\(\int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx\) [1773]

Optimal result

Integrand size = 40, antiderivative size = 119 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {4 (5+c) \sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{15 a}-\frac {4 \left (15+3 b+5 c-2 c^2+3 a x\right ) \sqrt {c+\sqrt {b+a x}}}{15 a}-\frac {4 \sqrt {-1-c} \arctan \left (\frac {\sqrt {-1-c} \sqrt {c+\sqrt {b+a x}}}{1+c}\right )}{a} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {527, 457, 90, 52, 65, 212} \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\frac {4 \sqrt {c+1} \text {arctanh}\left (\frac {\sqrt {\sqrt {a x+b}+c}}{\sqrt {c+1}}\right )}{a}-\frac {4 \left (\sqrt {a x+b}+c\right )^{5/2}}{5 a}-\frac {4 (1-c) \left (\sqrt {a x+b}+c\right )^{3/2}}{3 a}-\frac {4 \sqrt {\sqrt {a x+b}+c}}{a} \]

[In]

Int[(Sqrt[b + a*x]*Sqrt[c + Sqrt[b + a*x]])/(1 - Sqrt[b + a*x]),x]

[Out]

(-4*Sqrt[c + Sqrt[b + a*x]])/a - (4*(1 - c)*(c + Sqrt[b + a*x])^(3/2))/(3*a) - (4*(c + Sqrt[b + a*x])^(5/2))/(
5*a) + (4*Sqrt[1 + c]*ArcTanh[Sqrt[c + Sqrt[b + a*x]]/Sqrt[1 + c]])/a

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(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]))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[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]]

Rule 527

Int[(u_)^(m_.)*((a_.) + (b_.)*(v_)^(n_))^(p_.)*((c_.) + (d_.)*(v_)^(n_))^(q_.), x_Symbol] :> Dist[u^m/(Coeffic
ient[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {c+\sqrt {x}} \sqrt {x}}{1-\sqrt {x}} \, dx,x,b+a x\right )}{a} \\ & = \frac {2 \text {Subst}\left (\int \frac {x^2 \sqrt {c+x}}{1-x} \, dx,x,\sqrt {b+a x}\right )}{a} \\ & = \frac {2 \text {Subst}\left (\int \left ((-1+c) \sqrt {c+x}+\frac {\sqrt {c+x}}{1-x}-(c+x)^{3/2}\right ) \, dx,x,\sqrt {b+a x}\right )}{a} \\ & = -\frac {4 (1-c) \left (c+\sqrt {b+a x}\right )^{3/2}}{3 a}-\frac {4 \left (c+\sqrt {b+a x}\right )^{5/2}}{5 a}+\frac {2 \text {Subst}\left (\int \frac {\sqrt {c+x}}{1-x} \, dx,x,\sqrt {b+a x}\right )}{a} \\ & = -\frac {4 \sqrt {c+\sqrt {b+a x}}}{a}-\frac {4 (1-c) \left (c+\sqrt {b+a x}\right )^{3/2}}{3 a}-\frac {4 \left (c+\sqrt {b+a x}\right )^{5/2}}{5 a}+\frac {(2 (1+c)) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {c+x}} \, dx,x,\sqrt {b+a x}\right )}{a} \\ & = -\frac {4 \sqrt {c+\sqrt {b+a x}}}{a}-\frac {4 (1-c) \left (c+\sqrt {b+a x}\right )^{3/2}}{3 a}-\frac {4 \left (c+\sqrt {b+a x}\right )^{5/2}}{5 a}+\frac {(4 (1+c)) \text {Subst}\left (\int \frac {1}{1+c-x^2} \, dx,x,\sqrt {c+\sqrt {b+a x}}\right )}{a} \\ & = -\frac {4 \sqrt {c+\sqrt {b+a x}}}{a}-\frac {4 (1-c) \left (c+\sqrt {b+a x}\right )^{3/2}}{3 a}-\frac {4 \left (c+\sqrt {b+a x}\right )^{5/2}}{5 a}+\frac {4 \sqrt {1+c} \text {arctanh}\left (\frac {\sqrt {c+\sqrt {b+a x}}}{\sqrt {1+c}}\right )}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\frac {-4 \sqrt {c+\sqrt {b+a x}} \left (15+3 b-2 c^2+3 a x+5 \sqrt {b+a x}+c \left (5+\sqrt {b+a x}\right )\right )+60 \sqrt {-1-c} \arctan \left (\frac {\sqrt {c+\sqrt {b+a x}}}{\sqrt {-1-c}}\right )}{15 a} \]

[In]

Integrate[(Sqrt[b + a*x]*Sqrt[c + Sqrt[b + a*x]])/(1 - Sqrt[b + a*x]),x]

[Out]

(-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]]/Sqrt[-1 - c]])/(15*a)

