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

Optimal. Leaf size=98 \[ \frac {2 c f \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}}-\frac {f \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )} \]

[Out]

-1/2*f*(b+2*c*(e*x+d)^2)/(-4*a*c+b^2)/e/(a+b*(e*x+d)^2+c*(e*x+d)^4)+2*c*f*arctanh((b+2*c*(e*x+d)^2)/(-4*a*c+b^
2)^(1/2))/(-4*a*c+b^2)^(3/2)/e

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Rubi [A]  time = 0.13, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1142, 1107, 614, 618, 206} \[ \frac {2 c f \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}}-\frac {f \left (b+2 c (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 99, normalized size = 1.01 \[ -\frac {f \left (\frac {4 c \tan ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {b+2 c (d+e x)^2}{a+b (d+e x)^2+c (d+e x)^4}\right )}{2 e \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(f*((b + 2*c*(d + e*x)^2)/(a + b*(d + e*x)^2 + c*(d + e*x)^4) + (4*c*ArcTan[(b + 2*c*(d + e*x)^2)/Sqrt[-b
^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c]))/((b^2 - 4*a*c)*e)

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fricas [B]  time = 0.66, size = 1066, normalized size = 10.88 \[ \left [-\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f x^{2} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e f x + 2 \, {\left (c^{2} e^{4} f x^{4} + 4 \, c^{2} d e^{3} f x^{3} + {\left (6 \, c^{2} d^{2} + b c\right )} e^{2} f x^{2} + 2 \, {\left (2 \, c^{2} d^{3} + b c d\right )} e f x + {\left (c^{2} d^{4} + b c d^{2} + a c\right )} f\right )} \sqrt {b^{2} - 4 \, a c} \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^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} f}{2 \, {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} e^{5} x^{4} + 4 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e^{4} x^{3} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 6 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{2}\right )} e^{3} x^{2} + 2 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{3} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d\right )} e^{2} x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d^{2}\right )} e\right )}}, -\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2} f x^{2} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e f x - 4 \, {\left (c^{2} e^{4} f x^{4} + 4 \, c^{2} d e^{3} f x^{3} + {\left (6 \, c^{2} d^{2} + b c\right )} e^{2} f x^{2} + 2 \, {\left (2 \, c^{2} d^{3} + b c d\right )} e f x + {\left (c^{2} d^{4} + b c d^{2} + a c\right )} f\right )} \sqrt {-b^{2} + 4 \, a c} \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^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} f}{2 \, {\left ({\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} e^{5} x^{4} + 4 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e^{4} x^{3} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2} + 6 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{2}\right )} e^{3} x^{2} + 2 \, {\left (2 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{3} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d\right )} e^{2} x + {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} d^{2}\right )} e\right )}}\right ] \]

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)^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(b^2*c - 4*a*c^2)*e^2*f*x^2 + 4*(b^2*c - 4*a*c^2)*d*e*f*x + 2*(c^2*e^4*f*x^4 + 4*c^2*d*e^3*f*x^3 + (6
*c^2*d^2 + b*c)*e^2*f*x^2 + 2*(2*c^2*d^3 + b*c*d)*e*f*x + (c^2*d^4 + b*c*d^2 + a*c)*f)*sqrt(b^2 - 4*a*c)*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^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*d^2)*
f)/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*e^5*x^4 + 4*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e^4*x^3 + (b^5 - 8*a*b
^3*c + 16*a^2*b*c^2 + 6*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^2)*e^3*x^2 + 2*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c
^3)*d^3 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d)*e^2*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2
+ 16*a^2*c^3)*d^4 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d^2)*e), -1/2*(2*(b^2*c - 4*a*c^2)*e^2*f*x^2 + 4*(b^2*c
- 4*a*c^2)*d*e*f*x - 4*(c^2*e^4*f*x^4 + 4*c^2*d*e^3*f*x^3 + (6*c^2*d^2 + b*c)*e^2*f*x^2 + 2*(2*c^2*d^3 + b*c*d
)*e*f*x + (c^2*d^4 + b*c*d^2 + a*c)*f)*sqrt(-b^2 + 4*a*c)*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^3 - 4*a*b*c + 2*(b^2*c - 4*a*c^2)*d^2)*f)/((b^4*c - 8*a*b^2*c^2 + 16*a^2*c^
3)*e^5*x^4 + 4*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e^4*x^3 + (b^5 - 8*a*b^3*c + 16*a^2*b*c^2 + 6*(b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*d^2)*e^3*x^2 + 2*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^3 + (b^5 - 8*a*b^3*c + 16*a^2*
b*c^2)*d)*e^2*x + (a*b^4 - 8*a^2*b^2*c + 16*a^3*c^2 + (b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^4 + (b^5 - 8*a*b^3*
c + 16*a^2*b*c^2)*d^2)*e)]

