\(\int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx\) [17]

Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=-\frac {\sqrt {x+\sqrt {1+x}}}{x}-\frac {1}{4} \arctan \left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )+\frac {3}{4} \text {arctanh}\left (\frac {1-3 \sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right ) \] Output:

-1/4*arctan(1/2*(3+(1+x)^(1/2))/(x+(1+x)^(1/2))^(1/2))+3/4*arctanh(1/2*(1- 
3*(1+x)^(1/2))/(x+(1+x)^(1/2))^(1/2))-(x+(1+x)^(1/2))^(1/2)/x
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=-\frac {\sqrt {x+\sqrt {1+x}}}{x}-\frac {1}{2} \arctan \left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-\frac {3}{2} \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right ) \] Input:

Integrate[Sqrt[x + Sqrt[1 + x]]/x^2,x]
 

Output:

-(Sqrt[x + Sqrt[1 + x]]/x) - ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x] 
]]/2 - (3*ArcTanh[1 - Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]])/2
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {7267, 1347, 27, 1366, 25, 1154, 217, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {x+\sqrt {x+1}}}{x^2} \, dx\)

\(\Big \downarrow \) 7267

\(\displaystyle 2 \int \frac {\sqrt {x+1} \sqrt {x+\sqrt {x+1}}}{x^2}d\sqrt {x+1}\)

\(\Big \downarrow \) 1347

\(\displaystyle 2 \left (\frac {1}{2} \int \frac {2 \sqrt {x+1}+1}{2 x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (-\frac {1}{4} \int -\frac {2 \sqrt {x+1}+1}{x \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 1366

\(\displaystyle 2 \left (\frac {1}{4} \left (-\frac {3}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {1}{2} \int -\frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{\left (\sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {3}{2} \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle 2 \left (\frac {1}{4} \left (3 \int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-\int \frac {1}{-x-5}d\left (-\frac {\sqrt {x+1}+3}{\sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 2 \left (\frac {1}{4} \left (3 \int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {1}{4} \left (\frac {3}{2} \text {arctanh}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {\sqrt {x+\sqrt {x+1}}}{2 x}\right )\)

Input:

Int[Sqrt[x + Sqrt[1 + x]]/x^2,x]
 

Output:

2*(-1/2*Sqrt[x + Sqrt[1 + x]]/x + (-1/2*ArcTan[(3 + Sqrt[1 + x])/(2*Sqrt[x 
 + Sqrt[1 + x]])] + (3*ArcTanh[(1 - 3*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x] 
])])/2)/4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1347

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f 
_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a*h - g*c*x)*(a + c*x^2)^(p + 1)*((d + 
 e*x + f*x^2)^q/(2*a*c*(p + 1))), x] + Simp[2/(4*a*c*(p + 1))   Int[(a + c* 
x^2)^(p + 1)*(d + e*x + f*x^2)^(q - 1)*Simp[g*c*d*(2*p + 3) - a*(h*e*q) + ( 
g*c*e*(2*p + q + 3) - a*(2*h*f*q))*x + g*c*f*(2*p + 2*q + 3)*x^2, x], x], x 
] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && LtQ[p, -1] & 
& GtQ[q, 0]
 

rule 1366

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + ( 
f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Simp[(h/2 + c*(g/(2*q 
)))   Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Simp[(h/2 - c*(g/( 
2*q)))   Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d 
, e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && PosQ[(-a)*c]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(59)=118\).

Time = 0.04 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.59

method	result	size
derivativedivides	\(-\frac {\left (\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (-1+\sqrt {1+x}\right )}+\frac {3 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\frac {3 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )}{4}+\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}-\frac {\left (\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (1+\sqrt {1+x}\right )}-\frac {\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}-\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\frac {\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )}{4}+\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}\)	\(298\)
default	\(-\frac {\left (\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (-1+\sqrt {1+x}\right )}+\frac {3 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\frac {3 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )}{4}+\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}-\frac {\left (\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (1+\sqrt {1+x}\right )}-\frac {\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}-\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\frac {\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )}{4}+\frac {\left (1+2 \sqrt {1+x}\right ) \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}\)	\(298\)

