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

Optimal. Leaf size=239 \[ \frac {663 b^{15/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{154 a^{21/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {663 b^3 \sqrt {a x+b \sqrt [3]{x}}}{77 a^5}+\frac {1989 b^2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{55 a^3}+\frac {17 x^2 \sqrt {a x+b \sqrt [3]{x}}}{5 a^2}-\frac {3 x^3}{a \sqrt {a x+b \sqrt [3]{x}}} \]

[Out]

-3*x^3/a/(b*x^(1/3)+a*x)^(1/2)-663/77*b^3*(b*x^(1/3)+a*x)^(1/2)/a^5+1989/385*b^2*x^(2/3)*(b*x^(1/3)+a*x)^(1/2)
/a^4-221/55*b*x^(4/3)*(b*x^(1/3)+a*x)^(1/2)/a^3+17/5*x^2*(b*x^(1/3)+a*x)^(1/2)/a^2+663/154*b^(15/4)*x^(1/6)*(c
os(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^(21/4)/(b*x^(1/3)+a*x)^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2018, 2022, 2024, 2011, 329, 220} \[ \frac {1989 b^2 x^{2/3} \sqrt {a x+b \sqrt [3]{x}}}{385 a^4}+\frac {663 b^{15/4} \sqrt [6]{x} \left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right ) \sqrt {\frac {a x^{2/3}+b}{\left (\sqrt {a} \sqrt [3]{x}+\sqrt {b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{154 a^{21/4} \sqrt {a x+b \sqrt [3]{x}}}-\frac {663 b^3 \sqrt {a x+b \sqrt [3]{x}}}{77 a^5}-\frac {221 b x^{4/3} \sqrt {a x+b \sqrt [3]{x}}}{55 a^3}+\frac {17 x^2 \sqrt {a x+b \sqrt [3]{x}}}{5 a^2}-\frac {3 x^3}{a \sqrt {a x+b \sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(b*x^(1/3) + a*x)^(3/2),x]

[Out]

(-3*x^3)/(a*Sqrt[b*x^(1/3) + a*x]) - (663*b^3*Sqrt[b*x^(1/3) + a*x])/(77*a^5) + (1989*b^2*x^(2/3)*Sqrt[b*x^(1/
3) + a*x])/(385*a^4) - (221*b*x^(4/3)*Sqrt[b*x^(1/3) + a*x])/(55*a^3) + (17*x^2*Sqrt[b*x^(1/3) + a*x])/(5*a^2)
+ (663*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)*Ellip
ticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/(154*a^(21/4)*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 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 2022

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*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(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] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

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 \frac {x^3}{\left (b \sqrt [3]{x}+a x\right )^{3/2}} \, dx &=3 \operatorname {Subst}\left (\int \frac {x^{11}}{\left (b x+a x^3\right )^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {51 \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{2 a}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}-\frac {(221 b) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{10 a^2}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}+\frac {\left (1989 b^2\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{110 a^3}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}+\frac {1989 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}-\frac {\left (1989 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{154 a^4}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {663 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^5}+\frac {1989 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}+\frac {\left (663 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{154 a^5}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {663 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^5}+\frac {1989 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}+\frac {\left (663 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 )}{154 a^5 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {663 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^5}+\frac {1989 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}+\frac {\left (663 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^5 \sqrt {b \sqrt [3]{x}+a x}}\\ &=-\frac {3 x^3}{a \sqrt {b \sqrt [3]{x}+a x}}-\frac {663 b^3 \sqrt {b \sqrt [3]{x}+a x}}{77 a^5}+\frac {1989 b^2 x^{2/3} \sqrt {b \sqrt [3]{x}+a x}}{385 a^4}-\frac {221 b x^{4/3} \sqrt {b \sqrt [3]{x}+a x}}{55 a^3}+\frac {17 x^2 \sqrt {b \sqrt [3]{x}+a x}}{5 a^2}+\frac {663 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 )}{154 a^{21/4} \sqrt {b \sqrt [3]{x}+a x}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 124, normalized size = 0.52 \[ \frac {\sqrt {a x+b \sqrt [3]{x}} \left (154 a^4 x^{8/3}-238 a^3 b x^2+442 a^2 b^2 x^{4/3}+3315 b^4 \sqrt {\frac {a x^{2/3}}{b}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {a x^{2/3}}{b}\right )-1326 a b^3 x^{2/3}-3315 b^4\right )}{385 a^5 \left (a x^{2/3}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^(1/3) + a*x]*(-3315*b^4 - 1326*a*b^3*x^(2/3) + 442*a^2*b^2*x^(4/3) - 238*a^3*b*x^2 + 154*a^4*x^(8/3)
+ 3315*b^4*Sqrt[1 + (a*x^(2/3))/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^(2/3))/b)]))/(385*a^5*(b + a*x^(2/
3)))

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fricas [F]  time = 1.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} x^{5} + 3 \, a^{2} b^{2} x^{\frac {11}{3}} - 2 \, a b^{3} x^{3} - {\left (2 \, a^{3} b x^{4} - b^{4} x^{2}\right )} x^{\frac {1}{3}}\right )} \sqrt {a x + b x^{\frac {1}{3}}}}{a^{6} x^{4} + 2 \, a^{3} b^{3} x^{2} + b^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^4*x^5 + 3*a^2*b^2*x^(11/3) - 2*a*b^3*x^3 - (2*a^3*b*x^4 - b^4*x^2)*x^(1/3))*sqrt(a*x + b*x^(1/3))/
(a^6*x^4 + 2*a^3*b^3*x^2 + b^6), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 260, normalized size = 1.09 \[ -\frac {-308 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{5} x^{3}+476 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{4} b \,x^{\frac {7}{3}}-884 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{3} b^{2} x^{\frac {5}{3}}+2652 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a^{2} b^{3} x +4320 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, a \,b^{4} x^{\frac {1}{3}}+2310 \sqrt {a x +b \,x^{\frac {1}{3}}}\, a \,b^{4} x^{\frac {1}{3}}-3315 \sqrt {\left (a \,x^{\frac {2}{3}}+b \right ) x^{\frac {1}{3}}}\, \sqrt {-a b}\, \sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (a \,x^{\frac {1}{3}}-\sqrt {-a b}\right )}{\sqrt {-a b}}}\, \sqrt {-\frac {a \,x^{\frac {1}{3}}}{\sqrt {-a b}}}\, b^{4} \EllipticF \left (\sqrt {\frac {a \,x^{\frac {1}{3}}+\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{770 \left (a \,x^{\frac {2}{3}}+b \right ) a^{6} x^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/770*(-884*x^(5/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^3*b^2+476*x^(7/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^4*b-331
5*((a*x^(2/3)+b)*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))*(-a*b)^(1
/2)*((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)*(-1/(-a*b)^
(1/2)*a*x^(1/3))^(1/2)*b^4+2652*x*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a^2*b^3-308*((a*x^(2/3)+b)*x^(1/3))^(1/2)*x^3*
a^5+4320*x^(1/3)*((a*x^(2/3)+b)*x^(1/3))^(1/2)*a*b^4+2310*x^(1/3)*(a*x+b*x^(1/3))^(1/2)*a*b^4)/x^(1/3)/(a*x^(2
/3)+b)/a^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x + b*x^(1/3))^(3/2),x)

[Out]

int(x^3/(a*x + b*x^(1/3))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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