Integrand size = 15, antiderivative size = 69 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 b}{4 a^2 \sqrt {a+b x^4}}-\frac {1}{4 a x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \] Output:
-3/4*b/a^2/(b*x^4+a)^(1/2)-1/4/a/x^4/(b*x^4+a)^(1/2)+3/4*b*arctanh((b*x^4+ a)^(1/2)/a^(1/2))/a^(5/2)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-a-3 b x^4}{4 a^2 x^4 \sqrt {a+b x^4}}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{4 a^{5/2}} \] Input:
Integrate[1/(x^5*(a + b*x^4)^(3/2)),x]
Output:
(-a - 3*b*x^4)/(4*a^2*x^4*Sqrt[a + b*x^4]) + (3*b*ArcTanh[Sqrt[a + b*x^4]/ Sqrt[a]])/(4*a^(5/2))
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (b x^4+a\right )^{3/2}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/2}}dx^4}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {\int \frac {1}{x^4 \sqrt {b x^4+a}}dx^4}{a}+\frac {2}{a \sqrt {a+b x^4}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {x^8}{b}-\frac {a}{b}}d\sqrt {b x^4+a}}{a b}+\frac {2}{a \sqrt {a+b x^4}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x^4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^4}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x^4 \sqrt {a+b x^4}}\right )\) |
Input:
Int[1/(x^5*(a + b*x^4)^(3/2)),x]
Output:
(-(1/(a*x^4*Sqrt[a + b*x^4])) - (3*b*(2/(a*Sqrt[a + b*x^4]) - (2*ArcTanh[S qrt[a + b*x^4]/Sqrt[a]])/a^(3/2)))/(2*a))/4
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 = 0.67 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(\frac {b \left (-\frac {\sqrt {b \,x^{4}+a}}{b \,x^{4}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{4}+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {2}{\sqrt {b \,x^{4}+a}}\right )}{4 a^{2}}\) | \(55\) |
default | \(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
risch | \(-\frac {\sqrt {b \,x^{4}+a}}{4 a^{2} x^{4}}-\frac {b}{2 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
elliptic | \(-\frac {1}{4 a \,x^{4} \sqrt {b \,x^{4}+a}}-\frac {3 b}{4 a^{2} \sqrt {b \,x^{4}+a}}+\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{4}+a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}\) | \(63\) |
Input:
int(1/x^5/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4*b/a^2*(-(b*x^4+a)^(1/2)/b/x^4+3*arctanh((b*x^4+a)^(1/2)/a^(1/2))/a^(1/ 2)-2/(b*x^4+a)^(1/2))
Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.46 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {a} \log \left (\frac {b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) - 2 \, {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}, -\frac {3 \, {\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{4} + a}}\right ) + {\left (3 \, a b x^{4} + a^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}\right ] \] Input:
integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
[1/8*(3*(b^2*x^8 + a*b*x^4)*sqrt(a)*log((b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2*(3*a*b*x^4 + a^2)*sqrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4), -1/4*(3*(b^2*x^8 + a*b*x^4)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^4 + a)) + ( 3*a*b*x^4 + a^2)*sqrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4)]
Time = 1.74 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=- \frac {1}{4 a \sqrt {b} x^{6} \sqrt {\frac {a}{b x^{4}} + 1}} - \frac {3 \sqrt {b}}{4 a^{2} x^{2} \sqrt {\frac {a}{b x^{4}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{4 a^{\frac {5}{2}}} \] Input:
integrate(1/x**5/(b*x**4+a)**(3/2),x)
Output:
-1/(4*a*sqrt(b)*x**6*sqrt(a/(b*x**4) + 1)) - 3*sqrt(b)/(4*a**2*x**2*sqrt(a /(b*x**4) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x**2))/(4*a**(5/2))
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} a^{2} - \sqrt {b x^{4} + a} a^{3}\right )}} - \frac {3 \, b \log \left (\frac {\sqrt {b x^{4} + a} - \sqrt {a}}{\sqrt {b x^{4} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} \] Input:
integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
-1/4*(3*(b*x^4 + a)*b - 2*a*b)/((b*x^4 + a)^(3/2)*a^2 - sqrt(b*x^4 + a)*a^ 3) - 3/8*b*log((sqrt(b*x^4 + a) - sqrt(a))/(sqrt(b*x^4 + a) + sqrt(a)))/a^ (5/2)
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x^{4} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} - \frac {3 \, {\left (b x^{4} + a\right )} b - 2 \, a b}{4 \, {\left ({\left (b x^{4} + a\right )}^{\frac {3}{2}} - \sqrt {b x^{4} + a} a\right )} a^{2}} \] Input:
integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
-3/4*b*arctan(sqrt(b*x^4 + a)/sqrt(-a))/(sqrt(-a)*a^2) - 1/4*(3*(b*x^4 + a )*b - 2*a*b)/(((b*x^4 + a)^(3/2) - sqrt(b*x^4 + a)*a)*a^2)
Time = 0.57 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^4+a}}{\sqrt {a}}\right )}{4\,a^{5/2}}-\frac {1}{4\,a\,x^4\,\sqrt {b\,x^4+a}}-\frac {3\,b}{4\,a^2\,\sqrt {b\,x^4+a}} \] Input:
int(1/(x^5*(a + b*x^4)^(3/2)),x)
Output:
(3*b*atanh((a + b*x^4)^(1/2)/a^(1/2)))/(4*a^(5/2)) - 1/(4*a*x^4*(a + b*x^4 )^(1/2)) - (3*b)/(4*a^2*(a + b*x^4)^(1/2))
Time = 0.22 (sec) , antiderivative size = 513, normalized size of antiderivative = 7.43 \[ \int \frac {1}{x^5 \left (a+b x^4\right )^{3/2}} \, dx=\frac {-\sqrt {a}\, \sqrt {b \,x^{4}+a}\, a^{2}-7 \sqrt {a}\, \sqrt {b \,x^{4}+a}\, a b \,x^{4}-12 \sqrt {a}\, \sqrt {b \,x^{4}+a}\, b^{2} x^{8}-9 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a b \,x^{6}-12 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{2} x^{10}+9 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a b \,x^{6}+12 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{2} x^{10}-3 \sqrt {b}\, \sqrt {a}\, a^{2} x^{2}-13 \sqrt {b}\, \sqrt {a}\, a b \,x^{6}-12 \sqrt {b}\, \sqrt {a}\, b^{2} x^{10}-3 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{2} b \,x^{4}-15 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a \,b^{2} x^{8}-12 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}-\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{3} x^{12}+3 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a^{2} b \,x^{4}+15 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) a \,b^{2} x^{8}+12 \,\mathrm {log}\left (\frac {\sqrt {b \,x^{4}+a}+\sqrt {a}+\sqrt {b}\, x^{2}}{\sqrt {a}}\right ) b^{3} x^{12}}{4 \sqrt {a}\, a^{2} x^{4} \left (3 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, a \,x^{2}+4 \sqrt {b}\, \sqrt {b \,x^{4}+a}\, b \,x^{6}+a^{2}+5 a b \,x^{4}+4 b^{2} x^{8}\right )} \] Input:
int(1/x^5/(b*x^4+a)^(3/2),x)
Output:
( - sqrt(a)*sqrt(a + b*x**4)*a**2 - 7*sqrt(a)*sqrt(a + b*x**4)*a*b*x**4 - 12*sqrt(a)*sqrt(a + b*x**4)*b**2*x**8 - 9*sqrt(b)*sqrt(a + b*x**4)*log((sq rt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a*b*x**6 - 12*sqrt(b)*sq rt(a + b*x**4)*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b* *2*x**10 + 9*sqrt(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) + sqrt(a) + sq rt(b)*x**2)/sqrt(a))*a*b*x**6 + 12*sqrt(b)*sqrt(a + b*x**4)*log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b**2*x**10 - 3*sqrt(b)*sqrt(a)* a**2*x**2 - 13*sqrt(b)*sqrt(a)*a*b*x**6 - 12*sqrt(b)*sqrt(a)*b**2*x**10 - 3*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a**2*b*x**4 - 1 5*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a*b**2*x**8 - 1 2*log((sqrt(a + b*x**4) - sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b**3*x**12 + 3* log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a**2*b*x**4 + 15* log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*a*b**2*x**8 + 12* log((sqrt(a + b*x**4) + sqrt(a) + sqrt(b)*x**2)/sqrt(a))*b**3*x**12)/(4*sq rt(a)*a**2*x**4*(3*sqrt(b)*sqrt(a + b*x**4)*a*x**2 + 4*sqrt(b)*sqrt(a + b* x**4)*b*x**6 + a**2 + 5*a*b*x**4 + 4*b**2*x**8))