Integrand size = 19, antiderivative size = 93 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {d x \sqrt [4]{a+b x^4}}{2 b}-\frac {(2 b c-a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{2 \sqrt {a} \sqrt {b} \left (a+b x^4\right )^{3/4}} \] Output:
1/2*d*x*(b*x^4+a)^(1/4)/b-1/2*(-a*d+2*b*c)*(1+a/b/x^4)^(3/4)*x^3*InverseJa cobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^(1/2)/b^(1/2)/(b*x^4+a)^ (3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {d x \left (a+b x^4\right )+(2 b c-a d) x \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{2 b \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(3/4),x]
Output:
(d*x*(a + b*x^4) + (2*b*c - a*d)*x*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1 [1/4, 3/4, 5/4, -((b*x^4)/a)])/(2*b*(a + b*x^4)^(3/4))
Time = 0.39 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {913, 768, 858, 807, 229}
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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {(2 b c-a d) \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{2 b}+\frac {d x \sqrt [4]{a+b x^4}}{2 b}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (2 b c-a d) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{2 b \left (a+b x^4\right )^{3/4}}+\frac {d x \sqrt [4]{a+b x^4}}{2 b}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {d x \sqrt [4]{a+b x^4}}{2 b}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (2 b c-a d) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{2 b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {d x \sqrt [4]{a+b x^4}}{2 b}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (2 b c-a d) \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{4 b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {d x \sqrt [4]{a+b x^4}}{2 b}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (2 b c-a d) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \sqrt {a} \sqrt {b} \left (a+b x^4\right )^{3/4}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(3/4),x]
Output:
(d*x*(a + b*x^4)^(1/4))/(2*b) - ((2*b*c - a*d)*(1 + a/(b*x^4))^(3/4)*x^3*E llipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqrt[a]*Sqrt[b]*(a + b*x^ 4)^(3/4))
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int egerQ[m]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
\[\int \frac {d \,x^{4}+c}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x\]
Input:
int((d*x^4+c)/(b*x^4+a)^(3/4),x)
Output:
int((d*x^4+c)/(b*x^4+a)^(3/4),x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
integral((d*x^4 + c)/(b*x^4 + a)^(3/4), x)
Result contains complex when optimal does not.
Time = 1.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(3/4),x)
Output:
c*x*gamma(1/4)*hyper((1/4, 3/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 3/4)*gamma(5/4)) + d*x**5*gamma(5/4)*hyper((3/4, 5/4), (9/4,), b*x**4*exp_ polar(I*pi)/a)/(4*a**(3/4)*gamma(9/4))
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(3/4), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(3/4), x)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{3/4}} \,d x \] Input:
int((c + d*x^4)/(a + b*x^4)^(3/4),x)
Output:
int((c + d*x^4)/(a + b*x^4)^(3/4), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{3/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(3/4),x)
Output:
int(x**4/(a + b*x**4)**(3/4),x)*d + int(1/(a + b*x**4)**(3/4),x)*c