Integrand size = 15, antiderivative size = 149 \[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\frac {4 a^3 x^2}{65 b^2 \sqrt [4]{a+b x^4}}-\frac {2 a^2 x^2 \left (a+b x^4\right )^{3/4}}{65 b^2}+\frac {a x^6 \left (a+b x^4\right )^{3/4}}{39 b}+\frac {1}{13} x^{10} \left (a+b x^4\right )^{3/4}-\frac {4 a^{7/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{65 b^{5/2} \sqrt [4]{a+b x^4}} \] Output:
4/65*a^3*x^2/b^2/(b*x^4+a)^(1/4)-2/65*a^2*x^2*(b*x^4+a)^(3/4)/b^2+1/39*a*x ^6*(b*x^4+a)^(3/4)/b+1/13*x^10*(b*x^4+a)^(3/4)-4/65*a^(7/2)*(1+b*x^4/a)^(1 /4)*EllipticE(sin(1/2*arctan(b^(1/2)*x^2/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4 +a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.64 \[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\frac {x^2 \left (a+b x^4\right )^{3/4} \left (\left (1+\frac {b x^4}{a}\right )^{3/4} \left (-2 a^2+a b x^4+3 b^2 x^8\right )+2 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^4}{a}\right )\right )}{39 b^2 \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:
Integrate[x^9*(a + b*x^4)^(3/4),x]
Output:
(x^2*(a + b*x^4)^(3/4)*((1 + (b*x^4)/a)^(3/4)*(-2*a^2 + a*b*x^4 + 3*b^2*x^ 8) + 2*a^2*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^4)/a)]))/(39*b^2*(1 + (b*x^4)/a)^(3/4))
Time = 0.41 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {807, 248, 262, 262, 227, 225, 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.
\(\displaystyle \int x^9 \left (a+b x^4\right )^{3/4} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int x^8 \left (b x^4+a\right )^{3/4}dx^2\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \int \frac {x^8}{\sqrt [4]{b x^4+a}}dx^2+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \int \frac {x^4}{\sqrt [4]{b x^4+a}}dx^2}{3 b}\right )+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2}{5 b}\right )}{3 b}\right )+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{13} a \left (\frac {2 x^6 \left (a+b x^4\right )^{3/4}}{9 b}-\frac {2 a \left (\frac {2 x^2 \left (a+b x^4\right )^{3/4}}{5 b}-\frac {2 a \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^4}}\right )}{3 b}\right )+\frac {2}{13} x^{10} \left (a+b x^4\right )^{3/4}\right )\) |
Input:
Int[x^9*(a + b*x^4)^(3/4),x]
Output:
((2*x^10*(a + b*x^4)^(3/4))/13 + (3*a*((2*x^6*(a + b*x^4)^(3/4))/(9*b) - ( 2*a*((2*x^2*(a + b*x^4)^(3/4))/(5*b) - (2*a*(1 + (b*x^4)/a)^(1/4)*((2*x^2) /(1 + (b*x^4)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a] ]/2, 2])/Sqrt[b]))/(5*b*(a + b*x^4)^(1/4))))/(3*b)))/13)/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[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, 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]
\[\int x^{9} \left (b \,x^{4}+a \right )^{\frac {3}{4}}d x\]
Input:
int(x^9*(b*x^4+a)^(3/4),x)
Output:
int(x^9*(b*x^4+a)^(3/4),x)
\[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{9} \,d x } \] Input:
integrate(x^9*(b*x^4+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)*x^9, x)
Result contains complex when optimal does not.
Time = 0.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.19 \[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\frac {a^{\frac {3}{4}} x^{10} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10} \] Input:
integrate(x**9*(b*x**4+a)**(3/4),x)
Output:
a**(3/4)*x**10*hyper((-3/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/10
\[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{9} \,d x } \] Input:
integrate(x^9*(b*x^4+a)^(3/4),x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(3/4)*x^9, x)
\[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\int { {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{9} \,d x } \] Input:
integrate(x^9*(b*x^4+a)^(3/4),x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/4)*x^9, x)
Timed out. \[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\int x^9\,{\left (b\,x^4+a\right )}^{3/4} \,d x \] Input:
int(x^9*(a + b*x^4)^(3/4),x)
Output:
int(x^9*(a + b*x^4)^(3/4), x)
\[ \int x^9 \left (a+b x^4\right )^{3/4} \, dx=\frac {-6 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a^{2} x^{2}+5 \left (b \,x^{4}+a \right )^{\frac {3}{4}} a b \,x^{6}+15 \left (b \,x^{4}+a \right )^{\frac {3}{4}} b^{2} x^{10}+12 \left (\int \frac {x}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}d x \right ) a^{3}}{195 b^{2}} \] Input:
int(x^9*(b*x^4+a)^(3/4),x)
Output:
( - 6*(a + b*x**4)**(3/4)*a**2*x**2 + 5*(a + b*x**4)**(3/4)*a*b*x**6 + 15* (a + b*x**4)**(3/4)*b**2*x**10 + 12*int(((a + b*x**4)**(3/4)*x)/(a + b*x** 4),x)*a**3)/(195*b**2)