3.5.62 \(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{7/2} \, dx\) [462]

Optimal. Leaf size=137 \[ \frac {3 a^2 \left (a+b \sqrt [3]{x}\right )^7 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{8 b^3}-\frac {2 a \left (a+b \sqrt [3]{x}\right )^8 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{3 b^3}+\frac {3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{10 b^3} \]

[Out]

3/8*a^2*(a+b*x^(1/3))^7*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3-2/3*a*(a+b*x^(1/3))^8*(a^2+2*a*b*x^(1/3)+b^2
*x^(2/3))^(1/2)/b^3+3/10*(a+b*x^(1/3))^9*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)/b^3

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Rubi [A]
time = 0.05, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 660, 45} \begin {gather*} \frac {3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^9}{10 b^3}-\frac {2 a \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^8}{3 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 )^7}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(7/2),x]

[Out]

(3*a^2*(a + b*x^(1/3))^7*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(8*b^3) - (2*a*(a + b*x^(1/3))^8*Sqrt[a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3)])/(3*b^3) + (3*(a + b*x^(1/3))^9*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])/(10*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*} \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2} \, dx &=3 \text {Subst}\left (\int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{7/2} \, 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 )^7 \, dx,x,\sqrt [3]{x}\right )}{b^7 \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 )^7}{b^2}-\frac {2 a \left (a b+b^2 x\right )^8}{b^3}+\frac {\left (a b+b^2 x\right )^9}{b^4}\right ) \, dx,x,\sqrt [3]{x}\right )}{b^7 \left (a+b \sqrt [3]{x}\right )}\\ &=\frac {3 a^2 \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}-\frac {2 a \left (a+b \sqrt [3]{x}\right )^8 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{3 b^3}+\frac {3 \left (a+b \sqrt [3]{x}\right )^9 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}{10 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 117, normalized size = 0.85 \begin {gather*} \frac {\left (\left (a+b \sqrt [3]{x}\right )^2\right )^{7/2} \left (120 a^7 x+630 a^6 b x^{4/3}+1512 a^5 b^2 x^{5/3}+2100 a^4 b^3 x^2+1800 a^3 b^4 x^{7/3}+945 a^2 b^5 x^{8/3}+280 a b^6 x^3+36 b^7 x^{10/3}\right )}{120 \left (a+b \sqrt [3]{x}\right )^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^(7/2),x]

[Out]

(((a + b*x^(1/3))^2)^(7/2)*(120*a^7*x + 630*a^6*b*x^(4/3) + 1512*a^5*b^2*x^(5/3) + 2100*a^4*b^3*x^2 + 1800*a^3
*b^4*x^(7/3) + 945*a^2*b^5*x^(8/3) + 280*a*b^6*x^3 + 36*b^7*x^(10/3)))/(120*(a + b*x^(1/3))^7)

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Maple [A]
time = 0.09, size = 109, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )^{\frac {7}{2}} x \left (36 b^{7} x^{\frac {7}{3}}+280 a \,b^{6} x^{2}+945 a^{2} b^{5} x^{\frac {5}{3}}+1800 a^{3} b^{4} x^{\frac {4}{3}}+2100 a^{4} b^{3} x +1512 a^{5} b^{2} x^{\frac {2}{3}}+630 a^{6} b \,x^{\frac {1}{3}}+120 a^{7}\right )}{120 \left (a +b \,x^{\frac {1}{3}}\right )^{7}}\)	\(98\)
default	\(\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (36 b^{7} x^{\frac {10}{3}}+945 a^{2} b^{5} x^{\frac {8}{3}}+1800 a^{3} b^{4} x^{\frac {7}{3}}+1512 a^{5} b^{2} x^{\frac {5}{3}}+630 a^{6} b \,x^{\frac {4}{3}}+280 a \,b^{6} x^{3}+2100 a^{4} b^{3} x^{2}+120 a^{7} x \right )}{120 a +120 b \,x^{\frac {1}{3}}}\)	\(109\)

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))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/120*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(36*b^7*x^(10/3)+945*a^2*b^5*x^(8/3)+1800*a^3*b^4*x^(7/3)+1512*a^5
*b^2*x^(5/3)+630*a^6*b*x^(4/3)+280*a*b^6*x^3+2100*a^4*b^3*x^2+120*a^7*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 {7}{2}} a^{2} x^{\frac {1}{3}}}{8 \, b^{2}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {7}{2}} a^{3}}{8 \, b^{3}} + \frac {3 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {9}{2}} x^{\frac {1}{3}}}{10 \, b^{2}} - \frac {11 \, {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{\frac {9}{2}} a}{30 \, 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))^(7/2),x, algorithm="maxima")

[Out]

3/8*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(7/2)*a^2*x^(1/3)/b^2 + 3/8*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(7/2)*
a^3/b^3 + 3/10*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^(9/2)*x^(1/3)/b^2 - 11/30*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^
2)^(9/2)*a/b^3

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Fricas [A]
time = 0.40, size = 84, normalized size = 0.61 \begin {gather*} \frac {7}{3} \, a b^{6} x^{3} + \frac {35}{2} \, a^{4} b^{3} x^{2} + a^{7} x + \frac {63}{40} \, {\left (5 \, a^{2} b^{5} x^{2} + 8 \, a^{5} b^{2} x\right )} x^{\frac {2}{3}} + \frac {3}{20} \, {\left (2 \, b^{7} x^{3} + 100 \, a^{3} b^{4} x^{2} + 35 \, a^{6} b x\right )} x^{\frac {1}{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))^(7/2),x, algorithm="fricas")

[Out]

7/3*a*b^6*x^3 + 35/2*a^4*b^3*x^2 + a^7*x + 63/40*(5*a^2*b^5*x^2 + 8*a^5*b^2*x)*x^(2/3) + 3/20*(2*b^7*x^3 + 100
*a^3*b^4*x^2 + 35*a^6*b*x)*x^(1/3)

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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 {7}{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))**(7/2),x)

[Out]

Integral((a**2 + 2*a*b*x**(1/3) + b**2*x**(2/3))**(7/2), x)

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Giac [A]
time = 4.66, size = 140, normalized size = 1.02 \begin {gather*} \frac {3}{10} \, b^{7} x^{\frac {10}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {7}{3} \, a b^{6} x^{3} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {63}{8} \, a^{2} b^{5} x^{\frac {8}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + 15 \, a^{3} b^{4} x^{\frac {7}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {35}{2} \, a^{4} b^{3} x^{2} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {63}{5} \, a^{5} b^{2} x^{\frac {5}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + \frac {21}{4} \, a^{6} b x^{\frac {4}{3}} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right ) + a^{7} 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))^(7/2),x, algorithm="giac")

[Out]

3/10*b^7*x^(10/3)*sgn(b*x^(1/3) + a) + 7/3*a*b^6*x^3*sgn(b*x^(1/3) + a) + 63/8*a^2*b^5*x^(8/3)*sgn(b*x^(1/3) +
a) + 15*a^3*b^4*x^(7/3)*sgn(b*x^(1/3) + a) + 35/2*a^4*b^3*x^2*sgn(b*x^(1/3) + a) + 63/5*a^5*b^2*x^(5/3)*sgn(b
*x^(1/3) + a) + 21/4*a^6*b*x^(4/3)*sgn(b*x^(1/3) + a) + a^7*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 )}^{7/2} \,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))^(7/2),x)

[Out]

int((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(7/2), x)

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