3.2.44 \(\int \frac {(b \sqrt [3]{x}+a x)^{3/2}}{x^2} \, dx\) [144]

Optimal. Leaf size=144 \[ 4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+\frac {4 a^{3/4} b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt {b \sqrt [3]{x}+a x}} \]

[Out]

-2*(b*x^(1/3)+a*x)^(3/2)/x+4*a*(b*x^(1/3)+a*x)^(1/2)+4*a^(3/4)*b^(3/4)*x^(1/6)*(cos(2*arctan(a^(1/4)*x^(1/6)/b
^(1/4)))^2)^(1/2)/cos(2*arctan(a^(1/4)*x^(1/6)/b^(1/4)))*EllipticF(sin(2*arctan(a^(1/4)*x^(1/6)/b^(1/4))),1/2*
2^(1/2))*(x^(1/3)*a^(1/2)+b^(1/2))*((b+a*x^(2/3))/(x^(1/3)*a^(1/2)+b^(1/2))^2)^(1/2)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2043, 2045, 2046, 2036, 335, 226} \begin {gather*} \frac {4 a^{3/4} b^{3/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt {a x+b \sqrt [3]{x}}}-\frac {2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{x}+4 a \sqrt {a x+b \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^(1/3) + a*x)^(3/2)/x^2,x]

[Out]

4*a*Sqrt[b*x^(1/3) + a*x] - (2*(b*x^(1/3) + a*x)^(3/2))/x + (4*a^(3/4)*b^(3/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqr
t[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/
Sqrt[b*x^(1/3) + a*x]

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2045

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

Rule 2046

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

Rubi steps

\begin {align*} \int \frac {\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^2} \, dx &=3 \text {Subst}\left (\int \frac {\left (b x+a x^3\right )^{3/2}}{x^4} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+(6 a) \text {Subst}\left (\int \frac {\sqrt {b x+a x^3}}{x} \, dx,x,\sqrt [3]{x}\right )\\ &=4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+(4 a b) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+\frac {\left (4 a b \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {b \sqrt [3]{x}+a x}}\\ &=4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+\frac {\left (8 a b \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{\sqrt {b \sqrt [3]{x}+a x}}\\ &=4 a \sqrt {b \sqrt [3]{x}+a x}-\frac {2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{x}+\frac {4 a^{3/4} b^{3/4} \left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right ) \sqrt {\frac {b+a x^{2/3}}{\left (\sqrt {b}+\sqrt {a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 60, normalized size = 0.42 \begin {gather*} -\frac {2 b \sqrt {b \sqrt [3]{x}+a x} \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {a x^{2/3}}{b}\right )}{\sqrt {1+\frac {a x^{2/3}}{b}} x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(1/3) + a*x)^(3/2)/x^2,x]

[Out]

(-2*b*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((a*x^(2/3))/b)])/(Sqrt[1 + (a*x^(2/3))/b]*x^(
2/3))

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Maple [A]
time = 0.35, size = 130, normalized size = 0.90

method	result	size
default	\(\frac {4 x^{\frac {1}{3}} \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, b +2 x^{\frac {4}{3}} a^{2}-2 b^{2}}{x^{\frac {1}{3}} \sqrt {x^{\frac {1}{3}} \left (b +a \,x^{\frac {2}{3}}\right )}}\)	\(130\)
derivativedivides	\(-\frac {2 b \sqrt {b \,x^{\frac {1}{3}}+a x}}{x^{\frac {2}{3}}}+2 a \sqrt {b \,x^{\frac {1}{3}}+a x}+\frac {4 b \sqrt {-a b}\, \sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x^{\frac {1}{3}}-\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}\, \sqrt {-\frac {x^{\frac {1}{3}} a}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x^{\frac {1}{3}}+\frac {\sqrt {-a b}}{a}\right ) a}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b \,x^{\frac {1}{3}}+a x}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/x^(1/3)*(2*x^(1/3)*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^(1/2))^
(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a
*b)^(1/2)*b+x^(4/3)*a^2-b^2)/(x^(1/3)*(b+a*x^(2/3)))^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + b*x^(1/3))^(3/2)/x^2, 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^{1/3}\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(1/3))^(3/2)/x^2,x)

[Out]

int((a*x + b*x^(1/3))^(3/2)/x^2, x)

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