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

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

[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)

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Rubi [A]
time = 0.03, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 = {1164, 399, 223, 212, 385, 214} \begin {gather*} \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}} \end {gather*}

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 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*} \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) \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*}

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Mathematica [A]
time = 0.07, size = 70, normalized size = 1.13 \begin {gather*} \frac {\sqrt {2} \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1355\) vs. \(2(46)=92\).
time = 0.24, size = 1356, normalized size = 21.87

method	result	size
default	\(\text {Expression too large to display}\)	\(1356\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 113 vs. \(2 (44) = 88\).
time = 0.36, size = 113, normalized size = 1.82 \begin {gather*} \frac {1}{4} \, {\left (\sqrt {2} e^{\frac {1}{2}} \log \left (\frac {17 \, x^{4} e^{2} + 14 \, d x^{2} e + 4 \, \sqrt {2} {\left (3 \, x^{3} e^{2} + d x e\right )} \sqrt {x^{2} e + d} e^{\left (-\frac {1}{2}\right )} + d^{2}}{x^{4} e^{2} - 2 \, d x^{2} e + d^{2}}\right ) + 2 \, e^{\frac {1}{2}} \log \left (-2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right )\right )} e^{\left (-1\right )} \end {gather*}

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)*e^(1/2)*log((17*x^4*e^2 + 14*d*x^2*e + 4*sqrt(2)*(3*x^3*e^2 + d*x*e)*sqrt(x^2*e + d)*e^(-1/2) + d
^2)/(x^4*e^2 - 2*d*x^2*e + d^2)) + 2*e^(1/2)*log(-2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) - d))*e^(-1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\sqrt {d + e x^{2}}}{- d + e x^{2}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (44) = 88\).
time = 2.84, size = 107, normalized size = 1.73 \begin {gather*} \frac {\sqrt {2} d e^{\left (-\frac {1}{2}\right )} \log \left (\frac {{\left | 2 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{2 \, {\left | d \right |}} + \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (e\,x^2+d\right )}^{3/2}}{d^2-e^2\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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