3.463 \(\int \frac{1}{\sqrt{d+e x+f \sqrt{a+\frac{e^2 x^2}{f^2}}}} \, dx\)

Optimal. Leaf size=147 \[ \frac{a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{3/2} e}-\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}+\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

[Out]

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

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Rubi [A]  time = 0.108602, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2117, 897, 1157, 388, 206} \[ \frac{a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{3/2} e}-\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}+\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]],x]

[Out]

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

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

Rubi steps

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

Mathematica [A]  time = 0.250127, size = 143, normalized size = 0.97 \[ \frac{\frac{a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 d^{3/2}}+\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 d \left (f \left (-\sqrt{a+\frac{e^2 x^2}{f^2}}\right )-e x\right )}+\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]],x]

[Out]

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

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Maple [F]  time = 0.011, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{d+ex+f\sqrt{a+{\frac{{e}^{2}{x}^{2}}{{f}^{2}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2),x)

[Out]

int(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(e*x + sqrt(e^2*x^2/f^2 + a)*f + d), x)

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Fricas [A]  time = 1.33248, size = 659, normalized size = 4.48 \begin{align*} \left [\frac{a \sqrt{d} f^{2} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} - 2 \,{\left (\sqrt{d} e x - \sqrt{d} f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) + 2 \,{\left (d e x - d f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \, d^{2} e}, -\frac{a \sqrt{-d} f^{2} \arctan \left (\frac{\sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt{-d}}{d}\right ) -{\left (d e x - d f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt{e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{2 \, d^{2} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d + e x + f \sqrt{a + \frac{e^{2} x^{2}}{f^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(d + e*x + f*sqrt(a + e**2*x**2/f**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e x + \sqrt{\frac{e^{2} x^{2}}{f^{2}} + a} f + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(e*x + sqrt(e^2*x^2/f^2 + a)*f + d), x)