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

Optimal. Leaf size=233 \[ -\frac{f^2 \left (4 a-\frac{b^2 f^2}{e^2}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{4 \sqrt{2} e^{5/2} \sqrt{2 d e-b f^2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{3 e} \]

[Out]

(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)/(3*e) - (f^2*(4*a - (b^2*f^2)/e^2)*Sqrt[d + e*x + f*Sqrt[a +
b*x + (e^2*x^2)/f^2]])/(4*(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]])/(4*Sqrt[2]*
e^(5/2)*Sqrt[2*d*e - b*f^2])

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Rubi [A]  time = 0.304588, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2116, 897, 1257, 1153, 208} \[ -\frac{f^2 \left (4 a-\frac{b^2 f^2}{e^2}\right ) \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{4 \left (2 e \left (f \sqrt{a+\frac{x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac{f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{4 \sqrt{2} e^{5/2} \sqrt{2 d e-b f^2}}+\frac{\left (f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)/(3*e) - (f^2*(4*a - (b^2*f^2)/e^2)*Sqrt[d + e*x + f*Sqrt[a +
b*x + (e^2*x^2)/f^2]])/(4*(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]])/(4*Sqrt[2]*
e^(5/2)*Sqrt[2*d*e - b*f^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 1257

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^
(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4
)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1153

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

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 \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}} \, dx &=2 \operatorname{Subst}\left (\int \frac{\sqrt{x} \left (d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2\right )}{\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{x^2 \left (d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4\right )}{\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-\frac{b^2 f^2}{e^2}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 \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{\operatorname{Subst}\left (\int \frac{-e f^2 \left (4 a e^2-b^2 f^2\right )+4 e^2 \left (2 d e-b f^2\right ) x^2-8 e^3 x^4}{-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 )}{4 e^3}\\ &=-\frac{f^2 \left (4 a-\frac{b^2 f^2}{e^2}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 \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{\operatorname{Subst}\left (\int \left (-4 e^2 x^2-\frac{e f^2 \left (4 a e^2-b^2 f^2\right )}{-2 d e+b f^2+2 e x^2}\right ) \, dx,x,\sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}\right )}{4 e^3}\\ &=\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2}}{3 e}-\frac{f^2 \left (4 a-\frac{b^2 f^2}{e^2}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 \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{1}{4} \left (f^2 \left (4 a-\frac{b^2 f^2}{e^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 )\\ &=\frac{\left (d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}\right )^{3/2}}{3 e}-\frac{f^2 \left (4 a-\frac{b^2 f^2}{e^2}\right ) \sqrt{d+e x+f \sqrt{a+b x+\frac{e^2 x^2}{f^2}}}}{4 \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 )}{4 \sqrt{2} e^{5/2} \sqrt{2 d e-b f^2}}\\ \end{align*}

Mathematica [A]  time = 0.401768, size = 223, normalized size = 0.96 \[ \frac{\frac{\left (b^2 e f^4-4 a e^3 f^2\right ) \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{2 e \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+e x\right )+b f^2}-\frac{\sqrt{e} f^2 \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{e} \sqrt{f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x}}{\sqrt{2 d e-b f^2}}\right )}{\sqrt{4 d e-2 b f^2}}+\frac{4}{3} e^2 \left (f \sqrt{a+x \left (b+\frac{e^2 x}{f^2}\right )}+d+e x\right )^{3/2}}{4 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int \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((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x)

[Out]

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

[Out]

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

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Fricas [A]  time = 2.61705, size = 1472, normalized size = 6.32 \begin{align*} \left [-\frac{3 \,{\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 (3 \, b^{2} e f^{4} - 2 \, b d e^{2} f^{2} - 8 \, d^{2} e^{3} + 10 \,{\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}}{48 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )}}, \frac{3 \,{\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 (3 \, b^{2} e f^{4} - 2 \, b d e^{2} f^{2} - 8 \, d^{2} e^{3} + 10 \,{\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}}{24 \,{\left (b e^{3} f^{2} - 2 \, d e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(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
)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) - 4*(3*b^2*e*f^4 - 2*b*d*e^2*f^2 - 8*d^2*e^3 + 10*(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*e^3*f^2 - 2*d*e^4), 1/24*(3*(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*(3*b^2*e*f
^4 - 2*b*d*e^2*f^2 - 8*d^2*e^3 + 10*(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*e^3*f^2 - 2*d*e^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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((d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**(1/2),x)

[Out]

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

[Out]

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