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giac [B]  time = 0.44, size = 211, normalized size = 2.15 \[ -\frac {2 \, c f \arctan \left (\frac {2 \, c d^{2} f + 2 \, {\left (f x^{2} e + 2 \, d f x\right )} c e + b f}{\sqrt {-b^{2} + 4 \, a c} f}\right ) e^{\left (-1\right )}}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c d^{2} f^{3} + 2 \, {\left (f x^{2} e + 2 \, d f x\right )} c f^{2} e + b f^{3}}{2 \, {\left (c d^{4} f^{2} + 2 \, {\left (f x^{2} e + 2 \, d f x\right )} c d^{2} f e + b d^{2} f^{2} + {\left (f x^{2} e + 2 \, d f x\right )}^{2} c e^{2} + {\left (f x^{2} e + 2 \, d f x\right )} b f e + a f^{2}\right )} {\left (b^{2} e - 4 \, a c e\right )}} \]

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)^2,x, algorithm="giac")

[Out]

-2*c*f*arctan((2*c*d^2*f + 2*(f*x^2*e + 2*d*f*x)*c*e + b*f)/(sqrt(-b^2 + 4*a*c)*f))*e^(-1)/((b^2 - 4*a*c)*sqrt
(-b^2 + 4*a*c)) - 1/2*(2*c*d^2*f^3 + 2*(f*x^2*e + 2*d*f*x)*c*f^2*e + b*f^3)/((c*d^4*f^2 + 2*(f*x^2*e + 2*d*f*x
)*c*d^2*f*e + b*d^2*f^2 + (f*x^2*e + 2*d*f*x)^2*c*e^2 + (f*x^2*e + 2*d*f*x)*b*f*e + a*f^2)*(b^2*e - 4*a*c*e))

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maple [C]  time = 0.02, size = 484, normalized size = 4.94 \[ \frac {c e f \,x^{2}}{\left (c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a \right ) \left (4 a c -b^{2}\right )}+\frac {2 c d f x}{\left (c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a \right ) \left (4 a c -b^{2}\right )}+\frac {c \,d^{2} f}{\left (c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a \right ) \left (4 a c -b^{2}\right ) e}+\frac {c f \left (\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right ) e +d \right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+x \right )}{\left (4 a c -b^{2}\right ) e \left (2 c \,e^{3} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{3}+6 c d \,e^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{2}+6 e c \,d^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+2 c \,d^{3}+b e \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+b d \right )}+\frac {b f}{2 \left (c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a \right ) \left (4 a c -b^{2}\right ) e} \]

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)^2,x)

[Out]

f/(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)*c*e/(4*a*c-b^2)*x^2+
2*f/(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)*c*d/(4*a*c-b^2)*x+
f/(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)/(4*a*c-b^2)/e*c*d^2+
1/2*f/(c*e^4*x^4+4*c*d*e^3*x^3+6*c*d^2*e^2*x^2+4*c*d^3*e*x+b*e^2*x^2+c*d^4+2*b*d*e*x+b*d^2+a)/(4*a*c-b^2)/e*b+
f*c/(4*a*c-b^2)/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(-_R+x),_R=Root
Of(_Z^4*c*e^4+4*_Z^3*c*d*e^3+c*d^4+b*d^2+(6*c*d^2*e^2+b*e^2)*_Z^2+(4*c*d^3*e+2*b*d*e)*_Z+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, c f \int -\frac {e x + d}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{4} x^{4} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{3} x^{3} + {\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + {\left (b^{3} - 4 \, a b c + 6 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{2} x^{2} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2} + 2 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} + {\left (b^{3} - 4 \, a b c\right )} d\right )} e x}\,{d x} - \frac {2 \, c e^{2} f x^{2} + 4 \, c d e f x + {\left (2 \, c d^{2} + b\right )} f}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{5} x^{4} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{4} x^{3} + {\left (b^{3} - 4 \, a b c + 6 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{3} x^{2} + 2 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} + {\left (b^{3} - 4 \, a b c\right )} d\right )} e^{2} x + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} e\right )}} \]

