,c^2\,d^2\,e^8+16\,b\,c^2\,d\,e^9\,x+8\,b\,c^2\,e^{10}\,x^2+16\,a\,c^2\,e^8\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}+8\,c^2\,d\,e^8\,f\,x\right )\,1{}\mathrm {i}}{2\,e\,\sqrt {b^2-4\,a\,c}}+\frac {f\,\left (4\,c^2\,d^2\,e^7\,f+4\,c^2\,e^9\,f\,x^2+\frac {f\,\left (8\,b\,c^2\,d^2\,e^8+16\,b\,c^2\,d\,e^9\,x+8\,b\,c^2\,e^{10}\,x^2+16\,a\,c^2\,e^8\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}+8\,c^2\,d\,e^8\,f\,x\right )\,1{}\mathrm {i}}{2\,e\,\sqrt {b^2-4\,a\,c}}}{\frac {f\,\left (4\,c^2\,d^2\,e^7\,f+4\,c^2\,e^9\,f\,x^2-\frac {f\,\left (8\,b\,c^2\,d^2\,e^8+16\,b\,c^2\,d\,e^9\,x+8\,b\,c^2\,e^{10}\,x^2+16\,a\,c^2\,e^8\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}+8\,c^2\,d\,e^8\,f\,x\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}-\frac {f\,\left (4\,c^2\,d^2\,e^7\,f+4\,c^2\,e^9\,f\,x^2+\frac {f\,\left (8\,b\,c^2\,d^2\,e^8+16\,b\,c^2\,d\,e^9\,x+8\,b\,c^2\,e^{10}\,x^2+16\,a\,c^2\,e^8\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}+8\,c^2\,d\,e^8\,f\,x\right )}{2\,e\,\sqrt {b^2-4\,a\,c}}}\right )\,1{}\mathrm {i}}{e\,\sqrt {b^2-4\,a\,c}} \]

[In]

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

[Out]

(f*atan(((f*(4*c^2*d^2*e^7*f + 4*c^2*e^9*f*x^2 - (f*(16*a*c^2*e^8 + 8*b*c^2*d^2*e^8 + 8*b*c^2*e^10*x^2 + 16*b*
c^2*d*e^9*x))/(2*e*(b^2 - 4*a*c)^(1/2)) + 8*c^2*d*e^8*f*x)*1i)/(2*e*(b^2 - 4*a*c)^(1/2)) + (f*(4*c^2*d^2*e^7*f
+ 4*c^2*e^9*f*x^2 + (f*(16*a*c^2*e^8 + 8*b*c^2*d^2*e^8 + 8*b*c^2*e^10*x^2 + 16*b*c^2*d*e^9*x))/(2*e*(b^2 - 4*
a*c)^(1/2)) + 8*c^2*d*e^8*f*x)*1i)/(2*e*(b^2 - 4*a*c)^(1/2)))/((f*(4*c^2*d^2*e^7*f + 4*c^2*e^9*f*x^2 - (f*(16*
a*c^2*e^8 + 8*b*c^2*d^2*e^8 + 8*b*c^2*e^10*x^2 + 16*b*c^2*d*e^9*x))/(2*e*(b^2 - 4*a*c)^(1/2)) + 8*c^2*d*e^8*f*
x))/(2*e*(b^2 - 4*a*c)^(1/2)) - (f*(4*c^2*d^2*e^7*f + 4*c^2*e^9*f*x^2 + (f*(16*a*c^2*e^8 + 8*b*c^2*d^2*e^8 + 8
*b*c^2*e^10*x^2 + 16*b*c^2*d*e^9*x))/(2*e*(b^2 - 4*a*c)^(1/2)) + 8*c^2*d*e^8*f*x))/(2*e*(b^2 - 4*a*c)^(1/2))))
*1i)/(e*(b^2 - 4*a*c)^(1/2))