3.408 \(\int \frac {1}{(\sqrt {a+b x}+\sqrt {c+b x})^2} \, dx\)

Optimal. Leaf size=63 \[ \frac {(a-c)^2}{8 b \left (\sqrt {a+b x}+\sqrt {b x+c}\right )^4}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {b x+c}}\right )}{2 b} \]

[Out]

1/2*arctanh((b*x+a)^(1/2)/(b*x+c)^(1/2))/b+1/8*(a-c)^2/b/((b*x+a)^(1/2)+(b*x+c)^(1/2))^4

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Rubi [A]  time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.81, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {6689, 50, 63, 217, 206} \[ \frac {b x^2}{(a-c)^2}-\frac {(a+b x)^{3/2} \sqrt {b x+c}}{b (a-c)^2}+\frac {\sqrt {a+b x} \sqrt {b x+c}}{2 b (a-c)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {b x+c}}\right )}{2 b}+\frac {x (a+c)}{(a-c)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x] + Sqrt[c + b*x])^(-2),x]

[Out]

((a + c)*x)/(a - c)^2 + (b*x^2)/(a - c)^2 + (Sqrt[a + b*x]*Sqrt[c + b*x])/(2*b*(a - c)) - ((a + b*x)^(3/2)*Sqr
t[c + b*x])/(b*(a - c)^2) + ArcTanh[Sqrt[a + b*x]/Sqrt[c + b*x]]/(2*b)

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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 6689

Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)^(n_.)])^(m_), x_Symbol] :> Dis
t[(a*e^2 - c*f^2)^m, Int[ExpandIntegrand[u/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /; FreeQ[{a,
b, c, d, e, f, n}, x] && ILtQ[m, 0] && EqQ[b*e^2 - d*f^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (\sqrt {a+b x}+\sqrt {c+b x}\right )^2} \, dx &=\frac {\int \left (a \left (1+\frac {c}{a}\right )+2 b x-2 \sqrt {a+b x} \sqrt {c+b x}\right ) \, dx}{(a-c)^2}\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}-\frac {2 \int \sqrt {a+b x} \sqrt {c+b x} \, dx}{(a-c)^2}\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}-\frac {(a+b x)^{3/2} \sqrt {c+b x}}{b (a-c)^2}+\frac {\int \frac {\sqrt {a+b x}}{\sqrt {c+b x}} \, dx}{2 (a-c)}\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}+\frac {\sqrt {a+b x} \sqrt {c+b x}}{2 b (a-c)}-\frac {(a+b x)^{3/2} \sqrt {c+b x}}{b (a-c)^2}+\frac {1}{4} \int \frac {1}{\sqrt {a+b x} \sqrt {c+b x}} \, dx\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}+\frac {\sqrt {a+b x} \sqrt {c+b x}}{2 b (a-c)}-\frac {(a+b x)^{3/2} \sqrt {c+b x}}{b (a-c)^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {-a+c+x^2}} \, dx,x,\sqrt {a+b x}\right )}{2 b}\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}+\frac {\sqrt {a+b x} \sqrt {c+b x}}{2 b (a-c)}-\frac {(a+b x)^{3/2} \sqrt {c+b x}}{b (a-c)^2}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{2 b}\\ &=\frac {(a+c) x}{(a-c)^2}+\frac {b x^2}{(a-c)^2}+\frac {\sqrt {a+b x} \sqrt {c+b x}}{2 b (a-c)}-\frac {(a+b x)^{3/2} \sqrt {c+b x}}{b (a-c)^2}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {c+b x}}\right )}{2 b}\\ \end {align*}

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Mathematica [B]  time = 0.56, size = 179, normalized size = 2.84 \[ \frac {2 b x \left (b x-\sqrt {a+b x} \sqrt {b x+c}\right )+a \left (2 b x-\sqrt {a+b x} \sqrt {b x+c}\right )+c \left (2 b x-\sqrt {a+b x} \sqrt {b x+c}\right )+\frac {\sqrt {b} (c-a)^3 \sqrt {\frac {b x+c}{c-a}} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {a+b x}}{\sqrt {b (c-a)}}\right )}{\sqrt {b (c-a)} \sqrt {b x+c}}+2 c^2}{2 b (a-c)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x] + Sqrt[c + b*x])^(-2),x]

[Out]

