Integrand size = 15, antiderivative size = 80 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x^3 \sqrt [4]{a+b x^4}}{4 b}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}} \]
1/4*x^3*(b*x^4+a)^(1/4)/b+3/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)- 3/8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(7/4)
Time = 0.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4}+3 a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-3 a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{7/4}} \]
(2*b^(3/4)*x^3*(a + b*x^4)^(1/4) + 3*a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4 )] - 3*a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(7/4))
Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {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.
\(\displaystyle \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx}{4 b}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )}{4 b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x^3 \sqrt [4]{a+b x^4}}{4 b}-\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )}{4 b}\) |
(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 )
3.12.26.3.1 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 = 4.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(-\frac {3 \left (-\frac {4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{3} b^{\frac {3}{4}}}{3}+\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a +2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a \right )}{16 b^{\frac {7}{4}}}\) | \(85\) |
-3/16*(-4/3*(b*x^4+a)^(1/4)*x^3*b^(3/4)+ln((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b ^(1/4)*x-(b*x^4+a)^(1/4)))*a+2*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4))*a)/b^(7 /4)
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.51 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {4 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{3} - 3 \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (\frac {3 \, {\left (b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right ) + 3 \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right ) + 3 i \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (i \, b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right ) - 3 i \, b \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (-i \, b^{2} x \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a\right )}}{x}\right )}{16 \, b} \]
1/16*(4*(b*x^4 + a)^(1/4)*x^3 - 3*b*(a^4/b^7)^(1/4)*log(3*(b^2*x*(a^4/b^7) ^(1/4) + (b*x^4 + a)^(1/4)*a)/x) + 3*b*(a^4/b^7)^(1/4)*log(-3*(b^2*x*(a^4/ b^7)^(1/4) - (b*x^4 + a)^(1/4)*a)/x) + 3*I*b*(a^4/b^7)^(1/4)*log(-3*(I*b^2 *x*(a^4/b^7)^(1/4) - (b*x^4 + a)^(1/4)*a)/x) - 3*I*b*(a^4/b^7)^(1/4)*log(- 3*(-I*b^2*x*(a^4/b^7)^(1/4) - (b*x^4 + a)^(1/4)*a)/x))/b
Result contains complex when optimal does not.
Time = 0.73 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.46 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \]
x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a* *(3/4)*gamma(11/4))
Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=-\frac {3 \, {\left (\frac {2 \, a \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )}}{16 \, b} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} + a\right )} b}{x^{4}}\right )} x} \]
-3/16*(2*a*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - a*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 - 1/4* (b*x^4 + a)^(1/4)*a/((b^2 - (b*x^4 + a)*b/x^4)*x)
\[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {x^{6}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx=\int \frac {x^6}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \]