Integrand size = 40, antiderivative size = 119 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {4 (5+c) \sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{15 a}-\frac {4 \left (15+3 b+5 c-2 c^2+3 a x\right ) \sqrt {c+\sqrt {b+a x}}}{15 a}-\frac {4 \sqrt {-1-c} \arctan \left (\frac {\sqrt {-1-c} \sqrt {c+\sqrt {b+a x}}}{1+c}\right )}{a} \]
-4/15*(5+c)*(a*x+b)^(1/2)*(c+(a*x+b)^(1/2))^(1/2)/a-4/15*(3*a*x-2*c^2+3*b+ 5*c+15)*(c+(a*x+b)^(1/2))^(1/2)/a-4*(-1-c)^(1/2)*arctan((-1-c)^(1/2)*(c+(a *x+b)^(1/2))^(1/2)/(1+c))/a
Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\frac {-4 \sqrt {c+\sqrt {b+a x}} \left (15+3 b-2 c^2+3 a x+5 \sqrt {b+a x}+c \left (5+\sqrt {b+a x}\right )\right )+60 \sqrt {-1-c} \arctan \left (\frac {\sqrt {c+\sqrt {b+a x}}}{\sqrt {-1-c}}\right )}{15 a} \]
(-4*Sqrt[c + Sqrt[b + a*x]]*(15 + 3*b - 2*c^2 + 3*a*x + 5*Sqrt[b + a*x] + c*(5 + Sqrt[b + a*x])) + 60*Sqrt[-1 - c]*ArcTan[Sqrt[c + Sqrt[b + a*x]]/Sq rt[-1 - c]])/(15*a)
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1014, 948, 99, 2009}
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.
\(\displaystyle \int \frac {\sqrt {a x+b} \sqrt {\sqrt {a x+b}+c}}{1-\sqrt {a x+b}} \, dx\) |
\(\Big \downarrow \) 1014 |
\(\displaystyle \frac {\int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}}d(b+a x)}{a}\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {2 \int \frac {(b+a x) \sqrt {c+\sqrt {b+a x}}}{-b-a x+1}d\sqrt {b+a x}}{a}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {2 \int \left (-\left (c+\sqrt {b+a x}\right )^{3/2}+(c-1) \sqrt {c+\sqrt {b+a x}}+\frac {\sqrt {c+\sqrt {b+a x}}}{-b-a x+1}\right )d\sqrt {b+a x}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \left (2 \sqrt {c+1} \text {arctanh}\left (\frac {\sqrt {\sqrt {a x+b}+c}}{\sqrt {c+1}}\right )-\frac {2}{5} \left (\sqrt {a x+b}+c\right )^{5/2}-\frac {2}{3} (1-c) \left (\sqrt {a x+b}+c\right )^{3/2}-2 \sqrt {\sqrt {a x+b}+c}\right )}{a}\) |
(2*(-2*Sqrt[c + Sqrt[b + a*x]] - (2*(1 - c)*(c + Sqrt[b + a*x])^(3/2))/3 - (2*(c + Sqrt[b + a*x])^(5/2))/5 + 2*Sqrt[1 + c]*ArcTanh[Sqrt[c + Sqrt[b + a*x]]/Sqrt[1 + c]]))/a
3.18.73.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^(n_))^(p_.)*((c_.) + (d_.)*(v_)^(n_))^(q _.), x_Symbol] :> Simp[u^m/(Coefficient[v, x, 1]*v^m) Subst[Int[x^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x, v], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && LinearPairQ[u, v, x]
Time = 0.16 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(-\frac {2 \left (\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+2 \sqrt {c +\sqrt {a x +b}}-2 \sqrt {1+c}\, \operatorname {arctanh}\left (\frac {\sqrt {c +\sqrt {a x +b}}}{\sqrt {1+c}}\right )\right )}{a}\) | \(85\) |
default | \(-\frac {2 \left (\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+2 \sqrt {c +\sqrt {a x +b}}-2 \sqrt {1+c}\, \operatorname {arctanh}\left (\frac {\sqrt {c +\sqrt {a x +b}}}{\sqrt {1+c}}\right )\right )}{a}\) | \(85\) |
