\(\int \frac {1}{\sqrt {c+d (a+b x)}} \, dx\) [40]

Optimal result

Integrand size = 13, antiderivative size = 21 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \sqrt {c+d (a+b x)}}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {33, 32} \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \sqrt {d (a+b x)+c}}{b d} \]

[In]

Int[1/Sqrt[c + d*(a + b*x)],x]

[Out]

(2*Sqrt[c + d*(a + b*x)])/(b*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 33

Int[((a_.) + (b_.)*(u_))^(m_), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(a + b*x)^m, x], x, u], x]
/; FreeQ[{a, b, m}, x] && LinearQ[u, x] && NeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {c+d x}} \, dx,x,a+b x\right )}{b} \\ & = \frac {2 \sqrt {c+d (a+b x)}}{b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \sqrt {c+a d+b d x}}{b d} \]

[In]

Integrate[1/Sqrt[c + d*(a + b*x)],x]

[Out]

(2*Sqrt[c + a*d + b*d*x])/(b*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {2 \sqrt {b d x +a d +c}}{d b}\)	\(20\)
derivativedivides	\(\frac {2 \sqrt {b d x +a d +c}}{d b}\)	\(20\)
default	\(\frac {2 \sqrt {b d x +a d +c}}{d b}\)	\(20\)
trager	\(\frac {2 \sqrt {b d x +a d +c}}{d b}\)	\(20\)
pseudoelliptic	\(\frac {2 \sqrt {c +d \left (b x +a \right )}}{b d}\)	\(20\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \, \sqrt {b d x + a d + c}}{b d} \]

[In]

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

[Out]

2*sqrt(b*d*x + a*d + c)/(b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (15) = 30\).

Time = 0.49 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\begin {cases} \frac {x}{\sqrt {a d + c}} & \text {for}\: b = 0 \\\frac {x}{\sqrt {c}} & \text {for}\: d = 0 \\\frac {2 \sqrt {c + d \left (a + b x\right )}}{b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x/sqrt(a*d + c), Eq(b, 0)), (x/sqrt(c), Eq(d, 0)), (2*sqrt(c + d*(a + b*x))/(b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \, \sqrt {{\left (b x + a\right )} d + c}}{b d} \]

[In]

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

[Out]

2*sqrt((b*x + a)*d + c)/(b*d)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2 \, \sqrt {b d x + a d + c}}{b d} \]

[In]

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

[Out]

2*sqrt(b*d*x + a*d + c)/(b*d)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {c+d (a+b x)}} \, dx=\frac {2\,\sqrt {c+d\,\left (a+b\,x\right )}}{b\,d} \]

[In]

int(1/(c + d*(a + b*x))^(1/2),x)

[Out]

(2*(c + d*(a + b*x))^(1/2))/(b*d)