3.5.82 \(\int \frac {1}{\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}} \, dx\) [482]

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

[Out]

1/4*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2)*e^(1/2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^(1
/2))/e^(3/2)/(-b*f^2+2*d*e)^(3/2)*2^(1/2)+(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/e-1/2*f^2*(4*a*e-b^2*f^2/e
)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))

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Rubi [A]
time = 0.22, antiderivative size = 244, normalized size of antiderivative = 1.00, 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 = {2141, 911, 1171, 396, 214} \begin {gather*} -\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} \end {gather*}

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 214

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

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 2141

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]

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 \text {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 \text {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 \text {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 ) \text {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*}

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Mathematica [A]
time = 1.58, size = 341, normalized size = 1.40 \begin {gather*} \frac {\frac {2 \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \left (-b^2 f^4-4 b e f^2 \left (-d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 e^2 \left (-a f^2+2 d e x+2 d f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}{\left (2 d e-b f^2\right ) \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}+\frac {4 \sqrt {2} a e^2 f^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\left (-2 d e+b f^2\right )^{3/2}}-\frac {\sqrt {2} b^2 f^4 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\left (-2 d e+b f^2\right )^{3/2}}}{4 e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\sqrt {d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.52, size = 620, normalized size = 2.54 \begin {gather*} \left [-\frac {{\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} \log \left (b^{2} f^{4} - 4 \, b d f^{2} e - 8 \, d x e^{3} + 4 \, {\left (b f^{2} x + a f^{2}\right )} e^{2} + 2 \, {\left (2 \, \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} e - \sqrt {-2 \, b f^{2} e + 4 \, d e^{2}} {\left (b f^{2} + 2 \, x e^{2}\right )}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d} - 4 \, {\left (b f^{3} e - 2 \, d f e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right ) - 4 \, {\left (b^{2} f^{4} e - 6 \, b d f^{2} e^{2} + 4 \, d x e^{4} - 2 \, {\left (b f^{2} x - 4 \, d^{2}\right )} e^{3} + 2 \, {\left (b f^{3} e^{2} - 2 \, d f e^{3}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{8 \, {\left (b^{2} f^{4} e^{2} - 4 \, b d f^{2} e^{3} + 4 \, d^{2} e^{4}\right )}}, \frac {{\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} \sqrt {2 \, b f^{2} e - 4 \, d e^{2}} \arctan \left (-\frac {\sqrt {2 \, b f^{2} e - 4 \, d e^{2}} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{b f^{2} - 2 \, d e}\right ) + 2 \, {\left (b^{2} f^{4} e - 6 \, b d f^{2} e^{2} + 4 \, d x e^{4} - 2 \, {\left (b f^{2} x - 4 \, d^{2}\right )} e^{3} + 2 \, {\left (b f^{3} e^{2} - 2 \, d f e^{3}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}}{4 \, {\left (b^{2} f^{4} e^{2} - 4 \, b d f^{2} e^{3} + 4 \, d^{2} e^{4}\right )}}\right ] \end {gather*}

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

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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 + b x + \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+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 = 0.00, 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+b*x+e^2*x^2/f^2)^(1/2))^(1/2),x, algorithm="giac")

[Out]

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

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