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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.91, size = 442, normalized size = 4.51 \[ \frac {\frac {f\,\left (2\,c\,d^2+b\right )}{2\,e\,\left (4\,a\,c-b^2\right )}+\frac {2\,c\,d\,f\,x}{4\,a\,c-b^2}+\frac {c\,e\,f\,x^2}{4\,a\,c-b^2}}{a+x^2\,\left (6\,c\,d^2\,e^2+b\,e^2\right )+b\,d^2+c\,d^4+x\,\left (4\,c\,e\,d^3+2\,b\,e\,d\right )+c\,e^4\,x^4+4\,c\,d\,e^3\,x^3}+\frac {2\,c\,f\,\mathrm {atan}\left (\frac {{\left (4\,a\,c-b^2\right )}^4\,\left (x\,\left (\frac {8\,c^4\,d\,e^7\,f^2}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}-\frac {8\,b\,c^2\,f^2\,\left (b^3\,c^2\,d\,e^9-4\,a\,b\,c^3\,d\,e^9\right )}{a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}\right )+x^2\,\left (\frac {4\,c^4\,e^8\,f^2}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}-\frac {4\,b\,c^2\,f^2\,\left (b^3\,c^2\,e^{10}-4\,a\,b\,c^3\,e^{10}\right )}{a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}\right )+\frac {4\,c^4\,d^2\,e^6\,f^2}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {4\,b\,c^2\,f^2\,\left (8\,a^2\,c^3\,e^8-2\,a\,b^2\,c^2\,e^8+4\,a\,b\,c^3\,d^2\,e^8-b^3\,c^2\,d^2\,e^8\right )}{a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}\right )}{8\,c^4\,e^6\,f^2}\right )}{e\,{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f*(b + 2*c*d^2))/(2*e*(4*a*c - b^2)) + (2*c*d*f*x)/(4*a*c - b^2) + (c*e*f*x^2)/(4*a*c - b^2))/(a + x^2*(b*e^
2 + 6*c*d^2*e^2) + b*d^2 + c*d^4 + x*(2*b*d*e + 4*c*d^3*e) + c*e^4*x^4 + 4*c*d*e^3*x^3) + (2*c*f*atan(((4*a*c
- b^2)^4*(x*((8*c^4*d*e^7*f^2)/(a*(4*a*c - b^2)^(7/2)) - (8*b*c^2*f^2*(b^3*c^2*d*e^9 - 4*a*b*c^3*d*e^9))/(a*e^
2*(4*a*c - b^2)^(11/2))) + x^2*((4*c^4*e^8*f^2)/(a*(4*a*c - b^2)^(7/2)) - (4*b*c^2*f^2*(b^3*c^2*e^10 - 4*a*b*c
^3*e^10))/(a*e^2*(4*a*c - b^2)^(11/2))) + (4*c^4*d^2*e^6*f^2)/(a*(4*a*c - b^2)^(7/2)) + (4*b*c^2*f^2*(8*a^2*c^
3*e^8 - 2*a*b^2*c^2*e^8 - b^3*c^2*d^2*e^8 + 4*a*b*c^3*d^2*e^8))/(a*e^2*(4*a*c - b^2)^(11/2))))/(8*c^4*e^6*f^2)
))/(e*(4*a*c - b^2)^(3/2))

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sympy [B]  time = 5.07, size = 525, normalized size = 5.36 \[ - \frac {c f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {- 16 a^{2} c^{3} f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c^{2} f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} c f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c f + 2 c^{2} d^{2} f}{2 c^{2} e^{2} f} \right )}}{e} + \frac {c f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {16 a^{2} c^{3} f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c^{2} f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} c f \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c f + 2 c^{2} d^{2} f}{2 c^{2} e^{2} f} \right )}}{e} + \frac {b f + 2 c d^{2} f + 4 c d e f x + 2 c e^{2} f x^{2}}{8 a^{2} c e - 2 a b^{2} e + 8 a b c d^{2} e + 8 a c^{2} d^{4} e - 2 b^{3} d^{2} e - 2 b^{2} c d^{4} e + x^{4} \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \left (8 a b c e^{3} + 48 a c^{2} d^{2} e^{3} - 2 b^{3} e^{3} - 12 b^{2} c d^{2} e^{3}\right ) + x \left (16 a b c d e^{2} + 32 a c^{2} d^{3} e^{2} - 4 b^{3} d e^{2} - 8 b^{2} c d^{3} e^{2}\right )} \]

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)**2,x)

[Out]

-c*f*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (-16*a**2*c**3*f*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**2*c*
*2*f*sqrt(-1/(4*a*c - b**2)**3) - b**4*c*f*sqrt(-1/(4*a*c - b**2)**3) + b*c*f + 2*c**2*d**2*f)/(2*c**2*e**2*f)
)/e + c*f*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (16*a**2*c**3*f*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**
2*c**2*f*sqrt(-1/(4*a*c - b**2)**3) + b**4*c*f*sqrt(-1/(4*a*c - b**2)**3) + b*c*f + 2*c**2*d**2*f)/(2*c**2*e**
2*f))/e + (b*f + 2*c*d**2*f + 4*c*d*e*f*x + 2*c*e**2*f*x**2)/(8*a**2*c*e - 2*a*b**2*e + 8*a*b*c*d**2*e + 8*a*c
**2*d**4*e - 2*b**3*d**2*e - 2*b**2*c*d**4*e + x**4*(8*a*c**2*e**5 - 2*b**2*c*e**5) + x**3*(32*a*c**2*d*e**4 -
8*b**2*c*d*e**4) + x**2*(8*a*b*c*e**3 + 48*a*c**2*d**2*e**3 - 2*b**3*e**3 - 12*b**2*c*d**2*e**3) + x*(16*a*b*
c*d*e**2 + 32*a*c**2*d**3*e**2 - 4*b**3*d*e**2 - 8*b**2*c*d**3*e**2))

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