Integrand size = 16, antiderivative size = 124 \[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\frac {2 x^5}{b \sqrt [4]{a-b x^2}}+\frac {8 a x \left (a-b x^2\right )^{3/4}}{3 b^3}+\frac {20 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac {16 a^{5/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a-b x^2}} \] Output:
2*x^5/b/(-b*x^2+a)^(1/4)+8/3*a*x*(-b*x^2+a)^(3/4)/b^3+20/9*x^3*(-b*x^2+a)^ (3/4)/b^2-16/3*a^(5/2)*(1-b*x^2/a)^(1/4)*EllipticE(sin(1/2*arcsin(b^(1/2)* x/a^(1/2))),2^(1/2))/b^(7/2)/(-b*x^2+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.50 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=-\frac {2 x \left (-12 a^2+2 a b x^2+b^2 x^4+12 a^2 \sqrt [4]{1-\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )\right )}{9 b^3 \sqrt [4]{a-b x^2}} \] Input:
Integrate[x^6/(a - b*x^2)^(5/4),x]
Output:
(-2*x*(-12*a^2 + 2*a*b*x^2 + b^2*x^4 + 12*a^2*(1 - (b*x^2)/a)^(1/4)*Hyperg eometric2F1[1/4, 1/2, 3/2, (b*x^2)/a]))/(9*b^3*(a - b*x^2)^(1/4))
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {252, 262, 262, 227, 226}
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.
\(\displaystyle \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {2 x^5}{b \sqrt [4]{a-b x^2}}-\frac {10 \int \frac {x^4}{\sqrt [4]{a-b x^2}}dx}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 x^5}{b \sqrt [4]{a-b x^2}}-\frac {10 \left (\frac {2 a \int \frac {x^2}{\sqrt [4]{a-b x^2}}dx}{3 b}-\frac {2 x^3 \left (a-b x^2\right )^{3/4}}{9 b}\right )}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 x^5}{b \sqrt [4]{a-b x^2}}-\frac {10 \left (\frac {2 a \left (\frac {2 a \int \frac {1}{\sqrt [4]{a-b x^2}}dx}{5 b}-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}\right )}{3 b}-\frac {2 x^3 \left (a-b x^2\right )^{3/4}}{9 b}\right )}{b}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {2 x^5}{b \sqrt [4]{a-b x^2}}-\frac {10 \left (\frac {2 a \left (\frac {2 a \sqrt [4]{1-\frac {b x^2}{a}} \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}}dx}{5 b \sqrt [4]{a-b x^2}}-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}\right )}{3 b}-\frac {2 x^3 \left (a-b x^2\right )^{3/4}}{9 b}\right )}{b}\) |
\(\Big \downarrow \) 226 |
\(\displaystyle \frac {2 x^5}{b \sqrt [4]{a-b x^2}}-\frac {10 \left (\frac {2 a \left (\frac {4 a^{3/2} \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 x \left (a-b x^2\right )^{3/4}}{5 b}\right )}{3 b}-\frac {2 x^3 \left (a-b x^2\right )^{3/4}}{9 b}\right )}{b}\) |
Input:
Int[x^6/(a - b*x^2)^(5/4),x]
Output:
(2*x^5)/(b*(a - b*x^2)^(1/4)) - (10*((-2*x^3*(a - b*x^2)^(3/4))/(9*b) + (2 *a*((-2*x*(a - b*x^2)^(3/4))/(5*b) + (4*a^(3/2)*(1 - (b*x^2)/a)^(1/4)*Elli pticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a - b*x^2)^(1/4))))/( 3*b)))/b
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] ))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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]
\[\int \frac {x^{6}}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int(x^6/(-b*x^2+a)^(5/4),x)
Output:
int(x^6/(-b*x^2+a)^(5/4),x)
\[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {x^{6}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((-b*x^2 + a)^(3/4)*x^6/(b^2*x^4 - 2*a*b*x^2 + a^2), x)
Result contains complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.23 \[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\frac {x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{7 a^{\frac {5}{4}}} \] Input:
integrate(x**6/(-b*x**2+a)**(5/4),x)
Output:
x**7*hyper((5/4, 7/2), (9/2,), b*x**2*exp_polar(2*I*pi)/a)/(7*a**(5/4))
\[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {x^{6}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate(x^6/(-b*x^2 + a)^(5/4), x)
\[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {x^{6}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:
integrate(x^6/(-b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate(x^6/(-b*x^2 + a)^(5/4), x)
Timed out. \[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\int \frac {x^6}{{\left (a-b\,x^2\right )}^{5/4}} \,d x \] Input:
int(x^6/(a - b*x^2)^(5/4),x)
Output:
int(x^6/(a - b*x^2)^(5/4), x)
\[ \int \frac {x^6}{\left (a-b x^2\right )^{5/4}} \, dx=\int \frac {x^{6}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \] Input:
int(x^6/(-b*x^2+a)^(5/4),x)
Output:
int(x**6/((a - b*x**2)**(1/4)*a - (a - b*x**2)**(1/4)*b*x**2),x)