\(\int \frac {(d+e x^2)^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx\) [215]

Optimal result

Integrand size = 39, antiderivative size = 86 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {(3 c d-b e) x}{c^2}+\frac {e x^3}{3 c}-\frac {(2 c d-b e)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{c^{5/2} \sqrt {e} \sqrt {c d-b e}} \]

[Out]

(-b*e+3*c*d)*x/c^2+1/3*e*x^3/c-(-b*e+2*c*d)^2*arctanh(x*c^(1/2)*e^(1/2)/(-b*e+c*d)^(1/2))/c^(5/2)/e^(1/2)/(-b*
e+c*d)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1163, 398, 214} \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=-\frac {(2 c d-b e)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{c^{5/2} \sqrt {e} \sqrt {c d-b e}}+\frac {x (3 c d-b e)}{c^2}+\frac {e x^3}{3 c} \]

[In]

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

[Out]

((3*c*d - b*e)*x)/c^2 + (e*x^3)/(3*c) - ((2*c*d - b*e)^2*ArcTanh[(Sqrt[c]*Sqrt[e]*x)/Sqrt[c*d - b*e]])/(c^(5/2
)*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 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 {\left (d+e x^2\right )^2}{\frac {-c d^2+b d e}{d}+c e x^2} \, dx \\ & = \int \left (\frac {3 c d-b e}{c^2}+\frac {e x^2}{c}+\frac {4 c^2 d^2-4 b c d e+b^2 e^2}{c^2 \left (-c d+b e+c e x^2\right )}\right ) \, dx \\ & = \frac {(3 c d-b e) x}{c^2}+\frac {e x^3}{3 c}+\frac {(2 c d-b e)^2 \int \frac {1}{-c d+b e+c e x^2} \, dx}{c^2} \\ & = \frac {(3 c d-b e) x}{c^2}+\frac {e x^3}{3 c}-\frac {(2 c d-b e)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {c d-b e}}\right )}{c^{5/2} \sqrt {e} \sqrt {c d-b e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=-\frac {(-3 c d+b e) x}{c^2}+\frac {e x^3}{3 c}+\frac {(-2 c d+b e)^2 \arctan \left (\frac {\sqrt {c} \sqrt {e} x}{\sqrt {-c d+b e}}\right )}{c^{5/2} \sqrt {e} \sqrt {-c d+b e}} \]

[In]

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

[Out]

-(((-3*c*d + b*e)*x)/c^2) + (e*x^3)/(3*c) + ((-2*c*d + b*e)^2*ArcTan[(Sqrt[c]*Sqrt[e]*x)/Sqrt[-(c*d) + b*e]])/
(c^(5/2)*Sqrt[e]*Sqrt[-(c*d) + b*e])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {-\frac {1}{3} c \,x^{3} e +b e x -3 c d x}{c^{2}}+\frac {\left (b^{2} e^{2}-4 b c d e +4 c^{2} d^{2}\right ) \arctan \left (\frac {x c e}{\sqrt {\left (b e -c d \right ) e c}}\right )}{c^{2} \sqrt {\left (b e -c d \right ) e c}}\)	\(81\)
risch	\(\frac {e \,x^{3}}{3 c}-\frac {b e x}{c^{2}}+\frac {3 d x}{c}-\frac {\ln \left (x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) b^{2} e^{2}}{2 c^{2} \sqrt {-\left (b e -c d \right ) e c}}+\frac {2 \ln \left (x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) b d e}{c \sqrt {-\left (b e -c d \right ) e c}}-\frac {2 \ln \left (x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) d^{2}}{\sqrt {-\left (b e -c d \right ) e c}}+\frac {\ln \left (-x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) b^{2} e^{2}}{2 c^{2} \sqrt {-\left (b e -c d \right ) e c}}-\frac {2 \ln \left (-x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) b d e}{c \sqrt {-\left (b e -c d \right ) e c}}+\frac {2 \ln \left (-x c e +\sqrt {-\left (b e -c d \right ) e c}\right ) d^{2}}{\sqrt {-\left (b e -c d \right ) e c}}\)	\(281\)

[In]

int((e*x^2+d)^3/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x,method=_RETURNVERBOSE)

[Out]

