\(\int \frac {1}{b x+c (d+e x)^2} \, dx\) [489]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c e (d+e x)}{\sqrt {b} \sqrt {b+4 c d e}}\right )}{\sqrt {b} \sqrt {b+4 c d e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2006, 632, 212} \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c e (d+e x)}{\sqrt {b} \sqrt {b+4 c d e}}\right )}{\sqrt {b} \sqrt {b+4 c d e}} \]

[In]

Int[(b*x + c*(d + e*x)^2)^(-1),x]

[Out]

(-2*ArcTanh[(b + 2*c*e*(d + e*x))/(Sqrt[b]*Sqrt[b + 4*c*d*e])])/(Sqrt[b]*Sqrt[b + 4*c*d*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 2006

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && QuadraticQ[u, x] &&  !QuadraticMatch
Q[u, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c e (d+e x)}{\sqrt {b} \sqrt {b+4 c d e}}\right )}{\sqrt {b} \sqrt {b+4 c d e}} \]

[In]

Integrate[(b*x + c*(d + e*x)^2)^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {2 \,\operatorname {arctanh}\left (\frac {2 c \,e^{2} x +2 d c e +b}{\sqrt {4 b c d e +b^{2}}}\right )}{\sqrt {4 b c d e +b^{2}}}\)	\(43\)
risch	\(\frac {\ln \left (-2 c \,e^{2} x -2 d c e +\sqrt {b \left (4 d c e +b \right )}-b \right )}{\sqrt {b \left (4 d c e +b \right )}}-\frac {\ln \left (2 c \,e^{2} x +2 d c e +\sqrt {b \left (4 d c e +b \right )}+b \right )}{\sqrt {b \left (4 d c e +b \right )}}\)	\(81\)

[In]

int(1/(b*x+c*(e*x+d)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.04 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=\left [\frac {\log \left (\frac {2 \, c^{2} e^{4} x^{2} + 2 \, c^{2} d^{2} e^{2} + 4 \, b c d e + b^{2} + 2 \, {\left (2 \, c^{2} d e^{3} + b c e^{2}\right )} x - \sqrt {4 \, b c d e + b^{2}} {\left (2 \, c e^{2} x + 2 \, c d e + b\right )}}{c e^{2} x^{2} + c d^{2} + {\left (2 \, c d e + b\right )} x}\right )}{\sqrt {4 \, b c d e + b^{2}}}, \frac {2 \, \sqrt {-4 \, b c d e - b^{2}} \arctan \left (\frac {\sqrt {-4 \, b c d e - b^{2}} {\left (2 \, c e^{2} x + 2 \, c d e + b\right )}}{4 \, b c d e + b^{2}}\right )}{4 \, b c d e + b^{2}}\right ] \]

[In]

integrate(1/(b*x+c*(e*x+d)^2),x, algorithm="fricas")

[Out]

[log((2*c^2*e^4*x^2 + 2*c^2*d^2*e^2 + 4*b*c*d*e + b^2 + 2*(2*c^2*d*e^3 + b*c*e^2)*x - sqrt(4*b*c*d*e + b^2)*(2
*c*e^2*x + 2*c*d*e + b))/(c*e^2*x^2 + c*d^2 + (2*c*d*e + b)*x))/sqrt(4*b*c*d*e + b^2), 2*sqrt(-4*b*c*d*e - b^2
)*arctan(sqrt(-4*b*c*d*e - b^2)*(2*c*e^2*x + 2*c*d*e + b)/(4*b*c*d*e + b^2))/(4*b*c*d*e + b^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (49) = 98\).

Time = 0.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.21 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=\sqrt {\frac {1}{b \left (b + 4 c d e\right )}} \log {\left (x + \frac {- b^{2} \sqrt {\frac {1}{b \left (b + 4 c d e\right )}} - 4 b c d e \sqrt {\frac {1}{b \left (b + 4 c d e\right )}} + b + 2 c d e}{2 c e^{2}} \right )} - \sqrt {\frac {1}{b \left (b + 4 c d e\right )}} \log {\left (x + \frac {b^{2} \sqrt {\frac {1}{b \left (b + 4 c d e\right )}} + 4 b c d e \sqrt {\frac {1}{b \left (b + 4 c d e\right )}} + b + 2 c d e}{2 c e^{2}} \right )} \]

[In]

integrate(1/(b*x+c*(e*x+d)**2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(b*x+c*(e*x+d)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c*d*e+b>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=\frac {2 \, \arctan \left (\frac {2 \, c e^{2} x + 2 \, c d e + b}{\sqrt {-4 \, b c d e - b^{2}}}\right )}{\sqrt {-4 \, b c d e - b^{2}}} \]

[In]

integrate(1/(b*x+c*(e*x+d)^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \frac {1}{b x+c (d+e x)^2} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {2\,c\,x\,e^2+2\,c\,d\,e+b}{\sqrt {b}\,\sqrt {b+4\,c\,d\,e}}\right )}{\sqrt {b}\,\sqrt {b+4\,c\,d\,e}} \]

[In]

int(1/(c*(d + e*x)^2 + b*x),x)

[Out]

-(2*atanh((b + 2*c*d*e + 2*c*e^2*x)/(b^(1/2)*(b + 4*c*d*e)^(1/2))))/(b^(1/2)*(b + 4*c*d*e)^(1/2))