Integrand size = 20, antiderivative size = 122 \[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {(5 b c-2 a d) x^2}{5 b \sqrt [4]{a+b x^4}}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {\sqrt {a} (5 b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a+b x^4}} \] Output:
1/5*(-2*a*d+5*b*c)*x^2/b/(b*x^4+a)^(1/4)+1/5*d*x^2*(b*x^4+a)^(3/4)/b-1/5*a ^(1/2)*(-2*a*d+5*b*c)*(1+b*x^4/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x ^2/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61 \[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {x^2 \left (2 d \left (a+b x^4\right )+(5 b c-2 a d) \sqrt [4]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{10 b \sqrt [4]{a+b x^4}} \] Input:
Integrate[(x*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
(x^2*(2*d*(a + b*x^4) + (5*b*c - 2*a*d)*(1 + (b*x^4)/a)^(1/4)*Hypergeometr ic2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(10*b*(a + b*x^4)^(1/4))
Time = 0.38 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {959, 807, 227, 225, 212}
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.
\(\displaystyle \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(5 b c-2 a d) \int \frac {x}{\sqrt [4]{b x^4+a}}dx}{5 b}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {(5 b c-2 a d) \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{10 b}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^4}{a}+1} (5 b c-2 a d) \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{10 b \sqrt [4]{a+b x^4}}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^4}{a}+1} (5 b c-2 a d) \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{10 b \sqrt [4]{a+b x^4}}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^4}{a}+1} (5 b c-2 a d) \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{10 b \sqrt [4]{a+b x^4}}+\frac {d x^2 \left (a+b x^4\right )^{3/4}}{5 b}\) |
Input:
Int[(x*(c + d*x^4))/(a + b*x^4)^(1/4),x]
Output:
(d*x^2*(a + b*x^4)^(3/4))/(5*b) + ((5*b*c - 2*a*d)*(1 + (b*x^4)/a)^(1/4)*( (2*x^2)/(1 + (b*x^4)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/ Sqrt[a]]/2, 2])/Sqrt[b]))/(10*b*(a + b*x^4)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
\[\int \frac {x \left (d \,x^{4}+c \right )}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x\]
Input:
int(x*(d*x^4+c)/(b*x^4+a)^(1/4),x)
Output:
int(x*(d*x^4+c)/(b*x^4+a)^(1/4),x)
\[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
integral((d*x^5 + c*x)/(b*x^4 + a)^(1/4), x)
Result contains complex when optimal does not.
Time = 3.68 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.49 \[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\frac {c x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a}} + \frac {d x^{6} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{6 \sqrt [4]{a}} \] Input:
integrate(x*(d*x**4+c)/(b*x**4+a)**(1/4),x)
Output:
c*x**2*hyper((1/4, 1/2), (3/2,), b*x**4*exp_polar(I*pi)/a)/(2*a**(1/4)) + d*x**6*hyper((1/4, 3/2), (5/2,), b*x**4*exp_polar(I*pi)/a)/(6*a**(1/4))
\[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x/(b*x^4 + a)^(1/4), x)
\[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x*(d*x^4+c)/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x/(b*x^4 + a)^(1/4), x)
Timed out. \[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\int \frac {x\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \] Input:
int((x*(c + d*x^4))/(a + b*x^4)^(1/4),x)
Output:
int((x*(c + d*x^4))/(a + b*x^4)^(1/4), x)
\[ \int \frac {x \left (c+d x^4\right )}{\sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {x^{5}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {x}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) c \] Input:
int(x*(d*x^4+c)/(b*x^4+a)^(1/4),x)
Output:
int(x**5/(a + b*x**4)**(1/4),x)*d + int(x/(a + b*x**4)**(1/4),x)*c