3.223 \(\int \frac{1}{\sqrt{d+e x^2} (-c d^2+b d e+b e^2 x^2+c e^2 x^4)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{x}{d \sqrt{d+e x^2} (2 c d-b e)}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} x \sqrt{2 c d-b e}}{\sqrt{d+e x^2} \sqrt{c d-b e}}\right )}{\sqrt{e} \sqrt{c d-b e} (2 c d-b e)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.11729, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {1149, 382, 377, 208} \[ -\frac{x}{d \sqrt{d+e x^2} (2 c d-b e)}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{e} x \sqrt{2 c d-b e}}{\sqrt{d+e x^2} \sqrt{c d-b e}}\right )}{\sqrt{e} \sqrt{c d-b e} (2 c d-b e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d + e*x^2]*(-(c*d^2) + b*d*e + b*e^2*x^2 + c*e^2*x^4)),x]

[Out]

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

Rule 1149

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*x^2)/e)^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]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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

Mathematica [C]  time = 0.800298, size = 418, normalized size = 3.94 \[ -\frac{x \left (-\frac{2 c e x^2 \left (\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{e (b e-2 c d) x^2}{(b e-c d) \left (e x^2+d\right )}\right )}{c d-b e}+2 \left (\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}\right )^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{e (b e-2 c d) x^2}{(b e-c d) \left (e x^2+d\right )}\right )+\frac{10 c e x^2 \sqrt{\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}}}{c d-b e}-15 \sqrt{\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}}-\frac{10 c e x^2 \tanh ^{-1}\left (\sqrt{\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}}\right )}{c d-b e}+15 \tanh ^{-1}\left (\sqrt{\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}}\right )\right )}{5 \left (d+e x^2\right )^{3/2} (c d-b e) \left (\frac{e x^2 (b e-2 c d)}{\left (d+e x^2\right ) (b e-c d)}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[d + e*x^2]*(-(c*d^2) + b*d*e + b*e^2*x^2 + c*e^2*x^4)),x]

[Out]

-(x*(-15*Sqrt[(e*(-2*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2))] + (10*c*e*x^2*Sqrt[(e*(-2*c*d + b*e)*x^2)/(
(-(c*d) + b*e)*(d + e*x^2))])/(c*d - b*e) + 15*ArcTanh[Sqrt[(e*(-2*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2)
)]] - (10*c*e*x^2*ArcTanh[Sqrt[(e*(-2*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2))]])/(c*d - b*e) + 2*((e*(-2*
c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2)))^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, (e*(-2*c*d + b*e)*x^2)/((-(
c*d) + b*e)*(d + e*x^2))] - (2*c*e*x^2*((e*(-2*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2)))^(5/2)*Hypergeomet
ric2F1[2, 5/2, 7/2, (e*(-2*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2))])/(c*d - b*e)))/(5*(c*d - b*e)*((e*(-2
*c*d + b*e)*x^2)/((-(c*d) + b*e)*(d + e*x^2)))^(3/2)*(d + e*x^2)^(3/2))

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Maple [B]  time = 0.018, size = 771, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x^2+d)^(1/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x)

[Out]

-1/2*c^2*e/((-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(-(-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(-(b*e-c*d)*c*e)^(
1/2)/(-(b*e-2*c*d)/c)^(1/2)*ln((-2*(b*e-2*c*d)/c+2*(-(b*e-c*d)*c*e)^(1/2)/c*(x-(-(b*e-c*d)*c*e)^(1/2)/c/e)+2*(
-(b*e-2*c*d)/c)^(1/2)*((x-(-(b*e-c*d)*c*e)^(1/2)/c/e)^2*e+2*(-(b*e-c*d)*c*e)^(1/2)/c*(x-(-(b*e-c*d)*c*e)^(1/2)
/c/e)-(b*e-2*c*d)/c)^(1/2))/(x-(-(b*e-c*d)*c*e)^(1/2)/c/e))-1/2*c/d/((-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(-
(-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(x+(-d*e)^(1/2)/e)*((x+(-d*e)^(1/2)/e)^2*e-2*(-d*e)^(1/2)*(x+(-d*e)^(1/
2)/e))^(1/2)-1/2*c/d/((-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(-(-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(x-(-d*e
)^(1/2)/e)*((x-(-d*e)^(1/2)/e)^2*e+2*(-d*e)^(1/2)*(x-(-d*e)^(1/2)/e))^(1/2)+1/2*c^2*e/((-d*e)^(1/2)*c+(-(b*e-c
*d)*c*e)^(1/2))/(-(-d*e)^(1/2)*c+(-(b*e-c*d)*c*e)^(1/2))/(-(b*e-c*d)*c*e)^(1/2)/(-(b*e-2*c*d)/c)^(1/2)*ln((-2*
(b*e-2*c*d)/c-2*(-(b*e-c*d)*c*e)^(1/2)/c*(x+(-(b*e-c*d)*c*e)^(1/2)/c/e)+2*(-(b*e-2*c*d)/c)^(1/2)*((x+(-(b*e-c*
d)*c*e)^(1/2)/c/e)^2*e-2*(-(b*e-c*d)*c*e)^(1/2)/c*(x+(-(b*e-c*d)*c*e)^(1/2)/c/e)-(b*e-2*c*d)/c)^(1/2))/(x+(-(b
*e-c*d)*c*e)^(1/2)/c/e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c e^{2} x^{4} + b e^{2} x^{2} - c d^{2} + b d e\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^(1/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="maxima")

