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

Optimal. Leaf size=97 \[ \frac {2 a+b (d+e x)^2}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {b \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]

[Out]

1/2*(2*a+b*(e*x+d)^2)/(-4*a*c+b^2)/e/(a+b*(e*x+d)^2+c*(e*x+d)^4)-b*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 = 0.14, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1142, 1114, 638, 618, 206} \[ \frac {2 a+b (d+e x)^2}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {b \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a + b*(d + e*x)^2)/(2*(b^2 - 4*a*c)*e*(a + b*(d + e*x)^2 + c*(d + e*x)^4)) - (b*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 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 638

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

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 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+e x)^3}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3}{\left (a+b x^2+c x^4\right )^2} \, dx,x,d+e x\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x}{\left (a+b x+c x^2\right )^2} \, dx,x,(d+e x)^2\right )}{2 e}\\ &=\frac {2 a+b (d+e x)^2}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac {b \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,(d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e}\\ &=\frac {2 a+b (d+e x)^2}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {b \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 {2 a+b (d+e x)^2}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac {b \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 = 0.13, size = 100, normalized size = 1.03 \[ \frac {\frac {2 a+b (d+e x)^2}{\left (b^2-4 a c\right ) \left (a+(d+e x)^2 \left (b+c (d+e x)^2\right )\right )}-\frac {2 b \tan ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.92, size = 1021, normalized size = 10.53 \[ \left [\frac {{\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + 2 \, a b^{2} - 8 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2} - {\left (b c e^{4} x^{4} + 4 \, b c d e^{3} x^{3} + b c d^{4} + {\left (6 \, b c d^{2} + b^{2}\right )} e^{2} x^{2} + b^{2} d^{2} + 2 \, {\left (2 \, b c d^{3} + b^{2} d\right )} e x + a b\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 )}{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 {{\left (b^{3} - 4 \, a b c\right )} e^{2} x^{2} + 2 \, {\left (b^{3} - 4 \, a b c\right )} d e x + 2 \, a b^{2} - 8 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2} - 2 \, {\left (b c e^{4} x^{4} + 4 \, b c d e^{3} x^{3} + b c d^{4} + {\left (6 \, b c d^{2} + b^{2}\right )} e^{2} x^{2} + b^{2} d^{2} + 2 \, {\left (2 \, b c d^{3} + b^{2} d\right )} e x + a b\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 )}{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*x+d)^3/(a+b*(e*x+d)^2+c*(e*x+d)^4)^2,x, algorithm="fricas")

[Out]

