3.5.9 \(\int \frac {1}{x^4 \sqrt {a+b x^3}} \, dx\) [409]

Optimal. Leaf size=50 \[ -\frac {\sqrt {a+b x^3}}{3 a x^3}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 44, 65, 214} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {\sqrt {a+b x^3}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a + b*x^3]),x]

[Out]

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

Rule 272

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]]

Rubi steps

\begin {align*} \int \frac {1}{x^4 \sqrt {a+b x^3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3}}{3 a x^3}-\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{6 a}\\ &=-\frac {\sqrt {a+b x^3}}{3 a x^3}-\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{3 a}\\ &=-\frac {\sqrt {a+b x^3}}{3 a x^3}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 50, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+b x^3}}{3 a x^3}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a + b*x^3]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 39, normalized size = 0.78

method	result	size
default	\(\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{3 a \,x^{3}}\)	\(39\)
risch	\(\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{3 a \,x^{3}}\)	\(39\)
elliptic	\(\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{3 a \,x^{3}}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 68, normalized size = 1.36 \begin {gather*} -\frac {\sqrt {b x^{3} + a} b}{3 \, {\left ({\left (b x^{3} + a\right )} a - a^{2}\right )}} - \frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{6 \, a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(b*x^3 + a)*b/((b*x^3 + a)*a - a^2) - 1/6*b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a
)))/a^(3/2)

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 107, normalized size = 2.14 \begin {gather*} \left [\frac {\sqrt {a} b x^{3} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, \sqrt {b x^{3} + a} a}{6 \, a^{2} x^{3}}, -\frac {\sqrt {-a} b x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + \sqrt {b x^{3} + a} a}{3 \, a^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 1.17, size = 49, normalized size = 0.98 \begin {gather*} - \frac {\sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 a x^{\frac {3}{2}}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.04, size = 51, normalized size = 1.02 \begin {gather*} -\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {b x^{3} + a} b}{a x^{3}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 1.28, size = 59, normalized size = 1.18 \begin {gather*} \frac {b\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{6\,a^{3/2}}-\frac {\sqrt {b\,x^3+a}}{3\,a\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^3)^(1/2)),x)

[Out]

(b*log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x^6))/(6*a^(3/2)) - (a + b*x^3)^(1/2)/(
3*a*x^3)

________________________________________________________________________________________