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

Optimal. Leaf size=62 \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}} \]

[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]

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Rubi [A]  time = 0.0442698, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1150, 402, 217, 206, 377, 208} \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{\sqrt{e}} \]

Antiderivative was successfully verified.

[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 1150

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

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 217

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 377

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 208

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

Rubi steps

\begin{align*} \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\\ &=(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) \operatorname{Subst}\left (\int \frac{1}{d-2 d e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )-\operatorname{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]  time = 0.0283889, size = 61, normalized size = 0.98 \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} x}{\sqrt{d+e x^2}}\right )-\log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{\sqrt{e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.054, size = 1442, normalized size = 23.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*e/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))/(-d*e)^(1/2)*((x+(-d*e)^(1/2)/e)^2*e-2*(-d*e)^(1
/2)*(x+(-d*e)^(1/2)/e))^(3/2)+1/4*e/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*((x+(-d*e)^(1/2)/e)
^2*e-2*(-d*e)^(1/2)*(x+(-d*e)^(1/2)/e))^(1/2)*x+1/4*e^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1
/2))*d*ln(((x+(-d*e)^(1/2)/e)*e-(-d*e)^(1/2))/e^(1/2)+((x+(-d*e)^(1/2)/e)^2*e-2*(-d*e)^(1/2)*(x+(-d*e)^(1/2)/e
))^(1/2))+1/6*e/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))/(-d*e)^(1/2)*((x-(-d*e)^(1/2)/e)^2*e+2*
(-d*e)^(1/2)*(x-(-d*e)^(1/2)/e))^(3/2)+1/4*e/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*((x-(-d*e)
^(1/2)/e)^2*e+2*(-d*e)^(1/2)*(x-(-d*e)^(1/2)/e))^(1/2)*x+1/4*e^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)
+(d*e)^(1/2))*d*ln(((x-(-d*e)^(1/2)/e)*e+(-d*e)^(1/2))/e^(1/2)+((x-(-d*e)^(1/2)/e)^2*e+2*(-d*e)^(1/2)*(x-(-d*e
)^(1/2)/e))^(1/2))+1/6*e/(d*e)^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*((x+(d*e)^(1/2)/e)
^2*e-2*(d*e)^(1/2)*(x+(d*e)^(1/2)/e)+2*d)^(3/2)-1/4*e/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*(
(x+(d*e)^(1/2)/e)^2*e-2*(d*e)^(1/2)*(x+(d*e)^(1/2)/e)+2*d)^(1/2)*x-5/4*e^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-
d*e)^(1/2)+(d*e)^(1/2))*d*ln(((x+(d*e)^(1/2)/e)*e-(d*e)^(1/2))/e^(1/2)+((x+(d*e)^(1/2)/e)^2*e-2*(d*e)^(1/2)*(x
+(d*e)^(1/2)/e)+2*d)^(1/2))+e/(d*e)^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*d*((x+(d*e)^(
1/2)/e)^2*e-2*(d*e)^(1/2)*(x+(d*e)^(1/2)/e)+2*d)^(1/2)-e/(d*e)^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)
+(d*e)^(1/2))*d^(3/2)*2^(1/2)*ln((4*d-2*(d*e)^(1/2)*(x+(d*e)^(1/2)/e)+2*2^(1/2)*d^(1/2)*((x+(d*e)^(1/2)/e)^2*e
-2*(d*e)^(1/2)*(x+(d*e)^(1/2)/e)+2*d)^(1/2))/(x+(d*e)^(1/2)/e))-1/6*e/(d*e)^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(
-(-d*e)^(1/2)+(d*e)^(1/2))*((x-(d*e)^(1/2)/e)^2*e+2*(d*e)^(1/2)*(x-(d*e)^(1/2)/e)+2*d)^(3/2)-1/4*e/((-d*e)^(1/
2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*((x-(d*e)^(1/2)/e)^2*e+2*(d*e)^(1/2)*(x-(d*e)^(1/2)/e)+2*d)^(1/2)*
x-5/4*e^(1/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*d*ln(((x-(d*e)^(1/2)/e)*e+(d*e)^(1/2))/e^
(1/2)+((x-(d*e)^(1/2)/e)^2*e+2*(d*e)^(1/2)*(x-(d*e)^(1/2)/e)+2*d)^(1/2))-e/(d*e)^(1/2)/((-d*e)^(1/2)+(d*e)^(1/
2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*d*((x-(d*e)^(1/2)/e)^2*e+2*(d*e)^(1/2)*(x-(d*e)^(1/2)/e)+2*d)^(1/2)+e/(d*e)^(1
/2)/((-d*e)^(1/2)+(d*e)^(1/2))/(-(-d*e)^(1/2)+(d*e)^(1/2))*d^(3/2)*2^(1/2)*ln((4*d+2*(d*e)^(1/2)*(x-(d*e)^(1/2
)/e)+2*2^(1/2)*d^(1/2)*((x-(d*e)^(1/2)/e)^2*e+2*(d*e)^(1/2)*(x-(d*e)^(1/2)/e)+2*d)^(1/2))/(x-(d*e)^(1/2)/e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}{e^{2} x^{4} - d^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.05638, size = 491, normalized size = 7.92 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{d + e x^{2}}}{- d + e x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.17519, size = 32, normalized size = 0.52 \begin{align*} \frac{1}{2} \, e^{\left (-\frac{1}{2}\right )} \log \left ({\left (x e^{\frac{1}{2}} - \sqrt{x^{2} e + d}\right )}^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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