\(\int \frac {(d+e x^2)^{3/2}}{d^2-e^2 x^4} \, dx\) [195]

Optimal result

Integrand size = 26, antiderivative size = 62 \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1164, 399, 223, 212, 385, 214} \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}} \]

[In]

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

[Out]

-(ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/Sqrt[e]) + (Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d + e*x^2]])/Sqrt[
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 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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

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

Rule 399

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

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {d-e x^2+\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {2} d}\right )+\log \left (-\sqrt {e} x+\sqrt {d+e x^2}\right )}{\sqrt {e}} \]

[In]

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

[Out]

(Sqrt[2]*ArcTanh[(d - e*x^2 + Sqrt[e]*x*Sqrt[d + e*x^2])/(Sqrt[2]*d)] + Log[-(Sqrt[e]*x) + Sqrt[d + e*x^2]])/S
qrt[e]

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, \sqrt {2}}{2 x \sqrt {e}}\right )-\operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}}{x \sqrt {e}}\right )}{\sqrt {e}}\)	\(50\)
default	\(\text {Expression too large to display}\)	\(1356\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 3.21 \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=\left [\frac {\sqrt {2} \sqrt {e} \log \left (\frac {17 \, e^{2} x^{4} + 14 \, d e x^{2} + d^{2} + \frac {4 \, \sqrt {2} {\left (3 \, e^{2} x^{3} + d e x\right )} \sqrt {e x^{2} + d}}{\sqrt {e}}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right ) + 2 \, \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right )}{4 \, e}, -\frac {\sqrt {2} e \sqrt {-\frac {1}{e}} \arctan \left (\frac {\sqrt {2} {\left (3 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \sqrt {-\frac {1}{e}}}{4 \, {\left (e x^{3} + d x\right )}}\right ) - 2 \, \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right )}{2 \, e}\right ] \]

[In]

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

[Out]

[1/4*(sqrt(2)*sqrt(e)*log((17*e^2*x^4 + 14*d*e*x^2 + d^2 + 4*sqrt(2)*(3*e^2*x^3 + d*e*x)*sqrt(e*x^2 + d)/sqrt(
e))/(e^2*x^4 - 2*d*e*x^2 + d^2)) + 2*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d))/e, -1/2*(sqrt(2)
*e*sqrt(-1/e)*arctan(1/4*sqrt(2)*(3*e*x^2 + d)*sqrt(e*x^2 + d)*sqrt(-1/e)/(e*x^3 + d*x)) - 2*sqrt(-e)*arctan(s
qrt(-e)*x/sqrt(e*x^2 + d)))/e]

Sympy [F]

\[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=- \int \frac {\sqrt {d + e x^{2}}}{- d + e x^{2}}\, dx \]

[In]

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

[Out]

-Integral(sqrt(d + e*x**2)/(-d + e*x**2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (46) = 92\).

Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.76 \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=\frac {\sqrt {2} d \log \left (\frac {{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{2 \, \sqrt {e} {\left | d \right |}} + \frac {\log \left ({\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2}\right )}{2 \, \sqrt {e}} \]

[In]

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

[Out]

1/2*sqrt(2)*d*log(abs(2*(sqrt(e)*x - sqrt(e*x^2 + d))^2 - 4*sqrt(2)*abs(d) - 6*d)/abs(2*(sqrt(e)*x - sqrt(e*x^
2 + d))^2 + 4*sqrt(2)*abs(d) - 6*d))/(sqrt(e)*abs(d)) + 1/2*log((sqrt(e)*x - sqrt(e*x^2 + d))^2)/sqrt(e)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2}}{d^2-e^2 x^4} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{3/2}}{d^2-e^2\,x^4} \,d x \]

[In]

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

[Out]

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