-2/a*(2/5*(c+(a*x+b)^(1/2))^(5/2)-2/3*c*(c+(a*x+b)^(1/2))^(3/2)+2/3*(c+(a* x+b)^(1/2))^(3/2)+2*(c+(a*x+b)^(1/2))^(1/2)-2*(1+c)^(1/2)*arctanh((c+(a*x+ b)^(1/2))^(1/2)/(1+c)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.68 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\left [\frac {2 \, {\left (2 \, {\left (2 \, c^{2} - 3 \, a x - \sqrt {a x + b} {\left (c + 5\right )} - 3 \, b - 5 \, c - 15\right )} \sqrt {c + \sqrt {a x + b}} + 15 \, \sqrt {c + 1} \log \left (\frac {a x + 2 \, {\left (\sqrt {a x + b} \sqrt {c + 1} + \sqrt {c + 1}\right )} \sqrt {c + \sqrt {a x + b}} + 2 \, \sqrt {a x + b} {\left (c + 1\right )} + b + 2 \, c + 1}{a x + b - 1}\right )\right )}}{15 \, a}, \frac {4 \, {\left ({\left (2 \, c^{2} - 3 \, a x - \sqrt {a x + b} {\left (c + 5\right )} - 3 \, b - 5 \, c - 15\right )} \sqrt {c + \sqrt {a x + b}} - 15 \, \sqrt {-c - 1} \arctan \left (\frac {\sqrt {c + \sqrt {a x + b}} \sqrt {-c - 1}}{c + 1}\right )\right )}}{15 \, a}\right ] \]
[2/15*(2*(2*c^2 - 3*a*x - sqrt(a*x + b)*(c + 5) - 3*b - 5*c - 15)*sqrt(c + sqrt(a*x + b)) + 15*sqrt(c + 1)*log((a*x + 2*(sqrt(a*x + b)*sqrt(c + 1) + sqrt(c + 1))*sqrt(c + sqrt(a*x + b)) + 2*sqrt(a*x + b)*(c + 1) + b + 2*c + 1)/(a*x + b - 1)))/a, 4/15*((2*c^2 - 3*a*x - sqrt(a*x + b)*(c + 5) - 3*b - 5*c - 15)*sqrt(c + sqrt(a*x + b)) - 15*sqrt(-c - 1)*arctan(sqrt(c + sqr t(a*x + b))*sqrt(-c - 1)/(c + 1)))/a]
Time = 1.63 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\begin {cases} \frac {2 \cdot \left (\frac {2 \left (c - 1\right ) \left (c + \sqrt {a x + b}\right )^{\frac {3}{2}}}{3} - \frac {2 \left (c + \sqrt {a x + b}\right )^{\frac {5}{2}}}{5} - 2 \sqrt {c + \sqrt {a x + b}} - \frac {2 \left (c + 1\right ) \operatorname {atan}{\left (\frac {\sqrt {c + \sqrt {a x + b}}}{\sqrt {- c - 1}} \right )}}{\sqrt {- c - 1}}\right )}{a} & \text {for}\: a \neq 0 \\\frac {\sqrt {b} x \sqrt {\sqrt {b} + c}}{1 - \sqrt {b}} & \text {otherwise} \end {cases} \]
Piecewise((2*(2*(c - 1)*(c + sqrt(a*x + b))**(3/2)/3 - 2*(c + sqrt(a*x + b ))**(5/2)/5 - 2*sqrt(c + sqrt(a*x + b)) - 2*(c + 1)*atan(sqrt(c + sqrt(a*x + b))/sqrt(-c - 1))/sqrt(-c - 1))/a, Ne(a, 0)), (sqrt(b)*x*sqrt(sqrt(b) + c)/(1 - sqrt(b)), True))
Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {2 \, {\left (6 \, {\left (c + \sqrt {a x + b}\right )}^{\frac {5}{2}} - 10 \, {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} {\left (c - 1\right )} + 15 \, \sqrt {c + 1} \log \left (\frac {\sqrt {c + \sqrt {a x + b}} - \sqrt {c + 1}}{\sqrt {c + \sqrt {a x + b}} + \sqrt {c + 1}}\right ) + 30 \, \sqrt {c + \sqrt {a x + b}}\right )}}{15 \, a} \]
-2/15*(6*(c + sqrt(a*x + b))^(5/2) - 10*(c + sqrt(a*x + b))^(3/2)*(c - 1) + 15*sqrt(c + 1)*log((sqrt(c + sqrt(a*x + b)) - sqrt(c + 1))/(sqrt(c + sqr t(a*x + b)) + sqrt(c + 1))) + 30*sqrt(c + sqrt(a*x + b)))/a
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {4 \, {\left (c + 1\right )} \arctan \left (\frac {\sqrt {c + \sqrt {a x + b}}}{\sqrt {-c - 1}}\right )}{a \sqrt {-c - 1}} - \frac {4 \, {\left (3 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {5}{2}} - 5 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} c + 5 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} + 15 \, a^{4} \sqrt {c + \sqrt {a x + b}}\right )}}{15 \, a^{5}} \]
-4*(c + 1)*arctan(sqrt(c + sqrt(a*x + b))/sqrt(-c - 1))/(a*sqrt(-c - 1)) - 4/15*(3*a^4*(c + sqrt(a*x + b))^(5/2) - 5*a^4*(c + sqrt(a*x + b))^(3/2)*c + 5*a^4*(c + sqrt(a*x + b))^(3/2) + 15*a^4*sqrt(c + sqrt(a*x + b)))/a^5
Timed out. \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\int \frac {\sqrt {c+\sqrt {b+a\,x}}\,\sqrt {b+a\,x}}{\sqrt {b+a\,x}-1} \,d x \]