3.5.64 \(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{3/2} \, dx\) [464]

Optimal. Leaf size=137 \[ \frac {3 a^2 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 b^3}-\frac {6 a \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{5 b^3}+\frac {\left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{2 b^3} \]

[Out]

3/4*a^2*(a+b*x^(1/3))^3*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3-6/5*a*(a+b*x^(1/3))^4*(a^2+2*a*b*x^(1/3)+b^2
*x^(2/3))^(1/2)/b^3+1/2*(a+b*x^(1/3))^5*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3

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Rubi [A]
time = 0.04, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 659} \begin {gather*} \frac {\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^5}{2 b^3}-\frac {6 a \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^4}{5 b^3}+\frac {3 a^2 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^3}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*(a + b*x^(1/3))^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(4*b^3) - (6*a*(a + b*x^(1/3))^4*Sqrt[a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3)])/(5*b^3) + ((a + b*x^(1/3))^5*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(2*b^3)

Rule 659

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 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*} \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{3/2} \, dx &=3 \text {Subst}\left (\int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, 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 \left (\frac {a^2 \left (a b+b^2 x\right )^3}{b^2}-\frac {2 a \left (a b+b^2 x\right )^4}{b^3}+\frac {\left (a b+b^2 x\right )^5}{b^4}\right ) \, dx,x,\sqrt [3]{x}\right )}{b^3 \left (a+b \sqrt [3]{x}\right )}\\ &=\frac {3 a^2 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{4 b^3}-\frac {6 a \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{5 b^3}+\frac {\left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 67, normalized size = 0.49 \begin {gather*} \frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^{3/2} \left (20 a^3 x+45 a^2 b x^{4/3}+36 a b^2 x^{5/3}+10 b^3 x^2\right )}{20 \left (a+b \sqrt [3]{x}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((a + b*x^(1/3))^2)^(3/2)*(20*a^3*x + 45*a^2*b*x^(4/3) + 36*a*b^2*x^(5/3) + 10*b^3*x^2))/(20*(a + b*x^(1/3))^
3)

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Maple [A]
time = 0.05, size = 65, normalized size = 0.47

method	result	size
derivativedivides	\(\frac {\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )^{\frac {3}{2}} x \left (10 b^{3} x +36 a \,b^{2} x^{\frac {2}{3}}+45 a^{2} b \,x^{\frac {1}{3}}+20 a^{3}\right )}{20 \left (a +b \,x^{\frac {1}{3}}\right )^{3}}\)	\(54\)
default	\(\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (36 a \,b^{2} x^{\frac {5}{3}}+45 a^{2} b \,x^{\frac {4}{3}}+10 b^{3} x^{2}+20 a^{3} x \right )}{20 a +20 b \,x^{\frac {1}{3}}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(36*a*b^2*x^(5/3)+45*a^2*b*x^(4/3)+10*b^3*x^2+20*a^3*x)/(a+b*x^(1/3
))

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Maxima [A]
time = 0.30, size = 114, normalized size = 0.83 \begin {gather*} \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{\frac {1}{3}}}{4 \, b^{2}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {3}{2}} a^{3}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} x^{\frac {1}{3}}}{2 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} a}{10 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(3/2)*a^2*x^(1/3)/b^2 + 3/4*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(3/2)*
a^3/b^3 + 1/2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(5/2)*x^(1/3)/b^2 - 7/10*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)
^(5/2)*a/b^3

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Fricas [A]
time = 0.34, size = 32, normalized size = 0.23 \begin {gather*} \frac {1}{2} \, b^{3} x^{2} + \frac {9}{5} \, a b^{2} x^{\frac {5}{3}} + \frac {9}{4} \, a^{2} b x^{\frac {4}{3}} + a^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x^2 + 9/5*a*b^2*x^(5/3) + 9/4*a^2*b*x^(4/3) + a^3*x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.75, size = 64, normalized size = 0.47 \begin {gather*} \frac {1}{2} \, b^{3} x^{2} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {9}{5} \, a b^{2} x^{\frac {5}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {9}{4} \, a^{2} b x^{\frac {4}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + a^{3} x \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x^2*sgn(b*x^(1/3) + a) + 9/5*a*b^2*x^(5/3)*sgn(b*x^(1/3) + a) + 9/4*a^2*b*x^(4/3)*sgn(b*x^(1/3) + a) +
a^3*x*sgn(b*x^(1/3) + a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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