Integrand size = 16, antiderivative size = 81 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \arctan \left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \] Output:
-1/4*(-b*x^4+a)^(3/4)/a/x^4+1/8*b*arctan((-b*x^4+a)^(1/4)/a^(1/4))/a^(5/4) -1/8*b*arctanh((-b*x^4+a)^(1/4)/a^(1/4))/a^(5/4)
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=-\frac {\left (a-b x^4\right )^{3/4}}{4 a x^4}+\frac {b \arctan \left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{8 a^{5/4}} \] Input:
Integrate[1/(x^5*(a - b*x^4)^(1/4)),x]
Output:
-1/4*(a - b*x^4)^(3/4)/(a*x^4) + (b*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)])/(8* a^(5/4)) - (b*ArcTanh[(a - b*x^4)^(1/4)/a^(1/4)])/(8*a^(5/4))
Time = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {798, 52, 73, 27, 827, 216, 219}
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.
\(\displaystyle \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \sqrt [4]{a-b x^4}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (\frac {b \int \frac {1}{x^4 \sqrt [4]{a-b x^4}}dx^4}{4 a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {b x^8}{a-x^{16}}d\sqrt [4]{a-b x^4}}{a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \int \frac {x^8}{a-x^{16}}d\sqrt [4]{a-b x^4}}{a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{a-b x^4}-\frac {1}{2} \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{a-b x^4}\right )}{a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \left (\frac {1}{2} \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{a-b x^4}-\frac {\arctan \left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}-\frac {\arctan \left (\frac {\sqrt [4]{a-b x^4}}{\sqrt [4]{a}}\right )}{2 \sqrt [4]{a}}\right )}{a}-\frac {\left (a-b x^4\right )^{3/4}}{a x^4}\right )\) |
Input:
Int[1/(x^5*(a - b*x^4)^(1/4)),x]
Output:
(-((a - b*x^4)^(3/4)/(a*x^4)) - (b*(-1/2*ArcTan[(a - b*x^4)^(1/4)/a^(1/4)] /a^(1/4) + ArcTanh[(a - b*x^4)^(1/4)/a^(1/4)]/(2*a^(1/4))))/a)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] :> 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Time = 0.76 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(-\frac {-2 \arctan \left (\frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b \,x^{4}+\ln \left (\frac {-\left (-b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (-b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b \,x^{4}+4 \left (-b \,x^{4}+a \right )^{\frac {3}{4}} a^{\frac {1}{4}}}{16 a^{\frac {5}{4}} x^{4}}\) | \(89\) |
Input:
int(1/x^5/(-b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/16*(-2*arctan((-b*x^4+a)^(1/4)/a^(1/4))*b*x^4+ln((-(-b*x^4+a)^(1/4)-a^( 1/4))/(-(-b*x^4+a)^(1/4)+a^(1/4)))*b*x^4+4*(-b*x^4+a)^(3/4)*a^(1/4))/a^(5/ 4)/x^4
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.46 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=-\frac {a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - i \, a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) + i \, a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) - a x^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {1}{4}} \log \left (-a^{4} \left (\frac {b^{4}}{a^{5}}\right )^{\frac {3}{4}} + {\left (-b x^{4} + a\right )}^{\frac {1}{4}} b^{3}\right ) + 4 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}}}{16 \, a x^{4}} \] Input:
integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/16*(a*x^4*(b^4/a^5)^(1/4)*log(a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)^(1/4)* b^3) - I*a*x^4*(b^4/a^5)^(1/4)*log(I*a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)^(1 /4)*b^3) + I*a*x^4*(b^4/a^5)^(1/4)*log(-I*a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)^(1/4)*b^3) - a*x^4*(b^4/a^5)^(1/4)*log(-a^4*(b^4/a^5)^(3/4) + (-b*x^4 + a)^(1/4)*b^3) + 4*(-b*x^4 + a)^(3/4))/(a*x^4)
Result contains complex when optimal does not.
Time = 0.79 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=\frac {e^{\frac {3 i \pi }{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a}{b x^{4}}} \right )}}{4 \sqrt [4]{b} x^{5} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate(1/x**5/(-b*x**4+a)**(1/4),x)
Output:
exp(3*I*pi/4)*gamma(5/4)*hyper((1/4, 5/4), (9/4,), a/(b*x**4))/(4*b**(1/4) *x**5*gamma(9/4))
Time = 0.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=\frac {b {\left (\frac {2 \, \arctan \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{16 \, a} - \frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} b}{4 \, {\left ({\left (b x^{4} - a\right )} a + a^{2}\right )}} \] Input:
integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/16*b*(2*arctan((-b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((-b*x^4 + a)^( 1/4) - a^(1/4))/((-b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4))/a - 1/4*(-b*x^4 + a)^(3/4)*b/((b*x^4 - a)*a + a^2)
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (61) = 122\).
Time = 0.12 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.79 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=-\frac {\frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {2 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (-b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {\sqrt {2} b^{2} \log \left (\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \log \left (-\sqrt {2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {-b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} + \frac {8 \, {\left (-b x^{4} + a\right )}^{\frac {3}{4}} b}{a x^{4}}}{32 \, b} \] Input:
integrate(1/x^5/(-b*x^4+a)^(1/4),x, algorithm="giac")
Output:
-1/32*(2*sqrt(2)*(-a)^(3/4)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2 *(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + 2*sqrt(2)*(-a)^(3/4)*b^2*arctan(-1/ 2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(-b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + sq rt(2)*b^2*log(sqrt(2)*(-b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + s qrt(-a))/((-a)^(1/4)*a) + sqrt(2)*(-a)^(3/4)*b^2*log(-sqrt(2)*(-b*x^4 + a) ^(1/4)*(-a)^(1/4) + sqrt(-b*x^4 + a) + sqrt(-a))/a^2 + 8*(-b*x^4 + a)^(3/4 )*b/(a*x^4))/b
Time = 0.51 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=\frac {b\,\mathrm {atan}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}}-\frac {{\left (a-b\,x^4\right )}^{3/4}}{4\,a\,x^4}-\frac {b\,\mathrm {atanh}\left (\frac {{\left (a-b\,x^4\right )}^{1/4}}{a^{1/4}}\right )}{8\,a^{5/4}} \] Input:
int(1/(x^5*(a - b*x^4)^(1/4)),x)
Output:
(b*atan((a - b*x^4)^(1/4)/a^(1/4)))/(8*a^(5/4)) - (a - b*x^4)^(3/4)/(4*a*x ^4) - (b*atanh((a - b*x^4)^(1/4)/a^(1/4)))/(8*a^(5/4))
\[ \int \frac {1}{x^5 \sqrt [4]{a-b x^4}} \, dx=\int \frac {1}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}} x^{5}}d x \] Input:
int(1/x^5/(-b*x^4+a)^(1/4),x)
Output:
int(1/((a - b*x**4)**(1/4)*x**5),x)