Integrand size = 15, antiderivative size = 149 \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {45 a^4 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}-\frac {45 a^4 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}} \]
45/2048*a^3*x*(b*x^4+a)^(3/4)/b^3-9/512*a^2*x^5*(b*x^4+a)^(3/4)/b^2+1/64*a *x^9*(b*x^4+a)^(3/4)/b+1/16*x^13*(b*x^4+a)^(3/4)-45/4096*a^4*arctan(b^(1/4 )*x/(b*x^4+a)^(1/4))/b^(13/4)-45/4096*a^4*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4 ))/b^(13/4)
Time = 0.58 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.73 \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4} \left (45 a^3-36 a^2 b x^4+32 a b^2 x^8+128 b^3 x^{12}\right )-45 a^4 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-45 a^4 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}} \]
(2*b^(1/4)*x*(a + b*x^4)^(3/4)*(45*a^3 - 36*a^2*b*x^4 + 32*a*b^2*x^8 + 128 *b^3*x^12) - 45*a^4*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - 45*a^4*ArcTanh [(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(4096*b^(13/4))
Time = 0.29 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {811, 843, 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.
| \(\displaystyle \int x^{12} \left (a+b x^4\right )^{3/4} \, dx\) |
| \(\Big \downarrow \) 811 |
| \(\displaystyle \frac {3}{16} a \int \frac {x^{12}}{\sqrt [4]{b x^4+a}}dx+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \int \frac {x^8}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\right )}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\right )}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3}{16} a \left (\frac {x^9 \left (a+b x^4\right )^{3/4}}{12 b}-\frac {3 a \left (\frac {x^5 \left (a+b x^4\right )^{3/4}}{8 b}-\frac {5 a \left (\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\right )}{8 b}\right )}{4 b}\right )+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}\) |
(x^13*(a + b*x^4)^(3/4))/16 + (3*a*((x^9*(a + b*x^4)^(3/4))/(12*b) - (3*a* ((x^5*(a + b*x^4)^(3/4))/(8*b) - (5*a*((x*(a + b*x^4)^(3/4))/(4*b) - (a*(A rcTan[(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)))/(4*b)))/16
3.11.31.3.1 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* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 = 4.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.98
| method | result | size |
| pseudoelliptic | \(\frac {512 b^{\frac {13}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{13}+128 a \,b^{\frac {9}{4}} x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}}-144 a^{2} b^{\frac {5}{4}} x^{5} \left (b \,x^{4}+a \right )^{\frac {3}{4}}+180 a^{3} x \,b^{\frac {1}{4}} \left (b \,x^{4}+a \right )^{\frac {3}{4}}+90 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{4}-45 \ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{4}}{8192 b^{\frac {13}{4}}}\) | \(146\) |
1/8192/b^(13/4)*(512*b^(13/4)*(b*x^4+a)^(3/4)*x^13+128*a*b^(9/4)*x^9*(b*x^ 4+a)^(3/4)-144*a^2*b^(5/4)*x^5*(b*x^4+a)^(3/4)+180*a^3*x*b^(1/4)*(b*x^4+a) ^(3/4)+90*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^4-45*ln((-b^(1/4)*x-(b*x^4 +a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))*a^4)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.64 \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=-\frac {45 \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} + \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) - 45 i \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} + i \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) + 45 i \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} - i \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) - 45 \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} - \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) - 4 \, {\left (128 \, b^{3} x^{13} + 32 \, a b^{2} x^{9} - 36 \, a^{2} b x^{5} + 45 \, a^{3} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{8192 \, b^{3}} \]
-1/8192*(45*(a^16/b^13)^(1/4)*b^3*log(91125*((b*x^4 + a)^(1/4)*a^12 + (a^1 6/b^13)^(3/4)*b^10*x)/x) - 45*I*(a^16/b^13)^(1/4)*b^3*log(91125*((b*x^4 + a)^(1/4)*a^12 + I*(a^16/b^13)^(3/4)*b^10*x)/x) + 45*I*(a^16/b^13)^(1/4)*b^ 3*log(91125*((b*x^4 + a)^(1/4)*a^12 - I*(a^16/b^13)^(3/4)*b^10*x)/x) - 45* (a^16/b^13)^(1/4)*b^3*log(91125*((b*x^4 + a)^(1/4)*a^12 - (a^16/b^13)^(3/4 )*b^10*x)/x) - 4*(128*b^3*x^13 + 32*a*b^2*x^9 - 36*a^2*b*x^5 + 45*a^3*x)*( b*x^4 + a)^(3/4))/b^3
Result contains complex when optimal does not.
Time = 18.99 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.26 \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \]
a**(3/4)*x**13*gamma(13/4)*hyper((-3/4, 13/4), (17/4,), b*x**4*exp_polar(I *pi)/a)/(4*gamma(17/4))
Time = 0.28 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.51 \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\frac {45 \, a^{4} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{8192 \, b^{3}} + \frac {\frac {15 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{4} b^{3}}{x^{3}} + \frac {239 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{4} b^{2}}{x^{7}} - \frac {171 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a^{4} b}{x^{11}} + \frac {45 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} a^{4}}{x^{15}}}{2048 \, {\left (b^{7} - \frac {4 \, {\left (b x^{4} + a\right )} b^{6}}{x^{4}} + \frac {6 \, {\left (b x^{4} + a\right )}^{2} b^{5}}{x^{8}} - \frac {4 \, {\left (b x^{4} + a\right )}^{3} b^{4}}{x^{12}} + \frac {{\left (b x^{4} + a\right )}^{4} b^{3}}{x^{16}}\right )}} \]
45/8192*a^4*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(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 + 1/2048*(15*(b*x^4 + a)^(3/4)*a^4*b^3/x^3 + 239*(b*x^4 + a)^(7/4)*a^4*b^2/x ^7 - 171*(b*x^4 + a)^(11/4)*a^4*b/x^11 + 45*(b*x^4 + a)^(15/4)*a^4/x^15)/( b^7 - 4*(b*x^4 + a)*b^6/x^4 + 6*(b*x^4 + a)^2*b^5/x^8 - 4*(b*x^4 + a)^3*b^ 4/x^12 + (b*x^4 + a)^4*b^3/x^16)
\[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{12} \,d x } \]
Timed out. \[ \int x^{12} \left (a+b x^4\right )^{3/4} \, dx=\int x^{12}\,{\left (b\,x^4+a\right )}^{3/4} \,d x \]