Integrand size = 22, antiderivative size = 86 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=-\frac {c}{a x \sqrt [4]{a+b x^2}}-\frac {(3 b c-2 a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}} \] Output:
-c/a/x/(b*x^2+a)^(1/4)-(-2*a*d+3*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^(3/2)/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 = 10.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\frac {-2 a c-6 b c x^2+4 a d x^2+(3 b c-2 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{2 a^2 x \sqrt [4]{a+b x^2}} \] Input:
Integrate[(c + d*x^2)/(x^2*(a + b*x^2)^(5/4)),x]
Output:
(-2*a*c - 6*b*c*x^2 + 4*a*d*x^2 + (3*b*c - 2*a*d)*x^2*(1 + (b*x^2)/a)^(1/4 )*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)])/(2*a^2*x*(a + b*x^2)^(1/ 4))
Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {359, 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.
\(\displaystyle \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(3 b c-2 a d) \int \frac {1}{\left (b x^2+a\right )^{5/4}}dx}{2 a}-\frac {c}{a x \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 213 |
\(\displaystyle -\frac {\sqrt [4]{\frac {b x^2}{a}+1} (3 b c-2 a d) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{2 a^2 \sqrt [4]{a+b x^2}}-\frac {c}{a x \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {\sqrt [4]{\frac {b x^2}{a}+1} (3 b c-2 a d) E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt {b} \sqrt [4]{a+b x^2}}-\frac {c}{a x \sqrt [4]{a+b x^2}}\) |
Input:
Int[(c + d*x^2)/(x^2*(a + b*x^2)^(5/4)),x]
Output:
-(c/(a*x*(a + b*x^2)^(1/4))) - ((3*b*c - 2*a*d)*(1 + (b*x^2)/a)^(1/4)*Elli pticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(a^(3/2)*Sqrt[b]*(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[((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]
\[\int \frac {x^{2} d +c}{x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x)
Output:
int((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*(d*x^2 + c)/(b^2*x^6 + 2*a*b*x^4 + a^2*x^2), x)
Result contains complex when optimal does not.
Time = 2.58 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.63 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}} x} + \frac {d x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}}} \] Input:
integrate((d*x**2+c)/x**2/(b*x**2+a)**(5/4),x)
Output:
-c*hyper((-1/2, 5/4), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(5/4)*x) + d*x *hyper((1/2, 5/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/4)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*x^2), x)
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} x^{2}} \,d x } \] Input:
integrate((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*x^2), x)
Time = 1.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\frac {d\,x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{4};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{5/4}}-\frac {2\,c\,{\left (\frac {a}{b\,x^2}+1\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{4},\frac {7}{4};\ \frac {11}{4};\ -\frac {a}{b\,x^2}\right )}{7\,x\,{\left (b\,x^2+a\right )}^{5/4}} \] Input:
int((c + d*x^2)/(x^2*(a + b*x^2)^(5/4)),x)
Output:
(d*x*((b*x^2)/a + 1)^(5/4)*hypergeom([1/2, 5/4], 3/2, -(b*x^2)/a))/(a + b* x^2)^(5/4) - (2*c*(a/(b*x^2) + 1)^(5/4)*hypergeom([5/4, 7/4], 11/4, -a/(b* x^2)))/(7*x*(a + b*x^2)^(5/4))
\[ \int \frac {c+d x^2}{x^2 \left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{4}}d x \right ) c +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d \] Input:
int((d*x^2+c)/x^2/(b*x^2+a)^(5/4),x)
Output:
int(1/((a + b*x**2)**(1/4)*a*x**2 + (a + b*x**2)**(1/4)*b*x**4),x)*c + int (1/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*d