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

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

[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)

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Rubi [A]  time = 0.0812464, antiderivative size = 137, normalized size of antiderivative = 1., 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 = {1341, 646, 43} \[ \frac{3 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^9}{10 b^3}-\frac{2 a \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^8}{3 b^3}+\frac{3 a^2 \sqrt{a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}} \left (a+b \sqrt [3]{x}\right )^7}{8 b^3} \]

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 1341

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]

Rule 646

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 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])

Rubi steps

\begin{align*} \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2} \, dx &=3 \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.049556, size = 56, normalized size = 0.41 \[ \frac{\left (a+b \sqrt [3]{x}\right )^7 \sqrt{\left (a+b \sqrt [3]{x}\right )^2} \left (a^2-8 a b \sqrt [3]{x}+36 b^2 x^{2/3}\right )}{120 b^3} \]

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))^7*Sqrt[(a + b*x^(1/3))^2]*(a^2 - 8*a*b*x^(1/3) + 36*b^2*x^(2/3)))/(120*b^3)

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Maple [A]  time = 0.012, size = 109, normalized size = 0.8 \begin{align*}{\frac{1}{120}\sqrt{{a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}}} \left ( 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 \right ) \left ( a+b\sqrt [3]{x} \right ) ^{-1}} \end{align*}

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)

[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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.13304, size = 197, normalized size = 1.44 \begin{align*} \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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.17033, size = 189, normalized size = 1.38 \begin{align*} \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{align*}

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)