3.628 \(\int \sqrt{a+b \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=56 \[ \frac{4 \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^2 d}-\frac{4 a \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^2 d} \]

[Out]

(-4*a*(a + b*Sqrt[c + d*x])^(3/2))/(3*b^2*d) + (4*(a + b*Sqrt[c + d*x])^(5/2))/(5*b^2*d)

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Rubi [A]  time = 0.0327106, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {247, 190, 43} \[ \frac{4 \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^2 d}-\frac{4 a \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c + d*x]],x]

[Out]

(-4*a*(a + b*Sqrt[c + d*x])^(3/2))/(3*b^2*d) + (4*(a + b*Sqrt[c + d*x])^(5/2))/(5*b^2*d)

Rule 247

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]

Rule 190

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 43

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])

Rubi steps

\begin{align*} \int \sqrt{a+b \sqrt{c+d x}} \, dx &=\frac{\operatorname{Subst}\left (\int \sqrt{a+b \sqrt{x}} \, dx,x,c+d x\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int x \sqrt{a+b x} \, dx,x,\sqrt{c+d x}\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (-\frac{a \sqrt{a+b x}}{b}+\frac{(a+b x)^{3/2}}{b}\right ) \, dx,x,\sqrt{c+d x}\right )}{d}\\ &=-\frac{4 a \left (a+b \sqrt{c+d x}\right )^{3/2}}{3 b^2 d}+\frac{4 \left (a+b \sqrt{c+d x}\right )^{5/2}}{5 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0241736, size = 43, normalized size = 0.77 \[ \frac{4 \left (a+b \sqrt{c+d x}\right )^{3/2} \left (3 b \sqrt{c+d x}-2 a\right )}{15 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a + b*Sqrt[c + d*x])^(3/2)*(-2*a + 3*b*Sqrt[c + d*x]))/(15*b^2*d)

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Maple [A]  time = 0.004, size = 41, normalized size = 0.7 \begin{align*} 4\,{\frac{1/5\, \left ( a+b\sqrt{dx+c} \right ) ^{5/2}-1/3\,a \left ( a+b\sqrt{dx+c} \right ) ^{3/2}}{{b}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/d/b^2*(1/5*(a+b*(d*x+c)^(1/2))^(5/2)-1/3*a*(a+b*(d*x+c)^(1/2))^(3/2))

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Maxima [A]  time = 1.10962, size = 58, normalized size = 1.04 \begin{align*} \frac{4 \,{\left (\frac{3 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{5}{2}}}{b^{2}} - \frac{5 \,{\left (\sqrt{d x + c} b + a\right )}^{\frac{3}{2}} a}{b^{2}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(3*(sqrt(d*x + c)*b + a)^(5/2)/b^2 - 5*(sqrt(d*x + c)*b + a)^(3/2)*a/b^2)/d

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Fricas [A]  time = 2.38948, size = 122, normalized size = 2.18 \begin{align*} \frac{4 \,{\left (3 \, b^{2} d x + 3 \, b^{2} c + \sqrt{d x + c} a b - 2 \, a^{2}\right )} \sqrt{\sqrt{d x + c} b + a}}{15 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(3*b^2*d*x + 3*b^2*c + sqrt(d*x + c)*a*b - 2*a^2)*sqrt(sqrt(d*x + c)*b + a)/(b^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sqrt{c + d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c + d*x)), x)

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Giac [B]  time = 1.11971, size = 159, normalized size = 2.84 \begin{align*} \frac{4 \,{\left (3 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )}^{2} b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right ) - 5 \, \sqrt{{\left (\sqrt{d x + c} b + a\right )} b^{2}}{\left (\sqrt{d x + c} b + a\right )} a b^{2} \mathrm{sgn}\left ({\left (\sqrt{d x + c} b + a\right )} b - a b\right )\right )}{\left | b \right |}}{15 \, b^{6} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(3*sqrt((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)^2*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b) - 5*sqr
t((sqrt(d*x + c)*b + a)*b^2)*(sqrt(d*x + c)*b + a)*a*b^2*sgn((sqrt(d*x + c)*b + a)*b - a*b))*abs(b)/(b^6*d)