3.629 \(\int \frac {\sqrt {a+b \sqrt {c+d x}}}{x} \, dx\)

Optimal. Leaf size=116 \[ 4 \sqrt {a+b \sqrt {c+d x}}-2 \sqrt {a-b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )-2 \sqrt {a+b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right ) \]

[Out]

-2*arctanh((a+b*(d*x+c)^(1/2))^(1/2)/(a-b*c^(1/2))^(1/2))*(a-b*c^(1/2))^(1/2)-2*arctanh((a+b*(d*x+c)^(1/2))^(1
/2)/(a+b*c^(1/2))^(1/2))*(a+b*c^(1/2))^(1/2)+4*(a+b*(d*x+c)^(1/2))^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {371, 1398, 825, 827, 1166, 207} \[ 4 \sqrt {a+b \sqrt {c+d x}}-2 \sqrt {a-b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )-2 \sqrt {a+b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c + d*x]]/x,x]

[Out]

4*Sqrt[a + b*Sqrt[c + d*x]] - 2*Sqrt[a - b*Sqrt[c]]*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a - b*Sqrt[c]]] - 2
*Sqrt[a + b*Sqrt[c]]*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]]

Rule 207

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 825

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g*(d + e*x)^m)/
(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d + c*e*f)*x, x])/(a + c*x^2), x], x] /
; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \sqrt {c+d x}}}{x} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b \sqrt {x}}}{-c+x} \, dx,x,c+d x\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x \sqrt {a+b x}}{-c+x^2} \, dx,x,\sqrt {c+d x}\right )\\ &=4 \sqrt {a+b \sqrt {c+d x}}+2 \operatorname {Subst}\left (\int \frac {b c+a x}{\sqrt {a+b x} \left (-c+x^2\right )} \, dx,x,\sqrt {c+d x}\right )\\ &=4 \sqrt {a+b \sqrt {c+d x}}+4 \operatorname {Subst}\left (\int \frac {-a^2+b^2 c+a x^2}{a^2-b^2 c-2 a x^2+x^4} \, dx,x,\sqrt {a+b \sqrt {c+d x}}\right )\\ &=4 \sqrt {a+b \sqrt {c+d x}}+\left (2 \left (a-b \sqrt {c}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a+b \sqrt {c}+x^2} \, dx,x,\sqrt {a+b \sqrt {c+d x}}\right )+\left (2 \left (a+b \sqrt {c}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a-b \sqrt {c}+x^2} \, dx,x,\sqrt {a+b \sqrt {c+d x}}\right )\\ &=4 \sqrt {a+b \sqrt {c+d x}}-2 \sqrt {a-b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )-2 \sqrt {a+b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 116, normalized size = 1.00 \[ 4 \sqrt {a+b \sqrt {c+d x}}-2 \sqrt {a-b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a-b \sqrt {c}}}\right )-2 \sqrt {a+b \sqrt {c}} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c+d x}}}{\sqrt {a+b \sqrt {c}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c + d*x]]/x,x]

[Out]

4*Sqrt[a + b*Sqrt[c + d*x]] - 2*Sqrt[a - b*Sqrt[c]]*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a - b*Sqrt[c]]] - 2
*Sqrt[a + b*Sqrt[c]]*ArcTanh[Sqrt[a + b*Sqrt[c + d*x]]/Sqrt[a + b*Sqrt[c]]]

