\(\int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx\) [403]

Optimal result

Integrand size = 21, antiderivative size = 47 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 (a+b x)^{3/2}}{3 b (a-c)}-\frac {2 (c+b x)^{3/2}}{3 b (a-c)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6821} \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 (a+b x)^{3/2}}{3 b (a-c)}-\frac {2 (b x+c)^{3/2}}{3 b (a-c)} \]

[In]

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

[Out]

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

Rule 6821

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*} \text {integral}& = \frac {\int \left (\sqrt {a+b x}-\sqrt {c+b x}\right ) \, dx}{a-c} \\ & = \frac {2 (a+b x)^{3/2}}{3 b (a-c)}-\frac {2 (c+b x)^{3/2}}{3 b (a-c)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 \left ((a+b x)^{3/2}-(c+b x)^{3/2}\right )}{3 b (a-c)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {2 \left (b x +a \right )^{\frac {3}{2}}}{3 b \left (a -c \right )}-\frac {2 \left (b x +c \right )^{\frac {3}{2}}}{3 b \left (a -c \right )}\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - {\left (b x + c\right )}^{\frac {3}{2}}\right )}}{3 \, {\left (a b - b c\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (32) = 64\).

Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.89 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\begin {cases} \frac {2 a}{3 b \sqrt {a + b x} + 3 b \sqrt {b x + c}} + \frac {4 b x}{3 b \sqrt {a + b x} + 3 b \sqrt {b x + c}} + \frac {2 c}{3 b \sqrt {a + b x} + 3 b \sqrt {b x + c}} + \frac {2 \sqrt {a + b x} \sqrt {b x + c}}{3 b \sqrt {a + b x} + 3 b \sqrt {b x + c}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a} + \sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a/(3*b*sqrt(a + b*x) + 3*b*sqrt(b*x + c)) + 4*b*x/(3*b*sqrt(a + b*x) + 3*b*sqrt(b*x + c)) + 2*c/(
3*b*sqrt(a + b*x) + 3*b*sqrt(b*x + c)) + 2*sqrt(a + b*x)*sqrt(b*x + c)/(3*b*sqrt(a + b*x) + 3*b*sqrt(b*x + c))
, Ne(b, 0)), (x/(sqrt(a) + sqrt(c)), True))

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=-\frac {2}{3} \, \sqrt {b x + c} {\left (\frac {{\left (b x + a\right )} b}{a b^{2} - b^{2} c} - \frac {a b - b c}{a b^{2} - b^{2} c}\right )} + \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}}}{3 \, {\left (a b - b c\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.85 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.68 \[ \int \frac {1}{\sqrt {a+b x}+\sqrt {c+b x}} \, dx=\frac {2\,x\,\sqrt {a+b\,x}}{3\,\left (a-c\right )}-\frac {2\,x\,\sqrt {c+b\,x}}{3\,\left (a-c\right )}+\frac {2\,a\,\sqrt {a+b\,x}}{3\,b\,\left (a-c\right )}-\frac {2\,c\,\sqrt {c+b\,x}}{3\,b\,\left (a-c\right )} \]

[In]

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

[Out]

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