\(\int \frac {(d+e x^2)^3}{d^2-e^2 x^4} \, dx\) [190]

Optimal result

Integrand size = 24, antiderivative size = 38 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=-3 d x-\frac {e x^3}{3}+\frac {4 d^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}} \]

[Out]

-3*d*x-1/3*e*x^3+4*d^(3/2)*arctanh(x*e^(1/2)/d^(1/2))/e^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1164, 398, 214} \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=\frac {4 d^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-3 d x-\frac {e x^3}{3} \]

[In]

Int[(d + e*x^2)^3/(d^2 - e^2*x^4),x]

[Out]

-3*d*x - (e*x^3)/3 + (4*d^(3/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[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 1164

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (d+e x^2\right )^2}{d-e x^2} \, dx \\ & = \int \left (-3 d-e x^2+\frac {4 d^2}{d-e x^2}\right ) \, dx \\ & = -3 d x-\frac {e x^3}{3}+\left (4 d^2\right ) \int \frac {1}{d-e x^2} \, dx \\ & = -3 d x-\frac {e x^3}{3}+\frac {4 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=-3 d x-\frac {e x^3}{3}+\frac {4 d^{3/2} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}} \]

[In]

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

[Out]

-3*d*x - (e*x^3)/3 + (4*d^(3/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e]

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {e \,x^{3}}{3}-3 d x +\frac {4 d^{2} \operatorname {arctanh}\left (\frac {e x}{\sqrt {e d}}\right )}{\sqrt {e d}}\)	\(31\)
risch	\(-\frac {e \,x^{3}}{3}-3 d x +\frac {2 \sqrt {e d}\, d \ln \left (\sqrt {e d}\, x +d \right )}{e}-\frac {2 \sqrt {e d}\, d \ln \left (-\sqrt {e d}\, x +d \right )}{e}\)	\(55\)

[In]

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

[Out]

-1/3*e*x^3-3*d*x+4*d^2/(e*d)^(1/2)*arctanh(e*x/(e*d)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.37 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=\left [-\frac {1}{3} \, e x^{3} + 2 \, d \sqrt {\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {\frac {d}{e}} + d}{e x^{2} - d}\right ) - 3 \, d x, -\frac {1}{3} \, e x^{3} - 4 \, d \sqrt {-\frac {d}{e}} \arctan \left (\frac {e x \sqrt {-\frac {d}{e}}}{d}\right ) - 3 \, d x\right ] \]

[In]

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

[Out]

[-1/3*e*x^3 + 2*d*sqrt(d/e)*log((e*x^2 + 2*e*x*sqrt(d/e) + d)/(e*x^2 - d)) - 3*d*x, -1/3*e*x^3 - 4*d*sqrt(-d/e
)*arctan(e*x*sqrt(-d/e)/d) - 3*d*x]

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=- 3 d x - \frac {e x^{3}}{3} - 2 \sqrt {\frac {d^{3}}{e}} \log {\left (x - \frac {\sqrt {\frac {d^{3}}{e}}}{d} \right )} + 2 \sqrt {\frac {d^{3}}{e}} \log {\left (x + \frac {\sqrt {\frac {d^{3}}{e}}}{d} \right )} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((e*x^2+d)^3/(-e^2*x^4+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>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=-\frac {4 \, d^{2} \arctan \left (\frac {e x}{\sqrt {-d e}}\right )}{\sqrt {-d e}} - \frac {e^{4} x^{3} + 9 \, d e^{3} x}{3 \, e^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\left (d+e x^2\right )^3}{d^2-e^2 x^4} \, dx=\frac {4\,d^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {e\,x^3}{3}-3\,d\,x \]

[In]

int((d + e*x^2)^3/(d^2 - e^2*x^4),x)

[Out]

(4*d^(3/2)*atanh((e^(1/2)*x)/d^(1/2)))/e^(1/2) - (e*x^3)/3 - 3*d*x