[Out]

integrate(1/((c*e^2*x^4 + b*e^2*x^2 - c*d^2 + b*d*e)*sqrt(e*x^2 + d)), x)

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Fricas [B]  time = 2.78176, size = 1440, normalized size = 13.58 \begin{align*} \left [-\frac{4 \,{\left (2 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}\right )} \sqrt{e x^{2} + d} x + \sqrt{2 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}}{\left (c d e x^{2} + c d^{2}\right )} \log \left (\frac{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2} +{\left (17 \, c^{2} d^{2} e^{2} - 24 \, b c d e^{3} + 8 \, b^{2} e^{4}\right )} x^{4} + 2 \,{\left (7 \, c^{2} d^{3} e - 11 \, b c d^{2} e^{2} + 4 \, b^{2} d e^{3}\right )} x^{2} + 4 \, \sqrt{2 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}}{\left ({\left (3 \, c d e - 2 \, b e^{2}\right )} x^{3} +{\left (c d^{2} - b d e\right )} x\right )} \sqrt{e x^{2} + d}}{c^{2} e^{2} x^{4} + c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2} - 2 \,{\left (c^{2} d e - b c e^{2}\right )} x^{2}}\right )}{4 \,{\left (4 \, c^{3} d^{5} e - 8 \, b c^{2} d^{4} e^{2} + 5 \, b^{2} c d^{3} e^{3} - b^{3} d^{2} e^{4} +{\left (4 \, c^{3} d^{4} e^{2} - 8 \, b c^{2} d^{3} e^{3} + 5 \, b^{2} c d^{2} e^{4} - b^{3} d e^{5}\right )} x^{2}\right )}}, -\frac{2 \,{\left (2 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}\right )} \sqrt{e x^{2} + d} x + \sqrt{-2 \, c^{2} d^{2} e + 3 \, b c d e^{2} - b^{2} e^{3}}{\left (c d e x^{2} + c d^{2}\right )} \arctan \left (-\frac{\sqrt{-2 \, c^{2} d^{2} e + 3 \, b c d e^{2} - b^{2} e^{3}}{\left (c d^{2} - b d e +{\left (3 \, c d e - 2 \, b e^{2}\right )} x^{2}\right )} \sqrt{e x^{2} + d}}{2 \,{\left ({\left (2 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + b^{2} e^{4}\right )} x^{3} +{\left (2 \, c^{2} d^{3} e - 3 \, b c d^{2} e^{2} + b^{2} d e^{3}\right )} x\right )}}\right )}{2 \,{\left (4 \, c^{3} d^{5} e - 8 \, b c^{2} d^{4} e^{2} + 5 \, b^{2} c d^{3} e^{3} - b^{3} d^{2} e^{4} +{\left (4 \, c^{3} d^{4} e^{2} - 8 \, b c^{2} d^{3} e^{3} + 5 \, b^{2} c d^{2} e^{4} - b^{3} d e^{5}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^(1/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="fricas")

[Out]

[-1/4*(4*(2*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*sqrt(e*x^2 + d)*x + sqrt(2*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*(
c*d*e*x^2 + c*d^2)*log((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + (17*c^2*d^2*e^2 - 24*b*c*d*e^3 + 8*b^2*e^4)*x^4
+ 2*(7*c^2*d^3*e - 11*b*c*d^2*e^2 + 4*b^2*d*e^3)*x^2 + 4*sqrt(2*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*((3*c*d*e -
2*b*e^2)*x^3 + (c*d^2 - b*d*e)*x)*sqrt(e*x^2 + d))/(c^2*e^2*x^4 + c^2*d^2 - 2*b*c*d*e + b^2*e^2 - 2*(c^2*d*e
- b*c*e^2)*x^2)))/(4*c^3*d^5*e - 8*b*c^2*d^4*e^2 + 5*b^2*c*d^3*e^3 - b^3*d^2*e^4 + (4*c^3*d^4*e^2 - 8*b*c^2*d^
3*e^3 + 5*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2), -1/2*(2*(2*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*sqrt(e*x^2 + d)*x + s
qrt(-2*c^2*d^2*e + 3*b*c*d*e^2 - b^2*e^3)*(c*d*e*x^2 + c*d^2)*arctan(-1/2*sqrt(-2*c^2*d^2*e + 3*b*c*d*e^2 - b^
2*e^3)*(c*d^2 - b*d*e + (3*c*d*e - 2*b*e^2)*x^2)*sqrt(e*x^2 + d)/((2*c^2*d^2*e^2 - 3*b*c*d*e^3 + b^2*e^4)*x^3
+ (2*c^2*d^3*e - 3*b*c*d^2*e^2 + b^2*d*e^3)*x)))/(4*c^3*d^5*e - 8*b*c^2*d^4*e^2 + 5*b^2*c*d^3*e^3 - b^3*d^2*e^
4 + (4*c^3*d^4*e^2 - 8*b*c^2*d^3*e^3 + 5*b^2*c*d^2*e^4 - b^3*d*e^5)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d + e x^{2}\right )^{\frac{3}{2}} \left (b e - c d + c e x^{2}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x**2+d)**(1/2)/(c*e**2*x**4+b*e**2*x**2+b*d*e-c*d**2),x)

[Out]

Integral(1/((d + e*x**2)**(3/2)*(b*e - c*d + c*e*x**2)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x^2+d)^(1/2)/(c*e^2*x^4+b*e^2*x^2+b*d*e-c*d^2),x, algorithm="giac")

[Out]

Timed out