Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{a+b x^4}}{8 a x^8}+\frac {7 b \sqrt [4]{a+b x^4}}{32 a^2 x^4}-\frac {21 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {21 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}} \]
-1/8*(b*x^4+a)^(1/4)/a/x^8+7/32*b*(b*x^4+a)^(1/4)/a^2/x^4-21/64*b^2*arctan ((b*x^4+a)^(1/4)/a^(1/4))/a^(11/4)-21/64*b^2*arctanh((b*x^4+a)^(1/4)/a^(1/ 4))/a^(11/4)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{a+b x^4} \left (-4 a+7 b x^4\right )}{32 a^2 x^8}-\frac {21 b^2 \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}}-\frac {21 b^2 \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{11/4}} \]
((a + b*x^4)^(1/4)*(-4*a + 7*b*x^4))/(32*a^2*x^8) - (21*b^2*ArcTan[(a + b* x^4)^(1/4)/a^(1/4)])/(64*a^(11/4)) - (21*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^( 1/4)])/(64*a^(11/4))
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {798, 52, 52, 73, 756, 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.
\(\displaystyle \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^{12} \left (b x^4+a\right )^{3/4}}dx^4\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \int \frac {1}{x^8 \left (b x^4+a\right )^{3/4}}dx^4}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \left (-\frac {3 b \int \frac {1}{x^4 \left (b x^4+a\right )^{3/4}}dx^4}{4 a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \left (-\frac {3 \int \frac {1}{\frac {x^{16}}{b}-\frac {a}{b}}d\sqrt [4]{b x^4+a}}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \left (-\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \int \frac {1}{x^8+\sqrt {a}}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \left (-\frac {3 \left (-\frac {b \int \frac {1}{\sqrt {a}-x^8}d\sqrt [4]{b x^4+a}}{2 \sqrt {a}}-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} \left (-\frac {7 b \left (-\frac {3 \left (-\frac {b \arctan \left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}-\frac {b \text {arctanh}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{2 a^{3/4}}\right )}{a}-\frac {\sqrt [4]{a+b x^4}}{a x^4}\right )}{8 a}-\frac {\sqrt [4]{a+b x^4}}{2 a x^8}\right )\) |
(-1/2*(a + b*x^4)^(1/4)/(a*x^8) - (7*b*(-((a + b*x^4)^(1/4)/(a*x^4)) - (3* (-1/2*(b*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/a^(3/4) - (b*ArcTanh[(a + b*x^ 4)^(1/4)/a^(1/4)])/(2*a^(3/4))))/a))/(8*a))/4
3.12.17.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /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]]
Time = 4.40 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {-21 \ln \left (\frac {-\left (b \,x^{4}+a \right )^{\frac {1}{4}}-a^{\frac {1}{4}}}{-\left (b \,x^{4}+a \right )^{\frac {1}{4}}+a^{\frac {1}{4}}}\right ) b^{2} x^{8}-42 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right ) b^{2} x^{8}+28 b \,x^{4} \left (b \,x^{4}+a \right )^{\frac {1}{4}} a^{\frac {3}{4}}-16 a^{\frac {7}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{128 a^{\frac {11}{4}} x^{8}}\) | \(108\) |
1/128*(-21*ln((-(b*x^4+a)^(1/4)-a^(1/4))/(-(b*x^4+a)^(1/4)+a^(1/4)))*b^2*x ^8-42*arctan((b*x^4+a)^(1/4)/a^(1/4))*b^2*x^8+28*b*x^4*(b*x^4+a)^(1/4)*a^( 3/4)-16*a^(7/4)*(b*x^4+a)^(1/4))/a^(11/4)/x^8
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.10 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {21 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (21 \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) + 21 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (21 i \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) - 21 i \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-21 i \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) - 21 \, a^{2} x^{8} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} \log \left (-21 \, a^{3} \left (\frac {b^{8}}{a^{11}}\right )^{\frac {1}{4}} + 21 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}\right ) - 4 \, {\left (7 \, b x^{4} - 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, a^{2} x^{8}} \]
-1/128*(21*a^2*x^8*(b^8/a^11)^(1/4)*log(21*a^3*(b^8/a^11)^(1/4) + 21*(b*x^ 4 + a)^(1/4)*b^2) + 21*I*a^2*x^8*(b^8/a^11)^(1/4)*log(21*I*a^3*(b^8/a^11)^ (1/4) + 21*(b*x^4 + a)^(1/4)*b^2) - 21*I*a^2*x^8*(b^8/a^11)^(1/4)*log(-21* I*a^3*(b^8/a^11)^(1/4) + 21*(b*x^4 + a)^(1/4)*b^2) - 21*a^2*x^8*(b^8/a^11) ^(1/4)*log(-21*a^3*(b^8/a^11)^(1/4) + 21*(b*x^4 + a)^(1/4)*b^2) - 4*(7*b*x ^4 - 4*a)*(b*x^4 + a)^(1/4))/(a^2*x^8)
Result contains complex when optimal does not.
Time = 1.71 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac {3}{4}} x^{11} \Gamma \left (\frac {15}{4}\right )} \]
-gamma(11/4)*hyper((3/4, 11/4), (15/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b* *(3/4)*x**11*gamma(15/4))
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2} - 11 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} a^{2} - 2 \, {\left (b x^{4} + a\right )} a^{3} + a^{4}\right )}} - \frac {21 \, {\left (\frac {2 \, b^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}} - \frac {b^{2} \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {3}{4}}}\right )}}{128 \, a^{2}} \]
1/32*(7*(b*x^4 + a)^(5/4)*b^2 - 11*(b*x^4 + a)^(1/4)*a*b^2)/((b*x^4 + a)^2 *a^2 - 2*(b*x^4 + a)*a^3 + a^4) - 21/128*(2*b^2*arctan((b*x^4 + a)^(1/4)/a ^(1/4))/a^(3/4) - b^2*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(3/4))/a^2
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (80) = 160\).
Time = 0.30 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.35 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=\frac {\frac {42 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} + \frac {42 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} + \frac {21 \, \sqrt {2} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}} a^{2}} + \frac {21 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{3}} + \frac {8 \, {\left (7 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{3} - 11 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{3}\right )}}{a^{2} b^{2} x^{8}}}{256 \, b} \]
1/256*(42*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a^2) + 42*sqrt(2)*b^3*arctan(-1/2*sqrt(2 )*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a^2) + 21*sqrt(2)*b^3*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a ) + sqrt(-a))/((-a)^(3/4)*a^2) + 21*sqrt(2)*(-a)^(1/4)*b^3*log(-sqrt(2)*(b *x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^3 + 8*(7*(b*x^4 + a)^(5/4)*b^3 - 11*(b*x^4 + a)^(1/4)*a*b^3)/(a^2*b^2*x^8))/b
Time = 6.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^9 \left (a+b x^4\right )^{3/4}} \, dx=\frac {7\,{\left (b\,x^4+a\right )}^{5/4}}{32\,a^2\,x^8}-\frac {11\,{\left (b\,x^4+a\right )}^{1/4}}{32\,a\,x^8}-\frac {21\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{11/4}}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,21{}\mathrm {i}}{64\,a^{11/4}} \]