\(\int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx\) [465]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {a \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} x}{a+b \sqrt [3]{x}}+\frac {3 b \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} x^{4/3}}{4 \left (a+b \sqrt [3]{x}\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 660, 45} \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {3 b x^{4/3} \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 \left (a+b \sqrt [3]{x}\right )}+\frac {a x \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{a+b \sqrt [3]{x}} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)],x]

[Out]

(a*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]*x)/(a + b*x^(1/3)) + (3*b*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]*x
^(4/3))/(4*(a + b*x^(1/3)))

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 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1355

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.49 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {\sqrt {\left (a+b \sqrt [3]{x}\right )^2} \left (4 a x+3 b x^{4/3}\right )}{4 \left (a+b \sqrt [3]{x}\right )} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)],x]

[Out]

(Sqrt[(a + b*x^(1/3))^2]*(4*a*x + 3*b*x^(4/3)))/(4*(a + b*x^(1/3)))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48

method	result	size
derivativedivides	\(\frac {\operatorname {csgn}\left (a +b \,x^{\frac {1}{3}}\right ) \left (a +b \,x^{\frac {1}{3}}\right )^{2} \left (3 b^{2} x^{\frac {2}{3}}-2 a b \,x^{\frac {1}{3}}+a^{2}\right )}{4 b^{3}}\)	\(42\)
default	\(\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (3 b \,x^{\frac {4}{3}}+4 a x \right )}{4 a +4 b \,x^{\frac {1}{3}}}\)	\(43\)

[In]

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.11 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {3}{4} \, b x^{\frac {4}{3}} + a x \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2),x, algorithm="fricas")

[Out]

3/4*b*x^(4/3) + a*x

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.66 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=3 \left (\begin {cases} \sqrt {a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} \sqrt [3]{x}}{12 b^{2}} + \frac {a x^{\frac {2}{3}}}{12 b} + \frac {x}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**(1/2),x)

[Out]

3*Piecewise((sqrt(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))*(a**3/(12*b**3) - a**2*x**(1/3)/(12*b**2) + a*x**(2/3
)/(12*b) + x/4), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x**(1/3))**(3/2)/3 - 2*a**2*(a**2 + 2*a*b*x**(1/3))**(5/2)
/5 + (a**2 + 2*a*b*x**(1/3))**(7/2)/7)/(4*a**3*b**3), Ne(a*b, 0)), (x*sqrt(a**2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.30 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {3 \, \sqrt {b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}} a^{2} x^{\frac {1}{3}}}{2 \, b^{2}} + \frac {3 \, \sqrt {b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}} a^{3}}{2 \, b^{3}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} x^{\frac {1}{3}}}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} a}{4 \, b^{3}} \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.30 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {3}{4} \, b x^{\frac {4}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + a x \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2),x, algorithm="giac")

[Out]

3/4*b*x^(4/3)*sgn(b*x^(1/3) + a) + a*x*sgn(b*x^(1/3) + a)

Mupad [B] (verification not implemented)

Time = 8.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \, dx=\frac {\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}\,\left (a^3-4\,a^2\,b\,x^{1/3}-5\,a\,b^2\,x^{2/3}+3\,b\,x^{1/3}\,\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )\right )}{4\,b^3} \]

[In]

int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(1/2),x)

[Out]

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