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

Optimal. Leaf size=84 \[ \frac {4 \left (3 a c^2 x+3 b c^2-2 d^2\right ) \sqrt {c \sqrt {a x+b}+d}}{15 a c^2}+\frac {4 d \sqrt {a x+b} \sqrt {c \sqrt {a x+b}+d}}{15 a c} \]

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 0.67, 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} \begin {gather*} \frac {4 \left (c \sqrt {a x+b}+d\right )^{5/2}}{5 a c^2}-\frac {4 d \left (c \sqrt {a x+b}+d\right )^{3/2}}{3 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d + c*Sqrt[b + a*x]],x]

[Out]

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

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

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.51 \begin {gather*} \frac {4 \left (c \sqrt {a x+b}+d\right )^{3/2} \left (3 c \sqrt {a x+b}-2 d\right )}{15 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 55, normalized size = 0.65 \begin {gather*} \frac {4 \sqrt {d+c \sqrt {b+a x}} \left (-2 d^2+c d \sqrt {b+a x}+3 c^2 (b+a x)\right )}{15 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 50, normalized size = 0.60 \begin {gather*} \frac {4 \, {\left (3 \, a c^{2} x + 3 \, b c^{2} + \sqrt {a x + b} c d - 2 \, d^{2}\right )} \sqrt {\sqrt {a x + b} c + d}}{15 \, a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 99, normalized size = 1.18 \begin {gather*} \frac {4 \, {\left (\frac {5 \, {\left ({\left (\sqrt {a x + b} c + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {\sqrt {a x + b} c + d} d\right )} d}{c} + \frac {3 \, {\left (\sqrt {a x + b} c + d\right )}^{\frac {5}{2}} - 10 \, {\left (\sqrt {a x + b} c + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {\sqrt {a x + b} c + d} d^{2}}{c}\right )}}{15 \, a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(5*((sqrt(a*x + b)*c + d)^(3/2) - 3*sqrt(sqrt(a*x + b)*c + d)*d)*d/c + (3*(sqrt(a*x + b)*c + d)^(5/2) - 1
0*(sqrt(a*x + b)*c + d)^(3/2)*d + 15*sqrt(sqrt(a*x + b)*c + d)*d^2)/c)/(a*c)

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maple [A]  time = 0.06, size = 41, normalized size = 0.49

method	result	size
derivativedivides	\(\frac {\frac {4 \left (d +c \sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {4 d \left (d +c \sqrt {a x +b}\right )^{\frac {3}{2}}}{3}}{a \,c^{2}}\)	\(41\)
default	\(\frac {\frac {4 \left (d +c \sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {4 d \left (d +c \sqrt {a x +b}\right )^{\frac {3}{2}}}{3}}{a \,c^{2}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 43, normalized size = 0.51 \begin {gather*} \frac {4 \, {\left (\frac {3 \, {\left (\sqrt {a x + b} c + d\right )}^{\frac {5}{2}}}{c^{2}} - \frac {5 \, {\left (\sqrt {a x + b} c + d\right )}^{\frac {3}{2}} d}{c^{2}}\right )}}{15 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.86, size = 44, normalized size = 0.52 \begin {gather*} \frac {4\,{\left (d+c\,\sqrt {b+a\,x}\right )}^{5/2}}{5\,a\,c^2}-\frac {4\,d\,{\left (d+c\,\sqrt {b+a\,x}\right )}^{3/2}}{3\,a\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + c*(b + a*x)^(1/2))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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