Integrand size = 22, antiderivative size = 81 \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}+\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \] Output:
-1/3*c*(b*x^4+a)^(3/4)/a/x^3+1/2*d*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/ 4)+1/2*d*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}+\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}} \] Input:
Integrate[(c + d*x^4)/(x^4*(a + b*x^4)^(1/4)),x]
Output:
-1/3*(c*(a + b*x^4)^(3/4))/(a*x^3) + (d*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/ 4)])/(2*b^(1/4)) + (d*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(2*b^(1/4))
Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {953, 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.
\(\displaystyle \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle d \int \frac {1}{\sqrt [4]{b x^4+a}}dx-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle d \int \frac {1}{1-\frac {b x^4}{b x^4+a}}d\frac {x}{\sqrt [4]{b x^4+a}}-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle d \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}\right )-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle d \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle d \left (\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 \sqrt [4]{b}}\right )-\frac {c \left (a+b x^4\right )^{3/4}}{3 a x^3}\) |
Input:
Int[(c + d*x^4)/(x^4*(a + b*x^4)^(1/4)),x]
Output:
-1/3*(c*(a + b*x^4)^(3/4))/(a*x^3) + 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)))
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15
method | result | size |
pseudoelliptic | \(-\frac {\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right ) a d \,x^{3}-\frac {\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a d \,x^{3}}{2}+\frac {2 \left (b \,x^{4}+a \right )^{\frac {3}{4}} c \,b^{\frac {1}{4}}}{3}}{2 b^{\frac {1}{4}} x^{3} a}\) | \(93\) |
Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/2/b^(1/4)*(arctan((b*x^4+a)^(1/4)/x/b^(1/4))*a*d*x^3-1/2*ln((x*b^(1/4)+ (b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))*a*d*x^3+2/3*(b*x^4+a)^(3/4) *c*b^(1/4))/x^3/a
Timed out. \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=\text {Timed out} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/4),x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 3.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=\frac {b^{\frac {3}{4}} c \left (\frac {a}{b x^{4}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 a \Gamma \left (\frac {1}{4}\right )} + \frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [4]{a} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((d*x**4+c)/x**4/(b*x**4+a)**(1/4),x)
Output:
b**(3/4)*c*(a/(b*x**4) + 1)**(3/4)*gamma(-3/4)/(4*a*gamma(1/4)) + d*x*gamm a(1/4)*hyper((1/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/4)*gam ma(5/4))
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=-\frac {1}{4} \, d {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} c}{3 \, a x^{3}} \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/4),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)) - 1/3*(b*x^ 4 + a)^(3/4)*c/(a*x^3)
\[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{4}} \,d x } \] Input:
integrate((d*x^4+c)/x^4/(b*x^4+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(1/4)*x^4), x)
Time = 3.91 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=\frac {d\,x\,{\left (\frac {b\,x^4}{a}+1\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (b\,x^4+a\right )}^{1/4}}-\frac {c\,{\left (b\,x^4+a\right )}^{3/4}}{3\,a\,x^3} \] Input:
int((c + d*x^4)/(x^4*(a + b*x^4)^(1/4)),x)
Output:
(d*x*((b*x^4)/a + 1)^(1/4)*hypergeom([1/4, 1/4], 5/4, -(b*x^4)/a))/(a + b* x^4)^(1/4) - (c*(a + b*x^4)^(3/4))/(3*a*x^3)
\[ \int \frac {c+d x^4}{x^4 \sqrt [4]{a+b x^4}} \, dx=\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{4}}d x \right ) c \] Input:
int((d*x^4+c)/x^4/(b*x^4+a)^(1/4),x)
Output:
int(1/(a + b*x**4)**(1/4),x)*d + int(1/((a + b*x**4)**(1/4)*x**4),x)*c