Integrand size = 49, antiderivative size = 246 \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=-\frac {\sqrt {1+\sqrt {1+\sqrt {b+a x}}}}{x}-\frac {\sqrt {b+a x} \sqrt {1+\sqrt {1+\sqrt {b+a x}}}}{x}+\frac {\sqrt {1+\sqrt {b+a x}} \sqrt {1+\sqrt {1+\sqrt {b+a x}}}}{x}+\frac {1}{8} a \text {RootSum}\left [b-4 \text {$\#$1}^4+4 \text {$\#$1}^6-\text {$\#$1}^8\&,\frac {2 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right )-9 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right ) \text {$\#$1}^2+5 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right ) \text {$\#$1}^4}{2 \text {$\#$1}^3-3 \text {$\#$1}^5+\text {$\#$1}^7}\&\right ] \]
Time = 0.00 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\frac {\sqrt {1+\sqrt {1+\sqrt {b+a x}}} \left (-1-\sqrt {b+a x}+\sqrt {1+\sqrt {b+a x}}\right )}{x}+a \text {RootSum}\left [b-4 \text {$\#$1}^4+4 \text {$\#$1}^6-\text {$\#$1}^8\&,\frac {-\log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right )+\log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-2 \text {$\#$1}^3+\text {$\#$1}^5}\&\right ]-\frac {1}{8} a \text {RootSum}\left [b-4 \text {$\#$1}^4+4 \text {$\#$1}^6-\text {$\#$1}^8\&,\frac {6 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right )-7 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right ) \text {$\#$1}^2+3 \log \left (\sqrt {1+\sqrt {1+\sqrt {b+a x}}}-\text {$\#$1}\right ) \text {$\#$1}^4}{2 \text {$\#$1}^3-3 \text {$\#$1}^5+\text {$\#$1}^7}\&\right ] \]
(Sqrt[1 + Sqrt[1 + Sqrt[b + a*x]]]*(-1 - Sqrt[b + a*x] + Sqrt[1 + Sqrt[b + a*x]]))/x + a*RootSum[b - 4*#1^4 + 4*#1^6 - #1^8 & , (-Log[Sqrt[1 + Sqrt[ 1 + Sqrt[b + a*x]]] - #1] + Log[Sqrt[1 + Sqrt[1 + Sqrt[b + a*x]]] - #1]*#1 ^2)/(-2*#1^3 + #1^5) & ] - (a*RootSum[b - 4*#1^4 + 4*#1^6 - #1^8 & , (6*Lo g[Sqrt[1 + Sqrt[1 + Sqrt[b + a*x]]] - #1] - 7*Log[Sqrt[1 + Sqrt[1 + Sqrt[b + a*x]]] - #1]*#1^2 + 3*Log[Sqrt[1 + Sqrt[1 + Sqrt[b + a*x]]] - #1]*#1^4) /(2*#1^3 - 3*#1^5 + #1^7) & ])/8
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.
| \(\displaystyle \int \frac {\sqrt {a x+b} \sqrt {\sqrt {a x+b}+1}}{x^2 \sqrt {\sqrt {\sqrt {a x+b}+1}+1}} \, dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 2 a \int \frac {(b+a x) \sqrt {\sqrt {b+a x}+1}}{a^2 x^2 \sqrt {\sqrt {\sqrt {b+a x}+1}+1}}d\sqrt {b+a x}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 4 a \int \frac {(-b-a x+1)^2 (b+a x)}{\left (b-(-b-a x+1)^2\right )^2 \sqrt {\sqrt {\sqrt {b+a x}+1}+1}}d\sqrt {\sqrt {b+a x}+1}\) |
| \(\Big \downarrow \) 2003 |
| \(\displaystyle 4 a \int \frac {(b+a x) \left (1-\sqrt {\sqrt {b+a x}+1}\right )^2 \left (\sqrt {\sqrt {b+a x}+1}+1\right )^{3/2}}{\left (b-(-b-a x+1)^2\right )^2}d\sqrt {\sqrt {b+a x}+1}\) |
| \(\Big \downarrow \) 2091 |
| \(\displaystyle 4 a \int \frac {(b+a x) \left (1-\sqrt {\sqrt {b+a x}+1}\right )^2 \left (\sqrt {\sqrt {b+a x}+1}+1\right )^{3/2}}{\left (-(b+a x)^2+2 (b+a x)+b-1\right )^2}d\sqrt {\sqrt {b+a x}+1}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle 8 a \int \frac {(b+a x)^2 \left ((b+a x)^2-3 (b+a x)+2\right )^2}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}\) |
| \(\Big \downarrow \) 2527 |
| \(\displaystyle 8 a \left (\frac {1}{3} \int \frac {-14 (b+a x)^5+43 (b+a x)^4-36 (b+a x)^3+(12-5 b) (b+a x)^2}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 2029 |
