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

Optimal. Leaf size=34 \[ \frac{3 (a+b x)^{10/3}}{10 b^2}-\frac{3 a (a+b x)^{7/3}}{7 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0087666, antiderivative size = 34, normalized size of antiderivative = 1., 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} \[ \frac{3 (a+b x)^{10/3}}{10 b^2}-\frac{3 a (a+b x)^{7/3}}{7 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^(4/3),x]

[Out]

(-3*a*(a + b*x)^(7/3))/(7*b^2) + (3*(a + b*x)^(10/3))/(10*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 x (a+b x)^{4/3} \, dx &=\int \left (-\frac{a (a+b x)^{4/3}}{b}+\frac{(a+b x)^{7/3}}{b}\right ) \, dx\\ &=-\frac{3 a (a+b x)^{7/3}}{7 b^2}+\frac{3 (a+b x)^{10/3}}{10 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0420025, size = 24, normalized size = 0.71 \[ \frac{3 (a+b x)^{7/3} (7 b x-3 a)}{70 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)^(4/3),x]

[Out]

(3*(a + b*x)^(7/3)*(-3*a + 7*b*x))/(70*b^2)

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Maple [A]  time = 0.003, size = 21, normalized size = 0.6 \begin{align*} -{\frac{-21\,bx+9\,a}{70\,{b}^{2}} \left ( bx+a \right ) ^{{\frac{7}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/70*(b*x+a)^(7/3)*(-7*b*x+3*a)/b^2

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Maxima [A]  time = 1.05417, size = 35, normalized size = 1.03 \begin{align*} \frac{3 \,{\left (b x + a\right )}^{\frac{10}{3}}}{10 \, b^{2}} - \frac{3 \,{\left (b x + a\right )}^{\frac{7}{3}} a}{7 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54511, size = 96, normalized size = 2.82 \begin{align*} \frac{3 \,{\left (7 \, b^{3} x^{3} + 11 \, a b^{2} x^{2} + a^{2} b x - 3 \, a^{3}\right )}{\left (b x + a\right )}^{\frac{1}{3}}}{70 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/70*(7*b^3*x^3 + 11*a*b^2*x^2 + a^2*b*x - 3*a^3)*(b*x + a)^(1/3)/b^2

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Sympy [A]  time = 1.94136, size = 80, normalized size = 2.35 \begin{align*} \begin{cases} - \frac{9 a^{3} \sqrt [3]{a + b x}}{70 b^{2}} + \frac{3 a^{2} x \sqrt [3]{a + b x}}{70 b} + \frac{33 a x^{2} \sqrt [3]{a + b x}}{70} + \frac{3 b x^{3} \sqrt [3]{a + b x}}{10} & \text{for}\: b \neq 0 \\\frac{a^{\frac{4}{3}} x^{2}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-9*a**3*(a + b*x)**(1/3)/(70*b**2) + 3*a**2*x*(a + b*x)**(1/3)/(70*b) + 33*a*x**2*(a + b*x)**(1/3)/
70 + 3*b*x**3*(a + b*x)**(1/3)/10, Ne(b, 0)), (a**(4/3)*x**2/2, True))

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Giac [B]  time = 1.24592, size = 92, normalized size = 2.71 \begin{align*} \frac{3 \,{\left (\frac{5 \,{\left (4 \,{\left (b x + a\right )}^{\frac{7}{3}} - 7 \,{\left (b x + a\right )}^{\frac{4}{3}} a\right )} a}{b} + \frac{14 \,{\left (b x + a\right )}^{\frac{10}{3}} - 40 \,{\left (b x + a\right )}^{\frac{7}{3}} a + 35 \,{\left (b x + a\right )}^{\frac{4}{3}} a^{2}}{b}\right )}}{140 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/140*(5*(4*(b*x + a)^(7/3) - 7*(b*x + a)^(4/3)*a)*a/b + (14*(b*x + a)^(10/3) - 40*(b*x + a)^(7/3)*a + 35*(b*x
+ a)^(4/3)*a^2)/b)/b