3.5.63 \(\int \frac {1}{\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}} \, dx\) [463]

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

[Out]

1/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))/d^(3/2)/e+(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/
2)/e-1/2*a*f^2*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))

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Rubi [A]
time = 0.08, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 = {2142, 911, 1171, 396, 212} \begin {gather*} \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} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 396

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 911

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2142

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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}} \, dx &=\frac {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 0.87, size = 141, normalized size = 0.96 \begin {gather*} \frac {\frac {\sqrt {d} \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \left (-a f^2+2 d \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}+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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[d]*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]*(-(a*f^2) + 2*d*(e*x + f*Sqrt[a + (e^2*x^2)/f^2])))/(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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 0.42, size = 296, normalized size = 2.01 \begin {gather*} \left [\frac {{\left (a \sqrt {d} f^{2} \log \left (a f^{2} - 2 \, d x e + 2 \, d f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} - 2 \, {\left (\sqrt {d} x e - \sqrt {d} f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right ) + 2 \, {\left (d x e - d f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-1\right )}}{4 \, d^{2}}, -\frac {{\left (a \sqrt {-d} f^{2} \arctan \left (\frac {\sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) - {\left (d x e - d f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt {x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-1\right )}}{2 \, d^{2}}\right ] \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}}} \,d x \end {gather*}

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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