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

Optimal. Leaf size=64 \[ -\frac{b}{a^2 \sqrt{a+b x^3}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{a^{5/2}}-\frac{1}{3 a x^3 \sqrt{a+b x^3}} \]

[Out]

-(b/(a^2*Sqrt[a + b*x^3])) - 1/(3*a*x^3*Sqrt[a + b*x^3]) + (b*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/a^(5/2)

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Rubi [A]  time = 0.0375877, antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{\sqrt{a+b x^3}}{a^2 x^3}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{2}{3 a x^3 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)^(3/2)),x]

[Out]

2/(3*a*x^3*Sqrt[a + b*x^3]) - Sqrt[a + b*x^3]/(a^2*x^3) + (b*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/a^(5/2)

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{2}{3 a x^3 \sqrt{a+b x^3}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^3\right )}{a}\\ &=\frac{2}{3 a x^3 \sqrt{a+b x^3}}-\frac{\sqrt{a+b x^3}}{a^2 x^3}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )}{2 a^2}\\ &=\frac{2}{3 a x^3 \sqrt{a+b x^3}}-\frac{\sqrt{a+b x^3}}{a^2 x^3}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{a^2}\\ &=\frac{2}{3 a x^3 \sqrt{a+b x^3}}-\frac{\sqrt{a+b x^3}}{a^2 x^3}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0071917, size = 37, normalized size = 0.58 \[ -\frac{2 b \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^3}{a}+1\right )}{3 a^2 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 57, normalized size = 0.9 \begin{align*} -{\frac{2\,b}{3\,{a}^{2}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{3\,{x}^{3}{a}^{2}}\sqrt{b{x}^{3}+a}}+{b{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*b/a^2/((x^3+1/b*a)*b)^(1/2)-1/3/a^2*(b*x^3+a)^(1/2)/x^3+b*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54105, size = 383, normalized size = 5.98 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt{a} \log \left (\frac{b x^{3} + 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) - 2 \,{\left (3 \, a b x^{3} + a^{2}\right )} \sqrt{b x^{3} + a}}{6 \,{\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac{3 \,{\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b x^{3} + a^{2}\right )} \sqrt{b x^{3} + a}}{3 \,{\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*(b^2*x^6 + a*b*x^3)*sqrt(a)*log((b*x^3 + 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) - 2*(3*a*b*x^3 + a^2)*s
qrt(b*x^3 + a))/(a^3*b*x^6 + a^4*x^3), -1/3*(3*(b^2*x^6 + a*b*x^3)*sqrt(-a)*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a)
+ (3*a*b*x^3 + a^2)*sqrt(b*x^3 + a))/(a^3*b*x^6 + a^4*x^3)]

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Sympy [A]  time = 4.22682, size = 75, normalized size = 1.17 \begin{align*} - \frac{1}{3 a \sqrt{b} x^{\frac{9}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{\sqrt{b}}{a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(3*a*sqrt(b)*x**(9/2)*sqrt(a/(b*x**3) + 1)) - sqrt(b)/(a**2*x**(3/2)*sqrt(a/(b*x**3) + 1)) + b*asinh(sqrt(a
)/(sqrt(b)*x**(3/2)))/a**(5/2)

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Giac [A]  time = 1.09936, size = 89, normalized size = 1.39 \begin{align*} -\frac{1}{3} \, b{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \, b x^{3} + a}{{\left ({\left (b x^{3} + a\right )}^{\frac{3}{2}} - \sqrt{b x^{3} + a} a\right )} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b*(3*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*b*x^3 + a)/(((b*x^3 + a)^(3/2) - sqrt(b*x^3 + a
)*a)*a^2))