Maple [A] (verified)

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	\(-\frac {2 \left (\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+2 \sqrt {c +\sqrt {a x +b}}-2 \sqrt {1+c}\, \operatorname {arctanh}\left (\frac {\sqrt {c +\sqrt {a x +b}}}{\sqrt {1+c}}\right )\right )}{a}\)	\(85\)

[In]

int((a*x+b)^(1/2)*(c+(a*x+b)^(1/2))^(1/2)/(1-(a*x+b)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

-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)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.68 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\left [\frac {2 \, {\left (2 \, {\left (2 \, c^{2} - 3 \, a x - \sqrt {a x + b} {\left (c + 5\right )} - 3 \, b - 5 \, c - 15\right )} \sqrt {c + \sqrt {a x + b}} + 15 \, \sqrt {c + 1} \log \left (\frac {a x + 2 \, {\left (\sqrt {a x + b} \sqrt {c + 1} + \sqrt {c + 1}\right )} \sqrt {c + \sqrt {a x + b}} + 2 \, \sqrt {a x + b} {\left (c + 1\right )} + b + 2 \, c + 1}{a x + b - 1}\right )\right )}}{15 \, a}, \frac {4 \, {\left ({\left (2 \, c^{2} - 3 \, a x - \sqrt {a x + b} {\left (c + 5\right )} - 3 \, b - 5 \, c - 15\right )} \sqrt {c + \sqrt {a x + b}} - 15 \, \sqrt {-c - 1} \arctan \left (\frac {\sqrt {c + \sqrt {a x + b}} \sqrt {-c - 1}}{c + 1}\right )\right )}}{15 \, a}\right ] \]

[In]

integrate((a*x+b)^(1/2)*(c+(a*x+b)^(1/2))^(1/2)/(1-(a*x+b)^(1/2)),x, algorithm="fricas")

[Out]

[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 + sqrt(a*x + b))*sqrt(-c - 1)/(c + 1)))/a]

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=\begin {cases} \frac {2 \cdot \left (\frac {2 \left (c - 1\right ) \left (c + \sqrt {a x + b}\right )^{\frac {3}{2}}}{3} - \frac {2 \left (c + \sqrt {a x + b}\right )^{\frac {5}{2}}}{5} - 2 \sqrt {c + \sqrt {a x + b}} - \frac {2 \left (c + 1\right ) \operatorname {atan}{\left (\frac {\sqrt {c + \sqrt {a x + b}}}{\sqrt {- c - 1}} \right )}}{\sqrt {- c - 1}}\right )}{a} & \text {for}\: a \neq 0 \\\frac {\sqrt {b} x \sqrt {\sqrt {b} + c}}{1 - \sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

integrate((a*x+b)**(1/2)*(c+(a*x+b)**(1/2))**(1/2)/(1-(a*x+b)**(1/2)),x)

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {2 \, {\left (6 \, {\left (c + \sqrt {a x + b}\right )}^{\frac {5}{2}} - 10 \, {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} {\left (c - 1\right )} + 15 \, \sqrt {c + 1} \log \left (\frac {\sqrt {c + \sqrt {a x + b}} - \sqrt {c + 1}}{\sqrt {c + \sqrt {a x + b}} + \sqrt {c + 1}}\right ) + 30 \, \sqrt {c + \sqrt {a x + b}}\right )}}{15 \, a} \]

[In]

integrate((a*x+b)^(1/2)*(c+(a*x+b)^(1/2))^(1/2)/(1-(a*x+b)^(1/2)),x, algorithm="maxima")

[Out]

-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 + sqrt(a*x + b)) + sqrt(c + 1))) + 30*sqrt(c + sqrt(a*x + b)))/a

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx=-\frac {4 \, {\left (c + 1\right )} \arctan \left (\frac {\sqrt {c + \sqrt {a x + b}}}{\sqrt {-c - 1}}\right )}{a \sqrt {-c - 1}} - \frac {4 \, {\left (3 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {5}{2}} - 5 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} c + 5 \, a^{4} {\left (c + \sqrt {a x + b}\right )}^{\frac {3}{2}} + 15 \, a^{4} \sqrt {c + \sqrt {a x + b}}\right )}}{15 \, a^{5}} \]

[In]

integrate((a*x+b)^(1/2)*(c+(a*x+b)^(1/2))^(1/2)/(1-(a*x+b)^(1/2)),x, algorithm="giac")

[Out]

-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

Mupad [F(-1)]

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

[In]

int(-((c + (b + a*x)^(1/2))^(1/2)*(b + a*x)^(1/2))/((b + a*x)^(1/2) - 1),x)

[Out]

-int(((c + (b + a*x)^(1/2))^(1/2)*(b + a*x)^(1/2))/((b + a*x)^(1/2) - 1), x)