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

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

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Rubi [A]  time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2002, 2016, 2014} \begin {gather*} -\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} \end {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.05, size = 70, normalized size = 0.64 \begin {gather*} \frac {2 \left (a \sqrt [3]{x}+b\right ) \sqrt {a x+b x^{2/3}} \left (35 a^3 x-30 a^2 b x^{2/3}+24 a b^2 \sqrt [3]{x}-16 b^3\right )}{105 a^4 \sqrt [3]{x}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.06, size = 74, normalized size = 0.68 \begin {gather*} \frac {2 \sqrt {a x+b x^{2/3}} \left (35 a^4 x^{4/3}+5 a^3 b x-6 a^2 b^2 x^{2/3}+8 a b^3 \sqrt [3]{x}-16 b^4\right )}{105 a^4 \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x^(2/3) + a*x]*(-16*b^4 + 8*a*b^3*x^(1/3) - 6*a^2*b^2*x^(2/3) + 5*a^3*b*x + 35*a^4*x^(4/3)))/(105*a^
4*x^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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giac [A]  time = 0.18, size = 143, normalized size = 1.31 \begin {gather*} \frac {32 \, b^{\frac {9}{2}}}{105 \, a^{4}} + \frac {2 \, {\left (\frac {9 \, {\left (5 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} - 21 \, {\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} - 35 \, \sqrt {a x^{\frac {1}{3}} + b} b^{3}\right )} b}{a^{3}} + \frac {35 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} - 180 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b + 378 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b^{2} - 420 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{3} + 315 \, \sqrt {a x^{\frac {1}{3}} + b} b^{4}}{a^{3}}\right )}}{105 \, a} \end {gather*}

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*(9*(5*(a*x^(1/3) + b)^(7/2) - 21*(a*x^(1/3) + b)^(5/2)*b + 35*(a*x^(1/3) + b)^(3/2)
*b^2 - 35*sqrt(a*x^(1/3) + b)*b^3)*b/a^3 + (35*(a*x^(1/3) + b)^(9/2) - 180*(a*x^(1/3) + b)^(7/2)*b + 378*(a*x^
(1/3) + b)^(5/2)*b^2 - 420*(a*x^(1/3) + b)^(3/2)*b^3 + 315*sqrt(a*x^(1/3) + b)*b^4)/a^3)/a

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maple [A]  time = 0.04, size = 57, normalized size = 0.52 \begin {gather*} -\frac {2 \sqrt {a x +b \,x^{\frac {2}{3}}}\, \left (a \,x^{\frac {1}{3}}+b \right ) \left (-35 a^{3} x +30 a^{2} b \,x^{\frac {2}{3}}-24 a \,b^{2} x^{\frac {1}{3}}+16 b^{3}\right )}{105 a^{4} x^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a x + b x^{\frac {2}{3}}}\,{d x} \end {gather*}

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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mupad [B]  time = 5.19, size = 40, normalized size = 0.37 \begin {gather*} \frac {3\,x\,\sqrt {a\,x+b\,x^{2/3}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},4;\ 5;\ -\frac {a\,x^{1/3}}{b}\right )}{4\,\sqrt {\frac {a\,x^{1/3}}{b}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(1/2),x)

[Out]

(3*x*(a*x + b*x^(2/3))^(1/2)*hypergeom([-1/2, 4], 5, -(a*x^(1/3))/b))/(4*((a*x^(1/3))/b + 1)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a x + b x^{\frac {2}{3}}}\, dx \end {gather*}

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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