Integrand size = 16, antiderivative size = 72 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=-\frac {a \sqrt {a-b x^4}}{8 x^8}+\frac {5 b \sqrt {a-b x^4}}{16 x^4}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}} \] Output:
-1/8*a*(-b*x^4+a)^(1/2)/x^8+5/16*b*(-b*x^4+a)^(1/2)/x^4-3/16*b^2*arctanh(( -b*x^4+a)^(1/2)/a^(1/2))/a^(1/2)
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\frac {\sqrt {a-b x^4} \left (-2 a+5 b x^4\right )}{16 x^8}-\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{16 \sqrt {a}} \] Input:
Integrate[(a - b*x^4)^(3/2)/x^9,x]
Output:
(Sqrt[a - b*x^4]*(-2*a + 5*b*x^4))/(16*x^8) - (3*b^2*ArcTanh[Sqrt[a - b*x^ 4]/Sqrt[a]])/(16*Sqrt[a])
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {798, 51, 51, 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.
\(\displaystyle \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\left (a-b x^4\right )^{3/2}}{x^{12}}dx^4\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (-\frac {3}{4} b \int \frac {\sqrt {a-b x^4}}{x^8}dx^4-\frac {\left (a-b x^4\right )^{3/2}}{2 x^8}\right )\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{4} \left (-\frac {3}{4} b \left (-\frac {1}{2} b \int \frac {1}{x^4 \sqrt {a-b x^4}}dx^4-\frac {\sqrt {a-b x^4}}{x^4}\right )-\frac {\left (a-b x^4\right )^{3/2}}{2 x^8}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {3}{4} b \left (\int \frac {1}{\frac {a}{b}-\frac {x^8}{b}}d\sqrt {a-b x^4}-\frac {\sqrt {a-b x^4}}{x^4}\right )-\frac {\left (a-b x^4\right )^{3/2}}{2 x^8}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4} \left (-\frac {3}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a-b x^4}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a-b x^4}}{x^4}\right )-\frac {\left (a-b x^4\right )^{3/2}}{2 x^8}\right )\) |
Input:
Int[(a - b*x^4)^(3/2)/x^9,x]
Output:
(-1/2*(a - b*x^4)^(3/2)/x^8 - (3*b*(-(Sqrt[a - b*x^4]/x^4) + (b*ArcTanh[Sq rt[a - b*x^4]/Sqrt[a]])/Sqrt[a]))/4)/4
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 0.65 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82
method | result | size |
risch | \(-\frac {\sqrt {-b \,x^{4}+a}\, \left (-5 b \,x^{4}+2 a \right )}{16 x^{8}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{16 \sqrt {a}}\) | \(59\) |
default | \(-\frac {a \sqrt {-b \,x^{4}+a}}{8 x^{8}}+\frac {5 b \sqrt {-b \,x^{4}+a}}{16 x^{4}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{16 \sqrt {a}}\) | \(66\) |
elliptic | \(-\frac {a \sqrt {-b \,x^{4}+a}}{8 x^{8}}+\frac {5 b \sqrt {-b \,x^{4}+a}}{16 x^{4}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}}{x^{2}}\right )}{16 \sqrt {a}}\) | \(66\) |
pseudoelliptic | \(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {-b \,x^{4}+a}}{\sqrt {a}}\right ) b^{2} x^{8}+5 b \,x^{4} \sqrt {-b \,x^{4}+a}\, \sqrt {a}-2 \sqrt {-b \,x^{4}+a}\, a^{\frac {3}{2}}}{16 x^{8} \sqrt {a}}\) | \(67\) |
Input:
int((-b*x^4+a)^(3/2)/x^9,x,method=_RETURNVERBOSE)
Output:
-1/16*(-b*x^4+a)^(1/2)*(-5*b*x^4+2*a)/x^8-3/16*b^2/a^(1/2)*ln((2*a+2*a^(1/ 2)*(-b*x^4+a)^(1/2))/x^2)
Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.08 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{8} \log \left (\frac {b x^{4} + 2 \, \sqrt {-b x^{4} + a} \sqrt {a} - 2 \, a}{x^{4}}\right ) + 2 \, {\left (5 \, a b x^{4} - 2 \, a^{2}\right )} \sqrt {-b x^{4} + a}}{32 \, a x^{8}}, -\frac {3 \, \sqrt {-a} b^{2} x^{8} \arctan \left (\frac {\sqrt {-b x^{4} + a} \sqrt {-a}}{b x^{4} - a}\right ) - {\left (5 \, a b x^{4} - 2 \, a^{2}\right )} \sqrt {-b x^{4} + a}}{16 \, a x^{8}}\right ] \] Input:
integrate((-b*x^4+a)^(3/2)/x^9,x, algorithm="fricas")
Output:
[1/32*(3*sqrt(a)*b^2*x^8*log((b*x^4 + 2*sqrt(-b*x^4 + a)*sqrt(a) - 2*a)/x^ 4) + 2*(5*a*b*x^4 - 2*a^2)*sqrt(-b*x^4 + a))/(a*x^8), -1/16*(3*sqrt(-a)*b^ 2*x^8*arctan(sqrt(-b*x^4 + a)*sqrt(-a)/(b*x^4 - a)) - (5*a*b*x^4 - 2*a^2)* sqrt(-b*x^4 + a))/(a*x^8)]
Result contains complex when optimal does not.