| \(\displaystyle 8 a \left (\frac {1}{3} \int \frac {(b+a x)^2 \left (-14 (b+a x)^3+43 (b+a x)^2-36 (b+a x)-5 b+12\right )}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 2527 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \int \frac {47 (b+a x)^4-124 (b+a x)^3+5 (12-5 b) (b+a x)^2+42 b (b+a x)}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 2029 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \int \frac {(b+a x) \left (47 (b+a x)^3-124 (b+a x)^2+5 (12-5 b) (b+a x)+42 b\right )}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 2527 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{7} \int -\frac {-72 (b+a x)^3+(175 b+144) (b+a x)^2-294 b (b+a x)+47 b}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}+\frac {47 \sqrt {\sqrt {\sqrt {a x+b}+1}+1}}{7 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {47 \sqrt {\sqrt {\sqrt {a x+b}+1}+1}}{7 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}-\frac {1}{7} \int \frac {-72 (b+a x)^3+(175 b+144) (b+a x)^2-294 b (b+a x)+47 b}{\left (b-(-b-a x+2)^2 (b+a x)^2\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}\right )-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {47 \sqrt {\sqrt {\sqrt {a x+b}+1}+1}}{7 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}-\frac {1}{7} \int \left (-\frac {72 (b+a x)^3}{\left ((b+a x)^4-4 (b+a x)^3+4 (b+a x)^2-b\right )^2}+\frac {(175 b+144) (b+a x)^2}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}-\frac {294 b (b+a x)}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}+\frac {47 b}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}\right )d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}\right )-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 8 a \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{7} \left (-47 b \int \frac {1}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}+294 b \int \frac {b+a x}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}-(175 b+144) \int \frac {(b+a x)^2}{\left (-(b+a x)^4+4 (b+a x)^3-4 (b+a x)^2+b\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}+72 \int \frac {(b+a x)^3}{\left ((b+a x)^4-4 (b+a x)^3+4 (b+a x)^2-b\right )^2}d\sqrt {\sqrt {\sqrt {b+a x}+1}+1}\right )+\frac {47 \sqrt {\sqrt {\sqrt {a x+b}+1}+1}}{7 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )-\frac {14 (a x+b)^{3/2}}{5 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )+\frac {(a x+b)^{5/2}}{3 \left (b-(-a x-b+2)^2 (a x+b)^2\right )}\right )\) |
3.28.5.3.1 Defintions of rubi rules used
Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : > Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} , x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !IntegerQ[n]))
Int[(Fx_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)* (x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r) + d*x^(q - r))^p*Fx, x] /; FreeQ[{a, b, c, d, r, s, t, q}, x] && IntegerQ[p ] && PosQ[s - r] && PosQ[t - r] && PosQ[q - r] && !(EqQ[p, 1] && EqQ[u, 1] )
Int[(Px_)*(u_)^(p_.)*(z_)^(q_.), x_Symbol] :> Int[Px*ExpandToSum[z, x]^q*Ex pandToSum[u, x]^p, x] /; FreeQ[{p, q}, x] && PolyQ[Px, x] && BinomialQ[z, x ] && TrinomialQ[u, x] && !(BinomialMatchQ[z, x] && TrinomialMatchQ[u, x])
Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x ]}, Simp[Coeff[Pm, x, m]*x^(m - n + 1)*(Qn^(p + 1)/((m + n*p + 1)*Coeff[Qn, x, n])), x] + Simp[1/((m + n*p + 1)*Coeff[Qn, x, n]) Int[ExpandToSum[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p, x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[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 = 0.22 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(2 a \left (\frac {\frac {\left (1+\sqrt {1+\sqrt {a x +b}}\right )^{\frac {5}{2}}}{2}-\frac {3 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{\frac {3}{2}}}{2}+\sqrt {1+\sqrt {1+\sqrt {a x +b}}}}{-\left (1+\sqrt {1+\sqrt {a x +b}}\right )^{4}+4 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{3}-4 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{2}+b}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+4 \textit {\_Z}^{4}-b \right )}{\sum }\frac {\left (-5 \textit {\_R}^{4}+9 \textit {\_R}^{2}-2\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {a x +b}}}-\textit {\_R} \right )}{-\textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )}{16}\right )\) | \(188\) |
| default | \(2 a \left (\frac {\frac {\left (1+\sqrt {1+\sqrt {a x +b}}\right )^{\frac {5}{2}}}{2}-\frac {3 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{\frac {3}{2}}}{2}+\sqrt {1+\sqrt {1+\sqrt {a x +b}}}}{-\left (1+\sqrt {1+\sqrt {a x +b}}\right )^{4}+4 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{3}-4 \left (1+\sqrt {1+\sqrt {a x +b}}\right )^{2}+b}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 \textit {\_Z}^{6}+4 \textit {\_Z}^{4}-b \right )}{\sum }\frac {\left (-5 \textit {\_R}^{4}+9 \textit {\_R}^{2}-2\right ) \ln \left (\sqrt {1+\sqrt {1+\sqrt {a x +b}}}-\textit {\_R} \right )}{-\textit {\_R}^{7}+3 \textit {\_R}^{5}-2 \textit {\_R}^{3}}\right )}{16}\right )\) | \(188\) |
int((a*x+b)^(1/2)*(1+(a*x+b)^(1/2))^(1/2)/x^2/(1+(1+(a*x+b)^(1/2))^(1/2))^ (1/2),x,method=_RETURNVERBOSE)
2*a*(4*(1/8*(1+(1+(a*x+b)^(1/2))^(1/2))^(5/2)-3/8*(1+(1+(a*x+b)^(1/2))^(1/ 2))^(3/2)+1/4*(1+(1+(a*x+b)^(1/2))^(1/2))^(1/2))/(-(1+(1+(a*x+b)^(1/2))^(1 /2))^4+4*(1+(1+(a*x+b)^(1/2))^(1/2))^3-4*(1+(1+(a*x+b)^(1/2))^(1/2))^2+b)+ 1/16*sum((-5*_R^4+9*_R^2-2)/(-_R^7+3*_R^5-2*_R^3)*ln((1+(1+(a*x+b)^(1/2))^ (1/2))^(1/2)-_R),_R=RootOf(_Z^8-4*_Z^6+4*_Z^4-b)))
Timed out. \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\text {Timed out} \]
integrate((a*x+b)^(1/2)*(1+(a*x+b)^(1/2))^(1/2)/x^2/(1+(1+(a*x+b)^(1/2))^( 1/2))^(1/2),x, algorithm="fricas")
Timed out. \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\text {Timed out} \]
integrate((a*x+b)**(1/2)*(1+(a*x+b)**(1/2))**(1/2)/x**2/(1+(1+(a*x+b)**(1/ 2))**(1/2))**(1/2),x)
Not integrable
Time = 1.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\int { \frac {\sqrt {a x + b} \sqrt {\sqrt {a x + b} + 1}}{x^{2} \sqrt {\sqrt {\sqrt {a x + b} + 1} + 1}} \,d x } \]
integrate((a*x+b)^(1/2)*(1+(a*x+b)^(1/2))^(1/2)/x^2/(1+(1+(a*x+b)^(1/2))^( 1/2))^(1/2),x, algorithm="maxima")
Timed out. \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\text {Timed out} \]
integrate((a*x+b)^(1/2)*(1+(a*x+b)^(1/2))^(1/2)/x^2/(1+(1+(a*x+b)^(1/2))^( 1/2))^(1/2),x, algorithm="giac")
Not integrable
Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {b+a x} \sqrt {1+\sqrt {b+a x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {b+a x}}}} \, dx=\int \frac {\sqrt {\sqrt {b+a\,x}+1}\,\sqrt {b+a\,x}}{x^2\,\sqrt {\sqrt {\sqrt {b+a\,x}+1}+1}} \,d x \]
int((((b + a*x)^(1/2) + 1)^(1/2)*(b + a*x)^(1/2))/(x^2*(((b + a*x)^(1/2) + 1)^(1/2) + 1)^(1/2)),x)