3.5.1 \(\int \frac {x}{\sqrt [3]{-a+b x}} \, dx\)

Optimal. Leaf size=38 \[ \frac {3 (b x-a)^{5/3}}{5 b^2}+\frac {3 a (b x-a)^{2/3}}{2 b^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} \frac {3 (b x-a)^{5/3}}{5 b^2}+\frac {3 a (b x-a)^{2/3}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(-a + b*x)^(1/3),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.68 \begin {gather*} \frac {3 (b x-a)^{2/3} (3 a+2 b x)}{10 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(-a + b*x)^(1/3),x]

[Out]

(3*(-a + b*x)^(2/3)*(3*a + 2*b*x))/(10*b^2)

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IntegrateAlgebraic [A]  time = 0.01, size = 26, normalized size = 0.68 \begin {gather*} \frac {3 (b x-a)^{2/3} (3 a+2 b x)}{10 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(-a + b*x)^(1/3),x]

[Out]

(3*(-a + b*x)^(2/3)*(3*a + 2*b*x))/(10*b^2)

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fricas [A]  time = 0.91, size = 22, normalized size = 0.58 \begin {gather*} \frac {3 \, {\left (2 \, b x + 3 \, a\right )} {\left (b x - a\right )}^{\frac {2}{3}}}{10 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(2*b*x + 3*a)*(b*x - a)^(2/3)/b^2

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giac [A]  time = 0.98, size = 29, normalized size = 0.76 \begin {gather*} \frac {3 \, {\left (2 \, {\left (b x - a\right )}^{\frac {5}{3}} + 5 \, {\left (b x - a\right )}^{\frac {2}{3}} a\right )}}{10 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 0.61 \begin {gather*} \frac {3 \left (2 b x +3 a \right ) \left (b x -a \right )^{\frac {2}{3}}}{10 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/10*(2*b*x+3*a)/b^2*(b*x-a)^(2/3)

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maxima [A]  time = 1.34, size = 30, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left (b x - a\right )}^{\frac {5}{3}}}{5 \, b^{2}} + \frac {3 \, {\left (b x - a\right )}^{\frac {2}{3}} a}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 29, normalized size = 0.76 \begin {gather*} \frac {15\,a\,{\left (b\,x-a\right )}^{2/3}+6\,{\left (b\,x-a\right )}^{5/3}}{10\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x - a)^(1/3),x)

[Out]

(15*a*(b*x - a)^(2/3) + 6*(b*x - a)^(5/3))/(10*b^2)

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sympy [C]  time = 1.26, size = 486, normalized size = 12.79 \begin {gather*} \begin {cases} - \frac {9 a^{\frac {11}{3}} \left (-1 + \frac {b x}{a}\right )^{\frac {2}{3}} e^{\frac {i \pi }{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} - \frac {9 a^{\frac {11}{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} + \frac {3 a^{\frac {8}{3}} b x \left (-1 + \frac {b x}{a}\right )^{\frac {2}{3}} e^{\frac {i \pi }{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} + \frac {9 a^{\frac {8}{3}} b x}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} + \frac {6 a^{\frac {5}{3}} b^{2} x^{2} \left (-1 + \frac {b x}{a}\right )^{\frac {2}{3}} e^{\frac {i \pi }{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {9 a^{\frac {11}{3}} \left (1 - \frac {b x}{a}\right )^{\frac {2}{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} - \frac {9 a^{\frac {11}{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} - \frac {3 a^{\frac {8}{3}} b x \left (1 - \frac {b x}{a}\right )^{\frac {2}{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} + \frac {9 a^{\frac {8}{3}} b x}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} - \frac {6 a^{\frac {5}{3}} b^{2} x^{2} \left (1 - \frac {b x}{a}\right )^{\frac {2}{3}}}{- 10 a^{2} b^{2} e^{\frac {i \pi }{3}} + 10 a b^{3} x e^{\frac {i \pi }{3}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x-a)**(1/3),x)

[Out]

Piecewise((-9*a**(11/3)*(-1 + b*x/a)**(2/3)*exp(I*pi/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp(I*pi/3))
- 9*a**(11/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp(I*pi/3)) + 3*a**(8/3)*b*x*(-1 + b*x/a)**(2/3)*exp(I
*pi/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp(I*pi/3)) + 9*a**(8/3)*b*x/(-10*a**2*b**2*exp(I*pi/3) + 10*
a*b**3*x*exp(I*pi/3)) + 6*a**(5/3)*b**2*x**2*(-1 + b*x/a)**(2/3)*exp(I*pi/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a
*b**3*x*exp(I*pi/3)), Abs(b*x/a) > 1), (9*a**(11/3)*(1 - b*x/a)**(2/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*
x*exp(I*pi/3)) - 9*a**(11/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp(I*pi/3)) - 3*a**(8/3)*b*x*(1 - b*x/a
)**(2/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp(I*pi/3)) + 9*a**(8/3)*b*x/(-10*a**2*b**2*exp(I*pi/3) + 1
0*a*b**3*x*exp(I*pi/3)) - 6*a**(5/3)*b**2*x**2*(1 - b*x/a)**(2/3)/(-10*a**2*b**2*exp(I*pi/3) + 10*a*b**3*x*exp
(I*pi/3)), True))

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