\(\int \frac {x^3}{(a+b x^{3/2})^{2/3}} \, dx\) [2271]

Optimal result

Integrand size = 17, antiderivative size = 139 \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac {x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac {10 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{a+b x^{3/2}}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a^2 \log \left (\sqrt [3]{b} \sqrt {x}-\sqrt [3]{a+b x^{3/2}}\right )}{9 b^{8/3}} \]

[Out]

-5/9*a*x*(a+b*x^(3/2))^(1/3)/b^2+1/3*x^(5/2)*(a+b*x^(3/2))^(1/3)/b-5/9*a^2*ln(-(a+b*x^(3/2))^(1/3)+b^(1/3)*x^(
1/2))/b^(8/3)-10/27*a^2*arctan(1/3*(1+2*b^(1/3)*x^(1/2)/(a+b*x^(3/2))^(1/3))*3^(1/2))/b^(8/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {348, 327, 337} \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=-\frac {10 a^2 \arctan \left (\frac {\frac {2 \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{a+b x^{3/2}}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a^2 \log \left (\sqrt [3]{b} \sqrt {x}-\sqrt [3]{a+b x^{3/2}}\right )}{9 b^{8/3}}-\frac {5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac {x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b} \]

[In]

Int[x^3/(a + b*x^(3/2))^(2/3),x]

[Out]

(-5*a*x*(a + b*x^(3/2))^(1/3))/(9*b^2) + (x^(5/2)*(a + b*x^(3/2))^(1/3))/(3*b) - (10*a^2*ArcTan[(1 + (2*b^(1/3
)*Sqrt[x])/(a + b*x^(3/2))^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(8/3)) - (5*a^2*Log[b^(1/3)*Sqrt[x] - (a + b*x^(3/2))
^(1/3)])/(9*b^(8/3))

Rule 327

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

Rule 337

Int[(x_)/((a_) + (b_.)*(x_)^3)^(2/3), x_Symbol] :> With[{q = Rt[b, 3]}, Simp[-ArcTan[(1 + 2*q*(x/(a + b*x^3)^(
1/3)))/Sqrt[3]]/(Sqrt[3]*q^2), x] - Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*q^2), x]] /; FreeQ[{a, b}, x]

Rule 348

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

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {x^7}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac {(5 a) \text {Subst}\left (\int \frac {x^4}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right )}{3 b} \\ & = -\frac {5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac {x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}+\frac {\left (10 a^2\right ) \text {Subst}\left (\int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx,x,\sqrt {x}\right )}{9 b^2} \\ & = -\frac {5 a x \sqrt [3]{a+b x^{3/2}}}{9 b^2}+\frac {x^{5/2} \sqrt [3]{a+b x^{3/2}}}{3 b}-\frac {10 a^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{a+b x^{3/2}}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {5 a^2 \log \left (\sqrt [3]{b} \sqrt {x}-\sqrt [3]{a+b x^{3/2}}\right )}{9 b^{8/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.40 \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^{3/2}} \left (-5 a x+3 b x^{5/2}\right )}{9 b^2}-\frac {10 a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt {x}}{\sqrt [3]{b} \sqrt {x}+2 \sqrt [3]{a+b x^{3/2}}}\right )}{9 \sqrt {3} b^{8/3}}-\frac {10 a^2 \log \left (-\sqrt [3]{b} \sqrt {x}+\sqrt [3]{a+b x^{3/2}}\right )}{27 b^{8/3}}+\frac {5 a^2 \log \left (b^{2/3} x+\sqrt [3]{b} \sqrt {x} \sqrt [3]{a+b x^{3/2}}+\left (a+b x^{3/2}\right )^{2/3}\right )}{27 b^{8/3}} \]

[In]

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

[Out]

((a + b*x^(3/2))^(1/3)*(-5*a*x + 3*b*x^(5/2)))/(9*b^2) - (10*a^2*ArcTan[(Sqrt[3]*b^(1/3)*Sqrt[x])/(b^(1/3)*Sqr
t[x] + 2*(a + b*x^(3/2))^(1/3))])/(9*Sqrt[3]*b^(8/3)) - (10*a^2*Log[-(b^(1/3)*Sqrt[x]) + (a + b*x^(3/2))^(1/3)
])/(27*b^(8/3)) + (5*a^2*Log[b^(2/3)*x + b^(1/3)*Sqrt[x]*(a + b*x^(3/2))^(1/3) + (a + b*x^(3/2))^(2/3)])/(27*b
^(8/3))

Maple [F]

\[\int \frac {x^{3}}{\left (a +b \,x^{\frac {3}{2}}\right )^{\frac {2}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.84 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.29 \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {2 x^{4} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{\frac {3}{2}} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {11}{3}\right )} \]

[In]

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

[Out]

2*x**4*gamma(8/3)*hyper((2/3, 8/3), (11/3,), b*x**(3/2)*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(11/3))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.32 \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\frac {10 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{\sqrt {x}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{27 \, b^{\frac {8}{3}}} + \frac {5 \, a^{2} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{\sqrt {x}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}}}{x}\right )}{27 \, b^{\frac {8}{3}}} - \frac {10 \, a^{2} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}}}{\sqrt {x}}\right )}{27 \, b^{\frac {8}{3}}} + \frac {\frac {8 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {1}{3}} a^{2} b}{\sqrt {x}} - \frac {5 \, {\left (b x^{\frac {3}{2}} + a\right )}^{\frac {4}{3}} a^{2}}{x^{2}}}{9 \, {\left (b^{4} - \frac {2 \, {\left (b x^{\frac {3}{2}} + a\right )} b^{3}}{x^{\frac {3}{2}}} + \frac {{\left (b x^{\frac {3}{2}} + a\right )}^{2} b^{2}}{x^{3}}\right )}} \]

[In]

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

[Out]

10/27*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^(3/2) + a)^(1/3)/sqrt(x))/b^(1/3))/b^(8/3) + 5/27*a^2*l
og(b^(2/3) + (b*x^(3/2) + a)^(1/3)*b^(1/3)/sqrt(x) + (b*x^(3/2) + a)^(2/3)/x)/b^(8/3) - 10/27*a^2*log(-b^(1/3)
+ (b*x^(3/2) + a)^(1/3)/sqrt(x))/b^(8/3) + 1/9*(8*(b*x^(3/2) + a)^(1/3)*a^2*b/sqrt(x) - 5*(b*x^(3/2) + a)^(4/
3)*a^2/x^2)/(b^4 - 2*(b*x^(3/2) + a)*b^3/x^(3/2) + (b*x^(3/2) + a)^2*b^2/x^3)

Giac [F]

\[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\int { \frac {x^{3}}{{\left (b x^{\frac {3}{2}} + a\right )}^{\frac {2}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\left (a+b x^{3/2}\right )^{2/3}} \, dx=\int \frac {x^3}{{\left (a+b\,x^{3/2}\right )}^{2/3}} \,d x \]

[In]

int(x^3/(a + b*x^(3/2))^(2/3),x)

[Out]

int(x^3/(a + b*x^(3/2))^(2/3), x)