Integrand size = 22, antiderivative size = 79 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^4}}{a x}-\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \] Output:
-c*(b*x^4+a)^(1/4)/a/x-1/2*d*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(3/4)+1/2 *d*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(3/4)
Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=-\frac {c \sqrt [4]{a+b x^4}}{a x}-\frac {d \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}} \] Input:
Integrate[(c + d*x^4)/(x^2*(a + b*x^4)^(3/4)),x]
Output:
-((c*(a + b*x^4)^(1/4))/(a*x)) - (d*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)]) /(2*b^(3/4)) + (d*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(2*b^(3/4))
Time = 0.34 (sec) , antiderivative size = 80, 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, 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.
\(\displaystyle \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle d \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx-\frac {c \sqrt [4]{a+b x^4}}{a x}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle d \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}-\frac {c \sqrt [4]{a+b x^4}}{a x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle d \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )-\frac {c \sqrt [4]{a+b x^4}}{a x}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle d \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )-\frac {c \sqrt [4]{a+b x^4}}{a x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle d \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )-\frac {c \sqrt [4]{a+b x^4}}{a x}\) |
Input:
Int[(c + d*x^4)/(x^2*(a + b*x^4)^(3/4)),x]
Output:
-((c*(a + b*x^4)^(1/4))/(a*x)) + d*(-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)))
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[(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]
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 = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.13
method | result | size |
pseudoelliptic | \(-\frac {c \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{\frac {3}{4}}-\frac {\arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{x \,b^{\frac {1}{4}}}\right ) a d x}{2}-\frac {\ln \left (\frac {x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{-x \,b^{\frac {1}{4}}+\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right ) a d x}{4}}{b^{\frac {3}{4}} x a}\) | \(89\) |
Input:
int((d*x^4+c)/x^2/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/b^(3/4)*(c*(b*x^4+a)^(1/4)*b^(3/4)-1/2*arctan((b*x^4+a)^(1/4)/x/b^(1/4) )*a*d*x-1/4*ln((x*b^(1/4)+(b*x^4+a)^(1/4))/(-x*b^(1/4)+(b*x^4+a)^(1/4)))*a *d*x)/x/a
Timed out. \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\text {Timed out} \] Input:
integrate((d*x^4+c)/x^2/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 2.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{b} c \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {1}{4}\right )}{4 a \Gamma \left (\frac {3}{4}\right )} + \frac {d x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((d*x**4+c)/x**2/(b*x**4+a)**(3/4),x)
Output:
b**(1/4)*c*(a/(b*x**4) + 1)**(1/4)*gamma(-1/4)/(4*a*gamma(3/4)) + d*x**3*g amma(3/4)*hyper((3/4, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/4)* gamma(7/4))
Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\frac {1}{4} \, d {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {3}{4}}}\right )} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} c}{a x} \] Input:
integrate((d*x^4+c)/x^2/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
1/4*d*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - 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*x^4 + a )^(1/4)*c/(a*x)
\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^4+c)/x^2/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/((b*x^4 + a)^(3/4)*x^2), x)
Timed out. \[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\int \frac {d\,x^4+c}{x^2\,{\left (b\,x^4+a\right )}^{3/4}} \,d x \] Input:
int((c + d*x^4)/(x^2*(a + b*x^4)^(3/4)),x)
Output:
int((c + d*x^4)/(x^2*(a + b*x^4)^(3/4)), x)
\[ \int \frac {c+d x^4}{x^2 \left (a+b x^4\right )^{3/4}} \, dx=\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{2}}d x \right ) c \] Input:
int((d*x^4+c)/x^2/(b*x^4+a)^(3/4),x)
Output:
int(x**2/(a + b*x**4)**(3/4),x)*d + int(1/((a + b*x**4)**(3/4)*x**2),x)*c