\(\int \frac {\sqrt {e x} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [457]

Optimal result

Integrand size = 26, antiderivative size = 99 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {(2 b c-3 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} b^{3/2} \sqrt [4]{a+b x^2}} \] Output:

d*(e*x)^(3/2)/b/e/(b*x^2+a)^(1/4)-(-3*a*d+2*b*c)*(1+a/b/x^2)^(1/4)*(e*x)^( 
1/2)*EllipticE(sin(1/2*arccot(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(1/2)/b^(3/2) 
/(b*x^2+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \sqrt {e x} \left (3 a d+(2 b c-3 a d) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 a b \sqrt [4]{a+b x^2}} \] Input:

Integrate[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(x*Sqrt[e*x]*(3*a*d + (2*b*c - 3*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric 
2F1[3/4, 5/4, 7/4, -((b*x^2)/a)]))/(3*a*b*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {363, 249, 858, 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.

\(\displaystyle \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {(2 b c-3 a d) \int \frac {\sqrt {e x}}{\left (b x^2+a\right )^{5/4}}dx}{2 b}+\frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 249

\(\displaystyle \frac {\sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-3 a d) \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 858

\(\displaystyle \frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {\sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-3 a d) \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{2 b^2 \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {d (e x)^{3/2}}{b e \sqrt [4]{a+b x^2}}-\frac {\sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (2 b c-3 a d) E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{\sqrt {a} b^{3/2} \sqrt [4]{a+b x^2}}\)

Input:

Int[(Sqrt[e*x]*(c + d*x^2))/(a + b*x^2)^(5/4),x]
 

Output:

(d*(e*x)^(3/2))/(b*e*(a + b*x^2)^(1/4)) - ((2*b*c - 3*a*d)*(1 + a/(b*x^2)) 
^(1/4)*Sqrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(Sqrt[a]*b^( 
3/2)*(a + b*x^2)^(1/4))
 

Defintions of rubi rules used

rule 212

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]
 

rule 249

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[Sqrt[c* 
x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b*x^2)^(1/4)))   Int[1/(x^2*(1 + a/(b*x^2 
))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {\sqrt {e x}\, \left (x^{2} d +c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Output:

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(b^2*x^4 + 2*a*b*x^2 + a^ 
2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.79 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c \sqrt {e} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {7}{4}\right )} + \frac {d \sqrt {e} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate((e*x)**(1/2)*(d*x**2+c)/(b*x**2+a)**(5/4),x)
 

Output:

c*sqrt(e)*x**(3/2)*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(5/4)*gamma(7/4)) + d*sqrt(e)*x**(7/2)*gamma(7/4)*hyper((5/4 
, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)*sqrt(e*x)/(b*x^2 + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \sqrt {e x}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)*sqrt(e*x)/(b*x^2 + a)^(5/4), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {\sqrt {e\,x}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(5/4),x)
 

Output:

int(((e*x)^(1/2)*(c + d*x^2))/(a + b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {\sqrt {e x} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx=\sqrt {e}\, \left (\left (\int \frac {\sqrt {x}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c +\left (\int \frac {\sqrt {x}\, x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d \right ) \] Input:

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)
 

Output:

sqrt(e)*(int(sqrt(x)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2), 
x)*c + int((sqrt(x)*x**2)/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x 
**2),x)*d)