\(\int \frac {A+C x^4}{(a+b x^2)^{5/4}} \, dx\) [200]

Optimal result

Integrand size = 19, antiderivative size = 113 \[ \int \frac {A+C x^4}{\left (a+b x^2\right )^{5/4}} \, dx=-\frac {14 a C x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac {2 C x \left (a+b x^2\right )^{3/4}}{5 b^2}+\frac {2 \left (5 A b^2+12 a^2 C\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {a} b^{5/2} \sqrt [4]{a+b x^2}} \] Output:

-14/5*a*C*x/b^2/(b*x^2+a)^(1/4)+2/5*C*x*(b*x^2+a)^(3/4)/b^2+2/5*(5*A*b^2+1 
2*C*a^2)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^ 
(1/2))/a^(1/2)/b^(5/2)/(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 = 10.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80 \[ \int \frac {A+C x^4}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (10 A b^2+2 a C \left (6 a+b x^2\right )-\left (5 A b^2+12 a^2 C\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 a b^2 \sqrt [4]{a+b x^2}} \] Input:

Integrate[(A + C*x^4)/(a + b*x^2)^(5/4),x]
 

Output:

(x*(10*A*b^2 + 2*a*C*(6*a + b*x^2) - (5*A*b^2 + 12*a^2*C)*(1 + (b*x^2)/a)^ 
(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*a*b^2*(a + b*x^2 
)^(1/4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.35, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1472, 27, 299, 227, 225, 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.

\(\displaystyle \int \frac {A+C x^4}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 1472

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {2 \int \frac {b \left (\frac {2 C a^2}{b^2}+A\right )-a C x^2}{2 b \sqrt [4]{b x^2+a}}dx}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {\int \frac {b \left (\frac {2 C a^2}{b^2}+A\right )-a C x^2}{\sqrt [4]{b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\left (12 a^2 C+5 A b^2\right ) \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}-\frac {2 a C x \left (a+b x^2\right )^{3/4}}{5 b}}{a b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (12 a^2 C+5 A b^2\right ) \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}-\frac {2 a C x \left (a+b x^2\right )^{3/4}}{5 b}}{a b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (12 a^2 C+5 A b^2\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}-\frac {2 a C x \left (a+b x^2\right )^{3/4}}{5 b}}{a b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x \left (\frac {a^2 C}{b^2}+A\right )}{a \sqrt [4]{a+b x^2}}-\frac {\frac {\sqrt [4]{\frac {b x^2}{a}+1} \left (12 a^2 C+5 A b^2\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}-\frac {2 a C x \left (a+b x^2\right )^{3/4}}{5 b}}{a b}\)

Input:

Int[(A + C*x^4)/(a + b*x^2)^(5/4),x]
 

Output:

(2*(A + (a^2*C)/b^2)*x)/(a*(a + b*x^2)^(1/4)) - ((-2*a*C*x*(a + b*x^2)^(3/ 
4))/(5*b) + ((5*A*b^2 + 12*a^2*C)*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2 
)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[ 
b]))/(5*b*(a + b*x^2)^(1/4)))/(a*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

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

rule 299

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

rule 1472

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wi 
th[{Qx = PolynomialQuotient[(a + c*x^4)^p, d + e*x^2, x], R = Coeff[Polynom 
ialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, Simp[(-R)*x*((d + e*x^2) 
^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1))   Int[(d + e*x^2)^(q + 1 
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, 
e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]
 

Maple [F]

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

Input:

int((C*x^4+A)/(b*x^2+a)^(5/4),x)
 

Output:

int((C*x^4+A)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

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

integrate((C*x^4+A)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((C*x^4 + A)*(b*x^2 + a)^(3/4)/(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 = 1.91 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.50 \[ \int \frac {A+C x^4}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {A x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{4}}} + \frac {C x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac {5}{4}}} \] Input:

integrate((C*x**4+A)/(b*x**2+a)**(5/4),x)
 

Output:

A*x*hyper((1/2, 5/4), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/4) + C*x**5* 
hyper((5/4, 5/2), (7/2,), b*x**2*exp_polar(I*pi)/a)/(5*a**(5/4))
 

Maxima [F]

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

integrate((C*x^4+A)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((C*x^4 + A)/(b*x^2 + a)^(5/4), x)
 

Giac [F]

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

integrate((C*x^4+A)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((C*x^4 + A)/(b*x^2 + a)^(5/4), x)
 

Mupad [F(-1)]

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

int((A + C*x^4)/(a + b*x^2)^(5/4),x)
 

Output:

int((A + C*x^4)/(a + b*x^2)^(5/4), x)
 

Reduce [F]

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

int((C*x^4+A)/(b*x^2+a)^(5/4),x)
 

Output:

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