Integrand size = 11, antiderivative size = 97 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {5}{12} a x \sqrt [4]{a+b x^4}+\frac {1}{6} x \left (a+b x^4\right )^{5/4}-\frac {5 a^{3/2} \sqrt {b} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{12 \left (a+b x^4\right )^{3/4}} \] Output:
5/12*a*x*(b*x^4+a)^(1/4)+1/6*x*(b*x^4+a)^(5/4)-5/12*a^(3/2)*b^(1/2)*(1+a/b /x^4)^(3/4)*x^3*InverseJacobiAM(1/2*arccot(b^(1/2)*x^2/a^(1/2)),2^(1/2))/( b*x^4+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.60 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {a x \sqrt [4]{a+b x^4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^4}{a}\right )}{\sqrt [4]{1+\frac {b x^4}{a}}} \] Input:
Integrate[(a + b*x^4)^(5/4),x]
Output:
(a*x*(a + b*x^4)^(1/4)*Hypergeometric2F1[-5/4, 1/4, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^(1/4)
Time = 0.35 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {748, 748, 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.
\(\displaystyle \int \left (a+b x^4\right )^{5/4} \, dx\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {5}{6} a \int \sqrt [4]{b x^4+a}dx+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 748 |
\(\displaystyle \frac {5}{6} a \left (\frac {1}{2} a \int \frac {1}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {5}{6} a \left (\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x^3}dx}{2 \left (a+b x^4\right )^{3/4}}+\frac {1}{2} x \sqrt [4]{a+b x^4}\right )+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {5}{6} a \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{3/4} x}d\frac {1}{x}}{2 \left (a+b x^4\right )^{3/4}}\right )+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {5}{6} a \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {a x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4}}d\frac {1}{x^2}}{4 \left (a+b x^4\right )^{3/4}}\right )+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {5}{6} a \left (\frac {1}{2} x \sqrt [4]{a+b x^4}-\frac {\sqrt {a} \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right ),2\right )}{2 \left (a+b x^4\right )^{3/4}}\right )+\frac {1}{6} x \left (a+b x^4\right )^{5/4}\) |
Input:
Int[(a + b*x^4)^(5/4),x]
Output:
(x*(a + b*x^4)^(5/4))/6 + (5*a*((x*(a + b*x^4)^(1/4))/2 - (Sqrt[a]*Sqrt[b] *(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/ (2*(a + b*x^4)^(3/4))))/6
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_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Simp[a*n*(p/(n*p + 1)) Int[(a + b*x^n)^(p - 1), x], x] /; Fre eQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || LtQ[Denominat or[p + 1/n], Denominator[p]])
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]
\[\int \left (b \,x^{4}+a \right )^{\frac {5}{4}}d x\]
Input:
int((b*x^4+a)^(5/4),x)
Output:
int((b*x^4+a)^(5/4),x)
\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x^4+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(5/4), x)
Result contains complex when optimal does not.
Time = 0.64 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {a^{\frac {5}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**4+a)**(5/4),x)
Output:
a**(5/4)*x*gamma(1/4)*hyper((-5/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a) /(4*gamma(5/4))
\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x^4+a)^(5/4),x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(5/4), x)
\[ \int \left (a+b x^4\right )^{5/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {5}{4}} \,d x } \] Input:
integrate((b*x^4+a)^(5/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(5/4), x)
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.38 \[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {x\,{\left (b\,x^4+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{5/4}} \] Input:
int((a + b*x^4)^(5/4),x)
Output:
(x*(a + b*x^4)^(5/4)*hypergeom([-5/4, 1/4], 5/4, -(b*x^4)/a))/((b*x^4)/a + 1)^(5/4)
\[ \int \left (a+b x^4\right )^{5/4} \, dx=\frac {7 \left (b \,x^{4}+a \right )^{\frac {1}{4}} a x}{12}+\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} b \,x^{5}}{6}+\frac {5 \left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}d x \right ) a^{2}}{12} \] Input:
int((b*x^4+a)^(5/4),x)
Output:
(7*(a + b*x**4)**(1/4)*a*x + 2*(a + b*x**4)**(1/4)*b*x**5 + 5*int((a + b*x **4)**(1/4)/(a + b*x**4),x)*a**2)/12