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

Optimal. Leaf size=81 \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{2 c e \sqrt {b^2-4 a c}}+\frac {\log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 c e} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1142, 1114, 634, 618, 206, 628} \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {b^2-4 a c}}\right )}{2 c e \sqrt {b^2-4 a c}}+\frac {\log \left (a+b (d+e x)^2+c (d+e x)^4\right )}{4 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.95 \[ \frac {\log \left (a+b (d+e x)^2+c (d+e x)^4\right )-\frac {2 b \tan ^{-1}\left (\frac {b+2 c (d+e x)^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}}{4 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-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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fricas [B]  time = 0.71, size = 434, normalized size = 5.36 \[ \left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} e^{4} x^{4} + 8 \, c^{2} d e^{3} x^{3} + 2 \, c^{2} d^{4} + 2 \, {\left (6 \, c^{2} d^{2} + b c\right )} e^{2} x^{2} + 2 \, b c d^{2} + 4 \, {\left (2 \, c^{2} d^{3} + b c d\right )} e x + b^{2} - 2 \, a c + {\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c e^{2} x^{2} + 4 \, c d e x + 2 \, c d^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c e^{4} x^{4} + 4 \, c d e^{3} x^{3} + c d^{4} + {\left (6 \, c d^{2} + b\right )} e^{2} x^{2} + b d^{2} + 2 \, {\left (2 \, c d^{3} + b d\right )} e x + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )} e}\right ] \]

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),x, algorithm="fricas")

[Out]

[1/4*(sqrt(b^2 - 4*a*c)*b*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)*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*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)*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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giac [A]  time = 0.40, size = 130, normalized size = 1.60 \[ -\frac {b \arctan \left (\frac {2 \, c d^{2} + 2 \, {\left (x^{2} e + 2 \, d x\right )} c e + b}{\sqrt {-b^{2} + 4 \, a c}}\right ) e^{\left (-1\right )}}{2 \, \sqrt {-b^{2} + 4 \, a c} c} + \frac {e^{\left (-1\right )} \log \left (c d^{4} + 2 \, {\left (x^{2} e + 2 \, d x\right )} c d^{2} e + {\left (x^{2} e + 2 \, d x\right )}^{2} c e^{2} + b d^{2} + {\left (x^{2} e + 2 \, d x\right )} b e + a\right )}{4 \, c} \]

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),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.00, size = 151, normalized size = 1.86 \[ \frac {\left (\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{3} e^{3}+3 \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{2} d \,e^{2}+3 \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right ) d^{2} e +d^{3}\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+x \right )}{2 e \left (2 c \,e^{3} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{3}+6 c d \,e^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )^{2}+6 e c \,d^{2} \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+2 c \,d^{3}+b e \RootOf \left (\textit {\_Z}^{4} c \,e^{4}+4 \textit {\_Z}^{3} c d \,e^{3}+c \,d^{4}+b \,d^{2}+\left (6 c \,d^{2} e^{2}+b \,e^{2}\right ) \textit {\_Z}^{2}+\left (4 c \,d^{3} e +2 d e b \right ) \textit {\_Z} +a \right )+b d \right )} \]

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),x)

[Out]

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

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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.76, size = 278, normalized size = 3.43 \[ \frac {4\,a\,c\,e\,\ln \left (c\,d^4+4\,c\,d^3\,e\,x+6\,c\,d^2\,e^2\,x^2+b\,d^2+4\,c\,d\,e^3\,x^3+2\,b\,d\,e\,x+c\,e^4\,x^4+b\,e^2\,x^2+a\right )}{16\,a\,c^2\,e^2-4\,b^2\,c\,e^2}-\frac {b^2\,e\,\ln \left (c\,d^4+4\,c\,d^3\,e\,x+6\,c\,d^2\,e^2\,x^2+b\,d^2+4\,c\,d\,e^3\,x^3+2\,b\,d\,e\,x+c\,e^4\,x^4+b\,e^2\,x^2+a\right )}{16\,a\,c^2\,e^2-4\,b^2\,c\,e^2}-\frac {b\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,d^2}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,e^2\,x^2}{\sqrt {4\,a\,c-b^2}}+\frac {4\,c\,d\,e\,x}{\sqrt {4\,a\,c-b^2}}\right )}{2\,c\,e\,\sqrt {4\,a\,c-b^2}} \]

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),x)

[Out]

(4*a*c*e*log(a + b*d^2 + c*d^4 + b*e^2*x^2 + c*e^4*x^4 + 2*b*d*e*x + 6*c*d^2*e^2*x^2 + 4*c*d^3*e*x + 4*c*d*e^3
*x^3))/(16*a*c^2*e^2 - 4*b^2*c*e^2) - (b^2*e*log(a + b*d^2 + c*d^4 + b*e^2*x^2 + c*e^4*x^4 + 2*b*d*e*x + 6*c*d
^2*e^2*x^2 + 4*c*d^3*e*x + 4*c*d*e^3*x^3))/(16*a*c^2*e^2 - 4*b^2*c*e^2) - (b*atan(b/(4*a*c - b^2)^(1/2) + (2*c
*d^2)/(4*a*c - b^2)^(1/2) + (2*c*e^2*x^2)/(4*a*c - b^2)^(1/2) + (4*c*d*e*x)/(4*a*c - b^2)^(1/2)))/(2*c*e*(4*a*
c - b^2)^(1/2))

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sympy [B]  time = 1.64, size = 280, normalized size = 3.46 \[ \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) \log {\left (\frac {2 d x}{e} + x^{2} + \frac {- 8 a c e \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) + 2 a + 2 b^{2} e \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) + b d^{2}}{b e^{2}} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) \log {\left (\frac {2 d x}{e} + x^{2} + \frac {- 8 a c e \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) + 2 a + 2 b^{2} e \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c e \left (4 a c - b^{2}\right )} + \frac {1}{4 c e}\right ) + b d^{2}}{b e^{2}} \right )} \]

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),x)

[Out]

(-b*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + 1/(4*c*e))*log(2*d*x/e + x**2 + (-8*a*c*e*(-b*sqrt(-4*a*c + b
**2)/(4*c*e*(4*a*c - b**2)) + 1/(4*c*e)) + 2*a + 2*b**2*e*(-b*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + 1/(
4*c*e)) + b*d**2)/(b*e**2)) + (b*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + 1/(4*c*e))*log(2*d*x/e + x**2 +
(-8*a*c*e*(b*sqrt(-4*a*c + b**2)/(4*c*e*(4*a*c - b**2)) + 1/(4*c*e)) + 2*a + 2*b**2*e*(b*sqrt(-4*a*c + b**2)/(
4*c*e*(4*a*c - b**2)) + 1/(4*c*e)) + b*d**2)/(b*e**2))

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