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

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

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Rubi [A]  time = 0.14, antiderivative size = 84, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 \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*}

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Mathematica [A]  time = 0.08, size = 63, normalized size = 0.75 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 4.54, size = 87, normalized size = 1.04 \begin {gather*} \frac {2 \left (x^{2/3} \left (a \sqrt [3]{x}+b\right )\right )^{3/2} \left (35 a^4 x^{4/3}+50 a^3 b x+3 a^2 b^2 x^{2/3}-4 a b^3 \sqrt [3]{x}+8 b^4\right )}{105 a^3 x \left (a \sqrt [3]{x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((b + a*x^(1/3))*x^(2/3))^(3/2)*(8*b^4 - 4*a*b^3*x^(1/3) + 3*a^2*b^2*x^(2/3) + 50*a^3*b*x + 35*a^4*x^(4/3))
)/(105*a^3*(b + a*x^(1/3))*x)

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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)^(3/2)/x,x, algorithm="fricas")

[Out]

Timed out

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giac [B]  time = 0.21, size = 265, normalized size = 3.15 \begin {gather*} -\frac {2}{35} \, b {\left (\frac {8 \, b^{\frac {7}{2}}}{a^{3}} - \frac {\frac {7 \, {\left (3 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} - 10 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b + 15 \, \sqrt {a x^{\frac {1}{3}} + b} b^{2}\right )} b}{a^{2}} + \frac {3 \, {\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 )}}{a^{2}}}{a}\right )} + \frac {2}{105} \, a {\left (\frac {16 \, b^{\frac {9}{2}}}{a^{4}} + \frac {\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}}}{a}\right )} \end {gather*}

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]

-2/35*b*(8*b^(7/2)/a^3 - (7*(3*(a*x^(1/3) + b)^(5/2) - 10*(a*x^(1/3) + b)^(3/2)*b + 15*sqrt(a*x^(1/3) + b)*b^2
)*b/a^2 + 3*(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)/a^2)/a) + 2/105*a*(16*b^(9/2)/a^4 + (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 = 0.05, size = 48, normalized size = 0.57 \begin {gather*} \frac {2 \left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} \left (a \,x^{\frac {1}{3}}+b \right ) \left (35 a^{2} x^{\frac {2}{3}}-20 a b \,x^{\frac {1}{3}}+8 b^{2}\right )}{105 a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x+b\,x^{2/3}\right )}^{3/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(3/2)/x,x)

[Out]

int((a*x + b*x^(2/3))^(3/2)/x, x)

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

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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