\(\int \frac {1}{x^4 (a+b x^3)^{3/2}} \, dx\) [426]

Optimal result

Integrand size = 15, antiderivative size = 64 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {b}{a^2 \sqrt {a+b x^3}}-\frac {1}{3 a x^3 \sqrt {a+b x^3}}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{5/2}}-\frac {b}{a^2 \sqrt {a+b x^3}}-\frac {1}{3 a x^3 \sqrt {a+b x^3}} \]

[In]

Int[1/(x^4*(a + b*x^3)^(3/2)),x]

[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)

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 53

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {-a-3 b x^3}{3 a^2 x^3 \sqrt {a+b x^3}}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.86 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}-\frac {2 b}{3 a^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\)	\(57\)
elliptic	\(-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}-\frac {2 b}{3 a^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\)	\(57\)
pseudoelliptic	\(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {b \,x^{3}+a}\, b \,x^{3}-3 \sqrt {a}\, b \,x^{3}-a^{\frac {3}{2}}}{3 x^{3} a^{\frac {5}{2}} \sqrt {b \,x^{3}+a}}\)	\(62\)
risch	\(-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}-\frac {b \left (-\frac {2}{3 \sqrt {b \,x^{3}+a}}+3 a \left (\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\right )\right )}{2 a^{2}}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.70 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (3 \, a b x^{3} + a^{2}\right )} \sqrt {b x^{3} + a}}{6 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac {3 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b x^{3} + a^{2}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=- \frac {1}{3 a \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {\sqrt {b}}{a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{a^{\frac {5}{2}}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{3} + a\right )} b - 2 \, a b}{3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{3} + a} a^{3}\right )}} - \frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} \]

[In]

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

[Out]

-1/3*(3*(b*x^3 + a)*b - 2*a*b)/((b*x^3 + a)^(3/2)*a^2 - sqrt(b*x^3 + a)*a^3) - 1/2*b*log((sqrt(b*x^3 + a) - sq
rt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{3} + a\right )} b - 2 \, a b}{3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{3} + a} a\right )} a^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {b\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{2\,a^{5/2}}-\frac {2\,b}{3\,a^2\,\sqrt {b\,x^3+a}}-\frac {\sqrt {b\,x^3+a}}{3\,a^2\,x^3} \]

[In]

int(1/(x^4*(a + b*x^3)^(3/2)),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))/(2*a^(5/2)) - (2*b)/(3*a^2*(a + b
*x^3)^(1/2)) - (a + b*x^3)^(1/2)/(3*a^2*x^3)