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]

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

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.52, size = 50, normalized size = 0.89 \[ \frac {4 \, {\left (3 \, b^{2} d x + 3 \, b^{2} c + \sqrt {d x + c} a b - 2 \, a^{2}\right )} \sqrt {\sqrt {d x + c} b + a}}{15 \, b^{2} d} \]

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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giac [B]  time = 0.36, size = 99, normalized size = 1.77 \[ \frac {4 \, {\left (\frac {5 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {\sqrt {d x + c} b + a} a\right )} a}{b} + \frac {3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}} - 10 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {\sqrt {d x + c} b + a} a^{2}}{b}\right )}}{15 \, b d} \]

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*(5*((sqrt(d*x + c)*b + a)^(3/2) - 3*sqrt(sqrt(d*x + c)*b + a)*a)*a/b + (3*(sqrt(d*x + c)*b + a)^(5/2) - 1
0*(sqrt(d*x + c)*b + a)^(3/2)*a + 15*sqrt(sqrt(d*x + c)*b + a)*a^2)/b)/(b*d)

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maple [A]  time = 0.00, size = 41, normalized size = 0.73 \[ \frac {-\frac {4 \left (a +\sqrt {d x +c}\, b \right )^{\frac {3}{2}} a}{3}+\frac {4 \left (a +\sqrt {d x +c}\, b \right )^{\frac {5}{2}}}{5}}{b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.91, size = 43, normalized size = 0.77 \[ \frac {4 \, {\left (\frac {3 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} a}{b^{2}}\right )}}{15 \, d} \]

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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mupad [B]  time = 3.39, size = 44, normalized size = 0.79 \[ \frac {4\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{5/2}}{5\,b^2\,d}-\frac {4\,a\,{\left (a+b\,\sqrt {c+d\,x}\right )}^{3/2}}{3\,b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*(c + d*x)^(1/2))^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sqrt {c + d x}}\, dx \]

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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