\(\int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx\) [217]

Optimal result

Integrand size = 37, antiderivative size = 49 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{\sqrt {c} \sqrt {e} \sqrt {c d-b e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {1163, 214} \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{\sqrt {c} \sqrt {e} \sqrt {c d-b e}} \]

[In]

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

[Out]

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

Rule 214

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

Rule 1163

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {-c d+b e}}\right )}{\sqrt {c} \sqrt {e} \sqrt {-c d+b e}} \]

[In]

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

[Out]

ArcTan[(Sqrt[c]*Sqrt[e]*x)/Sqrt[-(c*d) + b*e]]/(Sqrt[c]*Sqrt[e]*Sqrt[-(c*d) + b*e])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {\arctan \left (\frac {x c e}{\sqrt {\left (b e -c d \right ) e c}}\right )}{\sqrt {\left (b e -c d \right ) e c}}\)	\(33\)
risch	\(-\frac {\ln \left (x c e +\sqrt {-\left (b e -c d \right ) e c}\right )}{2 \sqrt {-\left (b e -c d \right ) e c}}+\frac {\ln \left (-x c e +\sqrt {-\left (b e -c d \right ) e c}\right )}{2 \sqrt {-\left (b e -c d \right ) e c}}\)	\(75\)

[In]

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

[Out]

1/((b*e-c*d)*e*c)^(1/2)*arctan(x*c*e/((b*e-c*d)*e*c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.73 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\left [\frac {\log \left (\frac {c e x^{2} + c d - b e - 2 \, \sqrt {c^{2} d e - b c e^{2}} x}{c e x^{2} - c d + b e}\right )}{2 \, \sqrt {c^{2} d e - b c e^{2}}}, -\frac {\sqrt {-c^{2} d e + b c e^{2}} \arctan \left (-\frac {\sqrt {-c^{2} d e + b c e^{2}} x}{c d - b e}\right )}{c^{2} d e - b c e^{2}}\right ] \]

[In]

integrate((e*x^2+d)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="fricas")

[Out]

[1/2*log((c*e*x^2 + c*d - b*e - 2*sqrt(c^2*d*e - b*c*e^2)*x)/(c*e*x^2 - c*d + b*e))/sqrt(c^2*d*e - b*c*e^2), -
sqrt(-c^2*d*e + b*c*e^2)*arctan(-sqrt(-c^2*d*e + b*c*e^2)*x/(c*d - b*e))/(c^2*d*e - b*c*e^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (44) = 88\).

Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.53 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=- \frac {\sqrt {- \frac {1}{c e \left (b e - c d\right )}} \log {\left (- b e \sqrt {- \frac {1}{c e \left (b e - c d\right )}} + c d \sqrt {- \frac {1}{c e \left (b e - c d\right )}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{c e \left (b e - c d\right )}} \log {\left (b e \sqrt {- \frac {1}{c e \left (b e - c d\right )}} - c d \sqrt {- \frac {1}{c e \left (b e - c d\right )}} + x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x^2+d)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*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(e*(b*e-c*d)>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {\arctan \left (\frac {c e x}{\sqrt {-c^{2} d e + b c e^{2}}}\right )}{\sqrt {-c^{2} d e + b c e^{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int \frac {d+e x^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {\mathrm {atan}\left (\frac {c\,e\,x}{\sqrt {b\,c\,e^2-c^2\,d\,e}}\right )}{\sqrt {b\,c\,e^2-c^2\,d\,e}} \]

[In]

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

[Out]

atan((c*e*x)/(b*c*e^2 - c^2*d*e)^(1/2))/(b*c*e^2 - c^2*d*e)^(1/2)