3.12.94 \(\int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx\) [1194]

3.12.94.1 Optimal result

Integrand size = 27, antiderivative size = 87 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=\frac {1}{2} \left (-5-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}+4 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right )+\frac {7}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]

output

1/2*(-5-2*(1+x)^(1/2))*(x+(1+x)^(1/2))^(1/2)-4*arctanh(-1+(1+x)^(1/2)-(x+( 
1+x)^(1/2))^(1/2))+7/4*ln(-1-2*(1+x)^(1/2)+2*(x+(1+x)^(1/2))^(1/2))
 

3.12.94.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=\frac {1}{2} \left (-5-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}+4 \text {arctanh}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right )+\frac {7}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]

input

Integrate[Sqrt[x + Sqrt[1 + x]]/(1 - Sqrt[1 + x]),x]
 

output

((-5 - 2*Sqrt[1 + x])*Sqrt[x + Sqrt[1 + x]])/2 + 4*ArcTanh[1 - Sqrt[1 + x] 
 + Sqrt[x + Sqrt[1 + x]]] + (7*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 
+ x]]])/4
 

3.12.94.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {7267, 1231, 27, 1269, 1092, 219, 1154, 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.

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

\(\Big \downarrow \) 7267

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

\(\Big \downarrow \) 1231

\(\displaystyle 2 \left (-\frac {1}{4} \int -\frac {7 \sqrt {x+1}+1}{2 \left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {1}{8} \int \frac {7 \sqrt {x+1}+1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 1269

\(\displaystyle 2 \left (\frac {1}{8} \left (8 \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-7 \int \frac {1}{\sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 1092

\(\displaystyle 2 \left (\frac {1}{8} \left (8 \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-14 \int \frac {1}{3-x}d\frac {2 \sqrt {x+1}+1}{\sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {1}{8} \left (8 \int \frac {1}{\left (1-\sqrt {x+1}\right ) \sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}-7 \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 1154

\(\displaystyle 2 \left (\frac {1}{8} \left (-16 \int \frac {1}{3-x}d\frac {1-3 \sqrt {x+1}}{\sqrt {x+\sqrt {x+1}}}-7 \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 2 \left (\frac {1}{8} \left (-8 \text {arctanh}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )-7 \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+5\right )\right )\)

input

Int[Sqrt[x + Sqrt[1 + x]]/(1 - Sqrt[1 + x]),x]
 

output

2*(-1/4*(Sqrt[x + Sqrt[1 + x]]*(5 + 2*Sqrt[1 + x])) + (-8*ArcTanh[(1 - 3*S 
qrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] - 7*ArcTanh[(1 + 2*Sqrt[1 + x])/(2* 
Sqrt[x + Sqrt[1 + x]])])/8)
 

3.12.94.3.1 Defintions of rubi rules used

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 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 1092

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

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 1231

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

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]]
 

3.12.94.4 Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}+\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}-3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}\right )+2 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}}\right )\)	\(127\)
default	\(-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}+\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}-3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}\right )+2 \,\operatorname {arctanh}\left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (\sqrt {1+x}-1\right )^{2}+3 \sqrt {1+x}-2}}\right )\)	\(127\)

input

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

output

-1/2*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)+5/4*ln(1/2+(1+x)^(1/2)+(x+(1+ 
x)^(1/2))^(1/2))-2*(((1+x)^(1/2)-1)^2+3*(1+x)^(1/2)-2)^(1/2)-3*ln(1/2+(1+x 
)^(1/2)+(((1+x)^(1/2)-1)^2+3*(1+x)^(1/2)-2)^(1/2))+2*arctanh(1/2*(-1+3*(1+ 
x)^(1/2))/(((1+x)^(1/2)-1)^2+3*(1+x)^(1/2)-2)^(1/2))
 

3.12.94.5 Fricas [A] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=-\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 5\right )} + \frac {7}{8} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 5\right ) + 2 \, \log \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} + 1\right )} + 3 \, x + 2 \, \sqrt {x + 1} + 2}{x}\right ) \]

input

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

output

-1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 5) + 7/8*log(4*sqrt(x + sqrt(x 
 + 1))*(2*sqrt(x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 5) + 2*log((2*sqrt(x + 
sqrt(x + 1))*(sqrt(x + 1) + 1) + 3*x + 2*sqrt(x + 1) + 2)/x)
 

3.12.94.6 Sympy [F]

\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=- \int \frac {\sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1} - 1}\, dx \]

input

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

output

-Integral(sqrt(x + sqrt(x + 1))/(sqrt(x + 1) - 1), x)
 

3.12.94.7 Maxima [F]

\[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=\int { -\frac {\sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1} - 1} \,d x } \]

input

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

output

-integrate(sqrt(x + sqrt(x + 1))/(sqrt(x + 1) - 1), x)
 

3.12.94.8 Giac [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=-\frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 5\right )} + \frac {7}{4} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) + 2 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) - 2 \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \]

input

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

output

-1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 5) + 7/4*log(-2*sqrt(x + sqrt( 
x + 1)) + 2*sqrt(x + 1) + 1) + 2*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 
1) + 2)) - 2*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))
 

3.12.94.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x+\sqrt {1+x}}}{1-\sqrt {1+x}} \, dx=-\int \frac {\sqrt {x+\sqrt {x+1}}}{\sqrt {x+1}-1} \,d x \]

input

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

output

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