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

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

[Out]

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

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Rubi [A]  time = 0.122284, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2117, 897, 1257, 1153, 206} \[ -\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{2 e \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x\right )}+\frac{\left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{3 e}-\frac{a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{2 \sqrt{d} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.329926, size = 139, normalized size = 0.95 \[ -\frac{\frac{a f^2 \sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{f \sqrt{a+\frac{e^2 x^2}{f^2}}+e x}-\frac{2}{3} \left (f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x\right )^{3/2}+\frac{a f^2 \tanh ^{-1}\left (\frac{\sqrt{f \sqrt{a+\frac{e^2 x^2}{f^2}}+d+e x}}{\sqrt{d}}\right )}{\sqrt{d}}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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Fricas [A]  time = 1.36777, size = 664, normalized size = 4.52 \begin{align*} \left [\frac{3 \, 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 (5 \, 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}}{12 \, d e}, \frac{3 \, 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 (5 \, 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}}{6 \, d e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*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*sq
rt((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*(5*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*e), 1/6*(3*a*sqrt(-d)*f^2*arctan(sqrt(e*
x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) + (5*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*e)]

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

[Out]

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

[Out]

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