Input:

int((x+(1+x)^(1/2))^(1/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

-1/2/(-1+(1+x)^(1/2))*((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(3/2)+3/4*((-1+ 
(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2)+1/2*ln(1/2+(1+x)^(1/2)+((-1+(1+x)^(1 
/2))^2+3*(1+x)^(1/2)-2)^(1/2))-3/4*arctanh(1/2*(-1+3*(1+x)^(1/2))/((-1+(1+ 
x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2))+1/4*(1+2*(1+x)^(1/2))*((-1+(1+x)^(1/2) 
)^2+3*(1+x)^(1/2)-2)^(1/2)-1/2/(1+(1+x)^(1/2))*((1+(1+x)^(1/2))^2-(1+x)^(1 
/2)-2)^(3/2)-1/4*((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2)-1/2*ln(1/2+(1+x)^ 
(1/2)+((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))+1/4*arctan(1/2*(-3-(1+x)^(1 
/2))/((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))+1/4*(1+2*(1+x)^(1/2))*((1+(1 
+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=\frac {x \arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + 3 \, x \log \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} + 1\right )} - 3 \, x - 2 \, \sqrt {x + 1} - 2}{x}\right ) - 4 \, \sqrt {x + \sqrt {x + 1}}}{4 \, x} \] Input:

integrate((x+(1+x)^(1/2))^(1/2)/x^2,x, algorithm="fricas")
 

Output:

1/4*(x*arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) + 3*x*log 
((2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) + 1) - 3*x - 2*sqrt(x + 1) - 2)/x) 
- 4*sqrt(x + sqrt(x + 1)))/x
 

Sympy [F]

\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=\int \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2}}\, dx \] Input:

integrate((x+(1+x)**(1/2))**(1/2)/x**2,x)
 

Output:

Integral(sqrt(x + sqrt(x + 1))/x**2, x)
 

Maxima [F]

\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=\int { \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2}} \,d x } \] Input:

integrate((x+(1+x)^(1/2))^(1/2)/x^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(x + sqrt(x + 1))/x^2, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (59) = 118\).

Time = 0.36 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.27 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=-\frac {2 \, {\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{3} - 3 \, {\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{2} - \sqrt {x + \sqrt {x + 1}} + \sqrt {x + 1} + 1}{{\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{4} - 2 \, {\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{2} + 4 \, \sqrt {x + \sqrt {x + 1}} - 4 \, \sqrt {x + 1}} + \frac {1}{2} \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - \frac {3}{4} \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + \frac {3}{4} \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \] Input:

integrate((x+(1+x)^(1/2))^(1/2)/x^2,x, algorithm="giac")
 

Output:

-(2*(sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^3 - 3*(sqrt(x + sqrt(x + 1)) - s 
qrt(x + 1))^2 - sqrt(x + sqrt(x + 1)) + sqrt(x + 1) + 1)/((sqrt(x + sqrt(x 
 + 1)) - sqrt(x + 1))^4 - 2*(sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^2 + 4*sq 
rt(x + sqrt(x + 1)) - 4*sqrt(x + 1)) + 1/2*arctan(sqrt(x + sqrt(x + 1)) - 
sqrt(x + 1) - 1) - 3/4*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) + 2)) + 
 3/4*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=\int \frac {\sqrt {x+\sqrt {x+1}}}{x^2} \,d x \] Input:

int((x + (x + 1)^(1/2))^(1/2)/x^2,x)
 

Output:

int((x + (x + 1)^(1/2))^(1/2)/x^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{x^2} \, dx=\frac {10 \mathit {atan} \left (\sqrt {\sqrt {x +1}+x}+\sqrt {x +1}+1\right ) x -20 \sqrt {\sqrt {x +1}+x}-15 \,\mathrm {log}\left (\frac {10 \sqrt {\sqrt {x +1}+x}+10 \sqrt {x +1}}{\sqrt {5}}\right ) x +15 \,\mathrm {log}\left (\frac {2 \sqrt {\sqrt {x +1}+x}+2 \sqrt {x +1}-4}{\sqrt {5}}\right ) x -16 x}{20 x} \] Input:

int((x+(1+x)^(1/2))^(1/2)/x^2,x)
 

Output:

(10*atan(sqrt(sqrt(x + 1) + x) + sqrt(x + 1) + 1)*x - 20*sqrt(sqrt(x + 1) 
+ x) - 15*log((10*sqrt(sqrt(x + 1) + x) + 10*sqrt(x + 1))/sqrt(5))*x + 15* 
log((2*sqrt(sqrt(x + 1) + x) + 2*sqrt(x + 1) - 4)/sqrt(5))*x - 16*x)/(20*x 
)