\(\int \frac {(d+e x^2)^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx\) [216]

Optimal result

Integrand size = 39, antiderivative size = 64 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {x}{c}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{c^{3/2} \sqrt {e} \sqrt {c d-b e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1163, 396, 214} \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {x}{c}-\frac {(2 c d-b e) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{c^{3/2} \sqrt {e} \sqrt {c d-b e}} \]

[In]

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

[Out]

x/c - ((2*c*d - b*e)*ArcTanh[(Sqrt[c]*Sqrt[e]*x)/Sqrt[c*d - b*e]])/(c^(3/2)*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 396

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {x}{c}-\frac {(-2 c d+b e) \arctan \left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {-c d+b e}}\right )}{c^{3/2} \sqrt {e} \sqrt {-c d+b e}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {x}{c}+\frac {\left (-b e +2 c d \right ) \arctan \left (\frac {x c e}{\sqrt {\left (b e -c d \right ) e c}}\right )}{c \sqrt {\left (b e -c d \right ) e c}}\)	\(51\)
risch	\(\frac {x}{c}-\frac {\ln \left (x c e -\sqrt {-\left (b e -c d \right ) e c}\right ) b e}{2 c \sqrt {-\left (b e -c d \right ) e c}}+\frac {\ln \left (x c e -\sqrt {-\left (b e -c d \right ) e c}\right ) d}{\sqrt {-\left (b e -c d \right ) e c}}+\frac {\ln \left (-x c e -\sqrt {-\left (b e -c d \right ) e c}\right ) b e}{2 c \sqrt {-\left (b e -c d \right ) e c}}-\frac {\ln \left (-x c e -\sqrt {-\left (b e -c d \right ) e c}\right ) d}{\sqrt {-\left (b e -c d \right ) e c}}\)	\(172\)

[In]

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

[Out]

x/c+(-b*e+2*c*d)/c/((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 = 210, normalized size of antiderivative = 3.28 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\left [-\frac {\sqrt {c^{2} d e - b c e^{2}} {\left (2 \, c d - b e\right )} \log \left (\frac {c e x^{2} + c d - b e + 2 \, \sqrt {c^{2} d e - b c e^{2}} x}{c e x^{2} - c d + b e}\right ) - 2 \, {\left (c^{2} d e - b c e^{2}\right )} x}{2 \, {\left (c^{3} d e - b c^{2} e^{2}\right )}}, -\frac {\sqrt {-c^{2} d e + b c e^{2}} {\left (2 \, c d - b e\right )} \arctan \left (-\frac {\sqrt {-c^{2} d e + b c e^{2}} x}{c d - b e}\right ) - {\left (c^{2} d e - b c e^{2}\right )} x}{c^{3} d e - b c^{2} e^{2}}\right ] \]

[In]

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

[Out]

[-1/2*(sqrt(c^2*d*e - b*c*e^2)*(2*c*d - b*e)*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)) - 2*(c^2*d*e - b*c*e^2)*x)/(c^3*d*e - b*c^2*e^2), -(sqrt(-c^2*d*e + b*c*e^2)*(2*c*d - b*e)*arcta
n(-sqrt(-c^2*d*e + b*c*e^2)*x/(c*d - b*e)) - (c^2*d*e - b*c*e^2)*x)/(c^3*d*e - b*c^2*e^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (54) = 108\).

Time = 0.24 (sec) , antiderivative size = 212, normalized size of antiderivative = 3.31 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {\sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right ) \log {\left (x + \frac {- b c e \sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right ) + c^{2} d \sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right )}{b e - 2 c d} \right )}}{2} - \frac {\sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right ) \log {\left (x + \frac {b c e \sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right ) - c^{2} d \sqrt {- \frac {1}{c^{3} e \left (b e - c d\right )}} \left (b e - 2 c d\right )}{b e - 2 c d} \right )}}{2} + \frac {x}{c} \]

[In]

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

[Out]

sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*d)*log(x + (-b*c*e*sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*d) + c**2
*d*sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*d))/(b*e - 2*c*d))/2 - sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*d)
*log(x + (b*c*e*sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*d) - c**2*d*sqrt(-1/(c**3*e*(b*e - c*d)))*(b*e - 2*c*
d))/(b*e - 2*c*d))/2 + x/c

Maxima [F(-2)]

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

[In]

integrate((e*x^2+d)^2/(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 = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {{\left (2 \, c d - b e\right )} \arctan \left (\frac {c e x}{\sqrt {-c^{2} d e + b c e^{2}}}\right )}{\sqrt {-c^{2} d e + b c e^{2}} c} + \frac {x}{c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \frac {\left (d+e x^2\right )^2}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {x}{c}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,e\,x}{\sqrt {b\,e^2-c\,d\,e}}\right )\,\left (b\,e-2\,c\,d\right )}{c^{3/2}\,\sqrt {b\,e^2-c\,d\,e}} \]

[In]

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

[Out]

x/c - (atan((c^(1/2)*e*x)/(b*e^2 - c*d*e)^(1/2))*(b*e - 2*c*d))/(c^(3/2)*(b*e^2 - c*d*e)^(1/2))