Time = 2.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.62 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\begin {cases} - \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{4}} - 1}}{8 x^{6}} + \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{4}} - 1}}{16 x^{2}} - \frac {3 b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{16 \sqrt {a}} & \text {for}\: \left |{\frac {a}{b x^{4}}}\right | > 1 \\\frac {i a^{2}}{8 \sqrt {b} x^{10} \sqrt {- \frac {a}{b x^{4}} + 1}} - \frac {7 i a \sqrt {b}}{16 x^{6} \sqrt {- \frac {a}{b x^{4}} + 1}} + \frac {5 i b^{\frac {3}{2}}}{16 x^{2} \sqrt {- \frac {a}{b x^{4}} + 1}} + \frac {3 i b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{2}} \right )}}{16 \sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate((-b*x**4+a)**(3/2)/x**9,x)
Output:
Piecewise((-a*sqrt(b)*sqrt(a/(b*x**4) - 1)/(8*x**6) + 5*b**(3/2)*sqrt(a/(b *x**4) - 1)/(16*x**2) - 3*b**2*acosh(sqrt(a)/(sqrt(b)*x**2))/(16*sqrt(a)), Abs(a/(b*x**4)) > 1), (I*a**2/(8*sqrt(b)*x**10*sqrt(-a/(b*x**4) + 1)) - 7 *I*a*sqrt(b)/(16*x**6*sqrt(-a/(b*x**4) + 1)) + 5*I*b**(3/2)/(16*x**2*sqrt( -a/(b*x**4) + 1)) + 3*I*b**2*asin(sqrt(a)/(sqrt(b)*x**2))/(16*sqrt(a)), Tr ue))
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\frac {3 \, b^{2} \log \left (\frac {\sqrt {-b x^{4} + a} - \sqrt {a}}{\sqrt {-b x^{4} + a} + \sqrt {a}}\right )}{32 \, \sqrt {a}} - \frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {-b x^{4} + a} a b^{2}}{16 \, {\left ({\left (b x^{4} - a\right )}^{2} + 2 \, {\left (b x^{4} - a\right )} a + a^{2}\right )}} \] Input:
integrate((-b*x^4+a)^(3/2)/x^9,x, algorithm="maxima")
Output:
3/32*b^2*log((sqrt(-b*x^4 + a) - sqrt(a))/(sqrt(-b*x^4 + a) + sqrt(a)))/sq rt(a) - 1/16*(5*(-b*x^4 + a)^(3/2)*b^2 - 3*sqrt(-b*x^4 + a)*a*b^2)/((b*x^4 - a)^2 + 2*(b*x^4 - a)*a + a^2)
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {-b x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (-b x^{4} + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {-b x^{4} + a} a b^{3}}{b^{2} x^{8}}}{16 \, b} \] Input:
integrate((-b*x^4+a)^(3/2)/x^9,x, algorithm="giac")
Output:
1/16*(3*b^3*arctan(sqrt(-b*x^4 + a)/sqrt(-a))/sqrt(-a) - (5*(-b*x^4 + a)^( 3/2)*b^3 - 3*sqrt(-b*x^4 + a)*a*b^3)/(b^2*x^8))/b
Time = 0.76 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\frac {3\,a\,\sqrt {a-b\,x^4}}{16\,x^8}-\frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a-b\,x^4}}{\sqrt {a}}\right )}{16\,\sqrt {a}}-\frac {5\,{\left (a-b\,x^4\right )}^{3/2}}{16\,x^8} \] Input:
int((a - b*x^4)^(3/2)/x^9,x)
Output:
(3*a*(a - b*x^4)^(1/2))/(16*x^8) - (3*b^2*atanh((a - b*x^4)^(1/2)/a^(1/2)) )/(16*a^(1/2)) - (5*(a - b*x^4)^(3/2))/(16*x^8)
Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a-b x^4\right )^{3/2}}{x^9} \, dx=\frac {-2 \sqrt {a}\, \sqrt {-b \,x^{4}+a}\, a +5 \sqrt {a}\, \sqrt {-b \,x^{4}+a}\, b \,x^{4}+3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}\right )}{2}\right )\right ) b^{2} x^{8}}{16 \sqrt {a}\, x^{8}} \] Input:
int((-b*x^4+a)^(3/2)/x^9,x)
Output:
( - 2*sqrt(a)*sqrt(a - b*x**4)*a + 5*sqrt(a)*sqrt(a - b*x**4)*b*x**4 + 3*l og(tan(asin((sqrt(b)*x**2)/sqrt(a))/2))*b**2*x**8)/(16*sqrt(a)*x**8)