Integrand size = 16, antiderivative size = 103 \[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\frac {2 x^3}{3 b \left (a-b x^2\right )^{3/4}}+\frac {4 x \sqrt [4]{a-b x^2}}{3 b^2}-\frac {8 a^{3/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{5/2} \left (a-b x^2\right )^{3/4}} \] Output:
2/3*x^3/b/(-b*x^2+a)^(3/4)+4/3*x*(-b*x^2+a)^(1/4)/b^2-8/3*a^(3/2)*(1-b*x^2 /a)^(3/4)*InverseJacobiAM(1/2*arcsin(b^(1/2)*x/a^(1/2)),2^(1/2))/b^(5/2)/( -b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.49 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.64 \[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=-\frac {2 \left (-2 a x+b x^3+2 a x \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {b x^2}{a}\right )\right )}{3 b^2 \left (a-b x^2\right )^{3/4}} \] Input:
Integrate[x^4/(a - b*x^2)^(7/4),x]
Output:
(-2*(-2*a*x + b*x^3 + 2*a*x*(1 - (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3 /4, 3/2, (b*x^2)/a]))/(3*b^2*(a - b*x^2)^(3/4))
Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {252, 262, 231, 230}
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.
\(\displaystyle \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {2 x^3}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 \int \frac {x^2}{\left (a-b x^2\right )^{3/4}}dx}{b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {2 x^3}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 \left (\frac {2 a \int \frac {1}{\left (a-b x^2\right )^{3/4}}dx}{3 b}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{b}\) |
\(\Big \downarrow \) 231 |
\(\displaystyle \frac {2 x^3}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 \left (\frac {2 a \left (1-\frac {b x^2}{a}\right )^{3/4} \int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}}dx}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{b}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle \frac {2 x^3}{3 b \left (a-b x^2\right )^{3/4}}-\frac {2 \left (\frac {4 a^{3/2} \left (1-\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{3 b^{3/2} \left (a-b x^2\right )^{3/4}}-\frac {2 x \sqrt [4]{a-b x^2}}{3 b}\right )}{b}\) |
Input:
Int[x^4/(a - b*x^2)^(7/4),x]
Output:
(2*x^3)/(3*b*(a - b*x^2)^(3/4)) - (2*((-2*x*(a - b*x^2)^(1/4))/(3*b) + (4* a^(3/2)*(1 - (b*x^2)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2]) /(3*b^(3/2)*(a - b*x^2)^(3/4))))/b
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2] ))*EllipticF[(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)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( a + b*x^2)^(3/4) Int[1/(1 + b*(x^2/a))^(3/4), 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]
\[\int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]
Input:
int(x^4/(-b*x^2+a)^(7/4),x)
Output:
int(x^4/(-b*x^2+a)^(7/4),x)
\[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\int { \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^4/(-b*x^2+a)^(7/4),x, algorithm="fricas")
Output:
integral((-b*x^2 + a)^(1/4)*x^4/(b^2*x^4 - 2*a*b*x^2 + a^2), x)
Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.28 \[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\frac {x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 a^{\frac {7}{4}}} \] Input:
integrate(x**4/(-b*x**2+a)**(7/4),x)
Output:
x**5*hyper((7/4, 5/2), (7/2,), b*x**2*exp_polar(2*I*pi)/a)/(5*a**(7/4))
\[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\int { \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^4/(-b*x^2+a)^(7/4),x, algorithm="maxima")
Output:
integrate(x^4/(-b*x^2 + a)^(7/4), x)
\[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\int { \frac {x^{4}}{{\left (-b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \] Input:
integrate(x^4/(-b*x^2+a)^(7/4),x, algorithm="giac")
Output:
integrate(x^4/(-b*x^2 + a)^(7/4), x)
Timed out. \[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\int \frac {x^4}{{\left (a-b\,x^2\right )}^{7/4}} \,d x \] Input:
int(x^4/(a - b*x^2)^(7/4),x)
Output:
int(x^4/(a - b*x^2)^(7/4), x)
\[ \int \frac {x^4}{\left (a-b x^2\right )^{7/4}} \, dx=\int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {3}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \] Input:
int(x^4/(-b*x^2+a)^(7/4),x)
Output:
int(x**4/((a - b*x**2)**(3/4)*a - (a - b*x**2)**(3/4)*b*x**2),x)