Integrand size = 15, antiderivative size = 99 \[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\frac {a x^3}{4 \sqrt [4]{a+b x^4}}+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}+\frac {a^{3/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{4 \sqrt {b} \sqrt [4]{a+b x^4}} \] Output:
1/4*a*x^3/(b*x^4+a)^(1/4)+1/6*x^3*(b*x^4+a)^(3/4)+1/4*a^(3/2)*(1+a/b/x^4)^ (1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/b^(1/2)/(b *x^4+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.52 \[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\frac {x^3 \left (a+b x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:
Integrate[x^2*(a + b*x^4)^(3/4),x]
Output:
(x^3*(a + b*x^4)^(3/4)*Hypergeometric2F1[-3/4, 3/4, 7/4, -((b*x^4)/a)])/(3 *(1 + (b*x^4)/a)^(3/4))
Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {811, 839, 813, 858, 807, 212}
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.
\(\displaystyle \int x^2 \left (a+b x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{2} a \int \frac {x^2}{\sqrt [4]{b x^4+a}}dx+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle \frac {1}{2} a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {1}{2} a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx\right )+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle \frac {1}{2} a \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b \sqrt [4]{a+b x^4}}\right )+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {1}{2} a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} a \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{2} a \left (\frac {\sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )+\frac {1}{6} x^3 \left (a+b x^4\right )^{3/4}\) |
Input:
Int[x^2*(a + b*x^4)^(3/4),x]
Output:
(x^3*(a + b*x^4)^(3/4))/6 + (a*(x^3/(2*(a + b*x^4)^(1/4)) + (Sqrt[a]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqrt[ b]*(a + b*x^4)^(1/4))))/2
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) )*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x*((1 + a/(b*x^4) )^(1/4)/(b*(a + b*x^4)^(1/4))) Int[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
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]
\[\int x^{2} \left (b \,x^{4}+a \right )^{\frac {3}{4}}d x\]
Input:
int(x^2*(b*x^4+a)^(3/4),x)
Output:
int(x^2*(b*x^4+a)^(3/4),x)
\[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)*x^2, x)
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.39 \[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(x**2*(b*x**4+a)**(3/4),x)
Output:
a**(3/4)*x**3*gamma(3/4)*hyper((-3/4, 3/4), (7/4,), b*x**4*exp_polar(I*pi) /a)/(4*gamma(7/4))
\[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(3/4)*x^2, x)
\[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{2} \,d x } \] Input:
integrate(x^2*(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/4)*x^2, x)
Timed out. \[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\int x^2\,{\left (b\,x^4+a\right )}^{3/4} \,d x \] Input:
int(x^2*(a + b*x^4)^(3/4),x)
Output:
int(x^2*(a + b*x^4)^(3/4), x)
\[ \int x^2 \left (a+b x^4\right )^{3/4} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{3}}{6}+\frac {\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) a}{2} \] Input:
int(x^2*(b*x^4+a)^(3/4),x)
Output:
((a + b*x**4)**(3/4)*x**3 + 3*int(((a + b*x**4)**(3/4)*x**2)/(a + b*x**4), x)*a)/6