Integrand size = 22, antiderivative size = 116 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=-\frac {c}{a x \left (a+b x^2\right )^{5/4}}-\frac {(7 b c-2 a d) x}{5 a^2 \left (a+b x^2\right )^{5/4}}-\frac {3 (7 b c-2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} \sqrt {b} \sqrt [4]{a+b x^2}} \] Output:
-c/a/x/(b*x^2+a)^(5/4)-1/5*(-2*a*d+7*b*c)*x/a^2/(b*x^2+a)^(5/4)-3/5*(-2*a* d+7*b*c)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^ (1/2))/a^(5/2)/b^(1/2)/(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 = 118, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\frac {-2 \left (21 b^2 c x^4+a^2 \left (5 c-8 d x^2\right )+a b \left (28 c x^2-6 d x^4\right )\right )+3 (7 b c-2 a d) x^2 \left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{10 a^3 x \left (a+b x^2\right )^{5/4}} \] Input:
Integrate[(c + d*x^2)/(x^2*(a + b*x^2)^(9/4)),x]
Output:
(-2*(21*b^2*c*x^4 + a^2*(5*c - 8*d*x^2) + a*b*(28*c*x^2 - 6*d*x^4)) + 3*(7 *b*c - 2*a*d)*x^2*(a + b*x^2)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)])/(10*a^3*x*(a + b*x^2)^(5/4))
Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {359, 215, 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.
\(\displaystyle \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(7 b c-2 a d) \int \frac {1}{\left (b x^2+a\right )^{9/4}}dx}{2 a}-\frac {c}{a x \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle -\frac {(7 b c-2 a d) \left (\frac {3 \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{5 a}+\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}\right )}{2 a}-\frac {c}{a x \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle -\frac {(7 b c-2 a d) \left (\frac {3 \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{5 a^2 \sqrt [4]{a+b x^2}}+\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}\right )}{2 a}-\frac {c}{a x \left (a+b x^2\right )^{5/4}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {(7 b c-2 a d) \left (\frac {6 \sqrt [4]{\frac {b x^2}{a}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}+\frac {2 x}{5 a \left (a+b x^2\right )^{5/4}}\right )}{2 a}-\frac {c}{a x \left (a+b x^2\right )^{5/4}}\) |
Input:
Int[(c + d*x^2)/(x^2*(a + b*x^2)^(9/4)),x]
Output:
-(c/(a*x*(a + b*x^2)^(5/4))) - ((7*b*c - 2*a*d)*((2*x)/(5*a*(a + b*x^2)^(5 /4)) + (6*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2 ])/(5*a^(3/2)*Sqrt[b]*(a + b*x^2)^(1/4))))/(2*a)
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_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
\[\int \frac {x^{2} d +c}{x^{2} \left (b \,x^{2}+a \right )^{\frac {9}{4}}}d x\]
Input:
int((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x)
Output:
int((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)/(b^3*x^8 + 3*a*b^2*x^6 + 3*a^2*b*x^ 4 + a^3*x^2), x)
Result contains complex when optimal does not.
Time = 6.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.47 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {9}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {9}{4}} x} + \frac {d x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {9}{4}}} \] Input:
integrate((d*x**2+c)/x**2/(b*x**2+a)**(9/4),x)
Output:
-c*hyper((-1/2, 9/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(9/4)*x) + d*x *hyper((1/2, 9/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(9/4)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*x^2), x)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*x^2), x)
Time = 1.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\frac {d\,x\,{\left (\frac {b\,x^2}{a}+1\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{9/4}}-\frac {2\,c\,{\left (\frac {a}{b\,x^2}+1\right )}^{9/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {9}{4},\frac {11}{4};\ \frac {15}{4};\ -\frac {a}{b\,x^2}\right )}{11\,x\,{\left (b\,x^2+a\right )}^{9/4}} \] Input:
int((c + d*x^2)/(x^2*(a + b*x^2)^(9/4)),x)
Output:
(d*x*((b*x^2)/a + 1)^(9/4)*hypergeom([1/2, 9/4], 3/2, -(b*x^2)/a))/(a + b* x^2)^(9/4) - (2*c*(a/(b*x^2) + 1)^(9/4)*hypergeom([9/4, 11/4], 15/4, -a/(b *x^2)))/(11*x*(a + b*x^2)^(9/4))
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{9/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2} x^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{4}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{6}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a^{2}+2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} a b \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b^{2} x^{4}}d x \right ) d \] Input:
int((d*x^2+c)/x^2/(b*x^2+a)^(9/4),x)
Output:
int(1/((a + b*x**2)**(1/4)*a**2*x**2 + 2*(a + b*x**2)**(1/4)*a*b*x**4 + (a + b*x**2)**(1/4)*b**2*x**6),x)*c + 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)*d