Integrand size = 15, antiderivative size = 64 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {b}{a^2 \sqrt {a+b x^3}}-\frac {1}{3 a x^3 \sqrt {a+b x^3}}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{5/2}} \] Output:
-b/a^2/(b*x^3+a)^(1/2)-1/3/a/x^3/(b*x^3+a)^(1/2)+b*arctanh((b*x^3+a)^(1/2) /a^(1/2))/a^(5/2)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {-a-3 b x^3}{3 a^2 x^3 \sqrt {a+b x^3}}+\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{5/2}} \] Input:
Integrate[1/(x^4*(a + b*x^3)^(3/2)),x]
Output:
(-a - 3*b*x^3)/(3*a^2*x^3*Sqrt[a + b*x^3]) + (b*ArcTanh[Sqrt[a + b*x^3]/Sq rt[a]])/a^(5/2)
Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {798, 52, 61, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^6 \left (b x^3+a\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 b \int \frac {1}{x^3 \left (b x^3+a\right )^{3/2}}dx^3}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 b \left (\frac {\int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3}{a}+\frac {2}{a \sqrt {a+b x^3}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{a b}+\frac {2}{a \sqrt {a+b x^3}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x^3}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )\) |
Input:
Int[1/(x^4*(a + b*x^3)^(3/2)),x]
Output:
(-(1/(a*x^3*Sqrt[a + b*x^3])) - (3*b*(2/(a*Sqrt[a + b*x^3]) - (2*ArcTanh[S qrt[a + b*x^3]/Sqrt[a]])/a^(3/2)))/(2*a))/3
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Time = 0.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86
method | result | size |
pseudoelliptic | \(\frac {b \left (-\frac {\sqrt {b \,x^{3}+a}}{x^{3} b}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2}{\sqrt {b \,x^{3}+a}}\right )}{3 a^{2}}\) | \(55\) |
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 | \(-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}-\frac {2 b}{3 a^{2} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}+\frac {b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}\) | \(57\) |
risch | \(-\frac {\sqrt {b \,x^{3}+a}}{3 a^{2} x^{3}}-\frac {b \left (-\frac {2}{3 \sqrt {b \,x^{3}+a}}+3 a \left (\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\right )\right )}{2 a^{2}}\) | \(78\) |
Input:
int(1/x^4/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/3*b/a^2*(-(b*x^3+a)^(1/2)/x^3/b+3*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(1/ 2)-2/(b*x^3+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.66 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (3 \, a b x^{3} + a^{2}\right )} \sqrt {b x^{3} + a}}{6 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac {3 \, {\left (b^{2} x^{6} + a b x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{3} + a}}\right ) + {\left (3 \, a b x^{3} + a^{2}\right )} \sqrt {b x^{3} + a}}{3 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}\right ] \] Input:
integrate(1/x^4/(b*x^3+a)^(3/2),x, algorithm="fricas")
Output:
[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)*sqrt(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(-a)/sqrt(b*x^3 + a)) + ( 3*a*b*x^3 + a^2)*sqrt(b*x^3 + a))/(a^3*b*x^6 + a^4*x^3)]
Time = 1.87 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=- \frac {1}{3 a \sqrt {b} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {\sqrt {b}}{a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{a^{\frac {5}{2}}} \] Input:
integrate(1/x**4/(b*x**3+a)**(3/2),x)
Output:
-1/(3*a*sqrt(b)*x**(9/2)*sqrt(a/(b*x**3) + 1)) - sqrt(b)/(a**2*x**(3/2)*sq rt(a/(b*x**3) + 1)) + b*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/a**(5/2)
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{3} + a\right )} b - 2 \, a b}{3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{3} + a} a^{3}\right )}} - \frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{2 \, a^{\frac {5}{2}}} \] Input:
integrate(1/x^4/(b*x^3+a)^(3/2),x, algorithm="maxima")
Output:
-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) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^ (5/2)
Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=-\frac {b \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{3} + a\right )} b - 2 \, a b}{3 \, {\left ({\left (b x^{3} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{3} + a} a\right )} a^{2}} \] Input:
integrate(1/x^4/(b*x^3+a)^(3/2),x, algorithm="giac")
Output:
-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)
Time = 0.55 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {b\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{2\,a^{5/2}}-\frac {2\,b}{3\,a^2\,\sqrt {b\,x^3+a}}-\frac {\sqrt {b\,x^3+a}}{3\,a^2\,x^3} \] Input:
int(1/(x^4*(a + b*x^3)^(3/2)),x)
Output:
(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)
Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.12 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {b \,x^{3}+a}\, a^{2}-6 \sqrt {b \,x^{3}+a}\, a b \,x^{3}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {a}\right ) a b \,x^{3}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {a}\right ) b^{2} x^{6}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {a}\right ) a b \,x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {a}\right ) b^{2} x^{6}}{6 a^{3} x^{3} \left (b \,x^{3}+a \right )} \] Input:
int(1/x^4/(b*x^3+a)^(3/2),x)
Output:
( - 2*sqrt(a + b*x**3)*a**2 - 6*sqrt(a + b*x**3)*a*b*x**3 - 3*sqrt(a)*log( sqrt(a + b*x**3) - sqrt(a))*a*b*x**3 - 3*sqrt(a)*log(sqrt(a + b*x**3) - sq rt(a))*b**2*x**6 + 3*sqrt(a)*log(sqrt(a + b*x**3) + sqrt(a))*a*b*x**3 + 3* sqrt(a)*log(sqrt(a + b*x**3) + sqrt(a))*b**2*x**6)/(6*a**3*x**3*(a + b*x** 3))