-1/c^2*(-1/3*c*x^3*e+b*e*x-3*c*d*x)+(b^2*e^2-4*b*c*d*e+4*c^2*d^2)/c^2/((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 = 0.27 (sec) , antiderivative size = 311, normalized size of antiderivative = 3.62 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\left [\frac {2 \, {\left (c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{3} + 3 \, {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {c^{2} d e - b c e^{2}} \log \left (\frac {c e x^{2} + c d - b e - 2 \, \sqrt {c^{2} d e - b c e^{2}} x}{c e x^{2} - c d + b e}\right ) + 6 \, {\left (3 \, c^{3} d^{2} e - 4 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} x}{6 \, {\left (c^{4} d e - b c^{3} e^{2}\right )}}, \frac {{\left (c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{3} - 3 \, {\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {-c^{2} d e + b c e^{2}} \arctan \left (-\frac {\sqrt {-c^{2} d e + b c e^{2}} x}{c d - b e}\right ) + 3 \, {\left (3 \, c^{3} d^{2} e - 4 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} x}{3 \, {\left (c^{4} d e - b c^{3} e^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/6*(2*(c^3*d*e^2 - b*c^2*e^3)*x^3 + 3*(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sqrt(c^2*d*e - b*c*e^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)) + 6*(3*c^3*d^2*e - 4*b*c^2*d*e^2 + b^2*c*e^
3)*x)/(c^4*d*e - b*c^3*e^2), 1/3*((c^3*d*e^2 - b*c^2*e^3)*x^3 - 3*(4*c^2*d^2 - 4*b*c*d*e + b^2*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)) + 3*(3*c^3*d^2*e - 4*b*c^2*d*e^2 + b^2*c*e^3)*x
)/(c^4*d*e - b*c^3*e^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (75) = 150\).

Time = 0.35 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.20 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=x \left (- \frac {b e}{c^{2}} + \frac {3 d}{c}\right ) - \frac {\sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2} \log {\left (x + \frac {- b c^{2} e \sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2} + c^{3} d \sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2}}{b^{2} e^{2} - 4 b c d e + 4 c^{2} d^{2}} \right )}}{2} + \frac {\sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2} \log {\left (x + \frac {b c^{2} e \sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2} - c^{3} d \sqrt {- \frac {1}{c^{5} e \left (b e - c d\right )}} \left (b e - 2 c d\right )^{2}}{b^{2} e^{2} - 4 b c d e + 4 c^{2} d^{2}} \right )}}{2} + \frac {e x^{3}}{3 c} \]

[In]

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

[Out]

x*(-b*e/c**2 + 3*d/c) - sqrt(-1/(c**5*e*(b*e - c*d)))*(b*e - 2*c*d)**2*log(x + (-b*c**2*e*sqrt(-1/(c**5*e*(b*e
- c*d)))*(b*e - 2*c*d)**2 + c**3*d*sqrt(-1/(c**5*e*(b*e - c*d)))*(b*e - 2*c*d)**2)/(b**2*e**2 - 4*b*c*d*e + 4
*c**2*d**2))/2 + sqrt(-1/(c**5*e*(b*e - c*d)))*(b*e - 2*c*d)**2*log(x + (b*c**2*e*sqrt(-1/(c**5*e*(b*e - c*d))
)*(b*e - 2*c*d)**2 - c**3*d*sqrt(-1/(c**5*e*(b*e - c*d)))*(b*e - 2*c*d)**2)/(b**2*e**2 - 4*b*c*d*e + 4*c**2*d*
*2))/2 + e*x**3/(3*c)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x^2+d)^3/(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 = 101, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=\frac {{\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {c e x}{\sqrt {-c^{2} d e + b c e^{2}}}\right )}{\sqrt {-c^{2} d e + b c e^{2}} c^{2}} + \frac {c^{2} e^{4} x^{3} + 9 \, c^{2} d e^{3} x - 3 \, b c e^{4} x}{3 \, c^{3} e^{3}} \]

[In]

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

[Out]

(4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*arctan(c*e*x/sqrt(-c^2*d*e + b*c*e^2))/(sqrt(-c^2*d*e + b*c*e^2)*c^2) + 1/3*
(c^2*e^4*x^3 + 9*c^2*d*e^3*x - 3*b*c*e^4*x)/(c^3*e^3)

Mupad [B] (verification not implemented)

Time = 8.01 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.31 \[ \int \frac {\left (d+e x^2\right )^3}{-c d^2+b d e+b e^2 x^2+c e^2 x^4} \, dx=x\,\left (\frac {2\,d}{c}-\frac {b\,e-c\,d}{c^2}\right )+\frac {e\,x^3}{3\,c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,e\,x\,{\left (b\,e-2\,c\,d\right )}^2}{\sqrt {b\,e^2-c\,d\,e}\,\left (b^2\,e^2-4\,b\,c\,d\,e+4\,c^2\,d^2\right )}\right )\,{\left (b\,e-2\,c\,d\right )}^2}{c^{5/2}\,\sqrt {b\,e^2-c\,d\,e}} \]

[In]

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

[Out]

x*((2*d)/c - (b*e - c*d)/c^2) + (e*x^3)/(3*c) + (atan((c^(1/2)*e*x*(b*e - 2*c*d)^2)/((b*e^2 - c*d*e)^(1/2)*(b^
2*e^2 + 4*c^2*d^2 - 4*b*c*d*e)))*(b*e - 2*c*d)^2)/(c^(5/2)*(b*e^2 - c*d*e)^(1/2))