Integrand size = 15, antiderivative size = 123 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {x^9}{b \sqrt [4]{a+b x^4}}-\frac {45 a x \left (a+b x^4\right )^{3/4}}{32 b^3}+\frac {9 x^5 \left (a+b x^4\right )^{3/4}}{8 b^2}+\frac {45 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}}+\frac {45 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}} \]
-x^9/b/(b*x^4+a)^(1/4)-45/32*a*x*(b*x^4+a)^(3/4)/b^3+9/8*x^5*(b*x^4+a)^(3/ 4)/b^2+45/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(13/4)+45/64*a^2*arct anh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(13/4)
Time = 0.86 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} x \left (-45 a^2-9 a b x^4+4 b^2 x^8\right )}{\sqrt [4]{a+b x^4}}+45 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+45 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{13/4}} \]
((2*b^(1/4)*x*(-45*a^2 - 9*a*b*x^4 + 4*b^2*x^8))/(a + b*x^4)^(1/4) + 45*a^ 2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + 45*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(13/4))
Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {817, 843, 843, 770, 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.
| \(\displaystyle \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {9 \int \frac {x^8}{\sqrt [4]{b x^4+a}}dx}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {9 \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\right )}{b}-\frac {x^9}{b \sqrt [4]{a+b x^4}}\) |
-(x^9/(b*(a + b*x^4)^(1/4))) + (9*((x^5*(a + b*x^4)^(3/4))/(8*b) - (5*a*(( x*(a + b*x^4)^(3/4))/(4*b) - (a*(ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2* b^(1/4)) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/(2*b^(1/4))))/(4*b)))/(8 *b)))/b
3.12.54.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Time = 4.90 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02
| method | result | size |
| pseudoelliptic | \(-\frac {45 \left (-\frac {8 b^{\frac {9}{4}} x^{9}}{45}+\frac {2 a \,b^{\frac {5}{4}} x^{5}}{5}+\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}-\frac {\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{2}+2 a^{2} x \,b^{\frac {1}{4}}\right )}{64 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {13}{4}}}\) | \(125\) |
-45/64/(b*x^4+a)^(1/4)*(-8/45*b^(9/4)*x^9+2/5*a*b^(5/4)*x^5+arctan(1/b^(1/ 4)/x*(b*x^4+a)^(1/4))*a^2*(b*x^4+a)^(1/4)-1/2*ln((-b^(1/4)*x-(b*x^4+a)^(1/ 4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))*a^2*(b*x^4+a)^(1/4)+2*a^2*x*b^(1/4))/b^(1 3/4)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.37 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {45 \, {\left (b^{4} x^{4} + a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (\frac {91125 \, {\left (b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (b^{4} x^{4} + a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (-i \, b^{4} x^{4} - i \, a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (i \, b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) - 45 \, {\left (i \, b^{4} x^{4} + i \, a b^{3}\right )} \left (\frac {a^{8}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-\frac {91125 \, {\left (-i \, b^{10} x \left (\frac {a^{8}}{b^{13}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6}\right )}}{x}\right ) + 4 \, {\left (4 \, b^{2} x^{9} - 9 \, a b x^{5} - 45 \, a^{2} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \]
1/128*(45*(b^4*x^4 + a*b^3)*(a^8/b^13)^(1/4)*log(91125*(b^10*x*(a^8/b^13)^ (3/4) + (b*x^4 + a)^(1/4)*a^6)/x) - 45*(b^4*x^4 + a*b^3)*(a^8/b^13)^(1/4)* log(-91125*(b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) - 45*(-I*b ^4*x^4 - I*a*b^3)*(a^8/b^13)^(1/4)*log(-91125*(I*b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) - 45*(I*b^4*x^4 + I*a*b^3)*(a^8/b^13)^(1/4)*log (-91125*(-I*b^10*x*(a^8/b^13)^(3/4) - (b*x^4 + a)^(1/4)*a^6)/x) + 4*(4*b^2 *x^9 - 9*a*b*x^5 - 45*a^2*x)*(b*x^4 + a)^(3/4))/(b^4*x^4 + a*b^3)
Result contains complex when optimal does not.
Time = 2.43 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.30 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {17}{4}\right )} \]
x**13*gamma(13/4)*hyper((5/4, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(5/4)*gamma(17/4))
Time = 0.29 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.40 \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {32 \, a^{2} b^{2} - \frac {81 \, {\left (b x^{4} + a\right )} a^{2} b}{x^{4}} + \frac {45 \, {\left (b x^{4} + a\right )}^{2} a^{2}}{x^{8}}}{32 \, {\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{5}}{x} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{4}}{x^{5}} + \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} b^{3}}{x^{9}}\right )}} - \frac {45 \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{128 \, b^{3}} \]
-1/32*(32*a^2*b^2 - 81*(b*x^4 + a)*a^2*b/x^4 + 45*(b*x^4 + a)^2*a^2/x^8)/( (b*x^4 + a)^(1/4)*b^5/x - 2*(b*x^4 + a)^(5/4)*b^4/x^5 + (b*x^4 + a)^(9/4)* b^3/x^9) - 45/128*a^2*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + l og(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(1/ 4))/b^3
\[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^{12}}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {x^{12}}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \]