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

Optimal. Leaf size=119 \[ -\frac {4 \left (3 a x+3 b-2 c^2+5 c+15\right ) \sqrt {\sqrt {a x+b}+c}}{15 a}-\frac {4 (c+5) \sqrt {a x+b} \sqrt {\sqrt {a x+b}+c}}{15 a}-\frac {4 \sqrt {-c-1} \tan ^{-1}\left (\frac {\sqrt {-c-1} \sqrt {\sqrt {a x+b}+c}}{c+1}\right )}{a} \]

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Rubi [A]  time = 0.12, antiderivative size = 106, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {513, 446, 88, 50, 63, 206} \begin {gather*} -\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}+\frac {4 \sqrt {c+1} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a x+b}+c}}{\sqrt {c+1}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[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 50

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 63

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 88

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 446

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 513

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*} \int \frac {\sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{1-\sqrt {b+a x}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {c+\sqrt {x}} \sqrt {x}}{1-\sqrt {x}} \, dx,x,b+a x\right )}{a}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x^2 \sqrt {c+x}}{1-x} \, dx,x,\sqrt {b+a x}\right )}{a}\\ &=\frac {2 \operatorname {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 \operatorname {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)) \operatorname {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)) \operatorname {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} \tanh ^{-1}\left (\frac {\sqrt {c+\sqrt {b+a x}}}{\sqrt {1+c}}\right )}{a}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 96, normalized size = 0.81 \begin {gather*} \frac {60 \sqrt {c+1} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a x+b}+c}}{\sqrt {c+1}}\right )-4 \sqrt {\sqrt {a x+b}+c} \left (c \left (\sqrt {a x+b}+5\right )+5 \sqrt {a x+b}+3 a x+3 b-2 c^2+15\right )}{15 a} \end {gather*}

Antiderivative was successfully verified.

[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]*ArcTanh[Sqrt[c + Sqrt[b + a*x]]/Sqrt[1 + c]])/(15*a)

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IntegrateAlgebraic [A]  time = 0.16, size = 108, normalized size = 0.91 \begin {gather*} \frac {4 \sqrt {c+\sqrt {b+a x}} \left (-15-5 c+2 c^2-5 \sqrt {b+a x}-c \sqrt {b+a x}-3 (b+a x)\right )}{15 a}-\frac {4 \sqrt {-1-c} \tan ^{-1}\left (\frac {\sqrt {-1-c} \sqrt {c+\sqrt {b+a x}}}{1+c}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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 - 5*c + 2*c^2 - 5*Sqrt[b + a*x] - c*Sqrt[b + a*x] - 3*(b + a*x)))/(15*a) - (4*
Sqrt[-1 - c]*ArcTan[(Sqrt[-1 - c]*Sqrt[c + Sqrt[b + a*x]])/(1 + c)])/a

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fricas [A]  time = 0.46, size = 200, normalized size = 1.68 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.15, size = 107, normalized size = 0.90 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.08, size = 85, normalized size = 0.71

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+2 \sqrt {c +\sqrt {a x +b}}-2 \sqrt {1+c}\, \arctanh \left (\frac {\sqrt {c +\sqrt {a x +b}}}{\sqrt {1+c}}\right )\right )}{a}\)	\(85\)
default	\(-\frac {2 \left (\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\frac {2 \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+2 \sqrt {c +\sqrt {a x +b}}-2 \sqrt {1+c}\, \arctanh \left (\frac {\sqrt {c +\sqrt {a x +b}}}{\sqrt {1+c}}\right )\right )}{a}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maxima [A]  time = 1.80, size = 95, normalized size = 0.80 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {\sqrt {c+\sqrt {b+a\,x}}\,\sqrt {b+a\,x}}{\sqrt {b+a\,x}-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 5.58, size = 90, normalized size = 0.76 \begin {gather*} \frac {2 \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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