Integrand size = 19, antiderivative size = 102 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {(b c-a d) x}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {(2 b c+a d) \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{3 a^{3/2} \sqrt {b} \left (a+b x^4\right )^{3/4}} \] Output:
1/3*(-a*d+b*c)*x/a/b/(b*x^4+a)^(3/4)-1/3*(a*d+2*b*c)*(1+a/b/x^4)^(3/4)*x^3 *InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/a^(3/2)/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 = 72, normalized size of antiderivative = 0.71 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {x \left (b c-a d+(2 b c+a d) \left (1+\frac {b x^4}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^4}{a}\right )\right )}{3 a b \left (a+b x^4\right )^{3/4}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(7/4),x]
Output:
(x*(b*c - a*d + (2*b*c + a*d)*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^4)/a)]))/(3*a*b*(a + b*x^4)^(3/4))
Time = 0.40 (sec) , antiderivative size = 102, 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 = {910, 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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {(a d+2 b c) \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx}{3 a b}+\frac {x (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (a d+2 b c) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{3 a b \left (a+b x^4\right )^{3/4}}+\frac {x (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {x (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (a d+2 b c) \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{3 a b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {x (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (a d+2 b c) \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{6 a b \left (a+b x^4\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {x (b c-a d)}{3 a b \left (a+b x^4\right )^{3/4}}-\frac {x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} (a d+2 b c) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{3 a^{3/2} \sqrt {b} \left (a+b x^4\right )^{3/4}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(7/4),x]
Output:
((b*c - a*d)*x)/(3*a*b*(a + b*x^4)^(3/4)) - ((2*b*c + a*d)*(1 + a/(b*x^4)) ^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(3*a^(3/2)*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[(-(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])
\[\int \frac {d \,x^{4}+c}{\left (b \,x^{4}+a \right )^{\frac {7}{4}}}d x\]
Input:
int((d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int((d*x^4+c)/(b*x^4+a)^(7/4),x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(1/4)*(d*x^4 + c)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
Result contains complex when optimal does not.
Time = 4.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{4}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{4}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(7/4),x)
Output:
c*x*gamma(1/4)*hyper((1/4, 7/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 7/4)*gamma(5/4)) + d*x**5*gamma(5/4)*hyper((5/4, 7/4), (9/4,), b*x**4*exp_ polar(I*pi)/a)/(4*a**(7/4)*gamma(9/4))
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(7/4), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(7/4),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(7/4), x)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{7/4}} \,d x \] Input:
int((c + d*x^4)/(a + b*x^4)^(7/4),x)
Output:
int((c + d*x^4)/(a + b*x^4)^(7/4), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{7/4}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}} a +\left (b \,x^{4}+a \right )^{\frac {3}{4}} b \,x^{4}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(7/4),x)
Output:
int(x**4/((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*d + int(1 /((a + b*x**4)**(3/4)*a + (a + b*x**4)**(3/4)*b*x**4),x)*c