\(\int \frac {c+d x^2}{(e x)^{3/2} (a+b x^2)^{5/4}} \, dx\) [458]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 c}{a e \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {2 (2 b c-a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} \sqrt {b} e^2 \sqrt [4]{a+b x^2}} \] Output:

-2*c/a/e/(e*x)^(1/2)/(b*x^2+a)^(1/4)+2*(-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^(3/2)/b^(1 
/2)/e^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 = 10.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.75 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (-6 a c+2 (-2 b c+a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 a^2 (e x)^{3/2} \sqrt [4]{a+b x^2}} \] Input:

Integrate[(c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)),x]
 

Output:

(x*(-6*a*c + 2*(-2*b*c + a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[ 
3/4, 5/4, 7/4, -((b*x^2)/a)]))/(3*a^2*(e*x)^(3/2)*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 105, 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 = {359, 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.

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 249

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

\(\Big \downarrow \) 858

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

\(\Big \downarrow \) 212

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

Input:

Int[(c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)),x]
 

Output:

(-2*c)/(a*e*Sqrt[e*x]*(a + b*x^2)^(1/4)) + (2*(2*b*c - a*d)*(1 + a/(b*x^2) 
)^(1/4)*Sqrt[e*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 2])/(a^(3/2)*Sq 
rt[b]*e^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 359

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]
 

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]

\[\int \frac {x^{2} d +c}{\left (e x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x)
 

Output:

int((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(3/2)/(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*e^2*x^6 + 2*a*b*e^2* 
x^4 + a^2*e^2*x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 15.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=- \frac {d {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac {5}{4}} e^{\frac {3}{2}} x} + \frac {c \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \] Input:

integrate((d*x**2+c)/(e*x)**(3/2)/(b*x**2+a)**(5/4),x)
 

Output:

-d*hyper((1/2, 5/4), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(b**(5/4)*e**(3/2 
)*x) + c*gamma(-1/4)*hyper((-1/4, 5/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/ 
(2*a**(5/4)*e**(3/2)*sqrt(x)*gamma(3/4))
 

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*(e*x)^(3/2)), x)
 

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*(e*x)^(3/2)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int((c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)),x)
 

Output:

int((c + d*x^2)/((e*x)^(3/2)*(a + b*x^2)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {c+d x^2}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {\sqrt {e}\, \left (\left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b^{2} x^{4}+2 a b \,x^{2}+a^{2}}d x \right ) d +\left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{b^{2} x^{6}+2 a b \,x^{4}+a^{2} x^{2}}d x \right ) c \right )}{e^{2}} \] Input:

int((d*x^2+c)/(e*x)^(3/2)/(b*x^2+a)^(5/4),x)
 

Output:

(sqrt(e)*(int((sqrt(x)*(a + b*x**2)**(3/4))/(a**2 + 2*a*b*x**2 + b**2*x**4 
),x)*d + int((sqrt(x)*(a + b*x**2)**(3/4))/(a**2*x**2 + 2*a*b*x**4 + b**2* 
x**6),x)*c))/e**2