[1/2*((b^3 - 4*a*b*c)*e^2*x^2 + 2*(b^3 - 4*a*b*c)*d*e*x + 2*a*b^2 - 8*a^2*c + (b^3 - 4*a*b*c)*d^2 - (b*c*e^4*x
^4 + 4*b*c*d*e^3*x^3 + b*c*d^4 + (6*b*c*d^2 + b^2)*e^2*x^2 + b^2*d^2 + 2*(2*b*c*d^3 + b^2*d)*e*x + a*b)*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^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*((b^3 - 4*a*b*c)*e^2*x^2 + 2*(b^3 - 4*a*b*c)*d*e*x + 2*a*b^2 - 8*a^2*c +
(b^3 - 4*a*b*c)*d^2 - 2*(b*c*e^4*x^4 + 4*b*c*d*e^3*x^3 + b*c*d^4 + (6*b*c*d^2 + b^2)*e^2*x^2 + b^2*d^2 + 2*(2
*b*c*d^3 + b^2*d)*e*x + a*b)*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^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 [A]  time = 0.57, size = 171, normalized size = 1.76 \[ \frac {b \arctan \left (\frac {2 \, c d^{2} + 2 \, {\left (x^{2} e + 2 \, d x\right )} c e + b}{\sqrt {-b^{2} + 4 \, a c}}\right ) e^{\left (-1\right )}}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {b d^{2} + {\left (x^{2} e + 2 \, d x\right )} b e + 2 \, a}{2 \, {\left (c d^{4} + 2 \, {\left (x^{2} e + 2 \, d x\right )} c d^{2} e + {\left (x^{2} e + 2 \, d x\right )}^{2} c e^{2} + b d^{2} + {\left (x^{2} e + 2 \, d x\right )} b e + a\right )} {\left (b^{2} e - 4 \, a c e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 276, normalized size = 2.85 \[ \frac {b \left (-\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right ) e -d \right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+x \right )}{2 \left (4 a c -b^{2}\right ) e \left (2 c \,e^{3} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{3}+6 c d \,e^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{2}+6 e c \,d^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+2 c \,d^{3}+b e \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+b d \right )}+\frac {-\frac {b e \,x^{2}}{2 \left (4 a c -b^{2}\right )}-\frac {b d x}{4 a c -b^{2}}-\frac {b \,d^{2}+2 a}{2 \left (4 a c -b^{2}\right ) e}}{c \,e^{4} x^{4}+4 c d \,e^{3} x^{3}+6 c \,d^{2} e^{2} x^{2}+4 c \,d^{3} e x +b \,e^{2} x^{2}+c \,d^{4}+2 b d e x +b \,d^{2}+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/2*b*e/(4*a*c-b^2)*x^2-b*d/(4*a*c-b^2)*x-1/2/e*(b*d^2+2*a)/(4*a*c-b^2))/(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)+1/2/(4*a*c-b^2)*b/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=RootOf(_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 \[ -b \int -\frac {e x + d}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{4} x^{4} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{3} x^{3} + {\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + {\left (b^{3} - 4 \, a b c + 6 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{2} x^{2} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2} + 2 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} + {\left (b^{3} - 4 \, a b c\right )} d\right )} e x}\,{d x} + \frac {b e^{2} x^{2} + 2 \, b d e x + b d^{2} + 2 \, a}{2 \, {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{5} x^{4} + 4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d e^{4} x^{3} + {\left (b^{3} - 4 \, a b c + 6 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{3} x^{2} + 2 \, {\left (2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} + {\left (b^{3} - 4 \, a b c\right )} d\right )} e^{2} x + {\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + a b^{2} - 4 \, a^{2} c + {\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*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*(b*e^2*x^2 + 2*b*d*e*x + b*d^2 + 2*a)/((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 = 1.77, size = 427, normalized size = 4.40 \[ \frac {b\,\mathrm {atan}\left (\frac {{\left (4\,a\,c-b^2\right )}^4\,\left (x\,\left (\frac {b^3\,\left (2\,b^3\,c^2\,d\,e^9-8\,a\,b\,c^3\,d\,e^9\right )}{a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}-\frac {2\,b^2\,c^2\,d\,e^7}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}\right )+x^2\,\left (\frac {b^3\,\left (2\,b^3\,c^2\,e^{10}-8\,a\,b\,c^3\,e^{10}\right )}{2\,a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}-\frac {b^2\,c^2\,e^8}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}\right )-\frac {b^3\,\left (16\,a^2\,c^3\,e^8-4\,a\,b^2\,c^2\,e^8+8\,a\,b\,c^3\,d^2\,e^8-2\,b^3\,c^2\,d^2\,e^8\right )}{2\,a\,e^2\,{\left (4\,a\,c-b^2\right )}^{11/2}}-\frac {b^2\,c^2\,d^2\,e^6}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}\right )}{2\,b^2\,c^2\,e^6}\right )}{e\,{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {\frac {b\,d^2+2\,a}{2\,e\,\left (4\,a\,c-b^2\right )}+\frac {b\,e\,x^2}{2\,\left (4\,a\,c-b^2\right )}+\frac {b\,d\,x}{4\,a\,c-b^2}}{a+x^2\,\left (6\,c\,d^2\,e^2+b\,e^2\right )+b\,d^2+c\,d^4+x\,\left (4\,c\,e\,d^3+2\,b\,e\,d\right )+c\,e^4\,x^4+4\,c\,d\,e^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*atan(((4*a*c - b^2)^4*(x*((b^3*(2*b^3*c^2*d*e^9 - 8*a*b*c^3*d*e^9))/(a*e^2*(4*a*c - b^2)^(11/2)) - (2*b^2*c
^2*d*e^7)/(a*(4*a*c - b^2)^(7/2))) + x^2*((b^3*(2*b^3*c^2*e^10 - 8*a*b*c^3*e^10))/(2*a*e^2*(4*a*c - b^2)^(11/2
)) - (b^2*c^2*e^8)/(a*(4*a*c - b^2)^(7/2))) - (b^3*(16*a^2*c^3*e^8 - 4*a*b^2*c^2*e^8 - 2*b^3*c^2*d^2*e^8 + 8*a
*b*c^3*d^2*e^8))/(2*a*e^2*(4*a*c - b^2)^(11/2)) - (b^2*c^2*d^2*e^6)/(a*(4*a*c - b^2)^(7/2))))/(2*b^2*c^2*e^6))
)/(e*(4*a*c - b^2)^(3/2)) - ((2*a + b*d^2)/(2*e*(4*a*c - b^2)) + (b*e*x^2)/(2*(4*a*c - b^2)) + (b*d*x)/(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)

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sympy [B]  time = 4.88, size = 495, normalized size = 5.10 \[ \frac {b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {- 16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} + 2 b c d^{2}}{2 b c e^{2}} \right )}}{2 e} - \frac {b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (\frac {2 d x}{e} + x^{2} + \frac {16 a^{2} b c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} + 2 b c d^{2}}{2 b c e^{2}} \right )}}{2 e} + \frac {- 2 a - b d^{2} - 2 b d e x - b e^{2} x^{2}}{8 a^{2} c e - 2 a b^{2} e + 8 a b c d^{2} e + 8 a c^{2} d^{4} e - 2 b^{3} d^{2} e - 2 b^{2} c d^{4} e + x^{4} \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \left (8 a b c e^{3} + 48 a c^{2} d^{2} e^{3} - 2 b^{3} e^{3} - 12 b^{2} c d^{2} e^{3}\right ) + x \left (16 a b c d e^{2} + 32 a c^{2} d^{3} e^{2} - 4 b^{3} d e^{2} - 8 b^{2} c d^{3} e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(-1/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (-16*a**2*b*c**2*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**3*c*sqr
t(-1/(4*a*c - b**2)**3) - b**5*sqrt(-1/(4*a*c - b**2)**3) + b**2 + 2*b*c*d**2)/(2*b*c*e**2))/(2*e) - b*sqrt(-1
/(4*a*c - b**2)**3)*log(2*d*x/e + x**2 + (16*a**2*b*c**2*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**3*c*sqrt(-1/(4*a*
c - b**2)**3) + b**5*sqrt(-1/(4*a*c - b**2)**3) + b**2 + 2*b*c*d**2)/(2*b*c*e**2))/(2*e) + (-2*a - b*d**2 - 2*
b*d*e*x - b*e**2*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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