\(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{5/2} \, dx\) [463]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\frac {a^2 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{2 b^3}-\frac {6 a \left (a+b \sqrt [3]{x}\right )^6 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{7 b^3}+\frac {3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{8 b^3} \]

[Out]

1/2*a^2*(a+b*x^(1/3))^5*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3-6/7*a*(a+b*x^(1/3))^6*(a^2+2*a*b*x^(1/3)+b^2
*x^(2/3))^(1/2)/b^3+3/8*(a+b*x^(1/3))^7*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 137, 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} \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\frac {3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^7}{8 b^3}-\frac {6 a \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^6}{7 b^3}+\frac {a^2 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^5}{2 b^3} \]

[In]

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(5/2),x]

[Out]

(a^2*(a + b*x^(1/3))^5*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(2*b^3) - (6*a*(a + b*x^(1/3))^6*Sqrt[a^2 + 2*
a*b*x^(1/3) + b^2*x^(2/3)])/(7*b^3) + (3*(a + b*x^(1/3))^7*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(8*b^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 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, 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 )^5 \, dx,x,\sqrt [3]{x}\right )}{b^5 \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 (\frac {a^2 \left (a b+b^2 x\right )^5}{b^2}-\frac {2 a \left (a b+b^2 x\right )^6}{b^3}+\frac {\left (a b+b^2 x\right )^7}{b^4}\right ) \, dx,x,\sqrt [3]{x}\right )}{b^5 \left (a+b \sqrt [3]{x}\right )} \\ & = \frac {a^2 \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}-\frac {6 a \left (a+b \sqrt [3]{x}\right )^6 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{7 b^3}+\frac {3 \left (a+b \sqrt [3]{x}\right )^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{8 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.68 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^{5/2} \left (56 a^5 x+210 a^4 b x^{4/3}+336 a^3 b^2 x^{5/3}+280 a^2 b^3 x^2+120 a b^4 x^{7/3}+21 b^5 x^{8/3}\right )}{56 \left (a+b \sqrt [3]{x}\right )^5} \]

[In]

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(5/2),x]

[Out]

(((a + b*x^(1/3))^2)^(5/2)*(56*a^5*x + 210*a^4*b*x^(4/3) + 336*a^3*b^2*x^(5/3) + 280*a^2*b^3*x^2 + 120*a*b^4*x
^(7/3) + 21*b^5*x^(8/3)))/(56*(a + b*x^(1/3))^5)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.55

method	result	size
derivativedivides	\(\frac {{\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )}^{\frac {5}{2}} x \left (21 b^{5} x^{\frac {5}{3}}+120 b^{4} a \,x^{\frac {4}{3}}+280 a^{2} b^{3} x +336 a^{3} b^{2} x^{\frac {2}{3}}+210 b \,a^{4} x^{\frac {1}{3}}+56 a^{5}\right )}{56 \left (a +b \,x^{\frac {1}{3}}\right )^{5}}\)	\(76\)
default	\(\frac {\left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{\frac {5}{2}} \left (21 b^{5} x^{\frac {8}{3}}+120 b^{4} a \,x^{\frac {7}{3}}+336 a^{3} b^{2} x^{\frac {5}{3}}+210 b \,a^{4} x^{\frac {4}{3}}+280 a^{2} b^{3} x^{2}+56 a^{5} x \right )}{56 \left (a +b \,x^{\frac {1}{3}}\right )^{5}}\)	\(87\)

[In]

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/56*((a+b*x^(1/3))^2)^(5/2)*x*(21*b^5*x^(5/3)+120*b^4*a*x^(4/3)+280*a^2*b^3*x+336*a^3*b^2*x^(2/3)+210*b*a^4*x
^(1/3)+56*a^5)/(a+b*x^(1/3))^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.45 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=5 \, a^{2} b^{3} x^{2} + a^{5} x + \frac {3}{8} \, {\left (b^{5} x^{2} + 16 \, a^{3} b^{2} x\right )} x^{\frac {2}{3}} + \frac {15}{28} \, {\left (4 \, a b^{4} x^{2} + 7 \, a^{4} b x\right )} x^{\frac {1}{3}} \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(5/2),x, algorithm="fricas")

[Out]

5*a^2*b^3*x^2 + a^5*x + 3/8*(b^5*x^2 + 16*a^3*b^2*x)*x^(2/3) + 15/28*(4*a*b^4*x^2 + 7*a^4*b*x)*x^(1/3)

Sympy [A] (verification not implemented)

Time = 1.43 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.49 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=3 \left (\begin {cases} \sqrt {a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}} \left (\frac {a^{7}}{168 b^{3}} - \frac {a^{6} \sqrt [3]{x}}{168 b^{2}} + \frac {a^{5} x^{\frac {2}{3}}}{168 b} + \frac {55 a^{4} x}{168} + \frac {155 a^{3} b x^{\frac {4}{3}}}{168} + \frac {181 a^{2} b^{2} x^{\frac {5}{3}}}{168} + \frac {33 a b^{3} x^{2}}{56} + \frac {b^{4} x^{\frac {7}{3}}}{8}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {7}{2}}}{7} - \frac {2 a^{2} \left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b \sqrt [3]{x}\right )^{\frac {11}{2}}}{11}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x \left (a^{2}\right )^{\frac {5}{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**(5/2),x)

[Out]

3*Piecewise((sqrt(a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))*(a**7/(168*b**3) - a**6*x**(1/3)/(168*b**2) + a**5*x*
*(2/3)/(168*b) + 55*a**4*x/168 + 155*a**3*b*x**(4/3)/168 + 181*a**2*b**2*x**(5/3)/168 + 33*a*b**3*x**2/56 + b*
*4*x**(7/3)/8), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x**(1/3))**(7/2)/7 - 2*a**2*(a**2 + 2*a*b*x**(1/3))**(9/2)/
9 + (a**2 + 2*a*b*x**(1/3))**(11/2)/11)/(4*a**3*b**3), Ne(a*b, 0)), (x*(a**2)**(5/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} a^{2} x^{\frac {1}{3}}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {5}{2}} a^{3}}{2 \, b^{3}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {7}{2}} x^{\frac {1}{3}}}{8 \, b^{2}} - \frac {27 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {7}{2}} a}{56 \, b^{3}} \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(5/2),x, algorithm="maxima")

[Out]

1/2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(5/2)*a^2*x^(1/3)/b^2 + 1/2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(5/2)*
a^3/b^3 + 3/8*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(7/2)*x^(1/3)/b^2 - 27/56*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2
)^(7/2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.74 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\frac {3}{8} \, b^{5} x^{\frac {8}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {15}{7} \, a b^{4} x^{\frac {7}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + 5 \, a^{2} b^{3} x^{2} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + 6 \, a^{3} b^{2} x^{\frac {5}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {15}{4} \, a^{4} b x^{\frac {4}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + a^{5} x \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) \]

[In]

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(5/2),x, algorithm="giac")

[Out]

3/8*b^5*x^(8/3)*sgn(b*x^(1/3) + a) + 15/7*a*b^4*x^(7/3)*sgn(b*x^(1/3) + a) + 5*a^2*b^3*x^2*sgn(b*x^(1/3) + a)
+ 6*a^3*b^2*x^(5/3)*sgn(b*x^(1/3) + a) + 15/4*a^4*b*x^(4/3)*sgn(b*x^(1/3) + a) + a^5*x*sgn(b*x^(1/3) + a)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{5/2} \, dx=\int {\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^{5/2} \,d x \]

[In]

int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(5/2),x)

[Out]

int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(5/2), x)