Integrand size = 15, antiderivative size = 78 \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}}-\frac {a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}} \] Output:
1/4*x*(b*x^4+a)^(3/4)/b-1/8*a*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)-1/ 8*a*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\frac {2 \sqrt [4]{b} x \left (a+b x^4\right )^{3/4}-a \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )-a \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{8 b^{5/4}} \] Input:
Integrate[x^4/(a + b*x^4)^(1/4),x]
Output:
(2*b^(1/4)*x*(a + b*x^4)^(3/4) - a*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] - a*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(8*b^(5/4))
Time = 0.32 (sec) , antiderivative size = 85, 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, 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.
\(\displaystyle \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{4 b}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{4 b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{4 b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {x \left (a+b x^4\right )^{3/4}}{4 b}-\frac {a \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{4 b}\) |
Input:
Int[x^4/(a + b*x^4)^(1/4),x]
Output:
(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)
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*(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 = 0.66 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01
method | result | size |
pseudoelliptic | \(-\frac {-4 \left (b \,x^{4}+a \right )^{\frac {3}{4}} x \,b^{\frac {1}{4}}-2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) a +\ln \left (\frac {b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-b^{\frac {1}{4}} x +\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a}{16 b^{\frac {5}{4}}}\) | \(79\) |
Input:
int(x^4/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/16/b^(5/4)*(-4*(b*x^4+a)^(3/4)*x*b^(1/4)-2*arctan(1/b^(1/4)/x*(b*x^4+a) ^(1/4))*a+ln((b^(1/4)*x+(b*x^4+a)^(1/4))/(-b^(1/4)*x+(b*x^4+a)^(1/4)))*a)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.58 \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=-\frac {b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - i \, b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) + i \, b \left (\frac {a^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b^{4} x \left (\frac {a^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}}{x}\right ) - 4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} x}{16 \, b} \] Input:
integrate(x^4/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
-1/16*(b*(a^4/b^5)^(1/4)*log((b^4*x*(a^4/b^5)^(3/4) + (b*x^4 + a)^(1/4)*a^ 3)/x) - b*(a^4/b^5)^(1/4)*log(-(b^4*x*(a^4/b^5)^(3/4) - (b*x^4 + a)^(1/4)* a^3)/x) - I*b*(a^4/b^5)^(1/4)*log((I*b^4*x*(a^4/b^5)^(3/4) + (b*x^4 + a)^( 1/4)*a^3)/x) + I*b*(a^4/b^5)^(1/4)*log((-I*b^4*x*(a^4/b^5)^(3/4) + (b*x^4 + a)^(1/4)*a^3)/x) - 4*(b*x^4 + a)^(3/4)*x)/b
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.47 \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate(x**4/(b*x**4+a)**(1/4),x)
Output:
x**5*gamma(5/4)*hyper((1/4, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a** (1/4)*gamma(9/4))
Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.38 \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\frac {a {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{16 \, b} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a}{4 \, {\left (b^{2} - \frac {{\left (b x^{4} + a\right )} b}{x^{4}}\right )} x^{3}} \] Input:
integrate(x^4/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
1/16*a*(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 - 1/4*(b* x^4 + a)^(3/4)*a/((b^2 - (b*x^4 + a)*b/x^4)*x^3)
\[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {x^{4}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x^4/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate(x^4/(b*x^4 + a)^(1/4), x)
Timed out. \[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^4}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:
int(x^4/(a + b*x^4)^(1/4),x)
Output:
int(x^4/(a + b*x^4)^(1/4), x)
\[ \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \] Input:
int(x^4/(b*x^4+a)^(1/4),x)
Output:
int(x**4/(a + b*x**4)**(1/4),x)