Integrand size = 19, antiderivative size = 86 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {(b c-a d) x}{a b \sqrt [4]{a+b x^4}}+\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}} \] Output:
(-a*d+b*c)*x/a/b/(b*x^4+a)^(1/4)+1/2*d*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b ^(5/4)+1/2*d*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(5/4)
Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {\frac {2 \sqrt [4]{b} (b c-a d) x}{a \sqrt [4]{a+b x^4}}+d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{5/4}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(5/4),x]
Output:
((2*b^(1/4)*(b*c - a*d)*x)/(a*(a + b*x^4)^(1/4)) + d*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)] + d*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(2*b^(5/4))
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {910, 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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {d \int \frac {1}{\sqrt [4]{b x^4+a}}dx}{b}+\frac {x (b c-a d)}{a b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {d \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}}{b}+\frac {x (b c-a d)}{a b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {d \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )}{b}+\frac {x (b c-a d)}{a b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {d \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{b}+\frac {x (b c-a d)}{a b \sqrt [4]{a+b x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {d \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )}{b}+\frac {x (b c-a d)}{a b \sqrt [4]{a+b x^4}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(5/4),x]
Output:
((b*c - a*d)*x)/(a*b*(a + b*x^4)^(1/4)) + (d*(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)) ))/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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.09
method | result | size |
pseudoelliptic | \(\frac {\frac {\left (-a d +c b \right ) x}{b \left (b \,x^{4}+a \right )^{\frac {1}{4}}}-\frac {\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right ) a d}{2 b^{\frac {5}{4}}}+\frac {\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a d}{4 b^{\frac {5}{4}}}}{a}\) | \(94\) |
Input:
int((d*x^4+c)/(b*x^4+a)^(5/4),x,method=_RETURNVERBOSE)
Output:
((-a*d+b*c)/b*x/(b*x^4+a)^(1/4)-1/2*arctan((b*x^4+a)^(1/4)/x/b^(1/4))*a/b^ (5/4)*d+1/4*ln((x*b^(1/4)+(b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))*a /b^(5/4)*d)/a
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.21 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {4 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} {\left (b c - a d\right )} x + {\left (a b^{2} x^{4} + a^{2} b\right )} \left (\frac {d^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {b^{4} x \left (\frac {d^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) - {\left (a b^{2} x^{4} + a^{2} b\right )} \left (\frac {d^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (-\frac {b^{4} x \left (\frac {d^{4}}{b^{5}}\right )^{\frac {3}{4}} - {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) + {\left (-i \, a b^{2} x^{4} - i \, a^{2} b\right )} \left (\frac {d^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {i \, b^{4} x \left (\frac {d^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right ) + {\left (i \, a b^{2} x^{4} + i \, a^{2} b\right )} \left (\frac {d^{4}}{b^{5}}\right )^{\frac {1}{4}} \log \left (\frac {-i \, b^{4} x \left (\frac {d^{4}}{b^{5}}\right )^{\frac {3}{4}} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} d^{3}}{x}\right )}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
1/4*(4*(b*x^4 + a)^(3/4)*(b*c - a*d)*x + (a*b^2*x^4 + a^2*b)*(d^4/b^5)^(1/ 4)*log((b^4*x*(d^4/b^5)^(3/4) + (b*x^4 + a)^(1/4)*d^3)/x) - (a*b^2*x^4 + a ^2*b)*(d^4/b^5)^(1/4)*log(-(b^4*x*(d^4/b^5)^(3/4) - (b*x^4 + a)^(1/4)*d^3) /x) + (-I*a*b^2*x^4 - I*a^2*b)*(d^4/b^5)^(1/4)*log((I*b^4*x*(d^4/b^5)^(3/4 ) + (b*x^4 + a)^(1/4)*d^3)/x) + (I*a*b^2*x^4 + I*a^2*b)*(d^4/b^5)^(1/4)*lo g((-I*b^4*x*(d^4/b^5)^(3/4) + (b*x^4 + a)^(1/4)*d^3)/x))/(a*b^2*x^4 + a^2* b)
Result contains complex when optimal does not.
Time = 6.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right )}{4 a^{\frac {5}{4}} \sqrt [4]{1 + \frac {b x^{4}}{a}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(5/4),x)
Output:
c*x*gamma(1/4)/(4*a**(5/4)*(1 + b*x**4/a)**(1/4)*gamma(5/4)) + d*x**5*gamm a(5/4)*hyper((5/4, 5/4), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gam ma(9/4))
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=-\frac {1}{4} \, d {\left (\frac {\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}}{b} + \frac {4 \, x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b}\right )} + \frac {c x}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a} \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
-1/4*d*((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 + 4*x/(( b*x^4 + a)^(1/4)*b)) + c*x/((b*x^4 + a)^(1/4)*a)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(5/4), x)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \] Input:
int((c + d*x^4)/(a + b*x^4)^(5/4),x)
Output:
int((c + d*x^4)/(a + b*x^4)^(5/4), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} a +\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(5/4),x)
Output:
int(x**4/((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*d + int(1 /((a + b*x**4)**(1/4)*a + (a + b*x**4)**(1/4)*b*x**4),x)*c