(2*c^2 + 2*b*x*(b*x - Sqrt[a + b*x]*Sqrt[c + b*x]) + a*(2*b*x - Sqrt[a + b*x]*Sqrt[c + b*x]) + c*(2*b*x - Sqrt
[a + b*x]*Sqrt[c + b*x]) + (Sqrt[b]*(-a + c)^3*Sqrt[(c + b*x)/(-a + c)]*ArcSinh[(Sqrt[b]*Sqrt[a + b*x])/Sqrt[b
*(-a + c)]])/(Sqrt[b*(-a + c)]*Sqrt[c + b*x]))/(2*b*(a - c)^2)

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fricas [B]  time = 0.46, size = 103, normalized size = 1.63 \[ \frac {4 \, b^{2} x^{2} - 2 \, {\left (2 \, b x + a + c\right )} \sqrt {b x + a} \sqrt {b x + c} + 4 \, {\left (a b + b c\right )} x - {\left (a^{2} - 2 \, a c + c^{2}\right )} \log \left (-2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b x + c} - a - c\right )}{4 \, {\left (a^{2} b - 2 \, a b c + b c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b^2*x^2 - 2*(2*b*x + a + c)*sqrt(b*x + a)*sqrt(b*x + c) + 4*(a*b + b*c)*x - (a^2 - 2*a*c + c^2)*log(-2*
b*x + 2*sqrt(b*x + a)*sqrt(b*x + c) - a - c))/(a^2*b - 2*a*b*c + b*c^2)

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giac [B]  time = 0.24, size = 189, normalized size = 3.00 \[ -\frac {1}{2} \, \sqrt {b x + a} \sqrt {b x + c} {\left (\frac {2 \, {\left (a b - b c\right )} {\left (b x + a\right )}}{a^{3} b^{2} - 3 \, a^{2} b^{2} c + 3 \, a b^{2} c^{2} - b^{2} c^{3}} - \frac {a^{2} b - 2 \, a b c + b c^{2}}{a^{3} b^{2} - 3 \, a^{2} b^{2} c + 3 \, a b^{2} c^{2} - b^{2} c^{3}}\right )} + \frac {{\left (b x + a\right )}^{2} - {\left (b x + a\right )} a + {\left (b x + a\right )} c}{a^{2} b - 2 \, a b c + b c^{2}} - \frac {\log \left ({\left | -\sqrt {b x + a} + \sqrt {b x + c} \right |}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(b*x + a)*sqrt(b*x + c)*(2*(a*b - b*c)*(b*x + a)/(a^3*b^2 - 3*a^2*b^2*c + 3*a*b^2*c^2 - b^2*c^3) - (a
^2*b - 2*a*b*c + b*c^2)/(a^3*b^2 - 3*a^2*b^2*c + 3*a*b^2*c^2 - b^2*c^3)) + ((b*x + a)^2 - (b*x + a)*a + (b*x +
a)*c)/(a^2*b - 2*a*b*c + b*c^2) - 1/2*log(abs(-sqrt(b*x + a) + sqrt(b*x + c)))/b

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maple [B]  time = 0.01, size = 377, normalized size = 5.98 \[ \frac {\sqrt {\left (b x +a \right ) \left (b x +c \right )}\, a^{2} \ln \left (\frac {b^{2} x +\frac {1}{2} a b +\frac {1}{2} b c}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+a c +\left (a b +b c \right ) x}\right )}{4 \left (a -c \right )^{2} \sqrt {b x +c}\, \sqrt {b x +a}\, \sqrt {b^{2}}}-\frac {\sqrt {\left (b x +a \right ) \left (b x +c \right )}\, a c \ln \left (\frac {b^{2} x +\frac {1}{2} a b +\frac {1}{2} b c}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+a c +\left (a b +b c \right ) x}\right )}{2 \left (a -c \right )^{2} \sqrt {b x +c}\, \sqrt {b x +a}\, \sqrt {b^{2}}}+\frac {b \,x^{2}}{\left (a -c \right )^{2}}+\frac {\sqrt {\left (b x +a \right ) \left (b x +c \right )}\, c^{2} \ln \left (\frac {b^{2} x +\frac {1}{2} a b +\frac {1}{2} b c}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+a c +\left (a b +b c \right ) x}\right )}{4 \left (a -c \right )^{2} \sqrt {b x +c}\, \sqrt {b x +a}\, \sqrt {b^{2}}}+\frac {a x}{\left (a -c \right )^{2}}+\frac {c x}{\left (a -c \right )^{2}}-\frac {\sqrt {b x +c}\, \sqrt {b x +a}\, a}{2 \left (a -c \right )^{2} b}+\frac {\sqrt {b x +c}\, \sqrt {b x +a}\, c}{2 \left (a -c \right )^{2} b}-\frac {\sqrt {b x +a}\, \left (b x +c \right )^{\frac {3}{2}}}{\left (a -c \right )^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(a-c)^2*a+x/(a-c)^2*c+b*x^2/(a-c)^2-1/(a-c)^2/b*(b*x+a)^(1/2)*(b*x+c)^(3/2)-1/2/(a-c)^2/b*(b*x+c)^(1/2)*(b*x
+a)^(1/2)*a+1/2/(a-c)^2/b*(b*x+c)^(1/2)*(b*x+a)^(1/2)*c+1/4/(a-c)^2*((b*x+a)*(b*x+c))^(1/2)/(b*x+c)^(1/2)/(b*x
+a)^(1/2)*ln((1/2*a*b+1/2*b*c+x*b^2)/(b^2)^(1/2)+(b^2*x^2+(a*b+b*c)*x+a*c)^(1/2))/(b^2)^(1/2)*a^2-1/2/(a-c)^2*
((b*x+a)*(b*x+c))^(1/2)/(b*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*b+1/2*b*c+x*b^2)/(b^2)^(1/2)+(b^2*x^2+(a*b+b*c)*
x+a*c)^(1/2))/(b^2)^(1/2)*a*c+1/4/(a-c)^2*((b*x+a)*(b*x+c))^(1/2)/(b*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*b+1/2*
b*c+x*b^2)/(b^2)^(1/2)+(b^2*x^2+(a*b+b*c)*x+a*c)^(1/2))/(b^2)^(1/2)*c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (\sqrt {b x + a} + \sqrt {b x + c}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((sqrt(b*x + a) + sqrt(b*x + c))^(-2), x)

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mupad [B]  time = 0.24, size = 110, normalized size = 1.75 \[ \frac {b\,x^2}{{\left (a-c\right )}^2}+\frac {x\,\left (a+c\right )}{{\left (a-c\right )}^2}+\frac {\ln \left (a+c+2\,\sqrt {a+b\,x}\,\sqrt {c+b\,x}+2\,b\,x\right )\,{\left (a\,b-b\,c\right )}^2}{4\,b^3\,{\left (a-c\right )}^2}-\frac {2\,\sqrt {a+b\,x}\,\sqrt {c+b\,x}\,\left (\frac {x}{2}+\frac {a\,b+b\,c}{4\,b^2}\right )}{{\left (a-c\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^2)/(a - c)^2 + (x*(a + c))/(a - c)^2 + (log(a + c + 2*(a + b*x)^(1/2)*(c + b*x)^(1/2) + 2*b*x)*(a*b - b*c
)^2)/(4*b^3*(a - c)^2) - (2*(a + b*x)^(1/2)*(c + b*x)^(1/2)*(x/2 + (a*b + b*c)/(4*b^2)))/(a - c)^2

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sympy [A]  time = 1.04, size = 388, normalized size = 6.16 \[ \begin {cases} \frac {2 a \log {\left (\sqrt {a + b x} + \sqrt {b x + c} \right )}}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {a}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {4 b x \log {\left (\sqrt {a + b x} + \sqrt {b x + c} \right )}}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {2 b x}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {2 c \log {\left (\sqrt {a + b x} + \sqrt {b x + c} \right )}}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {c}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} + \frac {4 \sqrt {a + b x} \sqrt {b x + c} \log {\left (\sqrt {a + b x} + \sqrt {b x + c} \right )}}{4 a b + 8 b^{2} x + 4 b c + 8 b \sqrt {a + b x} \sqrt {b x + c}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\sqrt {a} + \sqrt {c}\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*log(sqrt(a + b*x) + sqrt(b*x + c))/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt(b*x + c))
+ a/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt(b*x + c)) + 4*b*x*log(sqrt(a + b*x) + sqrt(b*x + c))/(
4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt(b*x + c)) + 2*b*x/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b
*x)*sqrt(b*x + c)) + 2*c*log(sqrt(a + b*x) + sqrt(b*x + c))/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt
(b*x + c)) + c/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt(b*x + c)) + 4*sqrt(a + b*x)*sqrt(b*x + c)*lo
g(sqrt(a + b*x) + sqrt(b*x + c))/(4*a*b + 8*b**2*x + 4*b*c + 8*b*sqrt(a + b*x)*sqrt(b*x + c)), Ne(b, 0)), (x/(
sqrt(a) + sqrt(c))**2, True))

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