Integrand size = 24, antiderivative size = 83 \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\frac {1}{96} (-21+8 x) \sqrt {x+\sqrt {1+x}}+\frac {1}{48} \sqrt {1+x} (-89+24 x) \sqrt {x+\sqrt {1+x}}-\frac {115}{64} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]
1/96*(-21+8*x)*(x+(1+x)^(1/2))^(1/2)+1/48*(1+x)^(1/2)*(-89+24*x)*(x+(1+x)^ (1/2))^(1/2)-115/64*ln(1+2*(1+x)^(1/2)-2*(x+(1+x)^(1/2))^(1/2))
Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\frac {1}{96} \sqrt {x+\sqrt {1+x}} \left (-29-226 \sqrt {1+x}+8 (1+x)+48 (1+x)^{3/2}\right )-\frac {115}{64} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \]
(Sqrt[x + Sqrt[1 + x]]*(-29 - 226*Sqrt[1 + x] + 8*(1 + x) + 48*(1 + x)^(3/ 2)))/96 - (115*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]])/64
Time = 0.42 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.39, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {7267, 25, 2192, 27, 1160, 1087, 1092, 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.
\(\displaystyle \int \frac {(x-1) \sqrt {x+\sqrt {x+1}}}{\sqrt {x+1}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle 2 \int -\left ((1-x) \sqrt {x+\sqrt {x+1}}\right )d\sqrt {x+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int (1-x) \sqrt {x+\sqrt {x+1}}d\sqrt {x+1}\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {1}{4} \int \frac {1}{2} \sqrt {x+\sqrt {x+1}} \left (5 \sqrt {x+1}+14\right )d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {1}{8} \int \sqrt {x+\sqrt {x+1}} \left (5 \sqrt {x+1}+14\right )d\sqrt {x+1}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {23}{2} \int \sqrt {x+\sqrt {x+1}}d\sqrt {x+1}-\frac {5}{3} \left (x+\sqrt {x+1}\right )^{3/2}\right )+\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {23}{2} \left (\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+1\right )-\frac {5}{8} \int \frac {1}{\sqrt {x+\sqrt {x+1}}}d\sqrt {x+1}\right )-\frac {5}{3} \left (x+\sqrt {x+1}\right )^{3/2}\right )+\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {23}{2} \left (\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+1\right )-\frac {5}{4} \int \frac {1}{3-x}d\frac {2 \sqrt {x+1}+1}{\sqrt {x+\sqrt {x+1}}}\right )-\frac {5}{3} \left (x+\sqrt {x+1}\right )^{3/2}\right )+\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{8} \left (-\frac {23}{2} \left (\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+1\right )-\frac {5}{8} \text {arctanh}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right )-\frac {5}{3} \left (x+\sqrt {x+1}\right )^{3/2}\right )+\frac {1}{4} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}\right )\) |
2*((Sqrt[1 + x]*(x + Sqrt[1 + x])^(3/2))/4 + ((-5*(x + Sqrt[1 + x])^(3/2)) /3 - (23*((Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sqrt[1 + x]))/4 - (5*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/8))/2)/8)
3.12.19.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
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]]
Time = 0.90 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(-\frac {23 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}+\frac {115 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}+\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}\) | \(68\) |
default | \(-\frac {23 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}+\frac {115 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}+\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}\) | \(68\) |
-23/32*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)+115/64*ln(1/2+(1+x)^(1/2)+( x+(1+x)^(1/2))^(1/2))+1/2*(1+x)^(1/2)*(x+(1+x)^(1/2))^(3/2)-5/12*(x+(1+x)^ (1/2))^(3/2)
Time = 0.64 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\frac {1}{96} \, {\left (2 \, {\left (24 \, x - 89\right )} \sqrt {x + 1} + 8 \, x - 21\right )} \sqrt {x + \sqrt {x + 1}} + \frac {115}{128} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + 8 \, x + 8 \, \sqrt {x + 1} + 5\right ) \]
1/96*(2*(24*x - 89)*sqrt(x + 1) + 8*x - 21)*sqrt(x + sqrt(x + 1)) + 115/12 8*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) + 8*x + 8*sqrt(x + 1) + 5)
Time = 0.49 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=2 \sqrt {x + \sqrt {x + 1}} \left (\frac {x}{24} + \frac {\left (x + 1\right )^{\frac {3}{2}}}{4} - \frac {113 \sqrt {x + 1}}{96} - \frac {7}{64}\right ) + \frac {115 \log {\left (2 \sqrt {x + 1} + 2 \sqrt {x + \sqrt {x + 1}} + 1 \right )}}{64} \]
2*sqrt(x + sqrt(x + 1))*(x/24 + (x + 1)**(3/2)/4 - 113*sqrt(x + 1)/96 - 7/ 64) + 115*log(2*sqrt(x + 1) + 2*sqrt(x + sqrt(x + 1)) + 1)/64
\[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\int { \frac {\sqrt {x + \sqrt {x + 1}} {\left (x - 1\right )}}{\sqrt {x + 1}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\frac {1}{96} \, {\left (2 \, {\left (4 \, \sqrt {x + 1} {\left (6 \, \sqrt {x + 1} + 1\right )} - 65\right )} \sqrt {x + 1} + 19\right )} \sqrt {x + \sqrt {x + 1}} - \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} - \frac {115}{64} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \]
1/96*(2*(4*sqrt(x + 1)*(6*sqrt(x + 1) + 1) - 65)*sqrt(x + 1) + 19)*sqrt(x + sqrt(x + 1)) - 1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) - 115/64*lo g(-2*sqrt(x + sqrt(x + 1)) + 2*sqrt(x + 1) + 1)
Timed out. \[ \int \frac {(-1+x) \sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx=\int \frac {\sqrt {x+\sqrt {x+1}}\,\left (x-1\right )}{\sqrt {x+1}} \,d x \]