3.179 \(\int \frac{(b x^{2/3}+a x)^{3/2}}{x} \, dx\)

Optimal. Leaf size=84 \[ \frac{16 b^2 \left (a x+b x^{2/3}\right )^{5/2}}{105 a^3 x^{5/3}}-\frac{8 b \left (a x+b x^{2/3}\right )^{5/2}}{21 a^2 x^{4/3}}+\frac{2 \left (a x+b x^{2/3}\right )^{5/2}}{3 a x} \]

[Out]

(16*b^2*(b*x^(2/3) + a*x)^(5/2))/(105*a^3*x^(5/3)) - (8*b*(b*x^(2/3) + a*x)^(5/2))/(21*a^2*x^(4/3)) + (2*(b*x^
(2/3) + a*x)^(5/2))/(3*a*x)

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Rubi [A]  time = 0.138779, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ \frac{16 b^2 \left (a x+b x^{2/3}\right )^{5/2}}{105 a^3 x^{5/3}}-\frac{8 b \left (a x+b x^{2/3}\right )^{5/2}}{21 a^2 x^{4/3}}+\frac{2 \left (a x+b x^{2/3}\right )^{5/2}}{3 a x} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2)/x,x]

[Out]

(16*b^2*(b*x^(2/3) + a*x)^(5/2))/(105*a^3*x^(5/3)) - (8*b*(b*x^(2/3) + a*x)^(5/2))/(21*a^2*x^(4/3)) + (2*(b*x^
(2/3) + a*x)^(5/2))/(3*a*x)

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x} \, dx &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}-\frac{(4 b) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^{4/3}} \, dx}{9 a}\\ &=-\frac{8 b \left (b x^{2/3}+a x\right )^{5/2}}{21 a^2 x^{4/3}}+\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}+\frac{\left (8 b^2\right ) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^{5/3}} \, dx}{63 a^2}\\ &=\frac{16 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{105 a^3 x^{5/3}}-\frac{8 b \left (b x^{2/3}+a x\right )^{5/2}}{21 a^2 x^{4/3}}+\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}\\ \end{align*}

Mathematica [A]  time = 0.0579474, size = 63, normalized size = 0.75 \[ \frac{2 \left (a \sqrt [3]{x}+b\right )^2 \left (35 a^2 x^{2/3}-20 a b \sqrt [3]{x}+8 b^2\right ) \sqrt{a x+b x^{2/3}}}{105 a^3 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(2/3) + a*x)^(3/2)/x,x]

[Out]

(2*(b + a*x^(1/3))^2*(8*b^2 - 20*a*b*x^(1/3) + 35*a^2*x^(2/3))*Sqrt[b*x^(2/3) + a*x])/(105*a^3*x^(1/3))

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Maple [A]  time = 0.004, size = 48, normalized size = 0.6 \begin{align*}{\frac{2}{105\,x{a}^{3}} \left ( b{x}^{{\frac{2}{3}}}+ax \right ) ^{{\frac{3}{2}}} \left ( b+a\sqrt [3]{x} \right ) \left ( 35\,{x}^{2/3}{a}^{2}-20\,\sqrt [3]{x}ab+8\,{b}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(b*x^(2/3)+a*x)^(3/2)*(b+a*x^(1/3))*(35*x^(2/3)*a^2-20*x^(1/3)*a*b+8*b^2)/x/a^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + b x^{\frac{2}{3}}\right )}^{\frac{3}{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(2/3)+a*x)^(3/2)/x,x, algorithm="maxima")

[Out]

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

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Fricas [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((b*x^(2/3)+a*x)^(3/2)/x,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a x + b x^{\frac{2}{3}}\right )^{\frac{3}{2}}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**(2/3)+a*x)**(3/2)/x,x)

[Out]

Integral((a*x + b*x**(2/3))**(3/2)/x, x)

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Giac [A]  time = 1.17726, size = 155, normalized size = 1.85 \begin{align*} -\frac{16 \, b^{\frac{9}{2}}}{105 \, a^{3}} + \frac{2 \,{\left (\frac{3 \,{\left (15 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} - 42 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} b + 35 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} b^{2}\right )} b}{a^{2}} + \frac{35 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{9}{2}} - 135 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} b + 189 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} b^{2} - 105 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} b^{3}}{a^{2}}\right )}}{105 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(2/3)+a*x)^(3/2)/x,x, algorithm="giac")

[Out]

-16/105*b^(9/2)/a^3 + 2/105*(3*(15*(a*x^(1/3) + b)^(7/2) - 42*(a*x^(1/3) + b)^(5/2)*b + 35*(a*x^(1/3) + b)^(3/
2)*b^2)*b/a^2 + (35*(a*x^(1/3) + b)^(9/2) - 135*(a*x^(1/3) + b)^(7/2)*b + 189*(a*x^(1/3) + b)^(5/2)*b^2 - 105*
(a*x^(1/3) + b)^(3/2)*b^3)/a^2)/a