3.2280 \(\int \frac{1}{x^4 (a+b x^{3/2})^{2/3}} \, dx\)

Optimal. Leaf size=148 \[ \frac{5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}-\frac{10 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^{3/2}}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3}}-\frac{5 b^2 \log (x)}{18 a^{8/3}}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3} \]

[Out]

-(a + b*x^(3/2))^(1/3)/(3*a*x^3) + (5*b*(a + b*x^(3/2))^(1/3))/(9*a^2*x^(3/2)) - (10*b^2*ArcTan[(a^(1/3) + 2*(
a + b*x^(3/2))^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)) - (5*b^2*Log[x])/(18*a^(8/3)) + (5*b^2*Log[a^(1/
3) - (a + b*x^(3/2))^(1/3)])/(9*a^(8/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0878129, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {266, 51, 57, 617, 204, 31} \[ \frac{5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}-\frac{10 b^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{a+b x^{3/2}}+\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3}}-\frac{5 b^2 \log (x)}{18 a^{8/3}}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^(3/2))^(2/3)),x]

[Out]

-(a + b*x^(3/2))^(1/3)/(3*a*x^3) + (5*b*(a + b*x^(3/2))^(1/3))/(9*a^2*x^(3/2)) - (10*b^2*ArcTan[(a^(1/3) + 2*(
a + b*x^(3/2))^(1/3))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)) - (5*b^2*Log[x])/(18*a^(8/3)) + (5*b^2*Log[a^(1/
3) - (a + b*x^(3/2))^(1/3)])/(9*a^(8/3))

Rule 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b x^{3/2}\right )^{2/3}} \, dx &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{2/3}} \, dx,x,x^{3/2}\right )\\ &=-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{2/3}} \, dx,x,x^{3/2}\right )}{9 a}\\ &=-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}+\frac{\left (10 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{2/3}} \, dx,x,x^{3/2}\right )}{27 a^2}\\ &=-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac{5 b^2 \log (x)}{18 a^{8/3}}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{7/3}}\\ &=-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac{5 b^2 \log (x)}{18 a^{8/3}}+\frac{5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}+\frac{\left (10 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}\right )}{9 a^{8/3}}\\ &=-\frac{\sqrt [3]{a+b x^{3/2}}}{3 a x^3}+\frac{5 b \sqrt [3]{a+b x^{3/2}}}{9 a^2 x^{3/2}}-\frac{10 b^2 \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{a+b x^{3/2}}}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{9 \sqrt{3} a^{8/3}}-\frac{5 b^2 \log (x)}{18 a^{8/3}}+\frac{5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^{3/2}}\right )}{9 a^{8/3}}\\ \end{align*}

Mathematica [C]  time = 0.0093286, size = 41, normalized size = 0.28 \[ -\frac{2 b^2 \sqrt [3]{a+b x^{3/2}} \, _2F_1\left (\frac{1}{3},3;\frac{4}{3};\frac{b x^{3/2}}{a}+1\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}} \left ( a+b{x}^{{\frac{3}{2}}} \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [C]  time = 20.7382, size = 42, normalized size = 0.28 \begin{align*} - \frac{2 \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{a e^{i \pi }}{b x^{\frac{3}{2}}}} \right )}}{3 b^{\frac{2}{3}} x^{4} \Gamma \left (\frac{11}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*x**(3/2))**(2/3),x)

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*x^(3/2))^(2/3),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError