Integrand size = 19, antiderivative size = 98 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 (b c-a d) x}{5 a b \left (a+b x^2\right )^{5/4}}+\frac {2 (3 b c+2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}} \] Output:
2/5*(-a*d+b*c)*x/a/b/(b*x^2+a)^(5/4)+2/5*(2*a*d+3*b*c)*(1+b*x^2/a)^(1/4)*E llipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(3/2)/b^(3/2)/(b*x^ 2+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {2 a^2 d x+6 b^2 c x^3+4 a b x \left (2 c+d x^2\right )-(3 b c+2 a d) x \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{5 a^2 b \left (a+b x^2\right )^{5/4}} \] Input:
Integrate[(c + d*x^2)/(a + b*x^2)^(9/4),x]
Output:
(2*a^2*d*x + 6*b^2*c*x^3 + 4*a*b*x*(2*c + d*x^2) - (3*b*c + 2*a*d)*x*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a) ])/(5*a^2*b*(a + b*x^2)^(5/4))
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {298, 213, 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.
\(\displaystyle \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {(2 a d+3 b c) \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{5 a b}+\frac {2 x (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} (2 a d+3 b c) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{5 a^2 b \sqrt [4]{a+b x^2}}+\frac {2 x (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {2 \sqrt [4]{\frac {b x^2}{a}+1} (2 a d+3 b c) E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} b^{3/2} \sqrt [4]{a+b x^2}}+\frac {2 x (b c-a d)}{5 a b \left (a+b x^2\right )^{5/4}}\) |
Input:
Int[(c + d*x^2)/(a + b*x^2)^(9/4),x]
Output:
(2*(b*c - a*d)*x)/(5*a*b*(a + b*x^2)^(5/4)) + (2*(3*b*c + 2*a*d)*(1 + (b*x ^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*b^(3/ 2)*(a + b*x^2)^(1/4))
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)^(-5/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a*(a + b*x^2)^(1/4)) Int[1/(1 + b*(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
\[\int \frac {x^{2} d +c}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]
Input:
int((d*x^2+c)/(b*x^2+a)^(9/4),x)
Output:
int((d*x^2+c)/(b*x^2+a)^(9/4),x)
\[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate((d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^ 2 + a^3), x)
Result contains complex when optimal does not.
Time = 4.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.57 \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\frac {c x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {9}{4}}} + \frac {d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {9}{4}}} \] Input:
integrate((d*x**2+c)/(b*x**2+a)**(9/4),x)
Output:
c*x*hyper((1/2, 9/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(9/4) + d*x**3* hyper((3/2, 9/4), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(9/4))
\[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate((d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)/(b*x^2 + a)^(9/4), x)
\[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}}} \,d x } \] Input:
integrate((d*x^2+c)/(b*x^2+a)^(9/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)/(b*x^2 + a)^(9/4), x)
Timed out. \[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\int \frac {d\,x^2+c}{{\left (b\,x^2+a\right )}^{9/4}} \,d x \] Input:
int((c + d*x^2)/(a + b*x^2)^(9/4),x)
Output:
int((c + d*x^2)/(a + b*x^2)^(9/4), x)
\[ \int \frac {c+d x^2}{\left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) c \] Input:
int((d*x^2+c)/(b*x^2+a)^(9/4),x)
Output:
int(x**2/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*d + int(1/((a + b*x**2)**(1/4)*a**2 + 2*(a + b*x**2)**(1/4)*a*b*x**2 + (a + b*x**2)**(1/4)*b**2*x**4),x)*c