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

Optimal. Leaf size=208 \[ \frac {4 b^{15/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 a^{9/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 b^3 \sqrt {a x+b \sqrt [3]{x}}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {a x+b \sqrt [3]{x}}+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2} \]

[Out]

2/5*x*(b*x^(1/3)+a*x)^(3/2)-8/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^2+24/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/2)/a+12/5
5*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)+4/77*b^(15/4)*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)/a^(9/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2004, 2018, 2021, 2024, 2011, 329, 220} \[ \frac {4 b^{15/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{77 a^{9/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {8 b^3 \sqrt {a x+b \sqrt [3]{x}}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {a x+b \sqrt [3]{x}}+\frac {2}{5} x \left (a x+b \sqrt [3]{x}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-8*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^2) + (24*b^2*x^(2/3)*Sqrt[b*x^(1/3) + a*x])/(385*a) + (12*b*x^(4/3)*Sqrt[
b*x^(1/3) + a*x])/55 + (2*x*(b*x^(1/3) + a*x)^(3/2))/5 + (4*b^(15/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqrt[(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])/(77*a^(9/4
)*Sqrt[b*x^(1/3) + a*x])

Rule 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 329

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 2004

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

Rule 2011

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 2018

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 2021

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]

Rule 2024

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

Rubi steps

\begin {align*} \int \left (b \sqrt [3]{x}+a x\right )^{3/2} \, dx &=\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {1}{5} (2 b) \int \sqrt [3]{x} \sqrt {b \sqrt [3]{x}+a x} \, dx\\ &=\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {1}{5} (6 b) \operatorname {Subst}\left (\int x^3 \sqrt {b x+a x^3} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {1}{55} \left (12 b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}-\frac {\left (12 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a}\\ &=-\frac {8 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {\left (4 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 a^2}\\ &=-\frac {8 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {\left (4 b^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{77 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {8 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {\left (8 b^4 \sqrt {b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{77 a^2 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {8 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^2}+\frac {24 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a}+\frac {12}{55} b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}+\frac {2}{5} x \left (b \sqrt [3]{x}+a x\right )^{3/2}+\frac {4 b^{15/4} \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 )}{77 a^{9/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 106, normalized size = 0.51 \[ \frac {2 \sqrt {a x+b \sqrt [3]{x}} \left (5 b^3 \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {a x^{2/3}}{b}\right )-\left (5 b-11 a x^{2/3}\right ) \left (a x^{2/3}+b\right )^2 \sqrt {\frac {a x^{2/3}}{b}+1}\right )}{55 a^2 \sqrt {\frac {a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(2*Sqrt[b*x^(1/3) + a*x]*(-((5*b - 11*a*x^(2/3))*(b + a*x^(2/3))^2*Sqrt[1 + (a*x^(2/3))/b]) + 5*b^3*Hypergeome
tric2F1[-3/2, 1/4, 5/4, -((a*x^(2/3))/b)]))/(55*a^2*Sqrt[1 + (a*x^(2/3))/b])

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fricas [F]  time = 1.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x + b*x^(1/3))^(3/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + b*x^(1/3))^(3/2), x)

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maple [A]  time = 0.11, size = 164, normalized size = 0.79 \[ \frac {\frac {2 a^{5} x^{3}}{5}+\frac {56 a^{4} b \,x^{\frac {7}{3}}}{55}+\frac {262 a^{3} b^{2} x^{\frac {5}{3}}}{385}-\frac {16 a^{2} b^{3} x}{385}-\frac {8 a \,b^{4} x^{\frac {1}{3}}}{77}+\frac {4 \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b^{4} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{77}}{\sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/385*(10*b^4*(-a*b)^(1/2)*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-a*x^(1/3)+(-a*b)^(1/2))/(-
a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*a*x^(1/3))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/
2*2^(1/2))+131*a^3*b^2*x^(5/3)+196*a^4*b*x^(7/3)-8*a^2*b^3*x+77*a^5*x^3-20*a*b^4*x^(1/3))/a^3/((a*x^(2/3)+b)*x
^(1/3))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a x + b x^{\frac {1}{3}}\right )}^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + b*x^(1/3))^(3/2), x)

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mupad [B]  time = 5.19, size = 40, normalized size = 0.19 \[ \frac {2\,x\,{\left (a\,x+b\,x^{1/3}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {9}{4};\ \frac {13}{4};\ -\frac {a\,x^{2/3}}{b}\right )}{3\,{\left (\frac {a\,x^{2/3}}{b}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(1/3))^(3/2),x)

[Out]

(2*x*(a*x + b*x^(1/3))^(3/2)*hypergeom([-3/2, 9/4], 13/4, -(a*x^(2/3))/b))/(3*((a*x^(2/3))/b + 1)^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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