3.2316 \(\int (a+b \sqrt [3]{x})^5 \, dx\)

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b \sqrt [3]{x}\right )^6}{2 b^3}+\frac{3 \left (a+b \sqrt [3]{x}\right )^8}{8 b^3}-\frac{6 a \left (a+b \sqrt [3]{x}\right )^7}{7 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0392421, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 43} \[ \frac{a^2 \left (a+b \sqrt [3]{x}\right )^6}{2 b^3}+\frac{3 \left (a+b \sqrt [3]{x}\right )^8}{8 b^3}-\frac{6 a \left (a+b \sqrt [3]{x}\right )^7}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^(1/3))^5,x]

[Out]

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

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

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

Mathematica [A]  time = 0.0227737, size = 41, normalized size = 0.69 \[ \frac{\left (a+b \sqrt [3]{x}\right )^6 \left (a^2-6 a b \sqrt [3]{x}+21 b^2 x^{2/3}\right )}{56 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^(1/3))^5,x]

[Out]

((a + b*x^(1/3))^6*(a^2 - 6*a*b*x^(1/3) + 21*b^2*x^(2/3)))/(56*b^3)

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Maple [A]  time = 0., size = 55, normalized size = 0.9 \begin{align*} x{a}^{5}+{\frac{3\,{b}^{5}}{8}{x}^{{\frac{8}{3}}}}+{\frac{15\,a{b}^{4}}{7}{x}^{{\frac{7}{3}}}}+5\,{a}^{2}{b}^{3}{x}^{2}+6\,{a}^{3}{b}^{2}{x}^{5/3}+{\frac{15\,{a}^{4}b}{4}{x}^{{\frac{4}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957892, size = 73, normalized size = 1.24 \begin{align*} \frac{3}{8} \, b^{5} x^{\frac{8}{3}} + \frac{15}{7} \, a b^{4} x^{\frac{7}{3}} + 5 \, a^{2} b^{3} x^{2} + 6 \, a^{3} b^{2} x^{\frac{5}{3}} + \frac{15}{4} \, a^{4} b x^{\frac{4}{3}} + a^{5} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47549, size = 140, normalized size = 2.37 \begin{align*} 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 1.5631, size = 68, normalized size = 1.15 \begin{align*} a^{5} x + \frac{15 a^{4} b x^{\frac{4}{3}}}{4} + 6 a^{3} b^{2} x^{\frac{5}{3}} + 5 a^{2} b^{3} x^{2} + \frac{15 a b^{4} x^{\frac{7}{3}}}{7} + \frac{3 b^{5} x^{\frac{8}{3}}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15102, size = 73, normalized size = 1.24 \begin{align*} \frac{3}{8} \, b^{5} x^{\frac{8}{3}} + \frac{15}{7} \, a b^{4} x^{\frac{7}{3}} + 5 \, a^{2} b^{3} x^{2} + 6 \, a^{3} b^{2} x^{\frac{5}{3}} + \frac{15}{4} \, a^{4} b x^{\frac{4}{3}} + a^{5} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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