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

Optimal. Leaf size=109 \[ -\frac{32 b^3 \left (a x+b x^{2/3}\right )^{3/2}}{105 a^4 x}+\frac{16 b^2 \left (a x+b x^{2/3}\right )^{3/2}}{35 a^3 x^{2/3}}-\frac{4 b \left (a x+b x^{2/3}\right )^{3/2}}{7 a^2 \sqrt [3]{x}}+\frac{2 \left (a x+b x^{2/3}\right )^{3/2}}{3 a} \]

[Out]

(2*(b*x^(2/3) + a*x)^(3/2))/(3*a) - (32*b^3*(b*x^(2/3) + a*x)^(3/2))/(105*a^4*x) + (16*b^2*(b*x^(2/3) + a*x)^(
3/2))/(35*a^3*x^(2/3)) - (4*b*(b*x^(2/3) + a*x)^(3/2))/(7*a^2*x^(1/3))

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Rubi [A]  time = 0.136737, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2002, 2016, 2014} \[ -\frac{32 b^3 \left (a x+b x^{2/3}\right )^{3/2}}{105 a^4 x}+\frac{16 b^2 \left (a x+b x^{2/3}\right )^{3/2}}{35 a^3 x^{2/3}}-\frac{4 b \left (a x+b x^{2/3}\right )^{3/2}}{7 a^2 \sqrt [3]{x}}+\frac{2 \left (a x+b x^{2/3}\right )^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(2/3) + a*x],x]

[Out]

(2*(b*x^(2/3) + a*x)^(3/2))/(3*a) - (32*b^3*(b*x^(2/3) + a*x)^(3/2))/(105*a^4*x) + (16*b^2*(b*x^(2/3) + a*x)^(
3/2))/(35*a^3*x^(2/3)) - (4*b*(b*x^(2/3) + a*x)^(3/2))/(7*a^2*x^(1/3))

Rule 2002

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

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 \sqrt{b x^{2/3}+a x} \, dx &=\frac{2 \left (b x^{2/3}+a x\right )^{3/2}}{3 a}-\frac{(2 b) \int \frac{\sqrt{b x^{2/3}+a x}}{\sqrt [3]{x}} \, dx}{3 a}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{3/2}}{3 a}-\frac{4 b \left (b x^{2/3}+a x\right )^{3/2}}{7 a^2 \sqrt [3]{x}}+\frac{\left (8 b^2\right ) \int \frac{\sqrt{b x^{2/3}+a x}}{x^{2/3}} \, dx}{21 a^2}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{3/2}}{3 a}+\frac{16 b^2 \left (b x^{2/3}+a x\right )^{3/2}}{35 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{3/2}}{7 a^2 \sqrt [3]{x}}-\frac{\left (16 b^3\right ) \int \frac{\sqrt{b x^{2/3}+a x}}{x} \, dx}{105 a^3}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{3/2}}{3 a}-\frac{32 b^3 \left (b x^{2/3}+a x\right )^{3/2}}{105 a^4 x}+\frac{16 b^2 \left (b x^{2/3}+a x\right )^{3/2}}{35 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{3/2}}{7 a^2 \sqrt [3]{x}}\\ \end{align*}

Mathematica [A]  time = 0.0395579, size = 70, normalized size = 0.64 \[ \frac{2 \left (a \sqrt [3]{x}+b\right ) \sqrt{a x+b x^{2/3}} \left (-30 a^2 b x^{2/3}+35 a^3 x+24 a b^2 \sqrt [3]{x}-16 b^3\right )}{105 a^4 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^(2/3) + a*x],x]

[Out]

(2*(b + a*x^(1/3))*Sqrt[b*x^(2/3) + a*x]*(-16*b^3 + 24*a*b^2*x^(1/3) - 30*a^2*b*x^(2/3) + 35*a^3*x))/(105*a^4*
x^(1/3))

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Maple [A]  time = 0.003, size = 57, normalized size = 0.5 \begin{align*} -{\frac{2}{105\,{a}^{4}}\sqrt{b{x}^{{\frac{2}{3}}}+ax} \left ( b+a\sqrt [3]{x} \right ) \left ( 30\,{x}^{2/3}{a}^{2}b-24\,\sqrt [3]{x}a{b}^{2}-35\,x{a}^{3}+16\,{b}^{3} \right ){\frac{1}{\sqrt [3]{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*(b*x^(2/3)+a*x)^(1/2)*(b+a*x^(1/3))*(30*x^(2/3)*a^2*b-24*x^(1/3)*a*b^2-35*x*a^3+16*b^3)/x^(1/3)/a^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(2/3)+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x + b*x^(2/3)), 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)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**(2/3)+a*x)**(1/2),x)

[Out]

Integral(sqrt(a*x + b*x**(2/3)), x)

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Giac [A]  time = 1.11866, size = 89, normalized size = 0.82 \begin{align*} \frac{32 \, b^{\frac{9}{2}}}{105 \, a^{4}} + \frac{2 \,{\left (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}\right )}}{105 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(2/3)+a*x)^(1/2),x, algorithm="giac")

[Out]

32/105*b^(9/2)/a^4 + 2/105*(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^4