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

Optimal. Leaf size=244 \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

[Out]

Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]]/e - (f^2*(4*a*e - (b^2*f^2)/e)*Sqrt[d + e*x + f*Sqrt[a + b*x +
(e^2*x^2)/f^2]])/(2*(2*d*e - b*f^2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (f^2*(4*a*e^
2 - b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e - b*f^2]])/(
2*Sqrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

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Rubi [A]  time = 0.291653, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2116, 897, 1157, 388, 208} \[ -\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{2 \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]]/e - (f^2*(4*a*e - (b^2*f^2)/e)*Sqrt[d + e*x + f*Sqrt[a + b*x +
(e^2*x^2)/f^2]])/(2*(2*d*e - b*f^2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) + (f^2*(4*a*e^
2 - b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e - b*f^2]])/(
2*Sqrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

Rule 2116

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[((g + h*x^n)^p*(d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2))/(-2*d*e + b*f^2 +
2*e*x)^2, x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, 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 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+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}} \, dx &=2 \operatorname{Subst}\left (\int \frac{d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2}{\sqrt{x} \left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )\\ &=4 \operatorname{Subst}\left (\int \frac{d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4}{\left (-2 d e+b f^2+2 e x^2\right )^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )\\ &=-\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{2 \left (2 d e-b f^2\right ) \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (-8 d^2 e+8 b d f^2-4 a e f^2-\frac{b^2 f^4}{e}\right )+\left (2 d e-b f^2\right ) x^2}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{2 d e-b f^2}\\ &=\frac{\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{e}-\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{2 \left (2 d e-b f^2\right ) \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac{\left (f^2 \left (4 a e^2-b^2 f^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{2 e \left (2 d e-b f^2\right )}\\ &=\frac{\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{e}-\frac{f^2 \left (4 a e-\frac{b^2 f^2}{e}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{2 \left (2 d e-b f^2\right ) \left (b f^2+2 e \left (e x+f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.590883, size = 238, normalized size = 0.98 \[ \frac{f^2 \left (b^2 f^2-4 a e^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (2 d e-b f^2\right ) \left (2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2\right )}+\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{2 \sqrt{2} e^{3/2} \left (2 d e-b f^2\right )^{3/2}}+\frac{\sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]/e + (f^2*(-4*a*e^2 + b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + x*(b +
(e^2*x)/f^2)]])/(2*e*(2*d*e - b*f^2)*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) + (f^2*(4*a*e^2 -
b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[2*d*e - b*f^2]])/(2*S
qrt[2]*e^(3/2)*(2*d*e - b*f^2)^(3/2))

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Maple [F]  time = 0.015, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt{d+ex+f\sqrt{a+bx+{\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+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x)

[Out]

int(1/(d+e*x+f*(a+b*x+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{b x + \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+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.68942, size = 1508, normalized size = 6.18 \begin{align*} \left [\frac{{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}} \log \left (-b^{2} f^{4} + 4 \,{\left (b d e - a e^{2}\right )} f^{2} - 4 \,{\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x + 2 \,{\left (2 \, \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}} e f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt{-2 \, b e f^{2} + 4 \, d e^{2}}{\left (b f^{2} + 2 \, e^{2} x\right )}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} + 4 \,{\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 4 \,{\left (b^{2} e f^{4} - 6 \, b d e^{2} f^{2} + 8 \, d^{2} e^{3} - 2 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x + 2 \,{\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{8 \,{\left (b^{2} e^{2} f^{4} - 4 \, b d e^{3} f^{2} + 4 \, d^{2} e^{4}\right )}}, \frac{{\left (b^{2} f^{4} - 4 \, a e^{2} f^{2}\right )} \sqrt{2 \, b e f^{2} - 4 \, d e^{2}} \arctan \left (\frac{\sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}{\left (\sqrt{2 \, b e f^{2} - 4 \, d e^{2}} f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt{2 \, b e f^{2} - 4 \, d e^{2}}{\left (e x + d\right )}\right )}}{2 \,{\left (a e f^{2} - d^{2} e +{\left (b e f^{2} - 2 \, d e^{2}\right )} x\right )}}\right ) + 2 \,{\left (b^{2} e f^{4} - 6 \, b d e^{2} f^{2} + 8 \, d^{2} e^{3} - 2 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )} x + 2 \,{\left (b e^{2} f^{3} - 2 \, d e^{3} f\right )} \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt{e x + f \sqrt{\frac{b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \,{\left (b^{2} e^{2} f^{4} - 4 \, b d e^{3} f^{2} + 4 \, d^{2} e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*((b^2*f^4 - 4*a*e^2*f^2)*sqrt(-2*b*e*f^2 + 4*d*e^2)*log(-b^2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 -
2*d*e^3)*x + 2*(2*sqrt(-2*b*e*f^2 + 4*d*e^2)*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(-2*b*e*f^2 + 4*
d*e^2)*(b*f^2 + 2*e^2*x))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sq
rt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 4*(b^2*e*f^4 - 6*b*d*e^2*f^2 + 8*d^2*e^3 - 2*(b*e^3*f^2 - 2*d*e^4)*x +
2*(b*e^2*f^3 - 2*d*e^3*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)
/f^2) + d))/(b^2*e^2*f^4 - 4*b*d*e^3*f^2 + 4*d^2*e^4), 1/4*((b^2*f^4 - 4*a*e^2*f^2)*sqrt(2*b*e*f^2 - 4*d*e^2)*
arctan(1/2*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*(sqrt(2*b*e*f^2 - 4*d*e^2)*f*sqrt((b*f^2*x
+ e^2*x^2 + a*f^2)/f^2) - sqrt(2*b*e*f^2 - 4*d*e^2)*(e*x + d))/(a*e*f^2 - d^2*e + (b*e*f^2 - 2*d*e^2)*x)) + 2*
(b^2*e*f^4 - 6*b*d*e^2*f^2 + 8*d^2*e^3 - 2*(b*e^3*f^2 - 2*d*e^4)*x + 2*(b*e^2*f^3 - 2*d*e^3*f)*sqrt((b*f^2*x +
e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(b^2*e^2*f^4 - 4*b*d*e^3*f^2
+ 4*d^2*e^4)]

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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 + b x + \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+b*x+e**2*x**2/f**2)**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(d + e*x + f*sqrt(a + b*x + 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{b x + \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+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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