\(\int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx\) [649]

Optimal result

Integrand size = 17, antiderivative size = 54 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=-\frac {4 a \sqrt {a+b \sqrt {c+d x}}}{b^2 d}+\frac {4 \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {253, 196, 45} \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4 \left (a+b \sqrt {c+d x}\right )^{3/2}}{3 b^2 d}-\frac {4 a \sqrt {a+b \sqrt {c+d x}}}{b^2 d} \]

[In]

Int[1/Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4 \left (-2 a+b \sqrt {c+d x}\right ) \sqrt {a+b \sqrt {c+d x}}}{3 b^2 d} \]

[In]

Integrate[1/Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {4 \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{3}-4 a \sqrt {a +b \sqrt {d x +c}}}{b^{2} d}\)	\(41\)
default	\(\frac {\frac {4 \left (a +b \sqrt {d x +c}\right )^{\frac {3}{2}}}{3}-4 a \sqrt {a +b \sqrt {d x +c}}}{b^{2} d}\)	\(41\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4 \, \sqrt {\sqrt {d x + c} b + a} {\left (\sqrt {d x + c} b - 2 \, a\right )}}{3 \, b^{2} d} \]

[In]

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

[Out]

4/3*sqrt(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b - 2*a)/(b^2*d)

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\begin {cases} \frac {2 \left (\begin {cases} \frac {2 \left (- a \sqrt {a + b \sqrt {c + d x}} + \frac {\left (a + b \sqrt {c + d x}\right )^{\frac {3}{2}}}{3}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {c + d x}{2 \sqrt {a}} & \text {otherwise} \end {cases}\right )}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {a + b \sqrt {c}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4 \, {\left (\frac {{\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}}}{b^{2}} - \frac {3 \, \sqrt {\sqrt {d x + c} b + a} a}{b^{2}}\right )}}{3 \, d} \]

[In]

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

[Out]

4/3*((sqrt(d*x + c)*b + a)^(3/2)/b^2 - 3*sqrt(sqrt(d*x + c)*b + a)*a/b^2)/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {\sqrt {d x + c} b + a} a\right )}}{3 \, b^{2} d} \]

[In]

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

[Out]

4/3*((sqrt(d*x + c)*b + a)^(3/2) - 3*sqrt(sqrt(d*x + c)*b + a)*a)/(b^2*d)

Mupad [B] (verification not implemented)

Time = 17.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {a+b \sqrt {c+d x}}} \, dx=\frac {4\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{3/2}}{3\,b^2\,d}-\frac {4\,a\,\sqrt {a+b\,\sqrt {c+d\,x}}}{b^2\,d} \]

[In]

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

[Out]

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