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fricas [B]  time = 0.54, size = 194, normalized size = 1.67 \[ -\sqrt {a + \sqrt {b^{2} c}} \log \left (2 \, \sqrt {\sqrt {d x + c} b + a} + 2 \, \sqrt {a + \sqrt {b^{2} c}}\right ) + \sqrt {a + \sqrt {b^{2} c}} \log \left (2 \, \sqrt {\sqrt {d x + c} b + a} - 2 \, \sqrt {a + \sqrt {b^{2} c}}\right ) - \sqrt {a - \sqrt {b^{2} c}} \log \left (2 \, \sqrt {\sqrt {d x + c} b + a} + 2 \, \sqrt {a - \sqrt {b^{2} c}}\right ) + \sqrt {a - \sqrt {b^{2} c}} \log \left (2 \, \sqrt {\sqrt {d x + c} b + a} - 2 \, \sqrt {a - \sqrt {b^{2} c}}\right ) + 4 \, \sqrt {\sqrt {d x + c} b + a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(a + sqrt(b^2*c))*log(2*sqrt(sqrt(d*x + c)*b + a) + 2*sqrt(a + sqrt(b^2*c))) + sqrt(a + sqrt(b^2*c))*log(
2*sqrt(sqrt(d*x + c)*b + a) - 2*sqrt(a + sqrt(b^2*c))) - sqrt(a - sqrt(b^2*c))*log(2*sqrt(sqrt(d*x + c)*b + a)
+ 2*sqrt(a - sqrt(b^2*c))) + sqrt(a - sqrt(b^2*c))*log(2*sqrt(sqrt(d*x + c)*b + a) - 2*sqrt(a - sqrt(b^2*c)))
+ 4*sqrt(sqrt(d*x + c)*b + a)

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giac [A]  time = 0.49, size = 150, normalized size = 1.29 \[ \frac {2 \, {\left (2 \, \sqrt {\sqrt {d x + c} b + a} b - \frac {{\left (b^{3} c - a^{2} b\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-a + \sqrt {b^{2} c}}}\right )}{{\left (b \sqrt {c} + a\right )} \sqrt {b \sqrt {c} - a}} + \frac {{\left (b^{3} c - a^{2} b\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-a - \sqrt {b^{2} c}}}\right )}{{\left (b \sqrt {c} - a\right )} \sqrt {-b \sqrt {c} - a}}\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)^(1/2))^(1/2)/x,x, algorithm="giac")

[Out]

2*(2*sqrt(sqrt(d*x + c)*b + a)*b - (b^3*c - a^2*b)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-a + sqrt(b^2*c)))/((
b*sqrt(c) + a)*sqrt(b*sqrt(c) - a)) + (b^3*c - a^2*b)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-a - sqrt(b^2*c)))
/((b*sqrt(c) - a)*sqrt(-b*sqrt(c) - a)))/b

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maple [B]  time = 0.04, size = 221, normalized size = 1.91 \[ \frac {2 b^{2} c \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a -\sqrt {b^{2} c}}}\right )}{\sqrt {b^{2} c}\, \sqrt {-a -\sqrt {b^{2} c}}}-\frac {2 b^{2} c \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a +\sqrt {b^{2} c}}}\right )}{\sqrt {b^{2} c}\, \sqrt {-a +\sqrt {b^{2} c}}}+\frac {2 a \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a -\sqrt {b^{2} c}}}\right )}{\sqrt {-a -\sqrt {b^{2} c}}}+\frac {2 a \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a +\sqrt {b^{2} c}}}\right )}{\sqrt {-a +\sqrt {b^{2} c}}}+4 \sqrt {a +\sqrt {d x +c}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(a+(d*x+c)^(1/2)*b)^(1/2)+2/(b^2*c)^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2)*arctan((a+(d*x+c)^(1/2)*b)^(1/2)/(-(b^2*c
)^(1/2)-a)^(1/2))*b^2*c+2/(-(b^2*c)^(1/2)-a)^(1/2)*arctan((a+(d*x+c)^(1/2)*b)^(1/2)/(-(b^2*c)^(1/2)-a)^(1/2))*
a-2/(b^2*c)^(1/2)/((b^2*c)^(1/2)-a)^(1/2)*arctan((a+(d*x+c)^(1/2)*b)^(1/2)/((b^2*c)^(1/2)-a)^(1/2))*b^2*c+2/((
b^2*c)^(1/2)-a)^(1/2)*arctan((a+(d*x+c)^(1/2)*b)^(1/2)/((b^2*c)^(1/2)-a)^(1/2))*a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sqrt {d x + c} b + a}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)^(1/2))^(1/2)/x,x, algorithm="maxima")

[Out]

integrate(sqrt(sqrt(d*x + c)*b + a)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,\sqrt {c+d\,x}}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \sqrt {c + d x}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)**(1/2))**(1/2)/x,x)

[Out]

Integral(sqrt(a + b*sqrt(c + d*x))/x, x)

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