Integrand size = 15, antiderivative size = 106 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {7 a x^3 \sqrt [4]{a+b x^4}}{32 b^2}+\frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {21 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}}+\frac {21 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}} \]
-7/32*a*x^3*(b*x^4+a)^(1/4)/b^2+1/8*x^7*(b*x^4+a)^(1/4)/b-21/64*a^2*arctan (b^(1/4)*x/(b*x^4+a)^(1/4))/b^(11/4)+21/64*a^2*arctanh(b^(1/4)*x/(b*x^4+a) ^(1/4))/b^(11/4)
Time = 0.60 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.84 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (-7 a+4 b x^4\right )-21 a^2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+21 a^2 \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 b^{11/4}} \]
(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(-7*a + 4*b*x^4) - 21*a^2*ArcTan[(b^(1/4) *x)/(a + b*x^4)^(1/4)] + 21*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(6 4*b^(11/4))
Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {843, 843, 854, 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.
\(\displaystyle \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \int \frac {x^6}{\left (b x^4+a\right )^{3/4}}dx}{8 b}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx}{4 b}\right )}{8 b}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\right )}{8 b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )}{4 b}\right )}{8 b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )}{8 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x^7 \sqrt [4]{a+b x^4}}{8 b}-\frac {7 a \left (\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\right )}{8 b}\) |
(x^7*(a + b*x^4)^(1/4))/(8*b) - (7*a*((x^3*(a + b*x^4)^(1/4))/(4*b) - (3*a *(-1/2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]/b^(3/4) + ArcTanh[(b^(1/4)*x) /(a + b*x^4)^(1/4)]/(2*b^(3/4))))/(4*b)))/(8*b)
3.12.25.3.1 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[(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]
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]
Time = 4.61 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02
method | result | size |
pseudoelliptic | \(\frac {16 b^{\frac {7}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{7}-28 a \,x^{3} b^{\frac {3}{4}} \left (b \,x^{4}+a \right )^{\frac {1}{4}}+21 \ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a^{2}+42 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a^{2}}{128 b^{\frac {11}{4}}}\) | \(108\) |
1/128/b^(11/4)*(16*b^(7/4)*(b*x^4+a)^(1/4)*x^7-28*a*x^3*b^(3/4)*(b*x^4+a)^ (1/4)+21*ln((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))*a^2+ 42*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a^2)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.14 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {21 \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (\frac {21 \, {\left (b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 21 \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {21 \, {\left (b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) - 21 i \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {21 \, {\left (i \, b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) + 21 i \, b^{2} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-\frac {21 \, {\left (-i \, b^{3} x \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}\right )}}{x}\right ) + 4 \, {\left (4 \, b x^{7} - 7 \, a x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, b^{2}} \]
1/128*(21*b^2*(a^8/b^11)^(1/4)*log(21*(b^3*x*(a^8/b^11)^(1/4) + (b*x^4 + a )^(1/4)*a^2)/x) - 21*b^2*(a^8/b^11)^(1/4)*log(-21*(b^3*x*(a^8/b^11)^(1/4) - (b*x^4 + a)^(1/4)*a^2)/x) - 21*I*b^2*(a^8/b^11)^(1/4)*log(-21*(I*b^3*x*( a^8/b^11)^(1/4) - (b*x^4 + a)^(1/4)*a^2)/x) + 21*I*b^2*(a^8/b^11)^(1/4)*lo g(-21*(-I*b^3*x*(a^8/b^11)^(1/4) - (b*x^4 + a)^(1/4)*a^2)/x) + 4*(4*b*x^7 - 7*a*x^3)*(b*x^4 + a)^(1/4))/b^2
Result contains complex when optimal does not.
Time = 1.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.35 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {15}{4}\right )} \]
x**11*gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(3/4)*gamma(15/4))
Time = 0.40 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.46 \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {\frac {11 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b}{x} - \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{2}}{x^{5}}}{32 \, {\left (b^{4} - \frac {2 \, {\left (b x^{4} + a\right )} b^{3}}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2} b^{2}}{x^{8}}\right )}} + \frac {21 \, {\left (\frac {2 \, a^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{2} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{2}} \]
1/32*(11*(b*x^4 + a)^(1/4)*a^2*b/x - 7*(b*x^4 + a)^(5/4)*a^2/x^5)/(b^4 - 2 *(b*x^4 + a)*b^3/x^4 + (b*x^4 + a)^2*b^2/x^8) + 21/128*(2*a^2*arctan((b*x^ 4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - a^2*log(-(b^(1/4) - (b*x^4 + a)^(1/4)/ x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(3/4))/b^2
\[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {x^{10}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx=\int \frac {x^{10}}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]