3.5.15 \(\int \frac {x}{(a+b x)^{4/3}} \, dx\)

Optimal. Leaf size=32 \[ \frac {3 a}{b^2 \sqrt [3]{a+b x}}+\frac {3 (a+b x)^{2/3}}{2 b^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {43} \begin {gather*} \frac {3 a}{b^2 \sqrt [3]{a+b x}}+\frac {3 (a+b x)^{2/3}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x)^(4/3),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.72 \begin {gather*} \frac {3 (3 a+b x)}{2 b^2 \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x)^(4/3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.01, size = 23, normalized size = 0.72 \begin {gather*} \frac {3 (3 a+b x)}{2 b^2 \sqrt [3]{a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(a + b*x)^(4/3),x]

[Out]

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

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fricas [A]  time = 0.83, size = 29, normalized size = 0.91 \begin {gather*} \frac {3 \, {\left (b x + 3 \, a\right )} {\left (b x + a\right )}^{\frac {2}{3}}}{2 \, {\left (b^{3} x + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x+a)^(4/3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.93, size = 30, normalized size = 0.94 \begin {gather*} \frac {3 \, {\left (\frac {{\left (b x + a\right )}^{\frac {2}{3}}}{b} + \frac {2 \, a}{{\left (b x + a\right )}^{\frac {1}{3}} b}\right )}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x+a)^(4/3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 20, normalized size = 0.62 \begin {gather*} \frac {\frac {3 b x}{2}+\frac {9 a}{2}}{\left (b x +a \right )^{\frac {1}{3}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x+a)^(4/3),x)

[Out]

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

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maxima [A]  time = 1.30, size = 26, normalized size = 0.81 \begin {gather*} \frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}}}{2 \, b^{2}} + \frac {3 \, a}{{\left (b x + a\right )}^{\frac {1}{3}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x+a)^(4/3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.03, size = 20, normalized size = 0.62 \begin {gather*} \frac {9\,a+3\,b\,x}{2\,b^2\,{\left (a+b\,x\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x)^(4/3),x)

[Out]

(9*a + 3*b*x)/(2*b^2*(a + b*x)^(1/3))

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sympy [A]  time = 0.72, size = 41, normalized size = 1.28 \begin {gather*} \begin {cases} \frac {9 a}{2 b^{2} \sqrt [3]{a + b x}} + \frac {3 x}{2 b \sqrt [3]{a + b x}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {4}{3}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x+a)**(4/3),x)

[Out]

Piecewise((9*a/(2*b**2*(a + b*x)**(1/3)) + 3*x/(2*b*(a + b*x)**(1/3)), Ne(b, 0)), (x**2/(2*a**(4/3)), True))

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